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WBBSE Solutions For Class 6 Maths Chapter 3 Logical Approximation Of Number

July 2, 2024October 11, 2023 by Haritha

Class 6 Math Solutions WBBSE Chapter 3 Logical Approximation Of Number

Example 1. Let us find the nearest whole number which is a multiple of 10, by the following numbers.

333
1275
2364

3726
5147

Solution: Nearest whole number = 80

WBBSE Solutions For Class 6 Maths Chapter 3 Logical Approximation Of Number Nearest Whole N umber 80 Multiple By 10

Example 2. Let us find the nearest whole number which is a multiple of 100, for the following numbers

Solution:

Read and Learn More WBBSE Solutions For Class 6 Maths

WBBSE Solutions For Class 6 Maths Chapter 3 Logical Approximation Of Number Nearet Whole Number Multiple Of 100

Example 3. Let us find the nearest whole numbers to the following numbers which will be multiples of 1000.

332
2235
5561

4325
7671
8888

8968

Solution:

WBBSE Solutions For Class 6 Maths Chapter 3 Logical Approximation Of Number Nearest Whole Number Multiple Of 1000

Class 6 Math Solutions WBBSE

Example 4. Convert the following numbers to their nearest whole numbers which are multiples of 10,100 and 1000 and calculate.

28 + 71

316 + 45
728 – 156
489 – 36

Solution:

1. 28 + 71

Nearest multiple of 10 = 30 + 70

Nearest multiple of 100 = 0 + 100

Nearest multiple of 1000 = 0 + 0

2. 316 + 45

Nearest multiple of 10 = 320 + 50 = 370

Nearest multiple of 100 = 300 + 0 = 300

Nearest multiple of 1000 = 0 + 0 =0 Ans

3. 728 – 156

Nearest multiple of 10 = 730 – 160 = 570

Nearest multiple of 100 = 700 – 200 = 500

Nearest multiple of 1000 = 1000 – 0 =1000

4. 489 – 36

Nearest multiple of 10 = 490- 40 = 450

Nearest multiple of 100 = 500-0 =500

Nearest multiple of 1000 = 0 -0 = 0

Class 6 WBBSE Math Solutions Chapter 3 Logical Approximation Of Number Exercise 3

Question 1. Let us convert the following numbers to their nearest whole number which are multiples of 10:

1324
5968

Solution:

Nearest Whole Number (Multiples of 10)

12: 10
347: 350

1324: 1320
5968: 5970

Question 2. Let us convert the following numbers to the nearest whole number which are multiples of 100:

621
483
6521

2178

Solution:

Nearest Whole Number (Multiples of 10)

621: 600
483: 500
6521: 6500

2178: 2200

Question 3. Let us convert the following numbers to their nearest whole numbers which are multiples of 1000:

827
6719
8394

Solution:

Nearest Whole Number (Multiples of 10)

346: 0

827: 1000
6719: 7000
8394: 8000

Class 6 WBBSE Math Solutions

Question 4. Let us change the following numbers to their nearest whole numbers which are multiples of 10 and calculate.

1. 37 + 54
Solution: For the nearest multiple of 10, 40 + 50 = 90

2. 73 – 78
Solution: For the nearest multiple of 10, 70 – 50 = 20

3. 24 + 59
Solution: For the nearest multiple of 10, 20 + 60 = 80

4. 97 – 38
Solution: For the nearest multiple of 10,100 – 40 = 60

5. 76 – 29
Solution: For the nearest multiple of 10, 80 – 30 = 50

6. 66 + 73
Solution: For the nearest multiple of 10, 70 + 70 = 140

7. 251 + 175
Solution: For the nearest multiple of 10, 250 + 180 = 430

8. 462 – 271
Solution: For the nearest multiple of 10, 460 – 270 = 190

Question 5. Let us change the following numbers to their nearest whole numbers which are multiples of 100 and calculate:

1. 426+ 589
Solution: For the nearest multiple of 100, 400 + 600 = 1000

2. 356+ 435
Solution: For the nearest multiple of 100, 400 + 400 = 800

3. 678 -125
Solution: For the nearest multiple of 100, 700 -100 = 600

4. 1248 + 4329
Solution: For the nearest multiple of 100,1200 + 4300 = 5500

5. 170 + 895
Solution: For the nearest multiple of 100,200 + 900 = 1100

Class 6 WBBSE Math Solutions

6. 947 + 448
Solution: For the nearest multiple of 100, 900 + 400 = 1300

7. 5612 + 2095
Solution: For the nearest multiple of 100, 5600 + 2100 = 7700

8. 4258 – 2436
Solution: For the nearest multiple of 100, 4300 – 2400 = 1900

Question 6. Let us change the following numbers 1o their nearest whole nunv bers which are multiples of 1000 and calculate.

1. 2836 + 7466
Solution: For the nearest multiple of thousand, 3000 + 7000 = 10,000

2. 3076 + 5731
Solution: For the nearest multiple of thousand, 3000 + 6000 = 9000

3. 7767 + 3685
Solution: For the nearest multiple of thousand, 8000 + 4000 = 12000

4. 8005 + 7483
Solution: For the nearest multiple of thousand, 8000 + 7000 = 15000

5. 1375 + 6307
Solution: For the nearest multiple of thousand, 1000 + 6000 = 7000

Class 6 WBBSE Math Solutions

6. 8643 + 5285
Solution: For the nearest multiple of thousand, 9000 + 5000 = 14000

Question 7. Match left side with right side:

WBBSE Solutions For Class 6 Maths Chapter 3 Logical Approximation Of Number Match The Left Side With Right Side

Solution: 1. C, 2. A, 3. B, 4. F, 5. D, 6. E

WBBSE Solutions For Class 6 Maths Chapter 4 Roman Numbers And Numbers Upto Hundred

July 3, 2024October 11, 2023 by Haritha

WBBSE Class 6 Maths Solutions Chapter 4 Roman Numbers And Numbers Upto Hundred Exercise 4

Question 1. Let us write the following numbers in Roman numerals:

Solution:

Numbers in Roman numerals

7 – VII
9 – IX

14 – XIV
25 – XXV
36 – XXXVI

54 – LIV
65 – LXV
89 – LXXXIX

90 – XC
98 – XCVIII

Question 2. Let us write the following Roman numbers in Hindu-Arabic numbers:

IX, VIII, VII, IV, VI
XXX, XXXIX, XL, XLIX, XU
LV, LIX, LX, XC, XCV

XXVI, XI, XXXVI, XLV, LXXV

Read and Learn More WBBSE Solutions For Class 6 Maths

Solution:

Roman numbers in Hindu-Arabic numbers

IX – 9; VIII – 8, VII – 7, IV – 4, VI – 6
XXX – 30; XXXIX – 39; XL – 40; XLIX – 49; XLI – 41 .

LV – 55; LIX – 59; LX – 60; XC – 90; XCV – 95
XXVI – 26; XI – 11; XXXVI – 36; XIV – 45; LXXV -75

Question 3. Let’s choose the right answer.

WBBSE Solutions For Class 6 Maths Chapter 4 Roman Numbers And Numbers Upto Hundred

Solution: 40 – XL; 49 – XLIX; 90 – XC

Class 6 WBBSE Math Solutions

Question 4. Let us put <, = or > signs in the boxes given below.

5 □ IV
XIV □ 14

XIX □ XXI
LXXVI □ LXXIV

Solution:

5 [>] IV;
XIV [=];
XIX

XXI
LXXVI [>] LXXIV

WBBSE Solutions For Class 6 Maths Chapter 5 Concept Of Algebraic Variables

July 3, 2024October 11, 2023 by Haritha

Class 6 Math Solutions WBBSE Chapter 5 Concept Of Algebraic Variables

Example: To make 15 such triangles, let us find the number of match sticks required without actually forming the triangles.

WBBSE Solutions For Class 6 Maths Chapter 5 Concept Of Algebraic Variables Match Sticks Forming The Triangles-1
Solution:

The number of match sticks required without actually forming the triangles Are

Total number of match sticks = 3 x Number of triangles.

If there are triangles, No. of match sticks required = 3n.

WBBSE Solutions For Class 6 Maths Chapter 5 Concept Of Algebraic Variables Match Sticks Froming The Triangles

WBBSE Class 6 Maths Solutions Chapter 5 Concept Of Algebraic Variables Exercise 5.1

Question 1. Let us find the perimeter of the triangle in terms of x, y, and z.
Solution:

The perimeter of the triangle = x cm + ycm + z cm = (x+y+z) cm.

Read and Learn More WBBSE Solutions For Class 6 Maths

WBBSE Solutions For Class 6 Maths Chapter 5 Concept Of Algebraic Variables Perimeter Of The Triangle

Question 2. Let me form the letters V, Z, E, D………with match sticks, as given below, and try to form a relation to find the total number of match sticks required for any of these figures.
Solution:

1. WBBSE Solutions For Class 6 Maths Chapter 5 Concept Of Algebraic Variables Number Of Match Sticks Is 2x

Total number of match sticks = 2x

2. WBBSE Solutions For Class 6 Maths Chapter 5 Concept Of Algebraic Variables Number Of Match Sticks Is 3x

Total number of match sticks = 3x

3. WBBSE Solutions For Class 6 Maths Chapter 5 Concept Of Algebraic Variables Number Of Match Sticks Is 4x

Total number of match sticks = 4x

4. WBBSE Solutions For Class 6 Maths Chapter 5 Concept Of Algebraic Variables Number Sticks Is 4x

Total number = 4x Where x = Number of sticks in V, Z, E, D.

Question 3. I am 4 years younger than my brother. If my brother’s age is x years, let me write my age in terms of my brother’s age.
Solution:

My brother’s age = x years

I am 4 years younger than my brother.

∴ My age = (x – 4) years.

My age in terms of my brother’s age is (x minus 4) years.

Class 6 Maths Solutions WBBSE

Question 4. Rafika Begum and Aajma Khatoon are making garlands. Aajma Khatoon made 6 garlands more than Rafika Begum. Let us form a relation to find the number of garlands Aajma Khatoon made.
Solution:

Let Rafika Begum made x garlands.

Aajma Khatoon made 6 garlands more than Rafika Begum.

∴ Aajma Khatoon made (x + 6) garlands.

Question 5. In our shop, there are many oil pastel color boxes. Each box has 12 pastel colours. Without counting, let us form a relation to find how many pastel colors are there in the shop.
Solution:

Let there are x number of colour boxes.

Each box contains 12 pastel colors.

∴ Number of total pastel colours in the shop = 12 x.

Question 6. Today at a program of our school we seated ourselves in many rows. In each row there are 15 students. Let us form a relation to find the total number of students seated in the program of our school.
Solution:

Let there are x number of rows in the school. In each row there are 15 students.

∴ Total number of students = 15x.

Question 7. Sitting beside a pond, Partha and Tirtha, both were trying to catch fish. Partha caught 5 fishes more than Tirtha. Let us express the number of fishes caught by Partha in terms of the number of fishes caught by Tirtha.
Solution:

Let Tirtha caught x number of fishes. Partha caught 5 fishes more than Tirtha.

∴ Partha caught (x – 5) fishes.

Class 6 Maths Solutions WBBSE

Question 8. Mita and Maya are making paper boats. Mita made 2 boats less than Maya. Let us form a relation to find the number of boats made by Mita.
Solution:

Let Maya made x number of boats. Mita made 2 boats less than Maya.

∴ Mita made (x – 2) boats.

Question 9. Father bought several packets of sweets. If each packet contains 5 sweets, let us form a relation to find the total number of sweets bought by the father.
Solution:

Let father buy x packets of sweets. Each packet contains 5 sweets.

∴ Total number of sweets father bought = 5x.

Question 9. Grandmother prepared coconut sweets today. She gave my brother 2 more sweets than she gave me and my sister 3 less. Let us form a relationship to find how many sweets my brother and sister got in terms of the unknown number of sweets I got.
Solution:

Let I get x number of sweets.

My brother got (x + 2) sweets.

My sister got (x – 3) sweets.

Question 10. In the expressions given below, let us identify the variables and constants:

1. x + 2
Solution: x + 2 — variable x Constant 2

2. 5
Solution: 5 – Constant 5

3. y – 6
Solution: y-6 – variable y Constant6

4. 2p + 8
Solution: 2p + 8 – variable p – Constant 2, 8

5. -4
Solution: – 4 – Constant 4

6. 5 (3x – 2) + 5
Solution: 5 (3x-2)+5 – variable x – Constant 5, 3, – 2

7. 10-2p
Solution: 10p-2 – variable p – Constant 10,-2

8. 15z
Solution: 15z – variable z – Constant 15

9. x + 2y
Solution: x + 2y – variable x,y – Constant 2

Class 6 Maths Solutions WBBSE

Question 11. Let us express in language the following expressions:

1. y + 6
Solution: Sum of y and 6

2. a-2
Solution: Difference of a and 2

3. 5y
Solution: 5 times of y or y is multiplied by 5

4. a/9
Solution: a is divided by 9

5. 3x
Solution: 3 times of x or x is multiplied by 3

6. 5x+2
Solution: Sum of 2 and 5 times of x

7. a+b
Solution: Sum of a and b

8. 6x-4
Solution: Difference of 6 times of x and 4

9. x>5
Solution: x is greater than 5

10. x≥7
Solution: x is greater than or equal to 7

11. y ≯ 10
Solution: y is not greater than 0

12. y≤10
Solution: y is less than or equal to 10

13. x ≮ 7
Solution: x is not less than 7

14. x<9
Solution: x is less than 9

15. x≥12
Solution: x is greater than or equal to 12

16. x≥5
Solution: x is not greater than 5

Forming expressions with signs and symbols

WBBSE Solutions For Class 6 Maths Chapter 5 Concept Of Algebraic Variables Forming Expressions With Signs And Symbol

Solution:

WBBSE Solutions For Class 6 Maths Chapter 5 Concept Of Algebraic Variables Forming Expressions With Signs And Symbols

Chapter 5 Concept Of Algebraic Variables, Exercise 5.2

Question 1. Let us write in language the following expressions.

1. 10a
Solution: Ten times of a or a is multiplied by 10

2. a + c
Solution: Sum of a and c

3. x – y
Solution: Difference of x and y

4. 9x + 2
Solution: Sum of 9 times of x and 2

5. x + y + z
Solution: Sum of x, y and z

6. 3 x – 7
Solution: Seven less than three times of x

7. p/3-4
Solution: Four less than one-third of p

8. x > 6
Solution: x is greater than 6

9. p ≯ 9
Solution: p is not greater than 9

Question 2. Let us form the algebraic expressions of the following with signs and symbols

1. Sum of a and 7
Solution: a + 7

2. Subtract 9 from y
Solution: y – 9

3. Multiply a by 3
Solution: 3a

4. x is greater than 13
Solution: x>13

5. y is not equal to 5
Solution: y ≠ 5

6. x is divided by 8
Solution: x/8

7. Sum of ten times p and 7
Solution: 10p + 7

8. 8 is subtracted from 3 times x
Solution: 3x – 8

9. Sum of x and y
Solution: x + y

10. Product of x and y
Solution: x X y

11. k is less than y
Solution: k < y

12. b is not less than 8
Solution: b ≮ 8

Question 3. Let us find the difference between the following sets of algebraic expressions:

x+5, x-5, 5x
2p+3, 2p-3, 3p+2, 3p-2

Solution:

1. x+5 → Sum of x and 5

x- 5 → 5 is subtracted from x

5x → 5 times of x

2. 2p + 3 → Sum of twice p and 3

2p – 3 → 3 less than twice of p

3p + 2 → Sum of times of p and 2

3p – 2 → 2 less than 3 times of p

3. Let us prepare four different algebraic expressions with the variable x and constant 6. Let us write in a statement the expressions formed and find the difference amongst them :
Solution:

Four algebraic expressions x + 6; x – 6, 6x + 2 ; 6x – 2

x + 6 → Sum of x and 6

x – 6 > Difference of x and 6

6x + 2 > Sum of six times of x and 2

6x – 2 > Difference of six times of x and 2

4. Let us prepare four different algebraic expressions with the variable x and two different constants 9 and
Solution:

Four different algebraic expressions with variable x and two different constants 9 and 11.

9X + 11 → Sum of nine times of x and 11

11X —9 → Difference of 11 times x and 9

9x X 11 → Product of 11 and nine times of x

9 ÷ 11x → Nine divided by 11 times of x

5. With two variables and one constant, let us prepare four different expressions and find the difference among them
Solution:

Four different algebra expression with two variables and one constant. 2x + y + 1, 2x + 1 – y, 2y + x + 1, 2y + 1 – x

2x + y + 1 → Sum of twice x and y and 1

2x + 1 – y → y is subtracted from the sum of twice x and 1

2y + x + 1 → Sum of twice y and x and 1

2y + 1 – x → x is subtracted from the sum of twice y and 1.

Problem 4. Let us match the two sides

WBBSE Solutions For Class 6 Maths Chapter 5 Concept Of Algebraic Variables Let Us Match The Two Sides

Solution: 1. B, 2. C, 3. D, 4. E, 5. A

Question 5. Let us read and fill in the gaps

1. Present age of Mukti is x years, 4 years later her age will be years.
Solution:

Present age of Mukti is x years, 4 years later her age will be [x + 4] years.

2. Present age of Firoz is y years; 7 years ago his age was years.
Solution:

Present age of Firoz is y years; 7 years ago his age was [y- 7] years.

3. In Kumud’s garden there are x number of saplings of marigold. But in Shahida’s garden, there are three times more saplings than Kumud’s garden. There are saplings in Shahid’s garden.
Solution:

In Kumud’s garden there are x number of saplings of marigolds. But in Shahida’s garden there are three times more saplings than Kumud’s garden. There are [3x] saplings in Shahida’s garden.

4. Funds were collected for repairing the roads of Durganagar. Habib paid Rs. x but Muskan paid Rs. 10 more than twice the money Habib paid. Muskan paid Rs .
Solution:

Funds were collected for repairing the roads of Durganagar. Habib paid Rs. x but Muskan paid Rs. 10 more than twice the money Habib paid. Muskan paid Rs. [2x + 10].

5. The height of David is 3 cm less than Aminur. If height of Aminur is ‘x’ cm, height of David is cm.
Solution:

Height of Devid is 3 cm less than Aminur. If the height of Aminur is ‘x’ cm, height of David is [x- 3] cm.

Problem 6. Let us prepare stories which will satisfy the following expressions

1. Rs. (x + 18)
Solution:

∴ price of my exercise book is Rs. x. Price of Mihir’s exercise book is Rs.18 more than mine. Hence, price of Mihir’s exercise book is Rs. (x + 18).

2.(y — 21) cm.
Solution:

∴ Length of my classroom is y cm, but the length of another room is 21 cm less than my classroom. So the length Of the other room is (y – 21) cm.

3. 5x years
Solution:

My sister’s age is x years and I am 5 times older than her.

∴ My age is 5x years.

4. p/4
Solution:

A man bought p cakes, his 4 boys, each boy got p/4.

5. (2x + 3) metres
Solution:

The length of a room is x m but the length of another room is 3 m more than the twice of the 1st room.

∴ The length of 2nd room is (2x + 3) metres.

6. (3 x – 5) hours
Solution: A boy took x hours to do all the sums of an exercise but his friend took 5 hours less than thrice of the time taken by the 1 st boy.

∴ So, the 2nd boy took (3x – 5) hours to do the sums.

WBBSE Solutions For Class 6 Maths Chapter 1 Fraction

July 1, 2024October 11, 2023 by Haritha

Class 6 Math Solutions WBBSE Fraction

Question 1. Let us find the value of (40 ÷ 5) ÷ 4 (5 – 3) If there is no ‘sign’ after a number, it means multiplication
Solution:

(40 ÷ 5) ÷ 4 (5 – 3)

⇒ (40 ÷ 5) ÷ 4 (5 – 3) = 8 ÷ 4 of 2 (4 of 2 = 4 * 2)

⇒ (40 ÷ 5) ÷ 4 (5 – 3) = 8 + 8

⇒ (40 ÷ 5) ÷ 4 (5 – 3) = 1

Question 2. Let me find the values of {25 – (4 ÷ 9)} ÷ 3 and 25 – 4 + 9 ÷ 3
Solution:

{25 – (4 + 9)} ÷ 3

⇒ {25 – (4 + 9)} ÷ 3 = {25-13}÷3

⇒ {25 – (4 + 9)} ÷ 3 = 12÷3

⇒ {25 – (4 + 9)} ÷ 3 = 4

and

Read and Learn More WBBSE Solutions For Class 6 Maths

25 – 4 + 9÷3

⇒ {25 – (4 + 9)} ÷ 3 = 25-4 + 3

⇒ {25 – (4 + 9)} ÷ 3 = 28-4

⇒ {25 – (4 + 9)} ÷ 3 = 24

Class 6 Maths Solutions WBBSE

Question 3. Shall we get same values for (16 – 4) (5 – 3) and 16 – 4(5 -3).
Solution:

(16 – 4) (5 – 3)

⇒ (16 – 4) (5 – 3) = 12×2

⇒ (16 – 4) (5 – 3) = 24 and

16-4(5-3)

⇒ (16 – 4) (5 – 3) = 16 – 4 of 2

⇒ (16 – 4) (5 – 3) = 16-8

⇒ (16 – 4) (5 – 3) = 8

Question 4. Let me find the values for 10 – 3 – 5 and 20 ÷ 5 ÷ 2.
Solution:

10 – 3 – 5

⇒ (10 – 3) – 5 = 7 – 5

⇒ 7 – 5 = 2 and

20 ÷ 5 ÷ 2

⇒ 20 ÷ 5 ÷ 2 = 4 ÷ 2 = 2

Class 6 Maths Solutions WBBSE Fraction Exercise 1.1

Question 1. Verify, if the values are same for all these cases

1. 20 + 8 ÷ (4 – 2)
Solution:

20 + 8 ÷ (4 – 2)

⇒ 20 + 8 ÷ (4 – 2) = 20 + 8 ÷ 2

⇒ 20 + 8 ÷ (4 – 2) = 20 + 4

⇒ 20 + 8 ÷ (4 – 2) = 24

2. (20 + 8) ÷ (4 – 2)
Solution:

(20 + 8) ÷ (4 – 2)

⇒ (20 + 8) ÷ (4 – 2) = (20 + 8) ÷ 2

⇒ (20 + 8) ÷ (4 – 2) = 28 ÷ 2

⇒ (20 + 8) ÷ (4 – 2) = 14

3. (20-8)(4-2)
Solution:

⇒ (20 – 8) (4 – 2)

⇒ (20 – 8) (4 – 2) = 12 x 2

⇒ (20 – 8) (4 – 2) = 24

4. 20 – 8 (4 – 2)
Solution:

20 – 8 (4 – 2)

⇒ 20 – 8 (4 – 2) = 20 – 8(2)

⇒ 20 – 8 (4 – 2) = 20 – 16 =4

5. (20 + 8) ÷ 4 – 2
Solution:

(20 + 8) ÷ 4 – 2

⇒ (20 + 8) ÷ 4 – 2 = 28 ÷ 4 – 2

⇒ (20 + 8) ÷ 4 – 2 = 7 – 2 = 5

Class 6 Maths Solutions WBBSE

Question 2. Let us form similar simplification sums with numbers 12, 6, 3 and 1 and then find their values:

1. 12 + 6 ÷ (3 – 1)
Solution:

⇒ 12 + 6 ÷ (3 – 1 ) = 12 + 6 ÷2

⇒ 12 + 6 ÷ (3 – 1 ) = 12 + 3

⇒ 12 + 6 ÷ (3 – 1 ) = 15

2. (12 + 6) ÷ (3 -1)
Solution:

⇒ 12 + 6 ÷ (3 – 1 ) = 18 ÷ 2

⇒ 12 + 6 ÷ (3 – 1 ) = 9

3. (12-6) (3-1)
Solution:

⇒ (12 – 6 ) (3 – 1) = 6 x 2

⇒ (12 – 6 ) (3 – 1) = 15

4. 12 – 6(3 – 1)
Solution:

⇒ 12 – 6 (3 – 1) = 12 – 6 x 2 =12-12 = 0

5. (12 + 6) ÷ 3-1
Solution:

⇒ (12 + 6) ÷ (3 – 1) = 18 ÷ 3-1

⇒ (12 + 6) ÷ (3 – 1) = 6-1 = 5

Class 6 WBBSE Math Solutions

Question 3. Let us find the value of simplification sums:

1. $256 \div \overline{16 \div 2} \div \overline{18 \div 9} \times 2$
Solution:

⇒ $256 \div \overline{16 \div 2} \div \overline{18 \div 9} \times 2$

⇒ = 256 ÷ 8 ÷ 2 x 2

⇒ = 32 ÷ 2 x 2

⇒ = 16 x 2

⇒ = 32

2. (72 ÷ 8 x 9)- (72 ÷ 8 of 9)
Solution:

(72 ÷ 5- 8 x 9) – (72 ÷ 5- 8 of 9)

⇒ (72 ÷ 5- 8 x 9) – (72 ÷ 5- 8 of 9) = (9 x 9) – (72 ÷ 72)

⇒ (72 ÷ 5- 8 x 9) – (72 ÷ 5- 8 of 9) = 81-1

⇒ (72 ÷ 5- 8 x 9) – (72 ÷ 5- 8 of 9) = 80

3. 76-4 -[6 + {19-(48- $\overline{57-17}$))]
Solution:

76-4 -[6 + {19-(48- $\overline{57-17}$))]

⇒ = 76- 4 [6+ {19 – (48 – 40)}]

⇒ = 76 – 4 – [6+{19 – 8}]

⇒ = 76-4-[6+11]

⇒ = 76-4-17 = 72-17 = 55

4. {25 x 16 ÷ (60 ÷15) – 4 x (77 – 62)} ÷ (20 x 6 ÷ 3)
Solution:

⇒ {25 x 16 ÷ (60 ÷15) – 4 x (77 – 62)} ÷ (20 x 6 ÷ 3)

⇒ = {25 x 16 ÷ 4 – 4 x 15} ÷ (120 ÷ 3)

= {25 x 4 – 60} ÷ 40

= {100 – 60} ÷ 40

= 40 ÷ 40

= 1

5. [16 ÷ {42 – 38 + 2}] 12 ÷ (24 ÷ 6) x 2 + 4
Solution:

[16 ÷ {42 – 38 + 2}] 12 ÷ (24 ÷ 6) x 2 + 4

= [16 ÷ {42 – 40}] 12 ÷ 4 x 2 + 4

= [16 ÷ 2] 12 ÷ 4 x 2 + 4

= 8 of 12 ÷ 4 x 2 + 4

= 96 ÷ 4 x 2 + 4

= 24 x 2 + 4

= 48 + 4

= 52

Class 6 WBBSE Math Solutions

6. 4 x [24-{110- $\overline{11+3}$ x4) ÷ 9}] ÷ 2 of 9
Solution:

4 x [24-{110- $\overline{11+3}$ x 4) ÷ 9}] ÷ 2 of 9

= 4 x [24 – {110-14×4) ÷ 9}] ÷18.

= 4 x [24 – {(110-56) ÷ 9}] ÷18.

= 4 x [24 – {54 ÷ 9}] ÷18 = 4 x [24 – 6] ÷18

= 4 x 18 ÷ 18

= 4 x 1

= 4

7. 200 ÷ [88-{(12×13)-3(40-9)}]
Solution:

200 ÷ [88 -{(12×13)- 3(40 – 9)}]

= 200 ÷ [88-{156-3×31}]

= 200 ÷ [88-{156-93}]

= 200 ÷ [88 – 63]

= 200 ÷ 25

= 8

8. (987- $\overline{43+25}$)-10[5 + {(999 ÷ $\overline{9×3}$+($\overline{8×9}$ ÷ 6)4}]
Solution:

(987- $\overline{43+25}$)-10[5 + {(999 ÷ $\overline{9×3}$+($\overline{8×9}$ ÷ 6)4}]

= (987 – 68) – 10 [5 + {999 ÷ 27 + (72 ÷ 6)4}]

= 919 -10 [5 + {37+12 of 4}]

= 919-10 [5+ {37+ 48}]

= 919-10 [5+ 85]

= 919-10 x 90

= 919 – 900

= 19

Class 6 WBBSE Math Solutions

Question 4. Let us form a story for the following simplification and then solve:

1. You bring 12 eggs from the market, but 2 eggs are rotten. Now you divide the rest of the eggs between 2 of your friends. Find how many eggs each will get.
Solution:

(12-2) ÷ 2

=10 ÷2 = 5

∴ Each will get 5 eggs.

2. You have 90 rupees, you spent Rs. 48 on books. Again, your father gives you Rs. 21. With this total money you buy 7 pens of equal price. What is the price of each pen?
Solution:

(90-48+ 21) ÷ 7

= (42 + 21) ÷ 7

= 63 ÷ 7

= 9

∴ Price of each pen = Rs. 9.

Question 4. Let us express in mathematical language and solve:

Rajdeep’s father sold 125 guavas from their guava orchard at Rs. 2 each to Baruipur market. With the money, he bought 2 pens for Rs. 5 each and 2 exercise books for Rs. 20 each and divided the remaining money among two brothers and sisters to buy sweets. Let us find out how much money Rajdeep got for sweets.
Solution:

The total amount of money Rajdeep’s father received by selling 125 guavas at Rs. 2 each = 125 x Rs. 2 = Rs. 250

Now, he bought 2 pens for Rs. 5 each = Rs. 5×2 = Rs. 10 and he bought 2 exercise books for 20 each = Rs. 20 x 2 = Rs 40

∴ Total amount he spent = Rs. 10 + Rs. 40 = Rs. 50

Rest amount = Rs. 250 -(40 + 10) = 200

Each will get = Rs. 200 ÷ 2 = Rs. 100.

∴ In mathematical language,

[125 x 2-{(5×2)+ (20×2)}] ÷ 2

= [250-{10+ 40}] ÷ 2

= [250-50] ÷ 2

= 200 ÷ 2

= 100

WBBSE Math Solutions Class 6 Fraction Excercise 1.2

Question 1. 6 multiples of 5 which do not not contain ‘O’.
Solution: 5,15,25, 35, 45,55.

Question 2. Let us find 3 multiples of 7 which are greater than 50.
Solution: 56, 63, 70.

Question 3. Let us think and find 2-digit numbers which are multiples of 4.
Solution: 12,16,20,24,28,32,36,40.

Question 4. Let’s find three numbers which are the factors of 4.
Solution: 1,2, 4.

Question 5. Let’s find two numbers whose L.C.M. is 12 and their sum is 10.
Solution: The required numbers are 4 and 6as4 + 6 = 10 and L.C.M of 4 and 6 = 12.

Question 6. What are the prime factors of 14?
Solution: Prime factors of 14: 2, 7.

Question 7. What is the least prime number?
Solution: The least prime number is 2.

Question 8. Which number is neither prime nor composite?
Solution: 1

Question 9.

1. 42 is a multiple of which of the following numbers?

Solution: 42 is a multiple of

1. 7,

4. 6.

2. Which of the following numbers has 11 as a factor?

Solution: 3. 121.

WBBSE Math Solutions Class 6

Question 10. Which of the following pairs of numbers are mutually prime?

5, 7
10, 21
10, 15
16, 15.

Solution: 1. 5, 7

Question 11. Let us find two pairs of composite numbers which are mutually prime.
Solution: (4,15) and (4,9).

Question 12.

Let’s write H.C.F. of mutually prime numbers,
Let’s write L.C.M. of mutually prime numbers.

Solution:

Let the two prime numbers are 15 and 16.

H.C.F of 15 and 16 is 1.
L.C.M of 1 5 and 1 6 is (15 x 16) = 24.

Question 13. Finding prime factors and let’s find H.C.F. of the following

22,44;
54, 72;
27,64;
36,30;
28, 35, 49;
30, 72, 96;
20, 40, 80.

Solution:

WBBSE Solutions For Class 6 Maths Chapter 1 Fraction Prime Factors And Let HCF Of The Following

1. 22 = 2 x 1 1 x 1; 44 = 2 x 2 x 1 1 x 1

∴ H.C.F. = 2×11 x 1 =22

2. 54 = 2 x 3 x 3 x 3 x 1; 72 = 2x2x2x3x3x1

∴ H.C.F. = 2 x 3 x 3x 1 =18

3. 27 = 3 x 3 x 3 x 1; 64=2x2x2x2x2x2x1

∴ H.C.F =1

4. 36 = 2 x 2 x 3 x 3 x 1; 30 = 2x3x5x1

∴ H.C.F. = 2x3x1 =6

5. 28 = 2x2x7x1; 35 = 5 x 7 x 1; 49 = 7x7x1

∴ H.C.F. = 7×1=7

6. 30 = 2 x 3 x 5 x 1; 72 = 2x2x2x3x3x1; 96=2x2x2x2x2x3x1

∴ H.C.F =2x3x1=6

7. 20 = 2 x 2 x 5 x 1; 40 = 2x2x2x5x1; 80=2x2x2x2x5x1

∴ H.C.F = 2x2x5x1=20

WBBSE Math Solutions Class 6

Question 14. Let’s find L.C.M. of the following by division method

28, 35;
54, 72;
27, 63;
25, 35, 45;
48, 72, 96.

Solution:

WBBSE Solutions For Class 6 Maths Chapter 1 Fraction LCM Of The Following Division Method

Question 15. Let’s find L.C.M. of the following by finding their prime factors

25, 80;
36, 39;
32, 56;
36, 48 and 72;
25, 35 and 45;
32,
40 and 84.

Solution:

WBBSE Solutions For Class 6 Maths Chapter 1 Fraction LCM Of The Following By Their Prime Factors

1. 25 = 5 x 5 x 1; 80=2x2x2x2x5x1

∴ L.C.M of 25, 35 & 45

2. 6 = 2 x 2 x 3 x 3 x 1; 39 = 3 x 13 x 1

∴ L.C.M. of 36 and 39 =2x2x3x3x13 = 468.

3. 32 = 2x2x2x2x2x1; 56 = 2x2x2x7

∴ L.C.M. of 32 and 56 =2x2x2x 2x2x7 = 224

4. 36 = 2 x 2 x 3 x 3 x 1; 48=2x2x2x2x3x1; 72 = 2x2x2x3x3x1

∴ L.C.M of 36, 48 and 72 = 2x2x2x2x3x3 = 144

5. 25 = 5 x 5 x 1; 35 = 5 x 7 x 1; 45 = 3x3x5x1

∴ L.C.M. of 25,35 and 45 = 3x3x5x5x7=1575

6. 32 = 2x2x2x2x2x1; 40 = 2 x 2 x 2 x 5 x 1; 84 = 2x2x3x7x1

∴ L.C.M. of 32, 40 and 84 =2x2x2x2x2x3x5x1 = 3360

WBBSE Math Solutions Class 6

Question 16. Let’s find which pairs of numbers are mutually prime

47, 23;
25, 9;
49,35;
36, 54.

Solution:

WBBSE Solutions For Class 6 Maths Chapter 1 Fraction Pairs Of Numbers Are Mutuaaly Prime Examples

1. 47 = 47×1; 23 = 23 x 1

There are no common factors between 47 and 23.

∴ They are mutually prime.

2. 25 = 5 x 5 x 1; 9 = 3 x 3 x1

There are no common factors between 25 and 9.

∴ They are mutually prime.

3. 49 = 7 x 7 x 1; 39 = 3x13x1

There are no common factors between 49 and 39.

∴ They are mutually prime.

4. 36 = 2 x 2 x 3 x 3 x 1; 54 = 2x3x3x3x1

There are three common factors (2,3,3).

∴ 36 and 54 are not mutually prime.

Question 18. Let’s find H.C.F, and L.C.M. of the following numbers by short division method:

33 and 132,
90 and 144,
32, 40 and 72,
28, 49, 70

Solution:

WBBSE Solutions For Class 6 Maths Chapter 1 Fraction HCF And LCM Of The Numbers By Short Division Method

Question 19. Let’s find the least number which is divisible by 18, 24 and 42.
Solution:

First we have to find the L.C.M. of the least numbers.

L.C.M. = 2x2x2x3x3x7 = 504

∴ The required L.C.M = 504

Question 20. Let’s find the greatest number which when divided 45 and 60 does not leave any remainders.
Solution:

To find the greatest number we have to find the H.C.F. of 45 and 60.

∴ The required greatest number 15

Question 21. Let’s find the product of two numbers whose L.C.M. and H.C.F are 252 and 6 respectively.
Solution:

Product of two numbers = L.C.M. x H.C.F. = 252 x 6 = 1512

Question 22. If the H.C.F. and L.C.M. of two numbers are 8 and 280 and one of the numbers is 56, let’s find the other number.
Solution:

Product of two numbers = L.C.M. x H.C.F.

One number = $\frac{\text { L.C.M } \times \text { H.C.F }}{\text { other number }}=\frac{280 \times 8}{56}=40$

Question 23. Let’s write two numbers whose H.C.F. is 1.
Solution:

H.C.F. of two prime numbers is one. Similarly the H.C.F. of one prime number and one another number is one.

Question 24. Let’s find the maximum number of people among whom 48 rasogollas and 64 sandeshes are to be distributed in equal numbers.
Solution:

To find the number of people we have to find the H.C.F of 48 and 64.

H.C.F = 16

∴ The required number of people = 16.

WBBSE Math Solutions Class 6

Question 25. Bibhash and his friends decided to form a drama group of either 8 or 10 people. Let’s find the minimum number of people who will be able to form both types of groups.
Solution:

To find the number of people, we have to find the L.C.M. of 8 and 10.

WBBSE Solutions For Class 6 Maths Chapter 1 Fraction LCM Of 8 And 10

L.C.M. =2x2x2x5 = 40

∴ The required number of people is 40

Question 26. Panchayat has sent flower saplings to the students of class 6 of Jodunath Vidya Mandir for their school garden. It was found that if the saplings can be put in rows of 20, 24 and 30, then the number of sap lings in each row will be equal. Let’s find the minimum number of saplings that Panchayat sent to school.
Solution:

To find the minimum number of saplings, we have to find the L.C.M. of 20, 24 and 30.

L.C.M. of 20, 24, 30

L.C.M. = 2x2x2x3x5=120

WBBSE Solutions For Class 6 Maths Chapter 1 Fraction LCM Of 20 24 And 30

∴ The minimum required of flower saplings = 120

Question 27. The perimeter of the front wheel and rear wheel of an engine are 14 dcm and 35 dcm respectively. Let’s find the minimum distance the engine can cover before both wheels take complete revolutions together.
Solution:

To find the minimum distance we have to find the L.C.M of 14 and 25

WBBSE Solutions For Class 6 Maths Chapter 1 Fraction LCM Of 2 5 7

∴ L.C.M = 2x5x7 = 70.

∴ The required minimum distance = 70 dcm.

Question 28. For each of the following cases, let’s write two numbers whose:

H.C.F. is 7,
L.C.M. is 12,
H.C.F. is (put 1 digit number),
L.C.M. is (Put 1 digit number.)

Solution:

The required numbers are 7 and 14 as H.C.F. of 7 and 14 is 7.
The required numbers are 3 and 12 as L.C.M of 3 and 12 is 12.
The required numbers are 4 and 8. H.C.F. = 4
The required numbers are 3 and 6. L.C.M. = 6

Fraction Exercise 1.3

Question 1. I took a white rectangular piece of paper. I divided it horizontally into 3 equal parts and coloured. In the colours of our national flag, find the proportion of each colour.
Solution: Portion of saffron colour = 1/3 part

WBBSE Solutions For Class 6 Maths Chapter 1 Fraction Rectangular Piece Of paper

Portion of white colour = 1/3 part

Portion of green colour = 1/3 part

Question 2. Seema took a paper, same as mine, but coloured it different. Let’s find how and what portion she coloured.
Solution:

Portion of saffron colour = 2/3 part

WBBSE Solutions For Class 6 Maths Chapter 1 Fraction Portion Of Saffron Colour 2 by 3 part

1/2 part is coloured.

WBBSE Solutions For Class 6 Maths Chapter 1 Fraction 1 by 2 part Is Colored

Let’s colour 1/4 part

WBBSE Solutions For Class 6 Maths Chapter 1 Fraction 1 by 4 Part

Portion of green colour = 1/3 part

WBBSE Solutions For Class 6 Maths Chapter 1 Fraction 3 By 5 part Is Colored

3/5 part is coloured

WBBSE Solutions For Class 6 Maths Chapter 1 Fraction 3 By 4 part Is Colored

3/4 part is coloured

WBBSE Solutions For Class 6 Maths Chapter 1 Fraction 2 By 5 is Colored

2/5 is coloured

WBBSE Solutions For Class 6 Maths Chapter 1 Fraction 1 By 6 Part Is Colored

1/6 is coloured

WBBSE Solutions For Class 6 Maths Chapter 1 Fraction 2 B y part Is Colored

2/6 is coloured

WBBSE Solutions For Class 6 Maths Chapter 1 Fraction 4 By 10 part Is Colored

4/10 is coloured

WBBSE Solutions For Class 6 Maths Chapter 1 Fraction 1 By 2 part Is Colored (2)

1/2 is coloured

Question 3. Let me find what part of the mangoes I disposed of.

1/3 part of the number of mangoes collected are rotten.

In the picture 1/3 part of mangoes are circled with red —>

It is found, rotten mangoes are 1/3 part = 4 mangoes.

WBBSE Solutions For Class 6 Maths Chapter 1 Fraction 2 By 3 Part 8 mangoes

From picture it is found 2/3 part of mangoes are fish.

∴ 2/3 part of total number of mangoes = 8 mangoes.

Hence, from picture it is found, that 4 mangoes are rotten and 8 mangoes are fresh.

WBBSE Math Solutions Class 6

Question 4. Let’s find out from my picture in which parts the different fruits are present.
Solution:

WBBSE Solutions For Class 6 Maths Chapter 1 Fraction 1 By 4 Part Fruits

In the picture, out of the 4 equal parts of the fruits, the apples are 1/4 part.

That is, in the picture number of apples are 1/4 part of the total number of furits.

In the picture, mango is 3 part of 4 equal parts.

WBBSE Solutions For Class 6 Maths Chapter 1 Fraction 3 By 4 Part Of Total Number Of Fruits

That is, in the picture mango is 3/4 part of the total number of fruits.

In the picture aside, red flowers are 2 part of 3 equal parts of flowers.

Thus, 10/15 part of flowers are red flowers.

White flowers are 1 part of 3 equal parts of flowers.

Thus, 5/15 part of flowers are white flowers.

WBBSE Solutions For Class 6 Maths Chapter 1 Fraction 10 By 15 part

∴ 2/5 parts = 2 parts out of 5 equal parts = 4 balls.

WBBSE Solutions For Class 6 Maths Chapter 1 Fraction 1 By 4 p[art Of Equal Parts

∴ 1/4 parts = 1 part out of 4 equal parts = 2 coins.

Question 5. I Painted ted one of the window panes and a part of the other pane. Let me find, how much has been painted.
Solution:

Coloured → $\left(2+\frac{1}{2}\right)=2 \frac{1}{2}=\frac{2 \times 2+1}{2}=\frac{5}{2}$

WBBSE Solutions For Class 6 Maths Chapter 1 Fraction 13 By 6 part Is Colored

Coloured → $\left(2+\frac{1}{6}\right)=2 \frac{1}{6}=\frac{2 \times 6+1}{6}=\frac{13}{6}$

Question 6. Let’s find four equivalent fractions for each of the fractions 2/3 and 5/9
Solution:

Four equivalent fractions of 2/3 are

4/6 [Multiply numerator and denominator by 2]
6/9 [Multiply numerator and denominator by 3]
8/12 [Multiply numerator and denominator by 4]
10/15 [Multiply numerator & denominator by 5]

Four equivalent fractions of 5/9 are

10/18 [Multiply numerator and denominator by 2]
15/27 [Multiply numerator and denominator by 3]
20/36 [Multiply numerator and denominator by 4]
25/45 [Multiply numerator and denominator by 5]

WBBSE Class 6 Maths Solutions Fraction Exercise 1.4

Question 1. Let us write in blank boxes:
Solution:

7/10 part is coloured.
5/6 part is not coloured.
2/5 part is coloured.

Question 2. Let us write in blank boxes:

1. When we say half bread, what part of the whole bread we see?
Solution:

When we say half bread, what part of the whole bread we see 1-1/2 = 1/2 part.

2. I have a big chocolate. Dividing it into 8 pieces, I gave 3 pieces to my sister, 2 pieces to my brother and rest I ate myself. Let’s find what parts of chocolate we individually ate.
Solution:

My sister got 2/8 part of the chocolate.

My brother got 2/8 part = 1/4 part

∴ Remaining part = $\left[1-\left(\frac{3}{8}+\frac{1}{4}\right)\right]$

= $\left[1-\left(\frac{3+2}{8}\right)\right]$

= $\left[1-\frac{5}{8}\right]=\frac{8-5}{8}=\frac{3}{8} \text { part }$

∴ Part of chocolate I received = 3/8 part.

3. Let us write all numbers from 1 to 10. Out of these numbers what fraction are the prime numbers?
Solution:

The prime numbers between 1 to 10 are 2,3,5, and 7.

Total number=4 Fractional part of the prime number = 4/10 = 2/5 part.

WBBSE Class 6 Maths Solutions

4. There are few oranges in a basket. After giving half of the oranges to my grandfather, 2 oranges are left. Let’s find how many oranges were there in the basket.
Solution:

Number of oranges = $2 \div \frac{1}{2}=2 \times \frac{2}{1}$ = 4

5. In two glasses of equal size equal amount of lemon drink was prepared. One glass contains 1/5 part of sugar and the other contains 2/7 part of sugar. Let’s find without taking the drinks, which glass of lemon drink tastes sweeter.
Solution:

WBBSE Solutions For Class 6 Maths Chapter 1 Fraction Lemon Drint Part 2 By 7

∴ $\frac{2}{7}$ part of sugar glass of lemon drink testes sweeter.

6. $\frac{5}{7}$ part of school gate has been painted. Let’s find out which part is still to be painted.
Solution:

⇒ $\frac{5}{7}$ part of school gate has been painted.

Rest part of the gate which is still to be painted = $\text { 1. }-\frac{5}{7}=\frac{7-5}{7}=\frac{2}{7} \text { part }$

7. I have Rs. 20.1 spent Rs. 5. Let’s find what part I spent and what part is left with me.
Solution:

Total amount = Rs. 20.

I spent Rs. 5 = $\frac{5}{20}$ = $\frac{1}{4}$ part.

Remaianing part = $\left(1-\frac{1}{4}\right) \text { part }=\left(\frac{4-1}{4}\right) \text { part }=\frac{3}{4} \text { part. }$

8. Rajia has 36 berries. She will give me — part of her berries. Let me find out how many berries she will give me.
Solution:

Rajia has 36 berries.

She will give me = $\frac{2}{3}$ of 36 berries = (2 x 12) = 24 berries.

Question 4. Let us mark proper fraction by ; improper fraction by and mixed fraction by Δ
Solution:

Let us mark

Proper fraction
Imoroper fraction
Mixed fraction Δ

WBBSE Solutions For Class 6 Maths Chapter 1 Fraction Proper Fractions

Proper fraction: $\frac{1}{5}, \frac{2}{7}, \frac{3}{8}, \frac{6}{13}, \frac{1}{9}, \frac{2}{5}, \frac{5}{9}, \frac{4}{17}, \frac{11}{12}, \frac{3}{7}$

Improper fraction: $\frac{15}{13}, \frac{29}{19}, \frac{23}{17}$

⇒ $9 \frac{14}{15}, 1 \frac{22}{25}, 11 \frac{1}{9}, 2 \frac{3}{4}, 3 \frac{5}{11}$

Mixed fraction: $9 \frac{14}{15}, 1 \frac{22}{25}, 11 \frac{1}{9}, 2 \frac{3}{4}, 3 \frac{5}{11}$

WBBSE Class 6 Maths Solutions

Question 5. Let us write three equivalent fractions for each of the following

$\frac{1}{5}$
$\frac{2}{5}$
$1 \frac{1}{3}$
$6 \frac{1}{6}$
$3 \frac{4}{5}$

Solution:

1. $\frac{1}{5}$

1st equivalent fraction of $\frac{1}{5}=\frac{2}{10}$

2nd equivalent fraction of $\frac{1}{5}=\frac{3}{15}$

3rd equivalent fraction of $\frac{1}{5}=\frac{4}{20}$

2. $\frac{2}{5}$

1st equivalent fraction of $\frac{2}{5}=\frac{4}{10}$

2nd equivalent fraction of $\frac{2}{5}=\frac{6}{15}$

3rd equivalent fraction of $\frac{2}{5}=\frac{8}{20}$

3. 1$\frac{1}{3}=\frac{4}{3}$

1st equivalent fraction of $\frac{4}{3}=\frac{8}{6}$

2nd equivalent fraction of $\frac{4}{3}=\frac{12}{9}$

3rd equivalent fraction of $\frac{4}{3}=\frac{16}{12}$

4. 6$\frac{1}{6}=\frac{37}{6}$

1st equivalent fraction of $\frac{37}{6}=\frac{74}{12}$

2nd equivalent fraction of $\frac{37}{6}=\frac{111}{18}$

3rd equivalent fraction of $\frac{37}{6}=\frac{148}{24}$

5. 3$\frac{4}{5}=\frac{19}{5}$

1st equivalent fraction of $\frac{19}{5}=\frac{38}{10}$

2nd equivalent fraction of $\frac{19}{5}=\frac{57}{15}$

3rd equivalent fraction of $\frac{19}{5}=\frac{76}{20}$

Question 6. Let us express the following in their lowest form:

$\frac{28}{49}$
$\frac{54}{81}$
$\frac{72}{108}$
$\frac{243}{405}$
$\frac{165}{180}$

Solution:

1. $\frac{28}{49}=\frac{4}{7}$

2. $\frac{54}{81}=\frac{18}{27}=\frac{2}{3}$

3. $\frac{72}{108}=\frac{2}{3}$

4. $\frac{243}{405}=\frac{3}{5}$

5. $\frac{165}{180}=\frac{11}{12}$

WBBSE Class 6 Maths Solutions

Question 7. Let us arrange the following fractions from smaller to bigger in ascending order

1. $\frac{7}{2}, \frac{4}{4}, \frac{7}{5}$
Solution:

L.C.M. of 2,4,5=20

⇒ $\frac{7}{2}=\frac{7 \times 10}{2 \times 10}=\frac{70}{20}$

⇒ $\frac{7}{4}=\frac{7 \times 5}{4 \times 5}=\frac{35}{20}$

⇒ $\frac{7}{5}=\frac{7 \times 4}{5 \times 4}=\frac{28}{20}$

∴ $\frac{28}{20}<\frac{35}{20}<\frac{70}{20}$

i.e., $\frac{7}{5}<\frac{7}{4}<\frac{7}{2}$

2. $5 \frac{3}{4}, 5 \frac{5}{9}, 5 \frac{8}{12}$
Solution:

⇒ $5 \frac{3}{4}, 5 \frac{5}{9}, 5 \frac{8}{12}$

or, $\frac{23}{4}, \frac{50}{9}, \frac{68}{12}$

L.C.M of 4,9 and 12=36

⇒ $\frac{23}{4}=\frac{23 \times 9}{4 \times 9}=\frac{207}{36}$

⇒ $\frac{50}{9}=\frac{50 \times 4}{9 \times 4}=\frac{200}{36}$

⇒ $\frac{68}{12}=\frac{68 \times 3}{12 \times 3}=\frac{204}{36}$

⇒ $\frac{200}{36}<\frac{204}{36}<\frac{207}{36}$

or, $\frac{200}{36}<\frac{204}{36}<\frac{207}{36}$

or, $\frac{50}{9}<\frac{68}{12}<\frac{23}{4}$

∴ $5 \frac{5}{9}<5 \frac{8}{12}<5 \frac{3}{4}$

3. $1 \frac{1}{5}, 1 \frac{1}{7}, 1 \frac{1}{8}$
Solution:

⇒ $1 \frac{1}{5}, 1 \frac{1}{7}, 1 \frac{1}{8}$

or, $\frac{6}{5}, \frac{8}{7}, \frac{9}{8}$

L.C.M. of 5, 7, 8 = 280

⇒ $\frac{6}{5}=\frac{6 \times 56}{5 \times 56}=\frac{336}{280}$

⇒ $\frac{8}{7}=\frac{8 \times 40}{7 \times 40}=\frac{320}{280}$

⇒ $\frac{9}{8}=\frac{9 \times 35}{8 \times 35}=\frac{315}{280}$

∴ $\frac{315}{280}<\frac{320}{280}<\frac{336}{280}$

i.e., $\frac{9}{8}<\frac{8}{7}<\frac{6}{5}$

⇒ $1 \frac{1}{8}<1 \frac{1}{7}<1 \frac{1}{5}$

Class 6 Math WBBSE Solutions

4. $\frac{1}{3}, \frac{4}{5}, \frac{7}{15}$.
Solution:

L.C.M of 3,5 and 15=15

⇒ $\frac{1}{3}=\frac{1 \times 5}{3 \times 5}=\frac{5}{15}$

⇒ $\frac{4}{5}=\frac{4 \times 3}{5 \times 3}=\frac{12}{15}$

⇒ $\frac{7}{15}=\frac{7 \times 1}{15 \times 1}=\frac{7}{15}$

∴ $\frac{5}{15}<\frac{7}{15}<\frac{12}{15}$

i.e., $\frac{1}{3}<\frac{7}{15}<\frac{4}{5}$

5. $\frac{5}{7}, \frac{3}{4}, \frac{1}{4}$
Solution:

L.C.M. of 7,4 and 4=28

⇒ $\frac{5}{7}=\frac{5 \times 4}{7 \times 4}=\frac{20}{28}$

⇒ $\frac{3}{4}=\frac{3 \times 7}{4 \times 7}=\frac{21}{28}$

⇒ $\frac{1}{4}=\frac{1 \times 7}{4 \times 7}=\frac{7}{28}$

∴ $\frac{7}{28}<\frac{20}{28}<\frac{21}{28}$

∴ $\frac{1}{4}<\frac{5}{7}<\frac{3}{4}$

6. $3 \frac{1}{2}, 7 \frac{5}{9}, 7 \frac{1}{5}$
Solution:

⇒ $3 \frac{1}{2}, 7 \frac{5}{9}, 7 \frac{1}{5}$

⇒ $\frac{7}{2}, \frac{68}{9}, \frac{36}{5}$

L.C.M. of 2,9 and 5=90

⇒ $\frac{7}{2}=\frac{7 \times 45}{2 \times 45}=\frac{315}{90}$

⇒ $\frac{68}{9}=\frac{68 \times 10}{9 \times 10}=\frac{680}{90}$

⇒ $\frac{36}{5}=\frac{36 \times 18}{5 \times 18}=\frac{648}{90}$

⇒ $\frac{315}{90}<\frac{648}{90}<\frac{680}{90}$

or, $\frac{7}{2}<\frac{36}{5}<\frac{68}{9}$

∴ $3 \frac{1}{2}<7 \frac{1}{5}<7 \frac{5}{9}$

Class 6 Math WBBSE Solutions

7. $\frac{1}{8}, \frac{7}{10}, \frac{3}{5}$
Solution:

L.C.M. of 8,10 and 5=40

⇒ $\frac{1}{8}=\frac{1 \times 5}{8 \times 5}=\frac{5}{40}$

⇒ $\frac{7}{10}=\frac{7 \times 4}{10 \times 4}=\frac{28}{40}$

⇒ $\frac{3}{5}=\frac{3 \times 8}{5 \times 8}=\frac{24}{40}$

⇒ $\frac{5}{40}<\frac{24}{40}<\frac{28}{40}$

∴ $\frac{1}{8}<\frac{3}{5}<\frac{7}{10}$

8. $3 \frac{1}{2}, 3 \frac{5}{9}, 3 \frac{1}{5}$
Solution:

⇒ $3 \frac{1}{2}, 3 \frac{5}{9}, 3 \frac{1}{5}$

or, $\frac{7}{2}, \frac{32}{9}, \frac{16}{5}$

L.C.M. of 2,9 and 5=90

⇒ $\frac{7}{2}=\frac{7 \times 45}{2 \times 45}=\frac{315}{90}$

⇒ $\frac{32}{9}=\frac{32 \times 10}{9 \times 10}=\frac{320}{90}$

⇒ $\frac{16}{5}=\frac{16 \times 18}{5 \times 18}=\frac{288}{90}$

⇒ $\frac{288}{90}<\frac{315}{90}<\frac{320}{90}$

or, $\frac{16}{5}<\frac{7}{2}<\frac{32}{9}$

∴ $3 \frac{1}{5}<3 \frac{1}{2}<3 \frac{5}{9}$

Question 8. Let us find the values of the following

1. $\frac{2}{7}+\frac{2}{3}+1 \frac{1}{2}$
Solution:

= $\frac{2}{7}+\frac{2}{3}+\frac{3}{2}$

= $\frac{2 \times 6+2 \times 14+3 \times 21}{42}$

= $\frac{12+28+63}{42}=\frac{103}{42}$

= $2 \frac{19}{42}$

WBBSE Solutions For Class 6 Maths Chapter 1 Fraction LCM Of 7x3x2

2. $1 \frac{2}{5}-\frac{3}{8}+\frac{1}{4}$
Solution:

L.C.M Of 5, 8 and 4

= $\frac{7}{5}-\frac{3}{8}+\frac{1}{4}$

= $\frac{7 \times 8-3 \times 5+1 \times 10}{40}$

= $\frac{56-15+10}{40}$

= $\frac{51}{40}$=1 $\frac{11}{40}$

Class 6 Math WBBSE Solutions

WBBSE Solutions For Class 6 Maths Chapter 1 Fraction LCM Of 4 5 2

3. $\frac{2}{5}+\frac{3}{8}-\frac{1}{4}$
Solution:

L.C.M Of 5, 8 and 4

= $\frac{2 \times 8+3 \times 5-1 \times 10}{40}$

= $\frac{16+15-10}{40}=\frac{21}{40}$

4. $7-3 \frac{1}{8}-2 \frac{1}{3}$
Solution:

L.C.M of 1, 8 and 3 = 24

= $\frac{7}{1}-\frac{25}{8}-\frac{7}{3}$

= $\frac{24 \times 7-25 \times 3-7 \times 8}{24}$

= $\frac{168-75-56}{24}$

= $\frac{37}{24}=1 \frac{13}{24}$

WBBSE Solutions For Class 6 Maths Chapter 1 Fraction LCM Of 2x2x2x3

5. $\frac{4}{5}+\frac{5}{8}-1 \frac{1}{3}$
Solution:

L.C.M of 5, 8 and 3

= $\frac{4}{5}+\frac{5}{8}-\frac{4}{3}$

= $\frac{4 \times 24+5 \times 15-4 \times 40}{120}$

= $\frac{96+75-160}{120}$

= $\frac{171-160}{120}$

= $\frac{11}{120}$

WBBSE Solutions For Class 6 Maths Chapter 1 Fraction 2x2x2x3x5

L.C.M = 2x2x2x3x5 =120

Class 6 Math WBBSE Solutions

6. $1 \frac{3}{10}+1 \frac{4}{5}-1 \frac{1}{4}$
Solution:

L.C.M of 10, 5 and 4

= $\frac{13}{10}+\frac{9}{5}-\frac{5}{4}$

= $\frac{13 \times 2+9 \times 4-5 \times 5}{20}$

L.C.M = 2 x2 x 5 = 20

= $\frac{26+36-25}{20}$

= $\frac{37}{20}=1 \frac{17}{20}$

7. $2 \frac{5}{6}-1 \frac{8}{9}+1 \frac{3}{4}$
Solution:

L.C.M of 6, 9, 4

= $\frac{17}{6}-\frac{17}{9}+\frac{7}{4}$

= $\frac{17 \times 6-17 \times 4+7 \times 9}{36}$

= $\frac{102-68+63}{36}$ L.C.M = 2 x 2x 3x 3 = 36

WBBSE Solutions For Class 6 Maths Chapter 1 Fraction LCM Of 6x9x4

= $\frac{97}{36}=2 \frac{25}{36}$

8. $4 \frac{1}{7}$+2 $\frac{2}{5}-5$
Solution:

L.C.M of 7, 5 and 1

= $\frac{29}{7}+\frac{12}{5}-\frac{5}{1}$ = 7 x 5x 1 = 35

= $\frac{29 \times 5+12 \times 7-5 \times 35}{35}$

= $\frac{145+84-175}{35}$

= $\frac{54}{35}$

= $1 \frac{19}{35}$

Question 9. Let us work out

1. During tiffin hours of school $\frac{1}{4}$ part of drinking water of a tank was used. Another $\frac{1}{3}$ part of water was used when school got over. Let’s find what part of water is still left in the tank.
Solution:

Let total amount of water = 1 part

Water used = $\left(\frac{1}{4}+\frac{1}{3}\right) \text { part }=\frac{3+4}{12}=\frac{7}{12} \text { part }$

∴ Remaining water = $1-\frac{7}{12}=\frac{12-7}{12}=\frac{5}{12}$ part of the thank.

2. Today, I took a piece of cake for tiffin. My two friends Ayush ar Sabana respectively ate $\frac{1}{3}$ part and $\frac{2}{5}$ part of the cake. I ate the remaining part. Let me find who ate the most part of cake.
Solution:

Let total part of cake = 1

Ayush and Sabana ate = $\frac{1}{3}+\frac{2}{5}$

= $\frac{5+6}{15}=\frac{11}{15}$

Remaining part = $1-\frac{11}{15}=\frac{15-11}{15}=\frac{4}{15}$ part

Now Ayush, Sabana and I get $\frac{1}{3}, \frac{2}{5} and \frac{4}{15}$ parts of the cake respectively.

⇒ $\frac{1}{3}=\frac{1 \times 5}{3 \times 5}=\frac{5}{15}$

⇒ $\frac{2}{5}=\frac{2 \times 3}{5 \times 3}=\frac{6}{15}$

⇒ $\frac{4}{15}=\frac{4 \times 1}{15 \times 1}=\frac{4}{15}$

∴ $\frac{2}{5}$ is the greatest part.

Sabana ate most part of the cake.

Class 6 Math WBBSE Solutions

3. Out of 25 bighas of land Ratanbabu cultivated jute in 16 bigha But, out of 15 bighas of land, Ushadebi cultivated jute in 8 bighas. Let work out to find in what part of their land Ratanbabu and Ushade cultivated jute and who cultivated jute more.
Solution:

Ratanbabu cultivated jute in $\frac{16}{25}$ part of his land. Ushadebi cultivated jute in $\frac{8}{15}$ part of her land.

⇒ $\frac{16}{25}=\frac{16 \times 3}{25 \times 3}=\frac{48}{75}$ L.C.M. of 25 and15

⇒ $\frac{8}{15}=\frac{8 \times 5}{15 \times 5}=\frac{40}{75}$

⇒ $\begin{array}{ccc}
3 & 25, & 15 \\
5 & 25, & 5 \\
5 & 5, & 1 \\
& 1, & 1
\end{array}$

∴ L.C.M. = 3x5x5 = 75

∴ Ratanbabu cultivated jute more.

4. I have a 15-metre long white ribbon. I cut off $\frac{1}{3}$ part from it. What part of ribbon is left? Let’s find the length of the remaining part of the ribbon.
Solution:

Total length of white ribbon = 15 metres.

I take = $\frac{1}{3}$ x 15 m = 5 metre.

Remaining part = $1-\frac{1}{3}=\frac{3-1}{3}=\frac{2}{3}$ part

And the length of the remaining part of the ribbon = (15 – 5) m = 10 metres.

Class 6 Math Solutions WBBSE Decimal Fractions

We decided to decorate the front portion of the dais of our village club with some coloured tiles of equal size. 2 red tiles were given by me, 3 yellow tiles by Reba, 4 green tiles by Samsur and 1 white tile by Tina.

We all gathered before the dais with a total of 10 tiles.

Out of 10 tiles I gave 2 tiles → $\frac{2}{10}$ part = 0.2 part = .2

Out of 10 tiles, Reba gave 3 tiles → $\frac{3}{10}$ part = 0.3 part = .3

Out of 10 tiles, Samsur gave 4 tiles →$\frac{4}{10}$part = 0.4 part = .4

Tena give 1 tile → $\frac{1}{10}$ part = 0.1 part = .1

Hence, $\begin{array}{cc}
\text { Decimal } & \text { Tenths }\left(\frac{1}{10}\right) \\
\text { number } & \\
0.2 & 2 \\
0.3 & 3 \\
0.4 & 4 \\
0.1 & 1
\end{array}$

Now, few tiles on floor of the room of our club are broken. We need to replace them.

First, we do arrange for 100 tiles.

I shall give 12 tiles out of those 100 tiles.

I shall give $\frac{12}{100}$ part = 0.12 part or point one two parts. .

Rohit will give 38 tiles. Thus, he will give $\frac{38}{100}$part = 0.38 part.

Mira will give $\frac{34}{100}$ part = 0.34 part.

Class 6 Math Solutions WBBSE Fraction Exercise 1.5

Question 1. Let us express the coloured portion in terms of decimal fraction

WBBSE Solutions For Class 6 Maths Chapter 1 Fraction Decimal Fraction
Solution:

1. $\frac{3}{10}$ part = 0.3

2. $\frac{36}{100}$ part = 0.36

3. $\frac{60}{100}$ = 0.6

4. $\frac{16}{100}$ part = 0.6

5. $\frac{27}{100}$ part 0.27

Question 2.

1. Let us colour 0.15 part green and 0.53 part yellow. Find total coloured part.

WBBSE Solutions For Class 6 Maths Chapter 1 Fraction Colored Part
Solution:

0.15 part green and 0.53 part yellow.

∴ Total coloured part = 0.15 + 0.53 = 0.68.

WBBSE Solutions For Class 6 Maths Chapter 1 Fraction

2. Let us colour 0.33 part blue first and 0.15 part red. Find the total uncoloured parts.

WBBSE Solutions For Class 6 Maths Chapter 1 Fraction Colored Part
Class 6 Math WBBSE Solution:

0.33 part blue and 0.1 5 part red

∴ Total coloured part = 0.33 + 0.1 5 = 0.48

Total uncoloured part = 1-0.48 = 0.52

WBBSE Solutions For Class 6 Maths Chapter 1 Fraction Uncolored Part 0.52

Question 3. Let’s prepare a place value table for the following numbers and write the numbers in words.

1. 27.9
Solution:

=27+ $\frac{9}{10}$= Twenty seven point nine

= Two tens seven unit and nine tenth

2. 1.28
Solution:

= $1+\frac{2}{10}+\frac{8}{100}$

= one point two eight = one unit and two-tenth eight hundredths.

3.65.134
Solution:

65.134 = $60+5+\frac{1}{10}+\frac{3}{100}+\frac{4}{1000}$= Sixty five point one three four

= Six tens five units one tenth three hundredth four thousandths.

4. 42.009
Solution:

42.009+40+2+$\frac{9}{1000}$= Forty two point zero zero nine

= Four tens two units nine thousandths.

Class 6 Math Solutions WBBSE

5. 38.205
Solution:

= $30+8+\frac{2}{10}+\frac{5}{1000}$

= Thirty-eight point two zero five

= Three tens eight units and two-tenth five thousandths.

6. 4003.08
Solution:

4003.08 = Four thousand three point zero eight

= Four thousand three units eight hundedths.

7. 712.5
Solution:

712.5 = 700+10+2+$\frac{5}{10}$

= Seven hunderd twelve point five

= Seven hundred one tenth two units five tenths

8. 45.06
Solution:

45.06 = 40+5+$\frac{6}{100}$

= Forty-five point zero six

= Four tens five units six hundredths.

Question 4. Let’s convert the following decimal numbers to fractions:

0.3,
0.21,
0.039,
5.4,
102.035.

Solution:

0.3 = $\frac{3}{10}$
0.21 = $\frac{21}{100}$
0.039 = $\frac{39}{1000}$
5.4 = $\frac{54}{10}$ = $\frac{27}{5}$
102.035 = $\frac{102035}{1000}$ = $\frac{20407}{200}$ = 102$\frac{7}{200}$

Question 5. Let’s arrange the following in increasing order (ascending order):

1. 0.534, 0.52, 5.34, 0.513
Solution: In ascending order 0. 513, 0.52; 0.534, 5.34

2. 0.536, 0.335, 0.3354, 0.52
Solution: In ascending order 0.335; 0.3354; 0.52; 0.536

3. 2.0, 2.005, 20.05, 2.5
Solution: In ascending order 2.0; 2.005; 2.5; 20.05

Question 6. Let’s arrange the following in decreasing order (descending order):

1. 13.3,11.3,1.33,2.31
Solution: In descending order 13.3,11.3, 2.31,1.33

2. 3.007, 3.07, 37.30, 7.13
Solution: In descending order 37.30; 7.13, 3.07, 3.007

3. 0.88, 0.45, 8.45, 0.8217
Solution: In descending order 8.45, 0.88, 0.8217, 0.45

Question 7.

$WBBSE Solutions For Class 6 Maths Chapter 1 Fraction Decimal fraction and Place Value Of 8$

Solution:

WBBSE Solutions For Class 6 Maths Chapter 1 Fraction Decimal Fraction And Place Value Of 8-1

Question 8.

WBBSE Solutions For Class 6 Maths Chapter 1 Fraction Place Value Of 5 And Decimal Number-1

Solution:

WBBSE Solutions For Class 6 Maths Chapter 1 Fraction Place Value Of 5 And Decimal Number

Question 9. Let’s put >, = or < in the boxes given below.

1. 5.0 □ 0.5
Solution: 5.0>0.5

2. 72.1 □ 72.10
Solution: 72.1 =72.10

3. 68.5 □ 68.52
Solution: 68.5 < 68.52

4. 72.93 □ 729.3
Solution: 72.93 <729.3

5. 42.6 □ 42.600
Solution: 42.6 = 42.600

6. 2.33 □ 3.22
Solution: 42.6 = 42.600

7. 924 □ 924.0
Solution: 924 = 924.00

8. □ □ □ (Let us write own decimal numbers)
Solution: 84.67 > 72.67

Question 10. Let’s express the following in decimal numbers:

1. Six tenths
Solution: $\frac{6}{10}$ = 0.6

2. Nine hundredths
Solution: $\frac{9}{100}$ = 0.09

3. Two thousandths
Solution: $\frac{2}{1000}$ = 0.02

4. Two hundred three point four-five
Solution: 203.45

5. Four thousand two units five thousandths
Solution: 4002.005

6. Six hundred twenty-nine point zero five
Solution: 629.005

7. 2 + $\frac{3}{10}$
Solution: 2 +0.3 = 2.3

8. 10+7+$\frac{8}{1000}$
Solution: 17.008

Question 11. I had Rs. 5. I bought a pen for Rs. 3.50. Let’s find out how much money is left.
Solution: Remaining money = Rs. 5 – Rs. 3.50 = Rs. 1.50.

Question 12. What must be added to 2.75 to get 3?
Solution: Required number = 3 – 2.75 = 0.25.

Question 13. From a string of length 12.5 cm, Mira cuts off 8.5 cm. Let’s find, what portion of the string is left?
Solution: Remaining portion = 12.5 cm – 8.5 cm = 4 cm.

Question 14. The length of my exercise book is □ cm and its breadth is □ cm. Let me find the perimeter of my exercise book.
Solution: Let the length of the exercise book = 18 cm and the breadth of the exercise book = 12 cm.

∴ Perimeter = 2 x (length + breadth)

= 2 x (18 cm + 12 cm) = 2 x 30 cm = 60 cm.

Question 15. For an occasion at our house, father bought rice for Rs. 200, pulses for Rs. 125.50 and vegetables for Rs. 242.50. Let’s calculate the total amount of money spent by father.
Solution: Total amount of money spent = Rs. (200 + 125.50 + 242.50) = Rs. 568.

Question 16. In a long-jump competition, Sahil jumped 182.88 cm and Munna jumped 179.25 cm of length. Let’s find how much more did Sahil jump.
Solution: Sahil jumped more than Munna

= 182.88 cm – 179.25 cm = 3.63 cm.

Question 17. What must be added to 2.172 to get 5?
Solution: Required number = 5-2.172 = 2.828.

Question 18. 2.647 is subtracted from 4.15. How much is to be added to the result to get 10?
Solution: Subtraction of 2.647 from 4.15 = 4.150-2.647=1.503.

∴ Required number = 10 -1.503 = 8.497.

Question 19. Let’s find the values of the following

1. 0.07 + 0.09
Solution: 0.07 + 0.09 = 0.16

⇒ $\begin{aligned}
& 0.07 \\
& +0.09 \\
& \hline 0.16 \\
& \hline
\end{aligned}$

2. 4.11 + 1.6
Solution: 4.11 +1.6 = 5.71

⇒ $\begin{array}{r}
4.11 \\
+1.6 \\
\hline 5.71 \\
\hline
\end{array}$

3. 312.61 + 276.72
Solution: 589.33

⇒ $\begin{aligned}
& 312.61 \\
& \underline{276.72} \\
& \underline{589.33} \\
& \hline
\end{aligned}$

4. 5 – 0.555
Solution: 4.445

5. 27.56 + 14.69
Solution: 42.25

⇒ $\begin{aligned}
& 27.56 \\
& 14.69 \\
& \hline 42.25 \\
& \hline
\end{aligned}$

6. 4.3 + 3 – 6.4
Solution: 7.3 – 6.4 = 0.9

= 11.3-6.4 = 0.0

7. 3.36 – 4.62 + 2.18
Solution: 3.36 + 2.18- 4.62

= 5.54 – 4.62 = 0.92

8. 2.67 – 3.727 + 4.2.
Solution: 6.870 – 3.727 = 3.143

Fraction Exercise 1.6

Question 1. Let us try to place the following angles under right heading: 12°, 22.5°, 180°, 179°, 100°,39°,90°,69°,910
Solution:

Actute angles: 12°, 22.5°, 39°, 69°

Obtuse angles: 179°, 100°, 91°

Right angles: 90°

Straight angles: 180°

Question 2. Straight angle = 2 x □, hence straight angle is twice a right angle.
Solution:

Straight angle = 180° = 2 x 90° = 2 x right angle.

Question 3. Let’s find with which set of the lengths of line segments, triangles can be drawn:

1. 2 cm, 3 cm, 4 cm,
Solution:

Sum of 2 smaller sides = 2 cm + 3 cm = 5 cm is greater than the 3rd side (4 cm).

∴ Triangle can be drawn.

2. 4 cm, 3 cm, 7 cm
Solution:

Here sum of 2 smaller sides = 4 cm + 3cm = 7 cm is equal to the 3rd side (7 cm).

∴ Triangle can not be drawn.

3. 1 cm, 3 cm, 2 cm
Solution:

Here sum of 2 smaller sides = 1 cm + 2 cm= 3 cm is equal to the 3rd side (3 cm).

∴ Triangle can not be drawn.

4. □ cm, □ cm, □ cm.
Solution:

6 cm, 8 cm, 10 cm (let)

Here sum of 2 smaller sides = 6 cm + 8 cm = 14 cm is greater than the 3rd side (10 cm).

∴ Triangle can be drawn.

Question 4. Let us define actute angled and obtuse angled triangles.
Solution:

Acute angled triangle: A triangle whose three angles are acute, i.e., less than 90° is called an acute angled triangle.

Obtuse angled triangle: A triangle whose one angle is obtuse, i.e., greater than 90° but less than 180° is called obtuse angled triangle.

Question 5. Let’s measure the perimeter of the square in the figure.
Solution:

Side = 4 cm

Perimeter = 4 x side

= 4 x 4 cm

= 16 cm

WBBSE Solutions For Class 6 Maths Chapter 1 Fraction Perimeter Of The Square

Question 6. The diameter of circle in the figure = □ cm. Radius = □□□ cm = □ cm.
Solution:

Diameter = 4 cm

Radius = 4 ÷ 2 cm = 2 cm.

Fraction Exercise 1.7

Question 1. To construct an embankment on a certain part of river Ichhamati, 40 labourers are required to work for 35 days. If the work has to be finished in 28 days, let us find how many labourers will be needed.
Solution:

To finish the work:

⇒ $\begin{array}{ll}
\text { No. of days } & \text { No. of Labourers } \\
35 & 40 \\
28 & x \text { (let) }
\end{array}$

Number of days is inversely proportional to number of labourers.

∴ $\frac{40}{x}$ = $\frac{28}{35}$

∴ 28 x = 40 x 35

X = $\frac{40 \times 35}{28}$ = 50

∴ Number of labourers = 50

To construct an embarkment.

To construct an embarkment in 35 days, number of labourers required 40

To construct an embarkment in 1 day number of labourers required is 40×35

To construct an embarkment in 28 days number of labourers required $\frac{40 \times 35}{28}$ = 50

∴ Number of labourers required = 50

Question 2. Rajeeb, Debangana, Masum and Tajmira can do 150 sums in 6 days. Let us find in how many days Rajeeb and Tajmira will finish 250 sums if each of them does equal number of sums everyday.
Solution:

WBBSE Solutions For Class 6 Maths Chapter 1 Fraction Number of Days Required

4 persons can do 150 sums in 6 days.

1 person can do 150 sums in 6 x 4 days

1 persons can do 1 sum in $\frac{6 \times 4}{150}$ days

2 persons can do 1 sum in $\frac{6 \times 4}{150 \times 2}$ days.

2 persons can do 250 sums in $\frac{6 \times 4 \times 250}{150 \times 2}$ = 20 days.

∴ Number of days required = 20.

Question 3. Two labourers can polish 1/3 of a door in 1 day. Let us find how many labourers are needed to polish 2/3 of a door in 2 days.
Solution:

WBBSE Solutions For Class 6 Maths Chapter 1 Fraction Number Of Labourers Required

In 1 day $\frac{1}{3}$ part of the door is polished by 2 labourers.

In 1 day full part of the door is polished by 2 x 3 labourers.

In 2 days full part of the door is polished by $\frac{2 \times 3}{2}$ labourers.

In 2 days $\frac{2}{3}$ part of the door is polished by $\frac{2 \times 3}{2} \times \frac{2}{3}$ labourers = 2 labourers.

∴ Number of labourers required = 2

Question 4. For a mid-day meal of 500 students, 175 kg of rice is required for a week. After 75 kg of rice has been used, let us find how long remaining rice will last for 400 students.
Solution:

WBBSE Solutions For Class 6 Maths Chapter 1 Fraction Number Of Days Required Is 5

500 students can eat 175 kg rice in 7 days.

1 student can eat 175 kg rice in 7 x 500 days.

1 student can eat 1 kg rice in $\frac{7 \times 500}{175}$ days.

400 student can eat 1 kg rice $\frac{7 \times 500}{400 \times 175}$ days

400 student can eat 100 kg rice in $\frac{7 \times 500 \times 400}{400 \times 175}$ =5 days.

∴ Number of days required = 5.

Question 5. 4 tractors are required to cultivate 360 bighas of land in 20 days. Let us find how many tractors are needed to cultivate 1800 bighas of land in 10 days.
Solution:

WBBSE Solutions For Class 6 Maths Chapter 1 Fraction Number Of Tractors Required Is 40

360 bighas of land are cultivated in 20 days by 4 tractors.

1 bigha of land are cultivated in 1 days by $\frac{4}{360}$ tractors.

1 bigha of land are cultivated in 1 day by $\frac{4 \times 20}{360}$ tractors.

180 bighas of land are cultivated in 1 day by $\frac{4 \times 20 \times 1800}{360}$

1800 bighas of land are cultivated in 10 days by $\frac{4 \times 20 \times 1800}{360 \times 10}$=40 tractors.

∴ Number of tractors required is 40.

Question 6. In a fair, if 12 generators work for 6 hours a day, then 7 litres of deposited oil is required. If generators run for 4 hours a day for 9 days, then let us find how many generators can run with that remaining amount of oil.
Solution:

WBBSE Solutions For Class 6 Maths Chapter 1 Fraction Number Of Generators Required Is 14

For 6 hours a day in 7 days oil deposited for 12 generators.

For 6 hours a day in 1 day oil deposited for 12×7 generators.

For 1 hour a day in 1 day oil deposited for 12 x 7 x 6 generators.

For 4 hours a days in 1-day oil deposited for $\frac{12 \times 7 \times 6}{4}$ generators.

For 4 hours a day in 9 days oil deposited for $\frac{12 \times 7 \times 6}{9 \times 4}$ =14 generators.

∴ 14 generators can run with that remaining amount of oil.

Question 7. 15 Vans can carry 75 quintals of vegetables in 40 mins. Let us find out how long will 20 vans take to carry 100 quintals of vegetables.
Solution:

WBBSE Solutions For Class 6 Maths Chapter 1 Fraction Time Required Is 20 Minutes

15 vans can carry 75 quintals of vegetables in 40 minutes.

1 van can carry 75 quintals of vegetables in 40 x 15 minutes.

1 van can carry 100 quintals of vegetables in $\frac{40 \times 15}{75}$minutes.

20 vans can carry 100 quintals of vegetables in $\frac{40 \times 15 \times 100}{20 \times 75}$ minutes

∴ Time required = 20 minutes.

Question 8. 150 kg of wheat is stored for 20 hostel boys. But 30 kg wheat was wasted and 5 boys went home. With the remaining amount of wheat, let us find for how long the remaining boys be fed.
Solution:

WBBSE Solutions For Class 6 Maths Chapter 1 Fraction Number Of Days Required Is 32 Days

150 kg of wheat is stored for 20 boys for 30 days.

150 kg of wheat is stored for 1 boy for 30 x 20 days.

1 kg of wheat is stored for 1 boy for $\frac{30 \times 20}{150}$ days.

120 kg of wheat is stored for 1 boy for $\frac{30 \times 20 \times 120}{150}$ days.

120 kg of wheat is stored for 15 boys for $\frac{30 \times 20 \times 120}{15 \times 150}$ =32 days.

NEET General Organic Chemistry Quantitative Analysis Notes

September 14, 2023 by Haritha

Quantitative Analysis

Estimation of carbon and hydrogen by Liebig’s combustion method

The quantitative analysis deals with the determination of percentage of various elements. The following methods are employed for the determination of the percentage composition of elements present in the organic compounds.

Read And Learn More: NEET General Organic Chemistry Notes, Question And Answers

Percentage of hydrogen in the compound = $\frac{2}{18} \times \frac{\text { mass of water } \times 100}{\text { mass of organic compound }}$

Percentage of carbon in the compound = $\frac{12 \times \text { mass of carbon dioxide } \times 100}{44 \times \text { mass of organic compound }}$

Depending upon the composition of organic compounds, the following modifications are made in the estimation of carbon and hydrogen by Liebigs method.

Estimation of nitrogen by Kjeldahl’s method

Nitrogen By Kjeldahl’s Method Principle is based on the quantitative conversion of nitrogen of the organic compound to ammonium sulphate by sulphuric acid. The reaction product is treated with an alkali and the ammonia released is determined. From this the amount of nitrogen in the organic compound is calculated.

⇒ $\underset{\text { organic compound }}{[\mathrm{C}, \mathrm{H}, \mathrm{N}, \mathrm{S}]} \underset{\text { heat }}{\stackrel{\text { cooc. } \mathrm{H}_2 \mathrm{SO}_4}{\longrightarrow}} \mathrm{CO}_2+\mathrm{H}_2 \mathrm{O}+\mathrm{SO}_2+\left(\mathrm{NH}_4\right)_2 \mathrm{SO}_4$

⇒ $\left(\mathrm{NH}_4\right)_2 \mathrm{SO}_4+2 \mathrm{NaOH} \stackrel{\text { heat }}{\longrightarrow} \mathrm{Na}_2 \mathrm{SO}_4+2 \mathrm{H}_2 \mathrm{O}+2 \mathrm{NH}_3$

This method is simple, convenient, and largely used for the estimation of nitrogen in foodstuffs, drugs, fertilizers, and many other organic compounds. However, this method cannot be employed for the estimation of nitrogen in the following types of organic compounds.

Organic compounds containing nitrogen in aromatic rings such as pyridine, quinoline, etc.,

Organic compounds containing nitro (-NO₂) and diazo (- N = N -) groups.

% of nitrogen in the compound = $\frac{\text { volume of the acid }\left(\mathrm{cm}^3\right) \times \text { normality of the acid } \times 1.4}{\text { mass of organic compound in } \mathrm{g}}$

Estimation of Nitrogen by Duma’s method

Nitrogen By Duma’s method Principle: The organic compound containing nitrogen when heated with excess of copper oxide in the atmosphere of carbon dioxide, yields nitrogen in addition to carbon dioxide and water.

⇒ $\underset{\text { Organic compound }}{\left[\mathrm{C}_{\mathrm{x}} \mathrm{H}_y \mathrm{~N}_z\right]}+\left(2 \mathrm{x}+\frac{\mathrm{y}}{2}\right) \mathrm{CuO} \rightarrow\left(2 \mathrm{x}+\frac{\mathrm{y}}{2}\right) \mathrm{Cu}+\mathrm{xCO}_2+\frac{\mathrm{y}}{2} \mathrm{H}_2 \mathrm{O}+\frac{\mathrm{z}}{2} \mathrm{~N}_2$

Traces of nitrogen oxides formed during combustion of organic compound are reduced to nitrogen by passing the gaseous mixture over a heated copper gauze. The percentage of nitrogen present in a given organic compound is calculated from the volume of nitrogen collected over potassium hydroxide solution from a known mass of organic compound.

Percentage of nitrogen = $\frac{28 \times V_0 \times 100}{22,400 \times m}$

where V₀ = volume of nitrogen at STP; m = mass of organic compound in grams.

Estimation of Halogen by Carius method

Halogen By Carius Method Principle: A known mass of an organic compound is heated with fuming nitric acid and a few crystals of silver nitrate in a sealed tube called the Carius tube. The silver halide is formed.

⇒ $\underset{\text { Organic compouad }}{[\mathrm{C} \mathrm{H} \mathrm{X}]}+[\mathrm{O}] \underset{\text { beat }}{\stackrel{\text { fuming } \mathrm{HNO}_3}{\longrightarrow}} \mathrm{CO}_2+\mathrm{H}_2 \mathrm{O}+\mathrm{HX}$

⇒ $\mathrm{HX}+\mathrm{AgNO}_3 \rightarrow \underset{\text { Preceipitate }}{\mathrm{AgX}}+\mathrm{HNO}_3$

Percentage of halogen = $\frac{\text { atomic mass of } \mathrm{X}}{\text { molecular mass of } \mathrm{AgX}} \times \frac{\text { mass of silver halide in grams }}{\text { mass of organic compound in grams }} \times 100$

Percentage of chlorine = $\frac{35.5}{(108+35.5)} \times \frac{\text { mass of silver chloride in grams }}{\text { mass of organic compound in grams }} \times 100$

Percentage of bromine = $\frac{80.0}{(108+80)} \times \frac{\text { mass of silver bromide in grams }}{\text { mass of organic compound in grams }} \times 100$

Percentage of iodine = $\frac{127}{(108+127)} \times \frac{\text { mass of silver iodide in grams }}{\text { mass of organic compound in grams }} \times 100$

This method does not give satisfactory results in the estimation of iodine as Agl is slightly soluble in nitric acid and iodide may oxidize to iodine to some extent.
The results of this method is not very accurate in case of polyhalogenated aromatic compounds

Estimation of fluorine can not be carried out as AgF is soluble in water.

Estimation Of Sulfur by Carius Method

Sulfur By Carius Method Principle: An organic compound is digested with fuming nitric acid in a sealed tube. The sulfur present in the compound is quantitatively oxidised into sulphuric acid. Sulphuric acid so formed is precipitated as barium sulfate by adding excess of barium chloride.

Percentage of sulphur = $=\frac{32}{233} \times \frac{\text { mass of barium sulphate in grams }}{\text { mass of organic compounds in grams }} \times 100$

Estimation of phosphorous by Carius method

Phosphorous By Carius Method Principle: The phosphorus present in the organic compound is oxidised to orthophosphoric acid by heating with fuming nitric acid. The phosphoric acid so obtained is precipitated as MgNH₄PO₄ which on ignition is converted into Mg₂P₂O₇.

⇒ $\underset{\text { Organic component }}{[\mathrm{C}, \mathrm{H}, \mathrm{P}]}+[\mathrm{O}] \underset{\text { nitric acid }}{\stackrel{\text { filming }}{\longrightarrow}} \mathrm{H}_3 \mathrm{PO}_4+\mathrm{H}_2 \mathrm{O}+\mathrm{CO}_2$

⇒ $\mathrm{H}_3 \mathrm{PO}_4+\left[\mathrm{NH}_4 \mathrm{Cl}+\underset{\text { magnetic mixture }}{\mathrm{NH}_4 \mathrm{OH}}+\mathrm{MgCl}_2\right] \rightarrow \underset{\text { white precipitate }}{\mathrm{MgNH}_4 \mathrm{PO}_4}$

⇒ $2 \mathrm{MgNH}_4 \mathrm{PO}_4 \stackrel{\text { ignite }}{\longrightarrow} \underset{\text { magnesium pyophloctuse }}{\mathrm{Mg}_2 \mathrm{P}_2 \mathrm{O}_7}+\mathrm{H}_2 \mathrm{O}+2 \mathrm{NH}_3$

⇒ $\text { Percentage of phosphorus }=\frac{62}{222} \times \frac{\text { mass of } \mathrm{Mg}_2 \mathrm{P}_2 \mathrm{O}_7 \text { in grams }}{\text { mass of organic compound in grams }} \times 100$

Estimation of oxygen: There is no direct method for the estimation of oxygen present in the organic compound. The percentage of oxygen in the compound is generally estimated by difference.

Percentage of oxygen = 100 – [sum of percentage of all other elements present in it] The percentage of oxygen in an organic compound is usually found by difference between the total percentage composition (100) and the sum of the percentages of all other elements. However, oxygen can also be estimated directly as follows:

A definite mass of an organic compound is decomposed by heating in a stream of nitrogen gas. The mixture of gaseous products containing oxygen is passed over red-hot coke when all the oxygen is converted to carbon monoxide. This mixture is passed through warm Iodine pentoxide (I₂O₅) when carbon monoxide is oxidised to carbon dioxide producing iodine.

⇒ $Compound \stackrel{\text { heat }}{\longrightarrow} \mathrm{O}_2$+ other gaseous products

⇒ $\left.2 \mathrm{C}+\mathrm{O}_2 \stackrel{1373 \mathrm{~K}}{\longrightarrow} 2 \mathrm{CO}\right]$ x 5 ……..(1)

⇒ $\left.\mathrm{I}_2 \mathrm{O}_5+5 \mathrm{CO} \longrightarrow \mathrm{I}_2+5 \mathrm{CO}_2\right]$ x 2…..(2)

On making the amount of CO produced in equation (1) equal to the amount of CO used in equation (2) by multiplying equations (1) and (2) by 5 and 2 respectively; we find that each mole of oxygen liberated from the compound will produce two moles of carbon dioxide.

Thus 88 g carbon dioxide is obtained if 32 g oxygen is liberated.

Let the mass of organic compound taken be mg

Mass of carbon dioxide produced be m₁g

∴ $\mathrm{m}_1 \mathrm{~g}$ carbon dioxide is obtained from $\frac{32 \times \mathrm{m}_1}{88} \mathrm{~g} \mathrm{O}_2$

∴ Percentage of oxygen = $\frac{32 \times \mathrm{m}_1 \times 100}{88 \times \mathrm{m}} \%$

The percentage of oxygen can be derived from the amount of iodine produced also.

Presently, the estimation of elements in an organic compound is carried out by using microquantities of substances and automatic experimental techniques. The elements, carbon, hydrogen and nitrogen present in a compound are determined by an apparatus known as CHN elemental analyser. The analyzer requires only a very small amount of the substance (1-3 mg) and displays the values on a screen within a short time.

Empirical and Molecular Formulae: We know that a chemical formula gives the representation of a molecule of a substance in terms of symbols of various elements present in it. The determination of a formula of the substance involves the chemical analysis to determine

The constituent elements present.
The relative number of elements of each type present.

The chemical formula may be of two types:

Empirical formula and
Molecular formula

1. Empirical formula: The formula that gives the simple whole number ratio of the atoms of various elements present in one molecule of the compound is called empirical formula. For example, empirical formula of benzene is CH which indicates that atomic ratio of C: H in benzene is 1:1.

2. Molecular formula: The formula which gives the actual number of atoms of various elements present in one molecule of the compound is called molecular formula.

For example, molecular formula of benzene is C₆H₆ which tell that one molecule of benzene contains 6 atoms of carbon and 6 atoms of hydrogen.

Relation between empirical and molecular formulae; Molecular formula and empirical formula are related as molecular formula = n(empirical formula)

where n is a simple whole number and may have values 1, 2, 3 … It is equal to n = $\frac{\text { molecular mass }}{\text { empirical formula mass }}$

For example, the molecular mass of benzene is 78. The empirical formula of benzene is CH and therefore, its empirical formula mass is 13.

Thus, n = $\frac{\text { molecular mass }}{\text { empirical formula mass }}=\frac{78}{13}=6$

Therefore, the molecular formula of benzene = (CH)₆ = C₆H₆.

It may be noted that in many cases, the value of ‘n’ comes out to be one and, therefore, empirical formula and molecular formula are same in these cases. For example, empirical and molecular formulae are same for methane (CH₄), propane (C₃H₈), sucrose (C₁₂H₂₂O₁₁) etc.

Determination of the empirical formula of a compound: Once the percentage composition is known, the ratio of the numbers of atoms of each element present in the compound can be calculated to get the empirical formula. The method is to divide the percentage composition of each element by its atomic mass and to factorise the resulting numbers to obtain simple whole numbers. The steps involved in determining the empirical formula are:

Step 1: The relative number of atoms (also called atomic ratio) of various elements in the molecule of the compound is given by the following relation.

Relative number of atoms = $\frac{\text { percentage of an element }}{\text { atomic mass of the element }}$

Step 2: The simplest ratio of the atoms of the elements is obtained by dividing the relative numbers of atoms by the least value.

Step 3: Round off the value to the nearest whole number and multiply, if necessary, by a suitable integer to make them whole numbers. This gives the simplest whole-number ratio.

Step 4: Write the chemical formula with the simplest ratio of the atoms. The formula, thus, obtained represents the empirical formula of the compound.

Determination of the molecular formula of a compound

Step 1: Determine the empirical formula.

Step 2: Calculate the empirical formula mass by adding the atomic masses of the atoms in the empirical formula.

Step 3: Determine the molecular mass by a suitable method.

Step 4: Determine the value of n, as ‘n’ = $\frac{\text { molecular mass }}{\text { empirical formula mass }}$

Round off ‘n’ to the nearest whole number.

Step 5: Multiply the empirical formula by ‘n’ to get the molecular formula.

For example, if a compound has 23.3% carbon, 4.85% hydrogen, 40.78% nitrogen, and the remainder is oxygen, the empirical formula is calculated as follows:

C = $\frac{23 \cdot 3}{12}$, H = $\frac{4 \cdot 85}{1}$, N = $\frac{40 \cdot 78}{14}$ , O = $\frac{31 \cdot 07}{16}$

C = 194, H = 45, N =2.95, O = 1.94

Dividing by the smallest number, we get

C = $\frac{1 \cdot 94}{1 \cdot 94}$ = 1; H = $=\frac{4.85}{1.94}$ = 2.5: N = $\frac{2.95}{1.94}$ = 1.5 and O =$\frac{1.94}{1.94}$ = 1

The simplest whole number ratio is C₂H₅N₃O_2.

From the empirical formula, the molecular formula is arrived at by dividing the molecular mass by the empirical formula mass and the resulting whole number is multiplied with the empirical formula to get the molecular formula.

For example, if the empirical formula is CH₂ and the molecular mass is 56, the molecular formula is 56/14 = 4, 4 x CH₂ = C₄H

NEET General Organic Chemistry Purification And Characteristics Of Organic Compounds Notes

September 14, 2023 by Haritha

Purification And Characteristics Of Organic Compounds

Purification Of Organic Compounds: The methods which are employed for the purification of organic compounds or separation of individual components of a mixture containing two or more compounds are summarized below.

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Crystallisation Of Organic Compounds: The method is based on the difference in the solubilities of the compound and the impurities in a suitable solvent. Pure compound crystallises out from the solution and highly soluble impurities remain in the solution. Impure organic compounds like glucose, urea, cinnamic acid, etc are purified. Fractional crystallisation is used for the separation of a mixture of two compounds which are soluble in the same solvent but to a different extent. For example, Separation of sugar and salt.

Sublimation Of Organic Compounds: The process of direct conversion of a solid into the gaseous state on heating without passing through the intervening liquid state and vice versa on cooling is known as sublimation.

NEET General Organic Chemistry Purification And Characteristics Of Organic Compounds Sublimation Of Organic Compounds

Only those substances whose vapour pressure becomes equal to the atmospheric pressure much before their respective melting points are capable of undergoing sublimation. There exists an equilibrium between the solid and its vapours. Camphor, naphthalene, anthracene, iodine, benzoic acid, salicylic acid, NH₂Cl, HgCl₂, etc. are purified by sublimation.

Distillation Of Organic Compounds: Distillation is a method used to separate constituents of a liquid mixture which differ in their boiling points.

Distillation is a process which involves two steps:

Vapourisation: Liquid is converted into vapours.

Condensation: Vapours are condensed again into liquid.

Depending upon the difference in the boiling points of the constituent liquids, different types of distillation methods are employed.

Simple distillation Of Organic Compounds: Simple distillation is applied only for volatile liquids which boil without decomposition at atmospheric pressure and contain non-volatile impurities. This method can also be used for separating liquids having sufficient differences in their boiling points. For example,

Benzene (boiling point 353 K) and aniline (boiling point 475 K)
Chloroform (boiling point 334 K) and aniline (boiling point 457 K)
Ether (boiling point 308 K) and toluene (boiling point 383 K)

Nitrobenzene prepared in the laboratory can also be purified by distillation.

Fractional Distillation of Organic Compounds: This method is used for the separation of two or more volatile liquids from a liquid mixture which has boiling points close to each other. Liquids forming a constant boiling mixture (azeotropic mixture) such as rectified spirit cannot be separated by this method. Fractional distillation is used these days in industries, especially, in the distillation of petroleum, coal tar and crude alcohol. A mixture of methanol (boiling point 338 K) and propanone (boiling point 330 K) or a mixture of benzene and toluene may be separated by fractional distillation.

Distillation under reduced pressure (Vacuum distillation) Of Organic Compounds: The compounds, which decompose at a temperature below their normal boiling points, cannot be purified by distillation under ordinary atmospheric pressure. Glycerine is one such compound which decomposes at its boiling point. The pressure is reduced by suction pump and the distillation is carried out at lower temperature as glycerine can be distilled at 453 K (normal boiling point 563 K) under a pressure of 10-12 mm. Cane juice can also be concentrated by this method. This technique can be used to separate glycerol from spent lye in soap industry.

Steam distillation Of Organic Compounds: This method is used to purify the organic compounds which

Are volatile in steam but are immiscible with water.
Possess high vapour pressure at the boiling point of water.

Contain non-volatile impurities.

The compound to be purified is distilled with steam and impurities being non-volatile remains in mother liquor. For example, o-nitrophenol (volatile) and p-nitrophenol (non-volatile) are separated by this method.

Differential extraction (or solvent extraction) Of Organic Compounds: The process of separation of an organic compound (solid or liquid) from its aqueous solution by shaking with a suitable organic solvent is termed as solvent extraction. This method is employed for non-volatile compounds. For example, benzoic acid is extracted from its aqueous solution using benzene as solvent.

Chromatography

Chromatography is the technique of separating the components of a mixture in which the separation is achieved by the differential movement of individual components through a stationary phase under the influence of a mobile phase. Depending upon the nature of the stationary phase (either a solid or a liquid tightly bound on a solid support) and the nature of the mobile phase (either a liquid or a gas), different types of chromatographic techniques are followed.

NEET General Organic Chemistry Purification And Characteristics Of Organic Compounds Chromatography Types

The various components on the developed TLC plate are identified through their retardation factor, i.e., R_f values.

⇒ $\mathrm{R}_{\mathrm{f}}=\frac{\text { distance moved by the substance from base line }}{\text { distance moved by the solvent from base line }}$

A component with highest value of R_f elute first. A component with greater tendency to adsorb on solid has lesser the R_f value.

Qualitative Analysis (Detection Of Elements)

The qualitative analysis of an organic compound involves the detection of all the elements present in it. Carbon and hydrogen are generally present in all organic compounds. Other elements which may be present in organic compounds are oxygen, nitrogen, sulphur, halogens, phosphorus, etc. These elements are detected by the following tests.

Detection of Carbon and Hydrogen by Copper oxide test

Copper Oxide Test Principle Organic compounds undergo oxidation in the presence of a suitable oxidizing agent. In this process, carbon is oxidized to CO₂ and hydrogen is oxidized to water.

Copper Oxide Test Procedure The compound is intimately mixed with dry cupric oxide. The mixture is strongly heated in a hard glass test tube fitted with a cork and a delivery tube. The liberated gases are passed into lime water.

⇒ $\underset{\text { organic compound }}{[\mathrm{C}]}+2 \mathrm{CuO} \stackrel{\text { heat }}{\longrightarrow} \mathrm{CO}_2+2 \mathrm{Cu}$

⇒ $\mathrm{CO}_2+\underset{\text { lime water }}{\mathrm{Ca}(\mathrm{OH})_2} \longrightarrow \underset{\text { (milky) }}{\mathrm{CaCO}_3}+\mathrm{H}_2 \mathrm{O}$

⇒ $\underset{\text { organic compound }}{[2 \mathrm{H}]}+\mathrm{CuO} \stackrel{\text { heat }}{\longrightarrow} \mathrm{H}_2 \mathrm{O}+\mathrm{Cu}$

⇒ $\underset{\text { (white) }}{\mathrm{CuSO}_4}+5 \mathrm{H}_2 \mathrm{O} \longrightarrow \underset{\text { (blue) }}{\mathrm{CuSO}_4 \cdot 5 \mathrm{H}_2 \mathrm{O}}$

Detection of Nitrogen, Halogen and Sulphur by Lassaigne’s test

Detection of Nitrogen, Halogen and Sulphur by Lassaigne’s test Principle This is the most reliable test for the detection of nitrogen, sulphur, halogens etc. The given organic compound is fused with dry, metallic sodium. In this process, carbon and nitrogen present in the organic compounds get converted into cyanide ion (CN^–) sulphur (S) into sulphide ion S^2- and halogens (X^–) into halide ion (X“). The product is extracted into water. The solution contains sodium salts (NaCN, Na₂S and NaX).

Lassaigne’s Test For Nitrogen: A few crystals of ferrous sulphate are added to the first part of the filtrate. The mixture is boiled and cooled. It is acidified with hydrochloric acid and a few drops of ferric chloride solution are added. Sodium cyanide in the filtrate reacts with ferrous sulphate to give sodium ferrocyanide. It further reacts with ferric chloride to give a blue coloured solution of ferric ferrocyanide.

⇒ $[\mathrm{C}, \mathrm{N}]+\mathrm{Na} \stackrel{\text { fuse }}{\longrightarrow} \mathrm{NaCN}$

⇒ $\mathrm{FeSO}_4+2 \mathrm{NaCN} \longrightarrow \mathrm{Fe}(\mathrm{CN})_2+\mathrm{Na}_2 \mathrm{SO}_4$

⇒ $\mathrm{Fe}(\mathrm{CN})_2+4 \mathrm{NaCN} \longrightarrow \underset{\text { sodium ferrocyanide }}{\mathrm{Na}_4\left[\mathrm{Fe}(\mathrm{CN})_6\right]}$

⇒ $3 \mathrm{Na}_4\left[\mathrm{Fe}(\mathrm{CN})_6\right]+4 \mathrm{FeCl}_3 \longrightarrow \mathrm{Fe}_4\left[\mathrm{Fe}(\mathrm{CN})_6\right]_3+12 \mathrm{NaCl}$

If an organic compound contains nitrogen and sulphur, sodium thiocyanate is formed in the sodium extract which gives blood red colouration with ferric chloride due to the formation of ferric thiocyanate.

⇒ $\underset{\text { organic compound }}{[\mathrm{C}, \mathrm{N}, \mathrm{S}]}+\mathrm{Na} \stackrel{\text { fuse }}{\longrightarrow} \mathrm{NaCNS}$

⇒ $\mathrm{NaCNS}+\mathrm{FeCl}_3 \longrightarrow \mathrm{Fe}(\mathrm{CNS}) \mathrm{Cl}_2+\mathrm{NaCl}$

Lassaigne’s Test For Sulphur

1. Lead acetate test: Second part of the filtrate is treated with excess of acetic acid and lead acetate solution. A black precipitate of lead sulphide is formed.

⇒ $[\mathrm{S}]+2 \mathrm{Na} \stackrel{\text { fuse }}{\longrightarrow} \mathrm{Na}_2 \mathrm{~S}$

⇒ $\mathrm{Na}_2 \mathrm{~S}+\mathrm{Pb}\left(\mathrm{CH}_3 \mathrm{COO}\right)_2 \longrightarrow 2 \mathrm{CH}_3 \mathrm{COONa}+\underset{\text { black precipitate }}{\mathrm{PbS} \downarrow}$

2. Sodium nitroprusside test: To the sodium fusion extract, a few drops of sodium nitroprusside are added. The appearance of violet colour indicates the presence of sulphur.

⇒ $\mathrm{Na}_2 \mathrm{~S}+\mathrm{Na}_2\left[\mathrm{Fe}(\mathrm{CN})_5 \mathrm{NO}\right] \longrightarrow \underset{\text { sodium sulpho nitroprusside (violet colour) }}{\mathrm{Na}_4\left[\mathrm{Fe}(\mathrm{CN})_5 \mathrm{NOS}\right]}$

Lassaigne’s Test For Halogens

1. Silver nitrate test A portion of the stock solution is boiled with dil. HN03, cooled and silver nitrate is added. A white precipitate soluble in ammonium hydroxide shows the presence of chlorine.A pale yellow precipitate slightly soluble in ammonium hydroxide shows the presence of bromine.A yellow precipitate insoluble in ammonium hydroxide shows the presence of iodine.

⇒ $\mathrm{NaCl}+\mathrm{AgNO}_3 \longrightarrow \underset{\text { white }}{\mathrm{AgCl} \downarrow}+\mathrm{NaNO}_3$

⇒ $\mathrm{AgCl}+\underset{\text { ammonium hydroxide }}{2 \mathrm{NH}_4 \mathrm{OH}} \longrightarrow \underset{\text { diamine silver (I) chloride (soluble) }}{\mathrm{Ag}\left(\mathrm{NH}_3\right)_2 \mathrm{Cl}}+2 \mathrm{H}_2 \mathrm{O}$

⇒ $\mathrm{NaBr}+\mathrm{AgNO}_3 \longrightarrow \underset{\text { pale yellow }}{\mathrm{AgBr} \downarrow}+\mathrm{NaNO}_3$

⇒ $\mathrm{NaI}+\mathrm{AgNO}_3 \longrightarrow \underset{\text { yellow }}{\mathrm{AgI} \downarrow}+\mathrm{NaNO}_3 \\$

If nitrogen sulphur or both are found to be present in the organic compound, the sodium extract must be boiled with dilute nitric acid to expel hydrogen cyanide or hydrogen sulphide. If cyanide and sulphide ions are not removed, they interfere with the test forming precipitate of AgCN (white) and Ag₂S (black).

2. Beilstein’s test: Beilstein’s test is used to detect the halogen in an organic compound. A copper wire is heated in a Bunsen flame till no colour is imparted to the flame. The copper wire is dipped in the given organic compound and exposed to the non-luminous zone of the Bunsen flame. A bluish-green-coloured flame indicates the presence of halogen.

Lassaigne’s Test For Phosphorus: Organic compound containing phosphrous is fused with sodium peroxide. The phosphorus of the compound is oxidised to phosphate. The fused mass is extracted with water and filtered. The filtrate containing sodium phosphate is boiled with nitric acid and then treated with ammonium molybdate. A yellow solution or precipitate indicates the presence of phosphorus.

⇒ $\mathrm{Na}_3 \mathrm{PO}_4+3 \mathrm{HNO}_3 \rightarrow \mathrm{H}_3 \mathrm{PO}_4+3 \mathrm{NaNO}_3$

⇒ $\mathrm{H}_3 \mathrm{PO}_4+12\left(\mathrm{NH}_4\right)_2 \mathrm{MoO}_4+2 \mathrm{HNO}_3 \rightarrow\left(\mathrm{NH}_4\right)_3 \mathrm{PO}_4 12 \mathrm{MoO}_3+21 \mathrm{NH}_4 \mathrm{NO}_3+12 \mathrm{H}_2 \mathrm{O}$

NEET General Organic Chemistry Types Of Organic Reactions Notes

September 14, 2023 by Haritha

Types Of Organic Reactions

Substitution Reaction

The reactions in which an atom or a group in a molecule is replaced by another are called substitution reactions. The incoming group gets attached to the same carbon atom to which leaving group was attached. The substituting species may be a nucleophile, an electrophile, or a free radical.

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1. Nucleophilic substitution (S_N) reactions: The substitution reactions which are brought about by the attack of nucleophiles are called nucleophilic substitution reactions.

NEET General Organic Chemistry Types Of Organic Reactions Nucleophilic Substitution Reactions

The common examples of reactions are as follows.

⇒ $\mathrm{R}-\mathrm{CH}_2-\mathrm{X}+\mathrm{NaOH} \stackrel{\text { aqueous }}{\longrightarrow} \mathrm{R}-\mathrm{CH}_2-\mathrm{OH}+\mathrm{X}^{-}$

∴ $(\mathrm{X}=\mathrm{Cl}, \mathrm{Br}, \mathrm{I})$

⇒ $\mathrm{R}-\mathrm{CH}_2-\mathrm{X}+\mathrm{KCN} \stackrel{\text { alcohol }}{\longrightarrow} \mathrm{R}-\mathrm{CH}_2 \mathrm{CN}+\mathrm{X}^{-}$

⇒ $\mathrm{R}-\mathrm{CH}_2 \mathrm{OH}+\mathrm{SOCl}_2 \longrightarrow \mathrm{R}-\mathrm{CH}_2 \mathrm{Cl}+\mathrm{HCl}+\mathrm{SO}_2$

2. Electrophilic substitution (S_E) reactions: The substitution reactions which are brought about by the attack of an electrophile are called electrophilic substitution reactions. The substitution reactions of aromatic compound such as chlorination, nitration of benzene are representatives of reactions.

Chlorination: $\mathrm{Cl}_2+\mathrm{FeCl}_3 \rightarrow \mathrm{FeCl}+\mathrm{Cl}^{+}$

NEET General Organic Chemistry Types Of Organic Reactions Electrophilic Substitution Reaction Chlorination

3. Free radical substitution The substitution reactions which are brought about by the attack of free radical are called free radical substitution reactions. The chlorination of aliphatic hydrocarbons in presence of diffused sunlight is common example of free radical substitution.

NEET General Organic Chemistry Types Of Organic Reactions Free Radical Substitution

Addition Reaction

“An addition reaction is defined as one in which an unsaturated molecule combines with reagent (addendum) to give a single saturated or nearly saturated compound”. A few examples for different types of addition reactions are given below

1. Electrophilic addition reactions: The addition of common reagents like, HX, H₂O, HOCl, etc. to alkenes and alkynes are common examples of electrophilic addition reactions. The addition of unsymmetrical molecules like HX, H₂O, HOCl, etc. to unsymmetrical alkenes or alkynes takes place according to Markownikoffs’ rule.

NEET General Organic Chemistry Types Of Organic Reactions Addition Reacion Of Organic Compound

2. Nucleophilic Addition Reactions: The electron-deficient carbonyl group of aldehydes or ketones is easily attacked by nucleophiles which can supply an electron pair. Thus the addition reactions of carbonyl compounds initiated by nucleophiles and known as nucleophilic addition reactions.

Addition of hydrogen cyanide

NEET General Organic Chemistry Types Of Organic Reactions AdditionOf Hydrogen Cyanide

R′=H or alkyl (–CH₃, –CH₂CH₃, etc.) or aryl (C₆H₅–) or aralkyl (C₆H₅CH₂–)

Addition to Grignard reagent

NEET General Organic Chemistry Types Of Organic Reactions Addition Og Grignard Reagent

Addition of sodium bisulphite

NEET General Organic Chemistry Types Of Organic Reactions Addition Of Sodium Bisulphite

Elimination Reactions

An elimination reaction is one in which a molecule loses two atoms or groups without being replaced by other atoms or groups. The elimination reactions are two types, β-elimination reactions and α-elimination reactions

β-elimination reactions: This type of reaction involves the loss of two atoms or groups from vicinal (adjacent) carbon atoms resulting in the formation of a π bond. Thus, it is the reverse of addition reactions. The most familiar example of (β-elimination reactions are dehydrohalogenation reactions of alkyl halides, dehalogenation of haloalkanes, dehydration of alcohols, pyrolysis of esters, Hofmann elimination of quaternary ammonium hydroxide.

Dehydrohalogenation: When alkylhalides are treated with alcoholic potassium hydroxide solution or sodamide the corresponding alkenes are formed with the elimination of hydrogen halide.

NEET General Organic Chemistry Types Of Organic Reactions Dehydrohalogenation Heat

Sodamide (NaNH₂) is a stronger dehydrohalogenating agent.
An example of 1,4-elimination (δ-elimination) is

NEET General Organic Chemistry Types Of Organic Reactions 1,4 Elimination

Dehalogenation: Dehalogenation involves the removal of halogen molecule (X₂) from vicinal dihalide by heating with zinc dust in the alcoholic medium.

NEET General Organic Chemistry Types Of Organic Reactions Dehalogenation Beta Elimination

Dehydration of alcohols: When alcohols are heated with dehydrating agent like concentrated sulphuric acid, the corresponding alkenes are formed with the elimination of water (β-elimination). The reaction is called dehydration of alcohols.

NEET General Organic Chemistry Types Of Organic Reactions Dehydration Of Alcohols

α-elimination: It involves loss of two atoms from same atom resulting in the formation of reaction intermediates like carbine

NEET General Organic Chemistry Types Of Organic Reactions Alpha Elimination

Rearrangement Reactions

The reactions which proceed by a rearrangement or reshuffling of atoms or groups in the molecule to produce a structural isomer of the original substance are called rearrangement reactions.

NEET General Organic Chemistry Types Of Organic Reactions Rearrangement Reactions

NEET General Organic Chemistry Concepts In Organic Reaction Mechanism Notes

September 14, 2023 by Haritha

Concepts In Organic Reaction Mechanism

The organic reactions involve the breaking of covalent bonds in the reacting molecules and formation of new bonds to give product molecules. “A study of the sequential account of each step, describing details of formation energetics during bond cleavage and bond electron movement and the rate of transformation of reactant into products (kinetics) is referred to as reaction mechanism”.

NEET General Organic Chemistry Concepts In Organic Reaction Mechanism

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The organic molecule which reacts with attacking reagent is called substrate. In multistep organic reactions, the substrate react with reagent and leads to the formation of one or more reaction intermediates. The general reaction path involving the formation of one reaction intermediate is depicted as follows.

Fission Of Covalent Bond

The fission of covalent bond can take place in two ways depending on the nature of covalent bond, nature of the attacking reagent, and conditions of the reaction.

Homolytic Cleavage (Homolytic fission): In this type of fission each fragment formed as a result of cleavage of covalent bond gets on electron from the shared bond pair electrons”. This cleavage results in the formation of specie(s) with unshared electron called free radicals. The homolytic fission of a σ bond is shown as follows.

NEET General Organic Chemistry Concepts In Organic Reaction Mechanism Homolytic Cleavage

The free radicals contain unpaired electron (with odd number of electrons), electrically neutral and paramagnetic.

Heterolytic Cleavage (Heterolytic Fission): In this type of cleavage both the bond pair electrons of the covalent bond are taken away by electron-withdrawing fragment which results in the formation of electron-deficient and electron-rich fragments. The electron-deficient fragment is called electrophile while electron rich fragment is known as a nucleophile. The heterolytic fission is shown below

NEET General Organic Chemistry Concepts In Organic Reaction Mechanism Heterolytic Cleavage

Both electrophiles and nucleophile contain even number of electrons, influenced by strong electrical fields as they possess positive or negative charge and they are diamagnetic.

Substrate + Reagent ⇔ [reaction intermediate] → Product

Reaction Intermediates

A highly reactive, short-lived, and energetic intermediate formed in multistep organic reactions (reactions involving more than one step) by the action of reagent on substrate and readily transformed into the product(s) is called a reaction intermediate.

1. Carbocations: A reaction intermediate formed by heterolytic fission of a covalent bond that contains one positively charged carbon with three bond pair electrons (sextet of electrons) is called carbocation. The heterolytic fission of bromomethane yields methyl carbocation and bromide ion as shown below

NEET General Organic Chemistry Concepts In Organic Reaction Mechanism Carbocation Structers

The carbocations are classified into different groups depending upon the nature of carbon-bearing the positive charge. Alkyl carbocations may be primary (1°), secondary (2°), or tertiary (3°) carbocations.

NEET General Organic Chemistry Concepts In Organic Reaction Mechanism Carbocation Reactions

Structure of carbocation: The carbocations are electron deficient and contain six electrons (three bond pair electrons). In strong electrical field, carbocations move toward cathode. The carbocations have a trigonal planar structure and the positively charged carbon in sp² hybridized.

Stability of carbocations: The carbocations are electron deficient and stabilised by electron releasing group. The participation of empty p-orbital in lateral overlap with completely filled orbital is the major contributing factor for stability to stability in which positive charge is delocalised.

Alkyl group directly bonded to the positively charged carbon stabilise the carbocations due to inductive and hyperconjugation effects. The dispersal of the positive charge due to hyperconjugation stabilise the carbocation. The observed order of carbocation stability is

NEET General Organic Chemistry Concepts In Organic Reaction Mechanism Stability Of Carbocations

The greater the number of a-hydrogen atoms higher is the stability of carbocations. The reaction intermediates are highly reactive, usually order of reactivity of such intermediates is reverse that of its stability. Therefore order of reactivity of carbocations follows the sequence:

∴ CH₃ >1°>20>3°

Hybridised state of carbon: Higher the s-character, lesser the stability. The order of stability is

NEET General Organic Chemistry Concepts In Organic Reaction Mechanism Hybridiserd States Of Carbon

Carbocations Resonance: The greater the delocalization of positive charge of carbocations, the higher is the stability. The stability of benzyl carbocation is comparable with allyl carbocation as resonance energies is nearly the same

NEET General Organic Chemistry Concepts In Organic Reaction Mechanism Resonances

Carbocations undergo rearrangement forming more stable carbocations.

NEET General Organic Chemistry Concepts In Organic Reaction Mechanism Carbocations

2. Carbanions: A reaction intermediate formed by heterolytic fission of a covalent bond which results in negatively charged carbon with eight electrons in its valence shell is called carbanion.

The heterolytic cleavage of a covalent bond as indicated in the following reactions gives carbanions. The carbanions are classified as primary (1°), secondary (2°), and tertiary (3°) depending upon the nature of carbon bearing negative charge.

NEET General Organic Chemistry Concepts In Organic Reaction Mechanism Carbanioons Structure

Structure of carbanion: The carbanions are electron-rich with complete octet configuration. In strong electrical field, they move toward anode. The shape of alkyl carbanion is usually pyramidal like ammonia. The carbon atom carrying the negative charge is sp³ hybridized. In contrast, carbanions which are stabilized by resonance are planar and the carbon atom carrying a negative charge is sp² hybridized. The vinyl carbanion, phenyl carbanion, and cyclopentadiene carbanion are sp² hybridised as acetylide carbanion is sp hybridised.

Stability of carbanions: Alkyl group bonded to negatively charged carbon increases the intensity of negative charge due to +1 effect and destabilise the carbanion. The order of stability of alkyl carbanions is:

NEET General Organic Chemistry Concepts In Organic Reaction Mechanism Stability Of Carbanions

The order of reactivity of carbanions is reversed that of its stability. Therefore order of reactivity of carbanions follows the sequence:

NEET General Organic Chemistry Concepts In Organic Reaction Mechanism Carbanions

Hybridised state of carbon: Higher the s character greater the stability of carbanion. Order of stability is

NEET General Organic Chemistry Concepts In Organic Reaction Mechanism Hybridised State Of Carbon

Carbanions Resonance: Greater the delocalisation of negative charge by resonance, higher the stability The stability of benzyl carbanion is comparable with allyl carbanions.

NEET General Organic Chemistry Concepts In Organic Reaction Mechanism Resonanc

3. Free radicals: A reaction intermediate formed by the homolytic cleavage of a covalent bond that contains an unpaired electron is called free radical. The homolytic cleavage is favoured for nonpolar covalent bonds. Such cleavage for the bond is initiated by the action of heat, light, or a reagent. For example,

NEET General Organic Chemistry Concepts In Organic Reaction Mechanism Free Radicals

Depending upon nature of carbon atom carrying the unpaired electron, free radicals are also classified as primary (1°), secondary (2°), and tertiary (3°) free radicals.

NEET General Organic Chemistry Concepts In Organic Reaction Mechanism Free Radicals Types

Structure of free radicals: The free radicals are electron deficient since they contain seven electrons on carbon atom. They are electrically neutral and paramagnetic. The structure of alkyl radicals is not known with certainty. For alkyl radicals two possible structures have been proposed. The first is a planar sp² hybridised radical similar to a carbocation. The second one is a pyramidal sp³ hybridized radical similar to a carbanion.

Resonance stabilised free radicals such as allyl radicals and benzyl radicals are planar sp² hybridised.
The bridgehead free radicals are pyramidal (sp³-hybridised) because they cannot assume planar geometry due to angle strain. Further, the free radicals in which carbon is bonded to highly electronegative atoms are pyramidal.

NEET General Organic Chemistry Concepts In Organic Reaction Mechanism Structre Of Free Radical

Stability of free radicals: The relative stability of alkyl free radicals is explained on the basis of hyperconjugation and inductive effects. Greater the number of alkyl groups attached to the carbon atom carrying unpaired electron, higher the delocalisation and hence more stable is the alkyl radical.

The order of stability of a few alkyl radicals is given below

NEET General Organic Chemistry Concepts In Organic Reaction Mechanism Stability Of Free Radical

Like carbocations and carbanions, free radicals are highly reactive and short-lived intermediates because of the strong tendency of the carbon atom carrying the unpaired electron to acquire one more electron from an atom or a group to complete its octet. The reactivity of alkyl radical is reverse the order of stability CH₃> 1 > 2 > 3 resonance. The stability of allyl and benzyl radicals is comparable.

NEET General Organic Chemistry Concepts In Organic Reaction Mechanism Allyl Or Benzyl Radicals

The allyl or benzyl radicals are more stable than alkyl radicals.

Carbenes: The carbenes are neutral and highly reactive species generally obtained by successive elimination of an electrophile and a nucleophile from the same carbon atom (a-elimination). The carbon atom of carbene has six electrons in valence shell, out of which two constitute unshared electrons and two bond pair electrons. So they are divalent carbon species containing two unshared electrons and electrically neutral. Examples of different types of carbenes are given below.

NEET General Organic Chemistry Concepts In Organic Reaction Mechanism Carbenes

Electrophiles And Nucleophiles

The organic reactions proceed by the attack of highly reactive reagents on the substrate molecule. These reagents are called attacking reagents which may be electron deficient or electron rich. They are classified into two groups.

Electrophiles: The electron-deficient molecules or positively charged ions which are capable of accepting an electron pair from substrate molecule are called electrophiles. These species act as Lewis acids and attack the electron-rich centre of the organic molecules.

NEET General Organic Chemistry Concepts In Organic Reaction Mechanism Electrophiles

Following are common examples

Positive electrophiles: H⁺, Cl⁺, Br⁺, N⁺O₂, N⁺ = O, (R⁺) carbocations, etc.

Neutral electrophiles: SO₃, BF₃, AlCl₃, etc.

It may be noted that all the positively charged species do not act as electrophiles. The positively charged species which can accept an electron pair can act as electrophiles. The positively charged ions such as H₃O⁺, NH⁺₄, Na⁺, Ca²⁺ do not act as electrophile as they cannot accept electron pair, since all the ions have an octet configuration.

Nucleophiles: The molecules or negatively charge ions which are capable of donating an electron pair to the electron-deficient centre of the substrate are called nucleophiles. These species act as Lewis bases and attack on the electron-deficient center of organic molecules.

NEET General Organic Chemistry Concepts In Organic Reaction Mechanism Nucleophiles

The common examples of nucleophiles are given below:

Negative nucleophiles:

NEET General Organic Chemistry Concepts In Organic Reaction Mechanism Negative Nucleophiles

Neutral nucleophiles:

NEET General Organic Chemistry Concepts In Organic Reaction Mechanism Neutral Necleophiles

Comparison between Nucleophiles and electrophiles

NEET General Organic Chemistry Concepts In Organic Reaction Mechanism Comparison between Nucleophiles And Electrophiles

Electron Displacement Effects in Covalent Bonds: The electron pair displacement in organic molecules takes place under the influence of a hetero atom/group or by the attacking reagent. The displacement or shift of electron pair in the organic molecule under the influence of substituent makes the molecule permanently polar.

1. Inductive effect (I effect): The inductive effect is defined as “the permanent displacement of sigma (σ) bond pair of electrons towards more electronegative atom or group and as a result molecule becomes permanently polar”. Larger the displacement of σ bond pair electrons greater the polarity. Consider the carbon chain in which terminal carbon is bonded to a chlorine atom. Since chlorine is more electronegative bond pair of electrons are displaced towards chlorine.

NEET General Organic Chemistry Concepts In Organic Reaction Mechanism Inductive Effect

Positive inductive effect (+I effect): In this effect the substituent (Y) releases electron pair away from itself. In other wordso bond pair of electrons are displaced away from the substituent.

NEET General Organic Chemistry Concepts In Organic Reaction Mechanism Positive Inductive Effect

The order of electron releasing or releasing ability of substituent is given below.

NEET General Organic Chemistry Concepts In Organic Reaction Mechanism PositiveInductive Effect Order Of Electron Realesing

Negative inductive effect (-I effect): In this effect, the sigma bond pair of electrons are displaced towards the electron-withdrawing substituent (X).

NEET General Organic Chemistry Concepts In Organic Reaction Mechanism Elecron Withdrwaing Substituent

The order of electron-withdrawing ability (intensity of -I effect) of a few substituents is given below.

NEET General Organic Chemistry Concepts In Organic Reaction Mechanism Negative Inductive Effect Electron Withdrawing Substitutent

The inductive effect is a permanent effect operating in the ground state of the organic molecules. It explains facts such as the effect on dipole moment, reactivity of alkyl halides, acidity of carboxylic acids, and basicity of amines and alcohols.

2. Resonance effect (R−effect) or Mesomeric effect: The presence of alternate single and double bond or heteroatom-containing one or more lone pair electrons linked to multiple bonded atom (with single bond in an open chain or cyclic system) is called a conjugate system. Examples are 1,3-butadiene, aniline, phenol, nitrobenzene, etc.

In such systems, π or lone pair electrons are delocalised and the molecule develops polarity. “The permanent polarity produced in the molecule by the shift of pi(π) or lone pair electrons in the conjugate system creating electron deficient and electron-rich centres called resonance effect (R-effect). The effect is transmitted through the entire conjugate system. Depending upon the direction of shift of electron pair in the conjugate system, the R-effect is classified into two types.

Positive resonance effect (+R effect / +M effect): In this effect, of electron pair (π or lone pair) moves away from the substituent attached to conjugate system. The +R effect in aniline as shown below.

NEET General Organic Chemistry Concepts In Organic Reaction Mechanism Positive Resonance Effect

The substituents (atom/group) that exert only +R effect are as follows: -OH, -OR, -OCOR, -NH₂, -NHR, -NR₂, -NHCOR, -X (halogen ; X = -Cl, Br, I, etc.

The substituent which exerts +R effect are called electron-releasing groups. The resonance effect in which resonance structure violates the octet rule should not be considered. For example, structure (2) cannot be considered as resonance structure since it violates the octet rule because oxygen has 10 electrons in the valence shell.

Negative resonance effect (-R effect / -M effect): In this effect, the shift of π or lone pair electrons is towards the substituent attached to the conjugate system. The electron displacement depicted in nitrobenzene represents -R effect.

NEET General Organic Chemistry Concepts In Organic Reaction Mechanism Negative Resonance Depicted Nitrobenzene

The substituents, which exert only -R effect are given below.

NEET General Organic Chemistry Concepts In Organic Reaction Mechanism Negative Resonance R Effect

The substituent which exert-R effect is called electron withdrawn group. Resonance effect provides explanation to least reactivity of haloalkenes and aryl halides towards nucleophilic substitution reactions, acidic nature of phenols and carboxylic acids, mechanism of electrophilic substitution reactions of benzene.

Substitutes like NEET General Organic Chemistry Concepts In Organic Reaction Mechanism Phenyl etc. exert both +R and -R effects.

3. Hyperconjugation: The phenomenon of hyperconjugation is also known as the Baker-Nathan effect as it was proposed by Baker and Nathan. This effect is an extension of resonance in which C-H sigma (σ) bond pair electrons are involved in delocalization. The electron release of alkyl group bonded to the unsaturated system in which delocalization of electrons takes place through overlap between C-H sigma (σ) orbital and Pi(π) bond orbital or vacant p-orbital is known as hyperconjugation.

The hyperconjugation is a permanent effect and a stabilizing interaction. The delocalization of electrons by hyperconjugation in propene molecule is depicted as in the figure.

NEET General Organic Chemistry Concepts In Organic Reaction Mechanism Hyperconjugatrion Effect

Propene molecule may be regarded as the resonance hybrid of the following hyperconjugative structures.

NEET General Organic Chemistry Concepts In Organic Reaction Mechanism Propene Molecule

Since there is no bond between carbon and hydrogen atoms in these structures (2 -4), hyperconjugation is also called no bond resonance. It may be noted that although a free proton (H⁺) has been shown in the above structures, it is still bound firmly to the π- cloud and hence is not free to move.
The hyperconjugation effect is much weaker compared to the resonance effect, yet it is quite useful is explaining relative stability, and physical and chemical properties of organic molecules. Larger the number of hyperconjugative structures higher the delocalization of electron pairs and greater the stability of alkene. Number of hyperconjugative structures is equal to number of α-hydrogen atoms plus one. The relative stability of a few alkenes is given.

NEET General Organic Chemistry Concepts In Organic Reaction Mechanism Hypercojugative Structures

This order of stability is because of greater number of hyperconjugative contributing structures causing larger delocalization of n-electrons solve and hence accounts for higher stability of alkene. To understand the hyperconjugation effect in carbocations, let us take an example of ethyl carbocation (CH₂-CH⁺₂), in which the positively charged carbon atom has an empty p orbital. One of the C-Hασ bond orbital of methyl group align in the plane of empty p orbital and this bond pair electrons delocalise into the empty p-orbital as shown in figure.

NEET General Organic Chemistry Concepts In Organic Reaction Mechanism Hyperconjugatrion Effect

The overlap of completely filled C-H σ bond orbital with empty p-orbital of carbocation causes dispersion of positive charge and stabilize the carbocation. The ethyl carbocation is resonance hybrid of following contributing structures.

NEET General Organic Chemistry Concepts In Organic Reaction Mechanism Alkyl Groups Attached To A Positively

In general, larger the number of a-hydrogen atoms of alkyl groups attached to a positively charged carbon atom greater the stability of carbocations.

4. Electromeric effect (E effect): The electromeric effect is a temporary effect shown by organic compounds containing multiple bonds. The complete transfer or shift of n electron pair of a multiple bond to one of the bonded atoms during the attack of positively or negatively charged reagent is called electromeric effect or E effect.

This temporary effect takes place only in the presence of an attacking reagent. As soon as the reagent is removed, the molecule reverts back to original position. Depending upon the direction of displacement, the E effect is also of two types.

-E effect: An attacking reagent is said to have -E effect when the direction of π electron pair transfer of multiple bond is away from the attacking reagent. The -E effect operates during nucleophilic addition reaction of aldehydes and ketones.

NEET General Organic Chemistry Concepts In Organic Reaction Mechanism -E Effect

+E effect: An attacking reagent is said to have +E effect when the direction of π electron pair transfer of a multiple bond is towards the attacking reagent. The +E effect is observed during electrophilic addition reaction of alkenes and alkynes

NEET General Organic Chemistry Concepts In Organic Reaction Mechanism Propene E Effect

NEET General Organic Chemistry Concepts In Organic Reaction Mechanism E Effect Electrophilic Addition Reaction Of Alkenes And Alkunes

NEET General Organic Chemistry Isomerism Notes

September 14, 2023 by Haritha

Isomerism

The existence of two or more compounds with the same molecular formula but different physical and chemical properties is known as isomerism and the molecules themselves are called as isomers.

NEET General Organic Chemistry Isomerism Notes Isomerism

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The term was given by Berzelius. The difference in properties of the two isomers is due to differences in the arrangement of atoms within their molecules.

Isomerism is mainly classified into structural isomerism and stereoisomerism.

Structural Isomerism

Structural isomerism is due to the differences in structures of the isomers. Structural isomerism is further classified into 5 types.

1. Chain isomerism (nuclear isomerism): Compounds with same molecular formula but differ in the arrangement and number of carbon atoms within the molecule are called chain isomers and the phenomenon as chain isomerism.

2. Butane (C₄H₁₀) has two isomers – normal butane and isobutene. One isomer has a straight chain and the other has a branched chain.

NEET General Organic Chemistry Isomerism Notes Butane And Isobutane

3. Cyclohexane and methyl cyclopentane are nuclear isomers.

NEET General Organic Chemistry Isomerism Notes Cyclohexane And Methylcyclopentane

4. C₅H₁₂ has three chain isomers.

NEET General Organic Chemistry Isomerism Notes Three Chain Isomers

5. C₄H₉NH₂ also shows two chain isomers.

NEET General Organic Chemistry Isomerism Notes C4H9NH2 has Two Chain Isomers

Solve 1: How many chain isomers does butane have?
Solution:

NEET General Organic Chemistry Isomerism Notes Chain Isomers Does Have Butane

Solve 2: How many chain isomers does propylbenzene have?
Solution:

Functional Isomerism

Compounds with same molecular formula but differ in functional groups are called functional isomers and the phenomenon is known as functional isomerism. For example

Diethyl ether butyl alcohol both have the molecular formation C₄H₆O but contains different functional groups. Thus, functional group in diethyl ether is (-O-), while is butyl alcohol it is (-OH).

C₂H₅-O-C₂H₅(diethyl ether); C₄H₉-OH (butyl alochol)

Acetone and propionaldehyde both with the molecular formula are functional isomers. In acetone the functional group is (-CO-) while in acetaldehyde it is (-CHO)

CH₃-CO-CH₃ (acetone); CH₃-CH₂-CHO (acetaldehyde)

Cyanides are isomeric with isocyanides

RCN (Alkyl cyanide); RNC (Alkyl isocyanide)

Carboxylic acids are isomeric with esters.

CH₃CH₂COOH (Propanoic acid); CH₃COOCH₃ (Methyl ethanoate)

Nitroalkanes are isomeric with alkyl nitrites:

NEET General Organic Chemistry Isomerism Notes Nitroalkane

R-O-N = O (Alkyl nitrite)

Sometimes a double bond-containing compound may be isomeric with a triple bond-containing compound. This also is called as functional isomerism. Thus, butyne is isomeric with butadiene (molecular formula C₆H₆).

CH₃-CH₆C ≡ CH (1-Butyne);

CH₆ = CH-CH = CH₂(1, 3-Butadiene)

Unsaturated alcohols are isomeric with aldehydes. Thus,

CH₂=CH-OH (Vinyl alochol); CH₃CHO (Acetaldehyde)

Unsaturated alcohols containing three or more carbon atoms are isomeric to aldehydes as well as ketones:

CH₂ = CH-CH₂OH (Alkul alochol);

CH₃CH₂CHO (Propionaldehyde);

CH₃COCH₃ (Acetone)

Aromatic alcohols may be isomeric with phenols

NEET General Organic Chemistry Isomerism Notes Aromatic Alochols

Primary, secondary, and tertiary amines of same molecular formula are also functional isomers.

NEET General Organic Chemistry Isomerism Notes Primary Secondary And Tertiary Amines

Alkenes are isomeric with cycloalkanes:

NEET General Organic Chemistry Isomerism Notes Alkenes Are Isomeric With Cycloalkanes

Such isomers in which one is cyclic and the other is open chain are called ring-chain isomers. Alkynes and alkadienes are isomeric with cycloalkanes.

CH₃CH₂C ≡ CH (1-Butyne);

CH₂ =CH-CH = CH₂ (1,3 -Butadinene)

NEET General Organic Chemistry Isomerism Notes Cyclobutene

Position isomerism

Compounds which have the same structure (arrangement) of carbon chain (carbon skeleton) but differ in the position of the multiple bond or the functional group are called position isomers and the phenomenon is known as position isomerism.

Thus the following compounds can exhibit position isomerism:

Alkenes
Alkynes
Arenes

Alkyl halides
Aryl halides
Alcohols

Amines and
Nitro compounds

Alkenes containing four or more carbon atoms can exhibit position isomerism due to the difference in the position of the double bond on the same carbon skeleton, For example:

CH₃CH₂CH = CH₂ (but-1-ene) ; CH₃CH =CHCH₃ (But-2-ene)
CH₂CH₂CH₂CH = CH₂ (Pent-l-ene); CH₃CH₂CH = CHCH₃ (Pent-2-ene)
CH₃CH₂CH₂CH = CH₂ (Hex-l-ene); CH₃CH₂CH₂CH = CHCH₃ (Hex-2-ene); CH₃CH₂CH = CHCH₂CH₃(Hex-3-ene)

Alkynes containing four or more carbon atoms can exhibit position isomerism due to the difference in the position of triple bond on the same carbon skeleton. For example:

CH₃CH₂C = CH₂ (But-l-yne); CH₃C = CCH₃ (But-2-yne)
CH₃CH₂CH₂C (Pent-l-ene) = CH; CH₃CH₂C = CCH₃

Arenes containing eight or more carbon atoms exhibit position isomerism due to the difference in the position of alkyl groups on the benzene ring. For example:

NEET General Organic Chemistry Isomerism Notes Arenes

Alkyl halides containing three or more carbon atoms can exhibit position isomerism due to the difference in the position of halogen atom on the same carbon skeleton. For example

1. C₃H₇Cl has two position isomers:

NEET General Organic Chemistry Isomerism Notes Alkyl Halides

2. n – Pentane on monochlorination gives three isomeric chloromethanes:

NEET General Organic Chemistry Isomerism Notes Chloropentanes

Polyhalogen derivatives containing two or more carbon atoms can also exhibit position isomerism. For example:

CH₃CHCl₂ (1, 1- Dichloroethane)

ClCH₂CH₂Cl (1, 2-Dichloroethane)

Aryl halides containing two or more benzene rings can exhibit position isomerism due to the difference in the position of halogen atom. For example:

NEET General Organic Chemistry Isomerism Notes Aryl Halides

Polyhalogen compounds containing just one benzene ring can also exhibit position isomerism. For example:

NEET General Organic Chemistry Isomerism Notes Polyhalogen Compounds

Alcohols containing three or more carbon atoms exhibit position isomerism due to the difference in the position of functional group (-OH). For example:

1. The molecular formula C₃H₈O represents two isomeric alcohols:

CH₃CH₂CH₂OH (Propan 1-ol)

NEET General Organic Chemistry Isomerism Notes Alochols

2. Four monohydric alcohols differing in the position of -OH group can be derived from isopentane:

NEET General Organic Chemistry Isomerism Notes Four Monohydric Alochols

Primary amines (RNH₂) containing three or more carbon atoms can exhibit position isomerism due to the difference in the position of amino group on the same carbon skeleton. For example:

NEET General Organic Chemistry Isomerism Notes Primary Amines

Nitro compounds (RNO₂) containing three or more carbon atoms can exhibit position isomerism due to the difference in the position of nitro group on the same carbon skeleton. For example:

CH₃CH₂CH₂CH₂NO₂ (1-Nitobutane)

NEET General Organic Chemistry Isomerism Notes Nitro Compounds

Metamerism

Here two or more different compounds having the same molecular formula but different number and arrangements of carbon atoms on either side of the functional group is called metamerism. Such compounds are known as metamers.

NEET General Organic Chemistry Isomerism Notes Metamerism

Metamerism is never possible in compounds possessing the univalent functional group. Metamerism is due to the difference in the nature of alkyl groups attached to the same polyvalent is functional group such is Metamerism is exhibited by compounds of the same homologous series.

Ethers, R-Q-R, exhibit metamerism due to differences in the nature of the alkyl groups attached to the oxygen atom. Thus, the molecular formula, C₄H₁₀O, represents the following metamers:

NEET General Organic Chemistry Isomerism Notes Ethers R-O-R

Ketones , exhibit metamerism due to the difference in the nature of the alkyl groups attached to the carbonyl group. Thus, the molecular formula, C₅H₁₀O, represents the following metamers:

NEET General Organic Chemistry Isomerism Notes Ketones Metamers

Thioethers, R-S-R’, exhibit metamerism due to the difference in the nature of the alkyl groups attached to the sulfur atom. Thus, the molecular formula, C₄H₁₀S, represents the following metamers:

NEET General Organic Chemistry Isomerism Notes Thioethers

Secondary and tertiary amines exhibit metamerism due to the difference in the nature of the alkyl groups attached to the – NH – group and the atom respectively. Thus the molecular formula, C₄H₁₁N, represents the following metamers:

Esters, exhibit metamerism due to the difference in the nature of the alkyl groups attached to the -C – 0 – group. Thus, the following esters are metamers:

NEET General Organic Chemistry Isomerism Notes Esters

Note – If same polyvalent functional group is present in two or more organic compounds, then instead of chain or position isomerism, treat the phenomenon as metamerism.

Pentan – 2- one and pentan – 3- one are metamers and not position isomers. They can be included in position isomerism, if metamerism is not mentioned.

Similarly, pentan – 2- one and 3 – methylbutan-2-one are metamers and not chain isomers.

Metamers may be considered as position isomers. For instance, pentan – 2- one and penta-3- one may be regarded as position isomers as well as metamers.

Tautomerism

Here a single compound exists in two readily interconvertible structures that differ in position of the hydrogen atom. Tautomer exhibits dynamic equilibrium with each other. A very common form of tautomerism is that between a carbonyl compound containing an a – hydrogen and its enol form. This type of isomerism is also known as keto-enol isomerism.

NEET General Organic Chemistry Isomerism Notes Tautomerism

The percentage of enol form increases in the order simple aldehydes and ketones <β-keto esters <β-diketones <β-diketones having phenyl group < phenols. This increase in the enol content is due to the fact that the enol form of the above type of compounds is increasingly stabilized by resonance and hydrogen bonding than the corresponding keto form.

Ring Chain Isomerism

The phenomenon of existence of two or more compounds having the same molecular formula but possessing open chain and closed chain (cyclic structure) is called ring-chain isomerism. This type of isomerism arises due to different modes of linking of carbon atoms. Thus ring-chain isomers possess open chain or closed-chain structures as illustrated by the following examples:

Two ring-chain isomers are possible corresponding to the molecular formula C₃H₆:

NEET General Organic Chemistry Isomerism Notes Propene And Cyclopropane

Six pairs of ring-chain isomers are possible for the molecular formula C₄H₈:

NEET General Organic Chemistry Isomerism Notes Six Pairs Of Ring Chain

The molecular formula C₃H₄ represents the two ring-chain isomers:

NEET General Organic Chemistry Isomerism Notes Two Ring Chain Isomers

Ring chain isomerism can be included in functional isomerism, if not considered separately.

NEET General Organic Chemistry Naming Of Organic Compounds Containing A Functional Group Notes

September 14, 2023 by Haritha

Naming Of Organic Compounds Containing A Functional Group

Prefix and suffix names of a few common functional groups and their decreasing order of priority are given in the table.

NEET General Organic Chemistry Naming Of Organic Compounds Containing A Functional Group Decreasing Order Of Priority

It may be noted that the groups -R, -OR, -NO₂, -X, etc. are considered as substituents and are indicated as prefixes.

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Longest chain– The parent chain is the one which has a functional group directly attached to it, even if it violates the longest chain rule.
Numbering of parent chain-The numbering should be done in such a way that the carbon linking to the functional group gets the lowest number even if it violates the lowest sum rule or locant rule. When the functional group itself contains carbon atom, then that carbon atom is assigned as number 1.

Naming of compounds with polyfunctional group- If there is more than one functional group present in a compound, then one of the functional group is chosen as the principal functional group (secondary suffix) and the remaining functional groups are treated as (substituents) subordinate functional groups.

NEET General Organic Chemistry Naming Of Organic Compounds Containing A Functional Group Naming Of Compound With Polyfunctional Group

The functional group with higher priority in the sequence given below is termed as principal functional group.

Nomenclature Of Substituted Benzene Compounds

Benzene is a six-membered cyclic compound with alternate single and double bonds. It is represented in any one of the following ways:

NEET General Organic Chemistry Naming Of Organic Compounds Containing A Functional Group Nomenclature Of Substituted benzene Compounds

1. Naming of monosubstituted benzene

It is derived by adding the name of the substituent with the word benzene. Substituent + benzene 23/11 substituted benzene

Naming of monosubstituted benzene Example:

NEET General Organic Chemistry Naming Of Organic Compounds Containing A Functional Group Naming Of Monosubstituated Benzene

Special names of some monosubstituted benzene compounds

NEET General Organic Chemistry Naming Of Organic Compounds Containing A Functional Group Special Named Of Some Monosubstituted Benzene Compounds

If the functional group is attached to the carbon chain connected to benzene ring, then benzene ring is considered as a substituent and is prefixed before the root word as phenyl.

NEET General Organic Chemistry Naming Of Organic Compounds Containing A Functional Group 2, 3 Phenylchloroethane

2. Naming of disubstituted benzene

If the substituents are the same: In such case, the relative position of the substituents must be indicated by adding the symbols o – (1, 2); m – (1, 3) ; p – (1, 4).

NEET General Organic Chemistry Naming Of Organic Compounds Containing A Functional GroupNaming Of Disubstituted Benzene

Naming of disubstituted benzene Example:

NEET General Organic Chemistry Naming Of Organic Compounds Containing A Functional Group The Substitutens Are Same In Disubstituted Benzene

If the two substituents are different: The substituents are named in alphabetical order.

NEET General Organic Chemistry Naming Of Organic Compounds Containing A Functional Group Two Substituents Are Different

If there are more than two substituents: The numbering is done in such a way that it satisfies the lowest sum rule.

NEET General Organic Chemistry Naming Of Organic Compounds Containing A Functional Group There Are More Than Two Substitutents

Nomenclature Of Bicyclo Compounds

Bicyclo compounds contain two fused rings with the help of a bridge. We use the name of the alkane corresponding to the total number of carbon atoms as the base name.

The carbon atoms common to both the rings are called bridgeheads, and each bond or chain of atoms connecting the bridgehead atoms is called a bridge.

While naming the bi-cycloalkane we write an expression between the word bicyclo and alkane (in square bracket), that denotes the number of carbon atoms in each bridge. The numerals are written in descending order and the numbers are separated by a point.
If substituents are present, we number the bridged ring system beginning at one bridgehead, proceeding first along the longest bridge to the other bridge head, then along the second next longest bridge back to the first bridge head. The shortest bridge is numbered in the last.

Nomenclature of Bicyclo compounds Example:

NEET General Organic Chemistry Naming Of Organic Compounds Containing A Functional Group Nomeclature Of Bicyclo Compounds