Isomerism Question And Answers – NEET General Organic Chemistry

Isomerism

Question 1. Increasing order of stability among the three main conformations i.e., Eclipse, Anti, Gauche. of 2-fluoroethanol is

1. Eclipse, Gauche, Anti
2. Gauche, Eclipse, Anti
3. Eclipse, Anti, Gauche
4. Anti, Gauche, Eclipse

Answer: 3. Eclipse, Anti, Gauche

Solution: HO – CH₂– CH₂ – F

Gauche conformation is comparatively more stable due to the hydrogen linkage in between F and H at O-atom. hence order is Eclipse, Anti staggere, Gauche.

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

Question 2. How many isomers will C₃H₆ have?

1. 1
2. 2
3. Zero
4. 4

Answer: 2. 2

Solution: One propene and one cyclopropane.

Question 3. The molecular formula of diphenylmethane is C₁₃H₁₂

How many structural isomers are possible when one of the hydrogens is replaced by a chlorine atom?

1. 6
2. 4
3. 8
4. 7

Answer: 2. 4

Solution:

Question 4. In butane, which of the following forms has the lowest energy?

1. Gauche form
2. Eclipsed form
3. Staggered form
4. None of these

Answer: 3. Staggered form

Solution:  The staggered form has lower energy than the eclipsed form because the repulsive interaction between the H-atoms attached to two carbon atoms are minimal due to the maximum distance between them.

Question 5. Which of the following does not exhibit tautomerism?

Answer: 1

Solution: The compounds that contain the active methylene group at the adjacent position of the carboxyl group show tautomerism.

This compound does not contain an active methylene group, and hence does not exhibit tautomerism. Moreover, this compound is highly stable due to extensive cross-conjugation.

Question 6. The structure exhibits______________________ isomerism

1. Geometrical isomerism
2. Optical isomerism
3. Geometrical and optical isomerism
4. Tautomerism

Answer: 2. Optical Isomerism

Solution: One asymmetric carbon atom is present.

Question 7. Propanal and propanone are

1. Functional isomers
2. Enantiomers
3. Chain isomers
4. Structural isomers

Answer: 1. Functional isomers

Solution: Propanal and propanone are functional isomers

Question 8. Maleic and fumaric acids are :

1. Tautomers
2. Geometrical isomers
3. Chain isomers
4. Functional isomers

Answer: 2. Geometrical Isomers

Solution: Maleic and fumaric acids are geometrical isomers.

Question 9. Tautomerism is exhibited by

Answer: 1. 1

Solution: Tautomerism is a functional isomerism in which the isomers are readily interchangeable and they maintain a dynamic equilibrium with each other.

Question 10. An optically active compound is:

1. 1-bromobutane
2. 2-bromobutane
3. 1-bromo-2-methylpropane
4. 2-bromo-2-methylpropane

Answer: 2. 2-bromobutane

Solution: 2-bromobutane has asymmetric carbon atoms.

Question 11. The number of isomers possible for the compound with the molecular formula C₂BrClFI is

1. 3
2. 4
3. 5
4. 6

Answer:  4. 6

Solution: Molecular formula C₂BrClFl six isomers are possible.

Question 12. Which is incorrect about enantiomorphs?

1. They rotate the plane of polarized light in different directions
2. They have mostly identical physical properties
3. They have the same configuration
4. They have different biological properties

Answer: 3. They have the same configuration

Solution: Enantiomers differ in their configuration R or S. at the stereogenic centre.

Question 13. The total number of alkenes possible by dehydrogenation of 3-bromo-3- cyclopentyl hexane using alcoholic KOH is

1. 1
2. 3
3. 5
4. 7

Answer:  3. 5

Solution: The substrate has three different types of B-H, therefore, first, three structural isomers of alkenes are expected as

The last two alkenes 2 and 3 are also capable of showing geometrical isomerism hence two geometrical isomers for each of them will be counted giving a total of five isomers.

Question 14. The following compounds differ in

1. Configuration
2. Conformation
3. Structure
4. Chirality

Answer: 3. Structure

Solution: 

Since in the above structures, the positions of CI is different, these are position isomers, which is a type of structural isomerism.

Question 15. The correct statement about the compounds A, B and C is

1. A and B are identical
2. A and B are diastereomers
3. A and C are enantiomers
4. A and B are enantiomers

Answer: 4. A and B are enantiomers

Solution:

1. Enantiomers are pairs of optical isomers that are related as non-superimposable mirror images of each other.

2. Diastereomers are pairs of optical isomers that cannot be related as non-superimposable mirror images of each other.

∴ The only correct statement about given structures is that A. and B. are enantiomers.

Question 16. Isomerism among compounds due to the migration of a proton is known as

1. Geometrical
2. Optical
3. Tautomerism
4. Position

Answer: 3. Tautomerism

Solution: Definition of tautomerism.

Question 17. What is the number of possible optical isomers of glucose?

1. 3
2. 4
3. 12
4. 1

Answer:  4. 1

Solution: Glucose contains four chiral carbon atoms hence number of possible optical isomers is 2⁴ 

2⁴ = 2×2×2×2

= 16.

Question 18. Benz aldoxime exists in how many forms?

1. 1
2. 2
3. 3
4. 4

Answer: 2. 2

Solution: Benz aldoxime can exist in two geometrical isomeric forms known cis and trans.

Question 19. In the following the most stable conformation of n-butane is?

Answer: 2. 2

Solution:

1. The anti-staggered conformation of n-butane is more stable than gauche-staggered and eclipsed conformations of n-butane.

2. In anti-staggered n-butane, the methyl groups are placed at a dihedral angle of 1800, and the steric hindrance is minimal in anti-form than in gauche form.

Question 20. Which one of the following monoenes does not exhibit geometric isomerism?

1. C₄H₈
2. C₃H₆
3. C₅H₁₀
4. C₈H₁₆

Answer: 2.  C₃H₆

Solution: The main conditions for exhibiting geometrical isomerism are

1. Presence of double bond.
2. Presence of different groups on the same double bonded carbon.
3. Presence of at least one similar group on adjacent double bonded carbon atoms.

does not exhibit geometric isomerism due to the presence of same group on double-bonded atom (C₁ )

Question 21. The number of optical isomers of CH₃CH(OH)CH(OH)CHO is

1. Zero
2. 2
3. 3
4. 4

Answer: 4. 4

Solution: Possible number of optical isomers = 2ⁿ

= 2²

= 4

Question 22. The isomeric cis-2-butene and trans-2-butene can be distinguished on the basis of their:

1. Their physical nature
2. Their reduction products
3. The products they give on ozonolysis
4. The products they give in addition to bromine

Answer:

Solution: The two forms give different products on the addition of Br₂ cis butene gives a racemic mixture whereas trans butene gives meso form of 2,3,3-dibromo butene.

Answer: 3.

Question 23. n- pentane, iso-pentane, and neo– pentane are examples for isomers of the type

1. Geometrical
2. Optical
3. Chain
4. Positional

Answer:  3. Chain

Solution: The compounds which differ in the nature of carbon chain are called chain isomers, for example,

Question 24. The number of isomers in C₄H₁₀O  are

1. 7
2. 8
3. 6
4. 5

Answer: 1. 7

Solution: There are seven isomers in C₄H₁₀O. Out of these seven isomers, three are of ether. four are of alcohol

Question 25. In a lactic acid molecule, the methyl group, a hydroxyl group, a carboxylic acid group and a hydrogen atom are attached to a central carbon atom showing optical isomerism due to the molecular geometry at the

1. The carbon atom of the methyl group
2. Carbon atom of the carboxylic acid group
3. Central carbon atom
4. Oxygen of the hydroxyl group

Answer: 3. Central carbon atom

Solution: The central carbon atom is chiral carbon.

In a lactic acid molecule, the methyl group, a hydroxyl group, a carboxylic acid group and a hydrogen atom are attached to a central carbon atom showing optical isomerism due to the molecular geometry at the Central carbon atom

Question 26. Geometrical isomerism is possible in case of

1. Pentene-2
2. Propane
3. Pentane
4. Ethene

Answer: 1. Pentene – 2

Solution:  Pentene-2 exhibits cis and frans-isomerism.

Question 27. Isopentane can form four isomeric mono bromo derivatives. How many of them are optically active?

1. 1
2. 2
3. 3
4. None of these

Answer: 2. 2

Solution:

Question 28. A cyclic stereoisomer having the molecular formula C₄H₇Cl is classified and tabulated. Find out the correct set of numbers.

Answer: 1.

Solution: The acyclic stereoisomers of C₄H₇Cl. are

1. Number of optical isomers = 2ⁿ = 2¹ = 2
2. Hence, the total number of geometrical isomers = 6
3. Total number of optical isomers =2.

Question 29. The optical isomers, which are not enantiomers are called

1. Conformer
2. Diastereomer
3. Mirror images
4. None of these

Answer: 2. Diastereomer

Solution: Diastereoisomers are a pair of optical isomers that cannot be related as non-superimposable mirror images of each Other.

Question 30. A racemic mixture is a mixture of:

1. Meso and its isomers
2. d-and its l-isomers in equal proportions
3. d-and its l-isomers in different proportions
4. Meso and d-isomers

Answer: 2. d-and its l-isomers in equal proportions

Solution: A mixture of 50-50% of d and it’s form is called a racemic mixture.

Question 31. 2-methyl pent-3-enoic acid shows:

1. Optical isomerism
2. Geometrical isomerism
3. Both 1 and 2
4. None of these

Answer: 3. Both 1 and 2

Solution: The acid exists in cis and trans forms:

Also, it has an asymmetric carbon atom

Question 32. The number of optical enantiomorphs of tartaric acid is

1. 3
2. 2
3. 4
4. 1

Answer: 2. 2

Solution: a = 2n-1; where n is no. of asymmetric carbon; when a molecule possesses symmetry.

Question 33. The optically active alkane with the lowest molecular weight is

Answer: 3

Solution: Only this is optically active due to the central carbon being asymmetric.

Question 34. Which of the following structures permits cis-trans isomerism?

1. X₂C = CY₂
2. XYC = CZ₂
3. X₂C = CXY
4. XYC = CXY

Solution: Molecules with two similar groups attached on either of the doubly bonded carbons do not show geometrical isomerism.

Question 35. Methoxy methane and ethanol are

1. Position isomers
2. Chain isomers
3. Functional isomers
4. Optical isomers

Answer: 3. Functional isomers

Solution:

CH₃OCH₃ – Methoxy methane ₂₃₅

C ₂H₅OH₆ – Ethanol ether. alcohol. functional group

(C ₂H6O) (C₂H₆O)molecular formula

In methoxy methane and ethanol both molecular formulas are the same but the functional groups are different, so they are functional isomers.

Question 36. The number of structural and configurational isomers of a bromo compound, C5HgBr, formed by the addition of HBr to 2-pentyne respectively.

1. 1 and 2
2. 2 and 4
3. 4 and 2
4. 2 and 1

Answer:  2 and 4

Solution: Draw the isomers.

Question 37. A compound contains 2 dissimilar asymmetric carbon atoms. The number of optically active isomers is:

1. 2
2. 3
3. 4
4. 5

Answer: 3. 4

Solution: a = 2; where n is no. of dissimilar asymmetric carbon atoms and a is no. of optically active isomers.

Question 38. Stereoisomers are geometrical or optical. which are neither superimposable nor mirror image to each other are called?

1. Enantiomers
2. Mesomers
3. Tautomers
4. Diastereomers

Answer: 4. Diastereomers

Solution: It is the definition of Diastereoisomers.

Question 39. α-D- (+)-glucose and P-D- (+)-glucose are?

1. Enantiomers
2. Conformers
3. Epimers
4. Anomers

Answer: 4. Anomers

Solution:  Alpha D glucose and beta D glucose are examples of anomers. Alpha D glucose can be written as a-D (+)-glucose, whereas beta D glucose can be represented as P-D (+)-glucose.

Question 40. Isomers of propionic acid are

1. HCOOC₂H₅  and CH₃COOCH₃
2. HCOOC₂H5  and C₃H₇COOH
3. CH₃COOCH₃  and  C₃H₇OH
4. C₃H₇OH and CH₃COCH₃

Answer:  2. HCOOC₂H₅and CH₃COOCH₃

Solution: Isomers of propionic acid are as

Question 41. Identify, which of the below does not possess any element of symmetry.

1. (+) and (-) tartaric acid
2. Carbon tetrachloride
3. Methane
4. Meso-tartaric acid

Answer: 1. (+) and (-) tartaric acid

Solution: (+) and (-) tartaric acid does not possess any element of symmetry.

Question 42. Cis-trans, isomers generally

1. Contain an asymmetric carbon atom
2. Rotate the plane of polarized light
3. Are enantiomorphs
4. Contains a double-bonded carbon atom

Answer: 4. Contains a double-bonded carbon atom

Solution: Cis-trans isomers generally contain double-bonded carbon atoms.

Question 43. How many primary amines are possible with the formula C₄H₁₁N?

1. 1
2. 2
3. 3
4. 4

Answer:  4. 4

Solution: The possible primary amine with the formula C₄H₁₁N are

Question 44. For which of the following parameters the structural isomers C₂H₅OH and CH₃OCH₃ would be expected to have the same values? Assume ideal behaviour.

1. Heat of evaporation
2. Vapour pressure at the same temperature
3. Boiling points
4. Gaseous densities at the same temperature and pressure

Answer: 4. Gaseous densities at the same temperature and pressure

Solution: In CH₃CH₂OH, there is intermolecular H-bonding, while it is absent in isomeric ether CH₃OCH₃

1. Larger heat is required to vaporise CH₃CH₂OH as compared to CH₃,OCH₃, thus 1  is incorrect.

2. CH₃CH₂OH is less volatile than CH₃OCH₃, thus vapour pressures are different, thus 2 is incorrect.

3. Boiling point of CH₃CH₂OH>CH₃OCH₃, thus c. is incorrect.

Density = \(\frac{\text { mass }}{\text { volume }}\) due to ideal behaviour at a given temperature and pressure volume and molar J volume or  mass are the same.

Hence, they have the same vapour density.

Question 45. A compound having a molecular formula C₄H₁₀ can exhibit.

1. Metamerism
2. Functional isomerism
3. Positional isomerism
4. All of these

Answer: 4. All of these

Solution: Alcohols show position isomerism; Ethers show metamerism; Alcohols and ethers show functional isomerism.

Question 46. Which one of the follow 2, 3-dibromopentane

1. 3, 3-dibromopentane
2. 4-bromo-2-pentanol
3. 3-bromo-2-pentanol
4. 2, 3-dibromopentane

Answer: 3-bromo-2-pentanol

Solution: Those compounds which contain two or more asymmetric carbon atoms but are optically inactive due to the presence of plane of symmetry, are called meso compounds. Meso compounds are optically inactive due to internal compensation.

Out of the given compounds only 2, 4-dibromopentane have a plane of symmetry, so it is a meso compound.

Question 47. Ethylene dichloride and ethylidene chloride are isomeric compounds. The false statement about these isomers is that
they

1. React with alcoholic potash and give the same product
2. Are position isomers
3. Contain the same percentage of chlorine
4. Are both hydrolysed to the same product

Answer:  4. Are both hydrolysed to the same product

Question 48. Which of the following acids does not exhibit optical isomerism?

1. Lactic acid
2. Tartaric acid
3. Maleic acid
4. α -amino acids

Answer: 3. Maleic acid

Solution:

It has no asymmetric carbon; however, it shows geometrical isomerism.

Question 49. Ethyl acetoacetate shows which type of isomerism

1. Chain
2. Optical
3. Metamerism
4. Tautomerism

Answer: 4. Tautomerism

Solution: Ethyl acetoacetate shows tautomerism.

Question 50. The number of isomeric hexanes is?

1. 5
2. 2
3. 3
4. 4

Answer: 1. 5

Question 51. Which among the following statements is correct with respect to the optical isomers?

1. Enantiomers are non-superimposable mirror images.
2. Diastereomers are superimposable mirror images.
3. Enantiomers are superimposable mirror images.
4. Meso forms have no plane of symmetry.

Answer: 1. Enantiomers are non-superimposable mirror images

Solution: Enantiomers are non-superimposable mirror images, for example, lactic acid

Diastereomers are non-superimposable and are not mirror images of each other. Moreover, the meso form has a plane of symmetry.

Question 52. Hydrogen cyanide and hydrogen isocyanide are:

1. Functional isomers
2. Positional isomers
3. Metamers
4. Chain isomers

Answers: 1. Functional isomers

Question 53. 2-pentanone and 3-methyl-2-butanone are a pair of isomers.

1. Functional
2. Chain
3. Positional
4. Stereo

Answer:  2. Chain

Solution: 2-pentanone and 3-methyl-2-butanone are chain isomers because they differ in carbon skeleton.

Question 54. Which of the following is a dynamic isomerism?

1. Metamerism
2. Geometrical isomerism
3. Tautomerism
4. Coordinate isomerism

Answer: 3. Tautomerism

Solution: Tautomerism is a dynamic isomerism because two forms keto and enol. of substance cannot be separated; they are in dynamic equilibrium with each other.

Question 55. Cyclobutane and butene-1 are?

1. Chain isomers
2. Position isomers
3. Ring-chain isomers
4. Metamers

Answer: 3. Ring-chain isomers

Solution: Both have different mode of linkage, i.e., chain and ring.

Question 56. Among the following which is the one that does not exhibit functional group isomerism.

1. C₂H₆O
2. C₃H₈P
3. C₄H₁₀
4. C₄H₁₀O

Answer: 3. C₄H₁₀

Solution: Alkanes never show functional isomerism, metamerism, tautomerism and geometrical isomerism.

Question 57. Vinyl alcohol and acetaldehyde are

1. Geometrical isomers
2. Keto-enol tautomers
3. Chain isomers
4. None of these

Answer: 2. Keto-enol tautomers

Solution: Isomers differ in the arrangement of atoms but exist in dynamic equilibrium with each other shows the tautomerism. Acetaldehyde and vinyl alcohol are keto-enol tautomers.

Question 58. The structures that do not actually exist are known as

1. Tautomers
2. Conformational isomers
3. Canonical structures
4. Optical isomers

Answer: 3. Canonical structures

Solution: Canonical structures proposed in resonance are not the real structure of compounds. The compound showing resonance has a definite structure which can however not be drawn on paper.

Question 59. Chiral molecules are those which:

1. Are not superimposable on their mirror images
2. Are superimposable on their mirror images
3. Show geometrical isomerism
4. Are unstable molecules

Answer: 1. Are not superimposable on their mirror images

Solution: Chiral molecules should not contain any kind of symmetry.

Question 60. d-tartaric acid and l-tartaric acid are which type of isomer?

1. Structural isomers
2. Diastereoisomers
3. Tautomers
4. Enantiomers

Answer: 4. Enantiomers

Solution: d and l configurations are non- superimposable on each other and are mirror images, so they are termed as enantiomers. Enantiomers are compounds that are mirror images but are non-super impossible on each other. The structures that are mirror images are called enantiomorphs.

Question 61. Which of the following shows geometrical isomerism?

1. C₂H₅Br
2. (CH₂) (COOH)₂
3. (CH)₂(COOH)₂
4. C₂H₆

Answer: 3. (CH)₂(COOH)₂

Solution: The compounds must fulfil two conditions to show geometrical isomerism.

1. The compound should have at least one C=C.
2. The two groups attached to the same carbon must be different.
3. Out of given choices only (3). fulfils both conditions and shows geometrical isomerism.

Question 62. α-D-glucose and β-D-glucose have a specific rotation of +112° and +19° respectively. In aqueous solution the  rotation becomes +52°. This process is called as?

1. Inversion
2. Racemization
3. Mutarotation
4. Enolization

Answer: 3. Mutarotation

Solution: The conversion of a-glucose to p-glucose is called mutarotation.

Question 63. Which of the following pairs of carbon skeletons is an example of isomerism?

Answer: 1. 1

Solution: In the rest of them, all carbon chains are the same.

Question 64. Which of the following is a chiral compound?

1. Hexane
2. n-butane
3. Methane
4. 2,3,4, trimethyl hexane

Answer: 2,3,4, trimethyl hexane

Solution: In hexane, all the C-atoms are symmetric, so no carbon atom is chiral.

In n-butane all C-atom present are symmetric, hence it is achiral In methane, all groups attached are the same, hence it is also achiral.

In 2,3,4-trimethyl hexane, there are two chiral centres making the compound asymmetric.

Question 65. Geometrical isomerism is shown by:

Answer: 2

Solution: A molecule having doubly bonded carbon atoms shows geometrical isomerism only if both the doubly bonded carbon has altogether different group, i.e., _baC =C_ab or _abC = C_ac or _dcC=C_ab.

Question 66. How many optically active forms are possible for a compound of the formula,

CHO.CHOH.CHOH.CHOH.CH₂OH?

1. 2
2. 4
3. 3
4. 8

Answer: 4. 8

Solution: a = 2ⁿ; n = 3.

Question 67. The lactic acid molecule has

1. One chiral carbon atom
2. Two chiral carbon atoms
3. No chiral carbon atom
4. Asymmetric molecule

Answer: 1. One chiral carbon atom

Solution: A chiral carbon atom has all four different groups attached to it.

∴ It has one asymmetric or chiral carbon atom.

Question 68. Which of the following is an optically active compound?

1. Lactic acid
2. Chloro acetic acid
3. Meso-tartaric acid
4. Acetic acid

Answer: 1. Lactic acid

Solution: One asymmetric carbon atom is present in a lactic acid molecule. Hence, it is an optically active compound.

Question 69. Geometrical isomerism is attributed to?
Answer:

1. By restricted rotation around C = C bond
2. By the presence of one asymmetric carbon atom
3. Due to different groups attached to the same functional group
4. By swing of hydrogen atom between two polyvalent atoms

Answer: 1. By restricted rotation around C = C bond

Solution: It is a fact.

Question 70. Among the following four structures I to IV it is true that,

1. All four are chiral compounds
2. Only 1 and 2 are chiral compounds
3. Only 3 is a chiral compound
4. Only 2 and 4 are chiral compounds

Answer:  2. Only 1 and 2 are chiral compounds

Solution: Chiral compounds which have one chiral centre. All four atoms or groups attached to carbon are different.

Question 71. The number of chiral centres in +. -glucose

1. 4
2. 3
3. 2
4. 1

Answer: 1. 4

Solution: A carbon atom which is attached by four different groups is called the chiral centre of asymmetric carbon atoms. +.-glucose has four chiral centres.

Question 72. Geometrical isomerism is possible in

1. Acetone-oxime
2. Isobutene
3. Acetophenone-oxime
4. Benzophenone-oxime

Answer: 3. Acetophenone-oxime

Solution: Acetophenone oxime can show geometrical isomerism.

Question 73. Which of the following oxime can show geometrical isomerism?

Answer: 2

Solution: Due to asymmetric carbon atoms.

Question 74. Lactic acid, CH₃CH(OH)COOH  molecule shows

1. Geometrical isomerism
2. Metamerism
3. Optical isomerism
4. Tautomerism

Answer: 3. Optical isomerism

Solution: It shows optical isomerism due to the presence of asymmetric carbon atoms.

Question 75. The number of structural isomers possible for an organic compound with molecular formula C₅H₁₂

1. 5
2. 3
3. 4
4. 2

Answer: 2. 3

Solution: These are isopentane, neopentane and n-pentane.

Question 76. Which is a chiral molecule?

1. CH₃Cl
2. CH₂Cl₂
3. CHBr₃
4. CHClBrI

Answer: 4. CHClBrI

Solution: Carbon is asymmetric as all its valencies are attached to different groups.

Question 77. How many isomers are possible for the alkane C₄H₁₀?

1. 3
2. 5
3. 2
4. 4

Answer: 3.

Solution: Two isomers

Question 78. Which of the following compounds exhibits rotamers?

1. 2-butene
2. Maleic acid
3. Butane
4. Fumaric acid

Answer: 2-butene

Solution: 2-butene exhibit rotamers. Rotamers are the isomers formed by restricted rotation.

Question 79. Nitroethane can exhibit one of the following kinds of isomerism

1. Metamerism
2. Optical activity
3. Tautomerism
4. Position isomerism

Answer: 3. Tautomerism

Solution: Nitroalkanes exhibit tautomerism. In it, a-H-atom is labile and forms nitrolic acid.

Question 80. CH₃-CHO-CH₂-CH₃ has a chiral centre. Which one of the following represent its R-configuration?

Answer: 2

Solution: Follow priority rules.

Question 81. Which of the following molecules is achiral?

Answer: 4

Solution: A molecule having an asymmetric carbon atom and is not superimposable on its mirror image then it is chiral while if it is superimposable on its mirror image, it is achiral.

Question 82. Which of the following statements is necessarily true in the case of isomeric organic compounds?

1. They are hydrocarbons
2. They are optically active
3. They yield the same products on complete combustion
4. They have same melting or boiling points

Answer: 3. They yield the same products on complete combustion

Solution: Due to the same molecular formula.

Question 83. What kind of isomerism is possible for 1-chloro-2-nitroethene?

1. Functional group isomerism
2. Position isomerism
3. E/Z isomerism
4. Optical isomerism

Answer: 3. E/Z isomerism

Solution: The structure of 1-chloro-2-nitroethene is as

Question 84. On monochlorination of n-pentane, the number of isomers formed is

1. 4
2. 3
3. 2
4. 1

Answer: 2. 3

Solution: CH₃CH₂CHClCH₂CH₃; CH₃CHClCH₂CH₂CH₃; CH₂ClCH₂CH₂CHCH₃.

Question 85. One of the following compounds exhibits geometrical isomerism

1. CH₃CH₂CH₂CH₃
2. CH₃-HCCH₃.-HC.CH₃-CH₃
3. CH₃-HCCH₃.-CH₃
4. CH₃CH = CH-CH₃

Answer: 4. CH₃CH = CH-CH₃

Solution: Geometrical isomers of CH₃CH=CH-CH₃ are

Question 86. Which types of isomerism is shown by 2, 3-dichlorobutane?

1. Structural
2. Geometric
3. Optical
4. Diastereomer

Answer: 3. Optical

Question 87. Optically active compound among the following is:

1. 2-ethylbutanol
2. n-butanol
3. 2,2-dimethylbutanol
4. 2-methylbutanol-1

Answer: 2-methylbutanol-1

Solution: Chiral carbon is present. Hence it is optically active.

Question 88. Which of the following is a pair of functional isomers?

1. CH₃COCH₃,CH₃CHO
2. C₂H₅CO₂H,CH₃CO₂CH₃
3. C₂H₅CO₂H,CHCO₂C₂H₅
4. CH₃CO₂H,CH₃CHO

Answer: 2. C₂H₅CO₂H,CH₃CO₂CH₃

Solution: Carboxylic acid and esters show functional group isomerism. When two compounds have the same molecular formula but different functional groups, then functional isomerism arises.
for example., C₂H₅COOH and CH₃COOCH₃

Question 89. Diastereomers can be separated by?

1. Fractional distillation
2. Simple disillation
3. Electrophoresis
4. All of these

Answer: 1. Fractional distillation

Solution: It is a fact.

Question 90. Geometrical isomerism is not shown by

1. 1, 1-dichloro-1-pentene
2. 1,2-dichloro-1-pentene
3. 1, 3-dichloro-2-pentene
4. 1, 4-dichloro-2-pentene

Answer: 1, 1-dichloro-1-pentene

Solution: 1, 1-dichloro-1-pentene does not exhibit geometrical isomerism.

Question 91. n- pentane and 2-methyl butane are a pair of

1. Enantiomers
2. Stereoisomers
3. Diastereomers
4. Constitutional isomers

Answer: 4. Constitutional isomers

Solution: n-pentane and 2-methyl butane are constitutional isomers or chain isomers or skeletal isomers.

Question 92. A racemic mixture is formed by mixing two

1. Isomeric compounds
2. Chiral compounds
3. meso compounds
4. Enantiomers with chiral carbon

Answer: 2. Chiral compounds

Solution: Racemic mixture is formed by mixing two chiral compounds.

Question 93. CH₅≡N and C₆H₅N≡C exhibit which type of isomerism?

1. Position
2. Functional
3. Metamerism
4. Dextro isomerism

Answer:  2. Functional

Solution: CH₅≡N and C₆H₅N≡C are functional isomers.

Question 94. The production of an optically active compound from a symmetric molecule without resolution is termed as:

1. Walden inversion
2. Partial racemisation
3. Asymmetric synthesis
4. BVartial resolution

Answer: 3. Asymmetric synthesis

Solution: It is the definition of asymmetric synthesis.

Question 95. Metamers of ethyl propionate are

1. C₄H₉COOH and HCOOC₄H₉
2. C₄H₉COOH and C₇3COOC3H₇
3. CH₃COOCH₇ and CH₄COOC3H₇
4. CH₃COOC₃H₇and CH₇COOCH₇

Answer:  4. CH₃COOC₃H₇and CH₇COOCH₇

Solution: Metamers of ethyl propionate are as CH₃COOC₃H₇,C₃H₇COOCH₃

Question 96. Geometrical isomerism is not shown by which of the following? 

Answer: 3. CH₂ =C(Cl)CH₃

Solution: A molecule having doubly bonded carbon atoms shows geometrical isomerism only if both the doubly bonded carbon has altogether different group, i.e., _baC ≡C_ab or C_ac or _abC= _dcC=C_ab

Question 97. The number of optically active isomers of tartaric acid are

1. 1
2. 3
3. 4
4. 2

Answer: 4. 2

Solution: Total number of optical isomers = (2)ⁿ Where n = number of asymmetric carbon atoms.

=(2)²ⁿ= 4

Out of these four optical isomers, two are meso structures which are optically inactive.

∴ Only two structures d and l tartaric acid are optically active.

Question 98. The number of 1°,2° and 3° carbon atoms present in isopentane are respectively.

1. 3, 2, 1
2. 2, 3, 1
3. 3, 1, 1
4. 2, 2,1

Answer: 3. 3,1,1

Solution: (CH₃)₂CHCH₂CH₃ is isopentane.

Question 99. How many optically active stereomers are possible for butan-2, 3-diol?

1. 1
2. 2
3. 3
4. 4

Answer: 2. 2

Solution:

Where C* = Asymmetric C atom

It is a symmetrical molecule, so the number of optically active stereomers= 2n-1.

n = number of asymmetric C atom.

= 2^(2-1).

= 2^(-1)

= 2

Question 100. Which among the following statements is correct with respect to the optical isomers?

1. Enantiomers are non-superimposable mirror images
2. Diastereomers are superimposable mirror images
3. Enantiomers are superimposable mirror images
4. Meso forms have no plane of symmetry

Answer: 1. Enantiomers are non-superimposable mirror images

Solution: Enantiomers are non-superimposable mirror images. example, lactic  acid

Question 101. How many structural isomers are possible for C₄H₉CI?

1. 2
2. 4
3. 8
4. 10

Answer: 2. 4

Question 102. The number of isomeric alkenes with molecular formula C₆H₁₂ are

1. 8
2. 10
3. 11
4. 13

Answer: 4. 13

Solution: The number of isomeric alkenes with molecular formula C₆H₁₂ are 13.

Question 103. Tautomerism is not exhibited by

Answer: 1

Solution: For keto-enol isomerism a compound should have at least one a-hydrogen atom with respect to ketone group or in other words for tautomerism presence of the a-hydrogen atom is essential.

Does not exhibit tautomerism due to the absence of a-hydrogn atom.

Question 104. The compound which exhibits optical isomerism among the following is.

1. CH₃CHOHCH₃
2. (CH₃) ₂CHCH₂CH₃
3. CH₃CHClCH₂CH₃
4. CH₃CCl₂CH₂CH₃

Answer: 3. CH₃CHClCH₂CH₃

Solution: Due to asymmetric carbon atoms in it.

Question 105. Which is true about the following?

1. Only 3 is a chiral compound
2. Only 2 and 4 are chiral compounds
3. All four are chiral compounds
4. Only 1 and 2 are chiral compounds

Answer: 4. Only 1 and 2 are chiral compounds

Solution: The central carbon in 1 and 2 is asymmetric.

Question 106. Racemic compound has

1. An equimolar mixture of enantiomers
2. 1:1 mixture of enantiomer and diastereomer
3. 1:1 mixture of diastereomers
4. 1:2 mixture of enantiomers

Answer: 1. An  equimolar mixture of enantiomers

Solution: An equimolar mixture of the enantiomers dextro or laevo forms. is called a racemic mixture. It is represented as dl-form or ± form and is optically inactive due to the external compensation. The separation of racemic mixture into d- and l-forms is called as resolution.

Question 107. The number of stereoisomers obtained by bromination of trans-2-butene is?

1. 1
2. 2
3. 3
4. 4

Answer: 1. 1

Solution: Meso 2,3-dibromo butane is the product obtained upon bromination.

Question 108. Lactic acid shows optical activity in which state?

1. Solution state
2. Liquid state
3. Crystalline state
4. In all states

Answer: 4. In all states

Solution: Asymmetry is present in all the states.

Question 109. Which one of the starred carbons is the asymmetric one?

Answer: 1

Solution: It has altogether different groups.

Question 110. The number of isomers possible for the aromatic compound with the formula C₇H₈O? 

1. 2
2. 3
3. 4
4. 5

Answer: 4. 5

Solution: For the compound having formula C₇H₈O, there are 5 aromatic isomers.

1. Benzyl alcohol
2. o-methyl phenol
3. m-methyl phenol
4. p-methyl phenol
5. Anisole

Question 111. The number of isomers of the compound with molecular formula C₂H₂Br₂ is

1. 4
2. 3
3. 5
4. 2

Answer: 2. 3

Solution: The structure of isomers from C₂H₂Br₂ are CH₂=CBr₂

Question 112. The number of possible enantiomeric pairs that can be produced during monochlorination of 2-methyl butane

1. 2
2. 3
3. 4
4. 1

Answer: 1.  2

Solution: Two enantiomeric pairs are produced- If monochlorination occurs in no.1 carbon and no.3 carbon i.e., 1-Chloro-2-methyl butane and 2-Chloro-3-methylbutane.

Question 113. An alkane forms isomers if the number of carbon atoms are

1. ≥1
2. ≥2
3. ≥3
4. ≥4

Answer: 4. ≥4

Solution: Butane and isobutane and all higher alkanes show isomerism.

Question 114. Ethoxy ethane and methoxy propane are

1. Geometrical isomers
2. Optical isomers
3. Functional group isomers
4. Metamers

Answer: 4. Metamers

Solution: Ethers show metamerism.

Question 115. How many chiral compounds are possible on monochlorination of 2-methyl butane?

1. 2
2. 4
3. 6
4. 8

Answer: 2. 4

Solution:

Thus, out of four isomers only two have chiral carbon. Each has two isomers.

Question 116. Optical isomerism is shown by

1. Propanol-2
2. Butanol-2
3. Ethanol
4. Methanol

Answer: 2. Butanol-2

Solution: Optical isomerism is shown by an asymmetric carbon atom which has a carbon atom

Question 117. Which one of the following compounds will show optical isomerism?

1. (CH₃) ₂-CH-CH₂-CH₃
2. CH₃-CHOH-CH₃
3. CH₃-CHCl-CH₂-CH₃
4. CH₃-CCl₂-CH₂-CH₃

Answer: 3. CH₃-CHCl-CH₂-CH₃

Solution: Compound CH₃-CHCl-CH₂-CH₃ shows optical isomerism due to the presence of chiral carbon atoms.

Question 118. A molecule of urea can show

1. Chain isomerism
2. Position isomerism
3. Geometrical isomerism
4. Tautomerism

Answer:  4. Tautomerism

Solution: Urea shows tautomerism as

Question 119. If there is no rotation of plane polarised light by a compound in a specific solvent, though to be chiral, it means that:

1. It is certainly meso
2. It is a racemic mixture
3. It is certainly not chiral
4. None of the above

Answer: 1. It is certainly meso

Solution: Meso form is optically inactive.

Question 120. Which of the following does not show stereoisomerism?

Answer: 3

Solution: A, B and C will show stereoisomerism.

Question 121. An enantiomerically pure acid is treated with a racemic mixture of an alcohol having one chiral carbon. The ester formed will be

1. Optically active mixture
2. Pure enantiomer
3. Meso compound
4. Racemic mixture

Answer: 1. Optically active mixture

Solution:

• When optically active acid reacts with a racemic mixture of an alcohol, it forms two types of isomeric esters.
• In each, the configuration of the chiral centre of acid will remain thesame.
• So, the mixture will be optically active.

An enantiomerically pure acid is treated with a racemic mixture of an alcohol having one chiral carbon. The ester formed will be Optically active mixture

Question 122. How many chiral carbon atoms are present in 2, 3, 4- trichloro pentane?

1. 4
2. 1
3. 2
4. 3

Answer: 3. 2

Question 123. The type of isomerism observed in urea molecules is

1. Chain
2. Position
3. Geometrical
4. Functional

Answer: 4. Functional

Solution: NH₄CNO is a functional isomer of urea.

Question 124. The number of possible isomers for glucose are

1. 10
2. 14
3. 16
4. 20

Answer: 3. 16

Solution: Glucose has four dissimilar asymmetric carbon atoms; a = 24.

Question 125. The number of geometrical isomers in the case of a compound with the structure, CH₃ — CH=CH—CH=CH—C₂H₅ are

1. Four
2. Three
3. Two
4. Five

Answer: 1. Four

Solution: Two pairs of cis and transforms.

Question 126. The structures,CH₃—CH(NH₂)—CH₂—CH₂CH₃ and CH₃—CH₂—CH(NH₂)—CH₂CH₃ represent:

1. Chain isomers
2. Position isomers
3. Stereoisomers
4. Members

Answer: 2. Position isomers

Solution: 2-aminopentane and 3-aminopentane; Position is different.

Question 127. The total number of cyclic isomers possible for a hydrocarbon with molecular formula C₄H₆ is

1. 1
2. 3
3. 5
4. 7

Answer:  3. 5

Solution: C₄H₆ can have five cyclic isomers.

Question 128. Which of the following compounds exhibits geometrical isomerism?

1. C₂H₅Br
2. (CH)₂(COOH)₂
3. CH₃CHO(CH₂)₂
4. (COOH)₂

Answer: 2. (CH)₂(COOH)₂

Solution: (CH)₂(COOH)₂ actually represents HOOC−CH=CH−COOH which shows geometrical isomerism.

Question 129. The maximum number of alkene isomers for an alkene with molecular formula C₄H₈ are

1. 2
2. 3
3. 6
4. 5

Answer: 3. 6

Solution: The maximum number of alkene isomers for an alkene with molecular formula C₄H₈  are 6 namely, 1- butene, isobutylene, Cyclobutane, methyl cyclopropane, cis 2-butene, trans 2-butene.

Question 130. In fructose, the possible optical isomers are

1. 12
2. 8
3. 16
4. 4

Answer: 2. 8

Solution: Fructose has three chiral carbon atoms, hence the number of optical isomerism =2³

= 2×2×2

= 8

Question 131. The number of optical isomers of pent-3-en-2-ol.

1. 2
2. 4
3. 8
4. 16

Answer: 1. 2

Solution: CH₃CH = CH₂CHOHCH₃ has one asymmetric carbon.

Question 132. Which of the following acids shows stereoisomerism?

1. Oxalic acid
2. Tartaric acid
3. Acetic acid
4. Formic acid

Answer: Tartaric acid

Solution: In the case of 2,3-dihydroxybutanedioic acid, known as tartaric acid, the two chiral centres have the same four substituents and are equivalent. As a result, two of the four possible stereoisomers of this compound are identical due to a plane of symmetry, so there are only three stereoisomeric tartaric acids.

Question 133. How many types of functional groups can be present in an amine with the formula C₃H₉N?

1. 1
2. 2
3. 3
4. 4

Answer: 4. 4

Solution: 4 namely, propylamine, isopropyl amine, N-methyl ethylamine, and trimethylamine.

Question 134. The maximum number of stereoisomers possible for 3-hydroxy-2-methyl butanoic acid are

1. 1
2. 2
3. 3
4. 4

Answer: 4. 4

Solution:

Has two asymmetric carbon atoms and the molecule has no symmetry. Thus, the number of optical isomers = 2ⁿ

= 2²

= 4.

Question 135. An important chemical method to resolve a racemic mixture makes use of the formation of?

1. Mesocompound
2. Enantiomer
3. Racemers
4. Diastereoisomers

Answer: 4. Diastereoisomers

Solution: The resolution of the racemic mixture involves the formation of Diastereoisomers.

Question 136. Which one of the following pairs represents stereoisomerism?

1. Structural and geometrical isomerism
2. Linkage and geometrical isomerism
3. Chain and rotational isomerism
4. Optical and geometrical isomerism

Answer: 4. Optical and geometrical isomerism

Solution: Stereoisomerism is of two types, geometrical and optical.

Question 137. How many asymmetric carbon atoms are present in?

1. 1, 2-methylcyclohexane

2.  3-methyl cyclopentane and

3.  3-methylcyclohexane?

1. Two, one, one
2. One, one, one
3. Two, none, two
4. Two, none One

Answer: 1. Two, one, one

Solution: They contain two,, asymmetric carbon atoms respectively.

Question 138. Which class of compounds can exhibit geometrical isomerism?

1. C₆H₅CH = NOH
2. CH₃CH = CHCH₃
3. HOOCCH- CH₂– CHCOOH
4. All of the above

Answer: 4. All of the above

Solution: All of them show geometrical isomerism.

Question 139. Which type of isomerism is most common among the ethers?

1. Metamerism
2. Functional
3. Chain
4. Position

Answer: 1. Metamerism

Solution: Ethers show metamerism. Metamerism arises when a polyvalent functional group (For example, -O-,>C=O., etc.) is attached to different alkyl groups but the molecular formula remains the same for example, C₂H₅-O- C₂H₅ and CH₃-O-C₃H₇

Question 140. How many chiral isomers can be drawn from 2-bromo, 3-chloro butane?

1. 2
2. 3
3. 4
4. 5

Answer: 3. 4

Solution: 

∴ Number of asymmetric carbon atoms=2

∴ Number of chiral isomers = 2ⁿ

= 2²

= 4

Question 141. Identify the compound that exhibits tautomerism

1. 2-butene
2. Lactic acid
3. 2-pentanone
4. Phenol

Answer: 3. 2-pentanone

Solution:

Question 142. Which of the following may exist in enantiomorphs?

Answer: 4

Solution: Due to the presence of asymmetric carbon atoms.

Question 143. The two optical isomers given below, namely are

1. Enantiomers
2. Geometrical isomers
3. Diastereomers
4. Structural isomers

Answer: 3. Diastereomers

Solution: Diastereomers are a type of stereoisomer. Diastereomers are defined as non-mirror image, non-identical stereoisomers. Hence, they occur when two or more stereoisomers of a compound have different configurations at one or more of the equivalent stereo centres and are not mirror images of each other.

Question 144. CH₃CH₂OH and CH₃OCH₃are the examples of

1. Chain isomerism
2. Functional isomerism
3. Position isomerism
4. Metamerism

Answer: 2. Functional isomerism

Solution: CH₃CH₂OH and CH₃OCH₃ have different functional groups. (ie., -OH in alcohol and –O-in ether), hence they are examples of functional isomerism.

Question 145. Which is optically active?

1. Isobutyric acid
2. β-chloropropionic acid
3. Propionic acid
4. α-chloropropionic acid

Answer: 4. α-chloropropionic acid

Solution: CH₃CHClCOOH contains an asymmetric carbon atom.

Question 146. Compounds that rotate plane polarised light in a clockwise direction are known as?

1. Dextrorotatory
2. Laevorotatory
3. Optically inactive compounds
4. Racemic

Answer: Dextrorotatory

Solution: A characteristic of dextrorotatory.

Question 147. Maleic acid and fumaric acid are

1. Position isomers
2. Geometric isomers
3. Enantiomers
4. Functional isomers

Answer: 2. Geometric isomers

Solution: The structures of maleic and fumaric acids are given below

The structures of fumaric and maleic acid suggest that they are geometrical isomers because they have the same molecular formula but different spatial arrangements of atoms around a double bond.

Question 148. The total number of isomeric carbocations possible for the formula C₄H₉⁺

1. 3
2. 4
3. 2
4. 5

Answer: 2. 4

Solution: The possible carbocations are (CH₃)₃C⁺, (CH₃)₃CHC⁺H₂,CH₃CH₂CH₂C⁺H₂ and CH₃CH₂C⁺HCH₃

Question 149. Glucose and fructose are

1. Chain isomers
2. Position isomers
3. Functional isomers
4. Optical isomers

Answer: 3. Functional isomers

Solution: Glucose has an aldehyde group and a fructose keto group. The general formula for both is C₆H₁₂O₆.

Question 150. The compound isomeric with acetone is

1. Propionaldehyde
2. Propionic acid
3. Ethoxy ethane
4. None of these

Answer: 2. Propionaldehyde

Solution: Both have the same molecular formula.

Question 151. How many structural formulae are possible for C₅H₁₁Cl?

1. 6
2. 8
3. 10
4. 12

Answer: 2. 8

Solution: 1-chloropentane, 2-chloropentane, 3-chloropentane,1-chloro-3-methylbutane, 1-chloro-2-methylbutane, 1-chloro-2,2 dimethylpropane,2-chloro3-methylbutane, 2-chloro-2-methylbutane

Question 152. Total number of isomeric aldehydes and ketones that can exist with the molecular formula C₅H₁₀O :

1. 5
2. 8
3. 6
4. 7

Answer: 4. 7

Solution: Draw all possible isomers.

Question 153. Isomerism exhibited by acetic acid and methyl formate is

1. Functional
2. Chain
3. Geometrical
4. Central

Answer: 1. Functional

Solution: Acid has —COOH group whereas, ester has —COOR group.

Question 154. The number of isomeric structures for C₂H₇N would be?

1. 4
2. 3
3. 2
4. 1

Answer: 3. 2

Solution: CH₃CH₂NH₂and CH₃NHCH₃.

Question 155. Which of the following will have a meso-isomer also?

1. 2-chloroquine
2. 2, 3-dichlorobutane
3. 2, 3-dichloromethane
4. 2-hydroxy propanoic acid

Answer: 2. 2, 3-dichlorobutane

Solution:

One asymmetric carbon atom, forms d, and l-optical and isomers.

Two asymmetric carbon atoms, forms d, l and meso forms

Meso due to internal compensation

Two asymmetric carbon atoms but does not have symmetry. Hence, the meso form is not formed.

Question 156. 

Compound can exhibit

1. Geometrical isomerism
2. Tautomerism
3. Optical isomerism
4. Geometrical and optical isomerism

Answer: 3. Optical isomerism

Solution: A compound could be optically active only when it contains, at least one asymmetric carbon atom or a chiral centre.

Question 157. Select R-isomers from the following

1. 1 and 3
2. 2, 4 and 5
3. 1,2, and 3
4. 2 and 3

Answer: 3. 1,2, and 3

Solution: Draw the orientations.

Question 158. A similarity between optical and geometrical isomerism is that

1. Each forms an equal number of isomers for a given compound
2. If in a compound, one is present then so is the other
3. Both are included in stereoisomerism
4. They have no similarity

Answer:  3. Both are included in stereoisomerism

Solution: Both geometrical and optical isomerism are included in stereoisomerism.

Question 159. Acetone and propen-2-ol are?

1. Positional isomers
2. Keto-enol tautomers
3. Geometrical isomers
4. Chain isomers

Answer: 2. Keto-enol tautomers

Solution: Acetone has a ketone functional group and propenal have an aldehyde functional group therefore, these are functional isomers.

Question 160. How many isomers are possible for the compound having a molecular formula C₃H₅Br₃?

1. 5
2. 4
3. 6
4. 8

Answer: 1. 5

Solution: Draw position and chain isomers.

Question 161. Which one of the following shows functional isomerism?

1. C₂H₄
2. C₃H₆
3. C₂H₅OH
4. CH₂Cl₂

Answer: C₂H₅OH

Solution: Ethyl alcohol shows functional isomerism with dimethyl ether.

C₂H₅OH – Alcohol

CH₃-O-CH₃ – Ether 

Question 162. How many isomers of C₅H₁₁OH will be primary alcohols?

1. 5
2. 4
3. 3
4. 2

Answer: 2. 4

Solution: A total of 8 structural isomers of pentanol i.e., 1-pentanol, 2-pentanol, 3-pentanol, 2-methylbutan-1-ol, 3-methylbutan-1-ol, 2-methylbutan-2-ol, 2-methylbutan-3-ol and 2,2-dimethylpropanol are possible. Out of these 8, 4 are primary alcohols.

Question 163. The molecular formula of a saturated compound is C₂H₄Br₂. This formula permits the existence of which isomers?

1. Functional isomers
2. Optical isomers
3. Positional isomers
4. Cis-trans isomers

Answer: Positional isomers

Solution: 1,1-dibromoethane and 1,2-dibromoethane.

Question 164. C₆H₁₂ on the addition of HBr in the presence and in the absence of peroxide gives _______product.

1. Hexene-3
2. 2,3-dimethyl butane-2
3. Symmetrical alkene
4. All of these

Answer: 4. All of these

Solution: There are two symmetrical hexenes as given in (1) and (2).

Question 165. Buta-1,3-diene and But-2-yne are:

1. Position isomers
2. Functional isomers
3. Chain isomers
4. Tautomers

Answer: 2. Functional isomers

Solution: But-2-yen and Buta -1, 3-diene both have the same molecular formula of C₄H₆. In both of them, the functional group is different but the formula is the same. – Hence, they show functional isomerism.

Question 166. Which of the following compounds are optically active?

1. (CH₃)₂CHCH₂OH
2. CH₃CH₂OH
3. CCl₂F₂ CH₃
4. CHOHC₂H₅

Answer: 4. CHOHC₂H₅

Solution: CH₃CHOHC₂H₅ is optically active because it has chiral C*-atom

Question 167. Compounds whose molecules are superimposable on their mirror images even though they contain asymmetric carbon atoms or chiral centres are known as

1. Enantiomers
2. Racemers
3. Mesomers
4. Conformers

Answer: 3. Mesomers

Solution: Mesoforms are optically inactive as they are superimposable to their mirror images.

Question 168. Position isomerism is shown by

1. O-nitrophenol and p-nitrophenol
2. Dimethyl ether and ethanol
3. Pentan-2-one and pentan-3-one
4. Acetaldehyde and acetone

Answer: O-nitrophenol and p-nitrophenol

Solution: O-,m-,p- isomers are position isomers.

Question 169. The number of stereoisomers possible for a compound of the molecular formula

CH₃-CH=CH-CH(OH)-Me is

1. 3
2. 2
3. 4
4. 6

Answer:

Solution: There are four stereoisomers cis-R cis-S trans-R trans-S

Question 170. Which of the following will exhibit cis-trans isomerism?

1. CH₂Br–CH₂Br
2. CBr₃–CH₃
3. CHBr=CHBr
4. CBr₂=CH₂

Answer: 3. CHBr=CHBr

Solution: Due to the restricted rotation of a double bond, the alkene shows geometrical isomerism because the relative position of atoms or groups attached to the carbon atoms of the double bond gets fixed.

If the same groups or atoms are attached to the double bond bearing carbon, then alkene doesn’t show geometrical isomerism.

Question 171. The total number of cyclic structural as well as stereoisomers possible for a compound with the molecular formula C₅H10 is

1. 2
2. 4
3. 6
4. 7

Answer: 3. 6

Solution: The total number of cyclic isomers is six as shown below.

Question 172. The maximum number of possible optical isomers in 1-bromo-2-methylcyclohexane is

1. 4
2. 2
3. 8
4. 16

Answer: 1. 4

Solution: The molecule contains two carbons. The number of optical isomers is given by 2n, where n=No. of chiral carbons.

∴ Optical isomers 2² = 4.

Question 173. How many stereoisomers does this molecule have? CH₃CH=CHCH₂CHBrCH₃

1. 6
2. 8
3. 4
4. 2

Answer: 4. 2

Solution:

It has one chiral centre (two enantiomers) and two geometrical isomers cis⎼d, trans⎼d, cis⎼ and trans⎼l.

Question 174. How many optically active stereoisomers are possible for butane-2, 3-diol?

1. 0
2. 1
3. 2
4. 3

Answer: 4. 3

Solution: The structure of butane-2, 3-idol is as

∵ Optical isomers in compounds have similar asymmetric carbon atoms, which are even in number =2n-1

Here, n = 2

∴ Total number of optically active stereoisomers =

= 2(2)-1

= 4- 1=  3

Question 175. The number of isomeric pentyl alcohols are

1. Two
2. Four
3. Six
4. Eight

Answer:  4. Eight

Solution: n-pentanol, 2-pentanol, 3-pentanol, 2-methylbutanol, 2-methylbutan-2-ol, 3-methylbutanol, 2, 2-dimethypropanol, and 3-methylbutan-2-ol (8 isomers)

Question 176. Which of the following is most likely to show optical isomerism?

Answer: 2

Solution: Optical isomerism is shown by compounds which have one or more chiral carbon atoms.

∵ It has asymmetric or chiral carbon atoms,

∴ It shows optical isomerism.

Question 177. Which type of isomerism is shown by propanal and propanone?

1. Functional group
2. Metamerism
3. Tautomerism
4. Chain isomerism

Answer: 1. Functional group

Solution: When two compounds have similar molecular formulas but differ in the functional group then the isomerism is called functional group isomerism i.e.,

Question 178. The number of isomers possible for C4H8O is

1. 3
2. 4
3. 5
4. 6

Answer: 4. 6

Solution: There are six isomers possible for the compounds having molecular formula C₄H₈O, which are as follows

Question 179. Which of the following statements are correct?

1. Desmotropism is another name for tautomerism
2. Allyl carbocation is less stable than isopropyl carbocation
3. -I effect is exhibited by -NH₃⁺
4. The formula CH₂Cl₂ is non-polar

Answer: 1. Desmotropism is another name for tautomerism

Solution: Desmo (bond), tropism (turn). Thus, desmotropism, i.e., isomerism arised due to the turning of bond was the name given to tautomerism.

Question 180. Which of the following structures are superimposable?

1. 1 and 2
2. 2 and 3
3. 1 and 4
4. 1 and 3

Answer: 4. 1 and 3

Solution: (1) and (3) are enantiomeric forms of each other.

Question 181. The alkene that exhibits geometrical isomerism is

1. Propene
2. 2-methylpropene
3. 2-butene
4. 2-methyl-2-butene

Answer: 2 -butene

Solution: 2-Butene may exist as cis and trans isomers. The cis-isomer has the two methyl groups on the same side and the trans-isomer has the two methyl groups on opposite sides. Due to restricted rotation around the double bond, it exhibits geometrical isomerism.

Question 182. The isomers which are interconverted through rotation around a single bond are

1. Conformers
2. Diastereomers
3. Enantiomers
4. Position isomers

Answer: 1. Conformers

Solution: The isomerism which arises due to rotation about a C-C is called conformational isomerism and the isomers are called conformational isomers or rotational isomers or conformers.

Question 183. The total number of acyclic isomers including the stereoisomers with the molecular formula C₄H₇Cl

1. 11
2. 12
3. 9
4. 10

Answer: 2. 12

Question 184. n-pentane and neopentane exhibit

1. Functional isomers
2. Geometrical isomers
3. Chain isomers
4. Position isomers

Answer: 3. Chain isomers

Solution: N-pentane and isopentane or 2-methylbutane are chain isomers since both have different hydrocarbon chains.

Question 185. Geometrical isomerism is shown by

1. – C –C –
2. >C=C<
3. C≡C
4. None of these

Answer: 2. >C=C<

Solution: Geometrical isomerism is shown by >C=C< only when identical groups are not present on the double bonded carbon atoms.

Question 186. Example of geometrical isomerism is

1. 2-butanol
2. 2-butene
3. Butanal
4. 2-butyne

Answer: 2. 2-butene

Solution: 2-Butene may exist as cis and trans isomers. The cis-isomer has the two methyl groups on the same side and the trans-isomer has the two methyl groups on opposite sides. Due to restricted rotation around the double bond, it exhibits geometrical isomerism.

Question 187. Two crystalline forms of a substance, one being a mirror image of the other are called?

1. Pentane
2. Chain isomers
3. Stereoisomers
4. Functional isomers

Answer: 3. Stereoisomers

Solution: The mirror-image isomerism is a class of stereoisomerism and is included in optical

isomerism.

Question 188. Which pair represents chain isomers?

1. CH₃CHCl₂ and ClCH₂CH₂Cl
2. Propyl alcohol and isopropyl alcohol
3. 2-methyl butane and neopentane
4. Diethyl ether and dipropyl ether

Answer: 3. 2-methyl butane and neopentane

Solution: Note that propyl (propan-l-ol) and isopropyl alcohol (propan-2-ol) are position isomers.

Question 189. Different structures generated due to rotation about, the C – C axis, of an organic molecule, are examples of

1. Geometrical isomerism
2. Conformational isomerism
3. Optical isomerism
4. Structural isomerism

Answer: 2. Conformational isomerism

Solution: The different arrangement of atoms in space that results from the carbon-carbon single bond free rotation by 360° are called conformations or conformational isomers and this phenomenon is called conformational isomerism.

Question 190. Which of the following compounds is optically active?

1. 1 – butanol
2. Isopropyl alcohol
3. Acetaldehyde
4. 2-butanol

Answer: 4. 2-butanol

Solution: 2-butanol is optically active as it contains a chiral carbon atom.

Question 191. Which of the following will have meso isomers also?

1. 2-hydroxy propanoic acid
2. 2,3-dichlorobutane
3. 2,3-dichloromethane
4. 2-chloroquine

Answer: 2. 2,3-dichlorobutane

Solution: 2,3-Dichlorobutane has a meso-isomer due to the presence of the plane of symmetry.

Question 192. The isomerism which exists between CH₃CHCl₂ and CH₂ClCH₂Cl is

1. Chain
2. Functional
3. Positional
4. Metamerism

Answer: 3. Positional

Solution: The positions of Cl are different.

Question 193. Alkyl cyanide R-C≡N and alkyl isocyanides R–N≡C are

1. Tautomers
2. Metamers
3. Functional isomers
4. Geometrical isomers

Answer: Functional isomers

Solution: Both have different functional groups, i.e., –CN and –NC.

Question 194. Out of the following, the alkene that exhibits optical isomerism is

1. 3-methyl-2-pentene
2. 4-methyl-1-pentene
3. 3-methyl-1-pentene
4. 2-methyl-2-pentene

Answer: 3-methyl-1-pentene

Solution: For a compound to show optical isomerism, the presence of chiral carbon atoms is a necessary condition. H₂C=HC−H|C∗|CH₃  −CH₂−CH₃ 3-methyl-1-pentene

Question 195. Which of the following compounds will show metamerism?

1. CH₃– CO-C₂H₅
2. C2H5-S-C₂H₅
3. CH₃-O-CH₃
4. CH₃-O-C₂H₅

Answer: 3. CH₃-O-CH₃

Solution: Compounds having bivalent functional groups (like C=O, –O–, –S – etc) with at least 4 carbon atoms (in the case of ether and thioether) or at least 5 carbon atoms (in the case of ketones) exhibit metamerism. Hence, C₂H₅-S-C₂H₅ will show metamerism.

Question 196. Which of the following compounds exhibit stereoisomerism?

1. 3-methyl butyne –1
2. 2-methyl butene –1
3. 2-methyl butanoic acid
4. 3-methyl butanoic acid

Answer: 3. 2-methyl butanoic acid

Solution: 2-methyl butanoic acid exhibits stereoisomerism.

It shows optical isomerism because it contains an asymmetric carbon atom.

Question 197. Who proposed the tetrahedral mirror image structures to a pair of enantiomers?

1. Kekule
2. Wohler
3. Van’t Hoff
4. None of these

Answer: 3. Van’t Hoff

Solution: It is a fact.

Question 198. Geometrical isomers differ in

1. Position of functional groups
2. Position of atoms
3. Spatial arrangement of atoms
4. Length of the carbon chain

Answer: 3. Spatial arrangement of atoms

Solution: It is a fact.

Question 199. The number of isomeric alkanes having the molecular formula C₂H₁₂ is

1. Three
2. Five
3. Nine
4. Thirty-two

Answer: 1. Three

Solution: The isomers alkanes having the molecular formula C₂H₁₂ are as

Question 200. Two crystalline forms of a substance, one being a mirror image of the other are called

1. Pentane
2. Chain isomers
3. Stereoisomers
4. Functional isomers

Answer: 3. Stereoisomers

Solution: is a form of isomerism in which molecules have the same molecular formula and sequence of bonded atoms, but differ in the three-dimensional orientations of their atoms in space.

Question 201. A molecule having three different chiral carbon atoms, how many stereoisomers will it have?

1. 8
2. 3
3. 9
4. 6

Answer: 1. 8

Solution: The number of stereoisomers=2’

(Here, n=chiral carbon atom) Thus, the number of stereoisomers

= 2³

= 2×2×2

= 8

The mirror-image isomerism is a class of stereoisomerism and is included in optical isomerism.