NEET Chemistry

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:

1. Empirical formula and
2. 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

 

 

 

 

 

 

 

 

 

 

 

 

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.

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

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.

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:

1. Vapourisation: Liquid is converted into vapours.
2. 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,

1. Benzene (boiling point 353 K) and aniline (boiling point 475 K)
2. Chloroform (boiling point 334 K) and aniline (boiling point 457 K)
3. 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.

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.

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

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

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.

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

1. Nucleophilic substitution (S_N) reactions: The substitution reactions which are brought about by the attack of nucleophiles are called 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: \(\mathrm{Cl}_2+\mathrm{FeCl}_3 \rightarrow \mathrm{FeCl}+\mathrm{Cl}^{+}\)

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.

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.

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

R′=H or alkyl (–CH₃, –CH₂CH₃, etc.) or aryl (C₆H₅–) or aralkyl (C₆H₅CH₂–)

Addition to Grignard reagent

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.

 

1. Sodamide (NaNH₂) is a stronger dehydrohalogenating agent.
2. An example of 1,4-elimination (δ-elimination) is

Dehalogenation: Dehalogenation involves the removal of halogen molecule (X₂) from vicinal dihalide by heating with zinc dust in the alcoholic medium.

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.

α-elimination: It involves loss of two atoms from same atom resulting in the formation of reaction intermediates like carbine

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.

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”.

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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.

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

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

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.

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

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

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

Carbocations undergo rearrangement forming more stable 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.

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:

The order of reactivity of carbanions is reversed that of its stability. Therefore order of reactivity of carbanions follows the sequence:

Hybridised state of carbon: Higher the s character greater the stability of carbanion. Order of stability is

Carbanions Resonance: Greater the delocalisation of negative charge by resonance, higher the stability The stability of benzyl carbanion is comparable with allyl carbanions.

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,

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.

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.

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

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.

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.

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.

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.

The common examples of nucleophiles are given below:

Negative nucleophiles:

Neutral nucleophiles:

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.

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.

The order of electron releasing or releasing ability of substituent is given below.

Negative inductive effect (-I effect): In this effect, the sigma bond pair of electrons are displaced towards the electron-withdrawing substituent (X).

The order of electron-withdrawing ability (intensity of -I effect) of a few substituents is given below.

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.

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.

The substituents, which exert only -R effect are given below.

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 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.

Propene molecule may be regarded as the resonance hybrid of the following hyperconjugative structures.

• 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.

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.

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.

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.

+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

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.

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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.

3. Cyclohexane and methyl cyclopentane are nuclear isomers.

4. C₅H₁₂ has three chain isomers.

5. C₄H₉NH₂ also shows two chain isomers.

Solve 1: How many chain isomers does butane have?
Solution:

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:

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

Primary, secondary, and tertiary amines of same molecular formula are also functional isomers.

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)

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:

1. Alkenes
2. Alkynes
3. Arenes
4. Alkyl halides
5. Aryl halides
6. Alcohols
7. Amines and
8. 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:

1. CH₃CH₂CH = CH₂ (but-1-ene) ; CH₃CH =CHCH₃ (But-2-ene)
2. CH₂CH₂CH₂CH = CH₂ (Pent-l-ene); CH₃CH₂CH = CHCH₃ (Pent-2-ene)
3. 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:

1. CH₃CH₂C = CH₂ (But-l-yne); CH₃C = CCH₃ (But-2-yne)
2. 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:

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:

2. n – Pentane on monochlorination gives three isomeric chloromethanes:

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:

Polyhalogen compounds containing just one benzene ring can also exhibit position isomerism. For example:

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)

2. Four monohydric alcohols differing in the position of -OH group can be derived from isopentane:

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:

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)

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.

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:

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:

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:

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:

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.

1. 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.
2. 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.

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:

Six pairs of ring-chain isomers are possible for the molecular formula C₄H₈:

The molecular formula C₃H₄ represents the two ring-chain isomers:

Ring chain isomerism can be included in functional isomerism, if not considered separately.

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.

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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1. 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.
2. 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.
3. 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.

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:

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:

Special names 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.

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).

Naming of disubstituted benzene Example:

If the two substituents are different: The substituents are named in alphabetical order.

If there are more than two substituents: The numbering is done in such a way that it satisfies the lowest sum rule.

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.

1. 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.
2. 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.
3. 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:

Nomenclature Of Organic Compounds

The nomenclature deals with the naming of millions of organic compounds. The following systems are employed.

1. Trivial system (common system): It is the oldest system of naming organic compounds. In the early stages of the development of organic chemistry, organic compounds were named after the source from which they were first isolated. Generally, the names chosen had Latin or Greek roots. The following illustrations justify the statement.

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2. IUPAC System – Saturated hydrocarbons are the parent compounds and other organic compounds as their derivatives obtained by substituting one or more hydrogen atoms with functional groups. IUPAC name of any organic compound may consist of three parts. “prefix-root word-suffix”.

Root word- is assigned to organic molecule based on the number of carbon atoms present in the main parent chain

Suffix – is based on the nature of bonds and functional group present in the molecules. It is of 2 types,

(1) Primary suffix: A primary suffix is added next to the root word to indicate whether the parent chain is saturated or unsaturated.

Note: A carbon-carbon double bond or triple bond must be included in parent chain even if it contains less number of carbon atoms.

(2) The secondary suffix added next to the primary suffix to indicate the presence of a functional group in organic compounds which determines the class of organic compounds.

Secondary suffixes of a few functional groups are given.

Note – While adding the secondary suffix, the letter ‘e’ of the primary suffix (i.e., ane, ene, and yne) is dropped if the secondary suffix begins with a vowel (a, e, i, o, or u). It is retained if a secondary suffix begins with a consonant.

3. Prefix (substituent): All the groups which are not names in parent chain and functional groups are called as substituents. Its name placed before the root word. Example: – Alkyl group, halo atoms, nitro group

Alkyl Groups

Alkyl groups  derived from an alkane, by removing a hydrogen atom bonded to carbon. These groups are named simply by dropping -ane from the name of the corresponding alkane and replacing it by -yl. R is a general symbol, general formula for an alkyl group is C_nH_2n+1, because it contains one less hydrogen atom than the parent alkane, C_nH_2n+2.

Among the alkyl groups we encounter the problem of isomerism.

• While only one alkyl group can be derived from methane (the methyl, CH₃-)
• Ethane (the ethyl, CH₃CH₂-)
• Two or more alkyl groups can be derived from higher alkanes.

Alkyl groups Examples

1. From propane (C₃H₈), two alkyl groups can be derived. Removal of one of the hydrogens from one of the end carbon atoms gives an alkyl group that is called the propyl group or n-propyl group.

Removal of one of the hydrogens from the middle carbon atom gives an alkyl group that is called the isopropyl group.

Both these alkyl groups contain the propane chain but differ in the point of attachment of the group to the rest of the molecule.

2. There are four butyl groups, two derived from the straight-chain n-butane, and two derived from the branched-chain isobutane. These are given the designations: n- (normal), sec¬secondary), iso-, and tert- (tertiary) as shown below:

An alkyl group is described as,

• Primary if the carbon at the point of attachment is bonded to only one other carbon,
• As secondaryif bonded to two other carbons,
• Tertiary if bonded to three other carbons. Thus, if r is any hydrocarbon radical, the different kinds of alkyl groups are

RCH₂ -Primary

R₂CH– Secondary

R₂C- Tertiary

Alkyl Group Problems

Problem 1: Classify each of the following alkyl groups as primary, secondary, or tertiary: Butyl, Isopropyl, Isobutyl, sec-butyl, and tert-Butyl.

Solution:

EPrefix n-(normal) is used for those alkanes in which all the carbon atoms form a continuous chain with no branching.

CH₃CH₂CH₂CH₃ (n-butene)

CH₃CH₂CH₂CH₂CH₃ (n-pentane)

EPrefix iso is used for those alkanes in which one methyl group is attached to the next-to-end carbon atom (second last) of the continuous chain.

EPrefix neo is used for those alkanes which have two methyl groups attached to the second last carbon atom of the continuous chain.

Rules for naming of open chain Organic compound.

Rule 1 – Longest chain rule (root word rule): The continuous carbon chain containing maximum carbon atoms including the function group is selected. It is called the parent chain. For one to four carbon atoms of the parent chain, special root words are used but for chains of five or more carbon atoms Greek number roots are employed.

The generic root word for any parent chain is ‘alk’.

Similarly, the following hydrocarbon will be regarded as a substituted hexane because the longest continuous carbon chain (parent chain/ root chain) contains six carbon atoms.

Note – If two different chains of equal length are possible, the chain with maximum number of side chains or alkyl groups is selected. The number used to specify the position of the substituents is called locant.

Rule 2 – Lowest sum rule (rule of locant): Parent carbon chain is numbered using Arabic numerals 1, 2, 3, 4, and 5 in such a way that functional groups or substituents containing carbon receive least number.

E When the parent chain contains two or more substituents, the numbering is done from the end where the sum of the locants is least.

E Parent chain with one substituent: Prefix the name of the substituent to the root word of parent chain and indicate its position. The name of the substituent is separated from its locant by a hyphen (-). The name of the organic compound is written as one word.

E Naming of different substituents: When two or more different substituents are present on the parent chain, they are named in alphabetical order along with their appropriate locants.

E Numbering of different substituents at equivalent positions: If two different alkyl or halo groups are present at equivalent positions, the numbering of the parent chain is done in such a way that the substituent which comes first in the alphabetical order gets the lower number.

E Presence of the same substituent more than once: If the same substituent occurs more than once, the prefixes di, tri, tetra are prefixed to the name of the substituent. It may be noted that the position and name of the substituent are separated by a hyphen (-) whereas the numerals representing the positions of the substituents are separated by commas.

E Naming a complex substituent: In case the substituent on the parent chain is complex (i.e., it has a branched chain), it is named as a substituted alkyl group by numbering the carbon atom of this group attached to the parent chain as 1. The name of complex substituent is enclosed in the bracket to avoid confusion with the numbers of parent chain.

If same complex substituent occurs more than once on the parent chain, prefixes bis, tris, tetrakis, etc., are used before the name of the complex substituent. The application of IUPAC rules to the structure of a molecule to arrive at IUPAC name is illustrated as follows.

• Parent chain has 5 carbon atoms, and it has more substituents. The root word is ‘pent’.
• The primary suffix is ‘ane’
• Sum of locants is 9, the substituent that comes first in alphabetical order gets the lower number.
• Prefixes to the rootword are bromo, ethyl, and methyl. Hence, IUPAC name is “2-bromo-3-ethyl-4-methyl pentane”

A few examples are given below:

 

Functional Groups

A functional group is defined as “an atom or group of atoms present in a molecule which decides the chemical properties of the organic compounds”. Double and triple bonds are also considered as functional groups. The organic compounds are further classified into different classes depending upon the nature of functional group. Classification of organic compounds based on functional group is summarised in the table.

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

Homologous Series

A series of structurally related organic compounds with a same functional group can be represented by a general formula and differ by -CH₂ group is called homologous series. The individual members of such series is called homolouge and this phenomenon is known as homology.

Characteristics of a homologous series are as follows.

• Each homologues series can be represented by general molecular formula.
• Exp – General molecular formula of alkane is C_nH_2n+2. Molecular formula C_nH_2nO₂ represents alkanoic acids (carboxylic acids) and alkyl alkanoates (esters).
• All the members of a given homologous series (homologues) possess the same functional group.
• The successive members of a homologous series differ by a -CH₂– group or by mass units.
• Homologues of a homologous series can be prepared by general methods.
• The physical properties such as boiling point, melting point, density, etc., of the members of a homologues series show a regular gradation with increase in molecular mass.
• Chemical properties of the members of a homologous series are similar though the first homologue may vary considerably from the rest of the homologues.

Classification Of Organic Compounds

The compounds with C and H are called hydrocarbons. Organic compounds are considered as the derivatives of hydrocarbons obtained by replacing H-atoms by other atoms or groups. Based on the C – skeleton, organic compounds are divided as below,

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

Acyclic Compounds

The open-chain organic compounds are called acyclic compounds. The carbon chains may be linear or branched. The open-chain compounds are classified into two groups.

Saturated compounds: Open-chain compounds in which carbon atoms of parent chain are bonded with single bonds are called saturated acyclic compounds.

Saturated compounds Example:

Unsaturated compounds: The open-chain organic compounds in which parent chain contains one or more carbon-carbon double bonds or triple bonds are called unsaturated acyclic compounds.

Unsaturated compounds Example:

Cyclic Compounds

Here carbon atoms are linked by covalent bonds to form ring structures are called cyclic compounds. Depending upon the constituents of ring, these compounds are further classified into two groups.

Carbocyclic compounds(Hoinocydiccoinpoimds): The cyclic compounds containing only carbon atoms in their ring are called carbocyclic compounds. They are further divided into two groups.

Alicyclic compounds

The carbocyclic compounds which resemble with aliphatic compounds in most of their properties are called alicyclic compounds.

Aromatic Compounds

Benzene is the parent aromatic hydrocarbon. The organic compounds containing one or more isolated or fused benzene rings and their functionalised derivatives are called aromatic compounds or benzenoid compounds.

Aromatic compounds example:

The aromatic compounds that resemble benzene in most of their properties are called non-benzenoid aromatic compounds

Non-Benzenoid Aromatic compounds For example:

Heterocyclic Compounds

The cyclic compounds containing one or more heteroatoms (atom other than C it may be O, N, S, etc.,) in their ring are called heterocyclic compounds. Depending upon their chemical behaviors, they are further classified into two groups.

Alicyclic heterocyclic compounds: The aliphatic cyclic compounds containing one or more hetero atoms in their ring are called alicyclic heterocyclic compounds.

Alicyclic heterocyclic compounds For example:

Aromatic heterocyclic compounds: The aromatic cyclic compounds containing one or more hetero atoms in their molecules are called aromatic heterocyclic compounds.

Aromatic heterocyclic compounds For example:

General Organic Chemistry Introduction

‘The branch of chemistry dealing with these compounds, which are widely distributed in nature and play an important role in our daily lives’, is called organic chemistry.

In the earlier period of development of chemistry, compounds were classified into two types:

1. Organic compounds derived from ‘living matter’ (plants and animals).
2. Inorganic compounds prepared from ‘non-living matter’ (mineral sources).

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

Berzelius, a Swedish chemist proposed the mistaken notion that a ‘vital force’ present in living matter was essential for the synthesis of organic compounds. However, the synthesis of urea an organic compound present in urine, from ammonium cyanate, an inorganic compound by Frederich Wohler in 1828 effectively destroyed the myth of organic compounds being associated with a ‘vital force’.

Soon afterwards the pioneering work of Herman Kolbe who synthesized acetic acid and of Berthelot who synthesized methane showed conclusively that organic compounds are essentially the compounds formed by carbon with itself and other elements and that they can be synthesized in a laboratory as easily as inorganic compounds.

The chemistry of hydrocarbons and their derivatives constitutes organic chemistry.

The number of organic compounds available today are more compare to total inorganic compounds of all elements except carbon. This is due to unique catenation property of Carbon.

Catenation: is the property of an element where a large number of its own atoms join together through covalent bonds. Due to which it forms single as well as multiple covalent bonds with other carbon atoms.

It is further supplemented by the fact that it also forms covalent bonds with atoms of other elements like hydrogen, oxygen, nitrogen, Sulphur, phosphorus, and halogens in a variety of ways (i.e., single and multiple bonds). This property gives a scope for the Chemists to synthesize new compounds.

Tetravalency Of Carbon Atom

The atomic number of carbon is 6 and it has four electrons in its valence shell. In order to acquire a stable noble gas configuration, it can share its 4 electrons with the electrons of its atom or electrons of other atoms to form four covalent bonds. The bonds can be sigma(σ) or pi(π).

The type of bond determines the hybridization of carbon atoms and the geometry of the molecules.

Hybridisation In Carbon Compounds

Hybridization is defined as the intermixing of degenerate orbitals (orbitals at nearly the same energy) to produce an entirely equivalent number of new orbitals of same energy, identical shapes and symmetrically disposed of in planes. The orbitals formed are called hybrid orbitals.

• The orbitals of an isolated atom can undergo hybridization.
• Number of hybrid orbitals generated are equal to number of contributing atomic orbitals.
• The hybrid orbitals orient in space providing definite geometry to molecules or ion.
• Like atomic orbital, a hybrid orbital cannot have more than two electrons of opposite spins.

There are three types of hybridization,

1. sp³ hybridization (contain saturated organic compounds with only single covalent bonds)
2. sp² hybridization (here organic compounds having carbon atoms linked by double bonds)
3. sp hybridization (here organic compounds having carbon atoms linked by a triple bonds).

Prediction of hybridization It can be done by two methods,

1. First Method: In this method hybridization can be known by the number of bonds present on that particular atom.

First Method Examples:

2. Second Method: (Electron pair method)

The hybridized state of an atom of a molecule or an ion or radical can be predicted by calculating number of orbitals or electron pairs involved in hybridization (H) which is evaluated as follows.

H = (number of a bonds formed with adjected atom/s + number of lone pairs of electrons)

∴ ep = bp + lp;

where, ep = electron pair present in hybrid orbitals,

bp = bond pair present in hybrid orbitals

Number of bp = Number of atoms attached to the central atom of the species (do not include π electron pairs).

Number of lp’s can be determined as follows,

1. If carbon has – bonds or positive charge or odd electron, thanlp on carbon will be zero.
2. If carbon has negative charge, then Ip will be equal to one.

Second Method Example:

Applications Of Hybridisation

1. Size of the hybrid orbitals: Since – orbitals are closer to the nucleus than – orbitals, it is reasonable to expect that greater the character of an orbital the smaller it is. Thus, the decreasing order of the size of the three hybrid orbitals is opposite to that of the decreasing order of orbital character in the three hybrid orbitals. sp³ > sp² > sp

2. Electronegativity of different orbitals

1. The electronegativity of s-orbital is maximum.
2. Electronegativity of hybrid orbital ∝ % s-character in hybrid orbitals
   • Thus Sp-hybrid carbon is electronegative and it is electropositive in character and sp³-hybrid carbon can behave as electropositive in character sp² -hybrid carbon can behave as electropositive (in carbocation) as well as electronegative (in carbanion) in character.
3. Electronegativities of different hybrid and nonhybrid orbitals in decreasing order is as follows

3. ond length variation in hydrocarbons

⇒ \(\% \text { of s orbital character } \propto \frac{1}{\mathrm{C}-\mathrm{C} \text { bond length }} \propto \frac{1}{\mathrm{C}-\mathrm{H} \text { bond length }}\)

4. Bond strength in hydrocarbons: The shorter is the bond length, the greater is the compression between atomic nuclei, and hence greater is the strength of that bond.

Structural Representation Of Organic Compounds

Structural formula (structure) is the sequence in which different atoms constituting the molecule are bonded to one another. Structures of molecules of organic compounds can be described in various ways.

The most common types of representations are:

1. Lewis structure (or electron dot structure) Here dots are used to represent all of the valence electrons of all the bonded atoms in the molecule:

Writing dot structure is tedious and time-consuming. The other representations are more convenient and are, therefore, more often used.

2. Dash structural formula The Lewis structure can be simplified by representing a shared electron pair by a ‘stick’ (dash, -) between the bonded atoms. When there is one dash between two atoms, the atoms are said to be bonded by a single covalent bond.

• A double covalent bond, in which two pairs of electrons are shared, is shown by two dashes between the atoms. A triple bond is represented by three dashes between the atoms. The valence electrons that are not included in covalent bonds are called nonbonding electrons (lone pairs). These are assigned to specific atoms and are represented by dots drawn next to the symbols for these atoms.
• Lone pair of electrons on hetero atoms (for example, oxygen, nitrogen, sulphur, phosphorus, halogens) may or may not be shown. Such structural formulae which focus only on the valence electrons involved in bond formation are called complete structural formulae.

3. Condensed structural formula Complete structural formulae can be shortened by leaving out some or all of the covalent bonds and by denoting the number of identical groups attached to an atom by a subscript. The resulting description of the molecule is called a condensed structural formula. Thus,

CH₃CH₃ or C₂H₆ Ethane

H₂C = CH₂ or C₂H₄ Ethene

HC ≡ CH or C₂H₂ Ethyne

CH₃CH₂Cl or C₂H₅Cl Ethyl Chloride

CH₃CH₂CH₂OH or CH₃(CH₂)₂OH Methanol

CH₃CH₂CH₂CH₂CO₂H or CH₃(CH₂)₃CO₂H Pentanoic Acid

Condensed structural formulae are easier to write than dash formulae. In condensed formulae, all of the hydrogen atoms that are attached to a particular carbon are usually written immediately after the carbon. In fully condensed formulae, all of the atoms that are attached to the carbon are usually written immediately after that carbon, listing hydrogens first.

For example, the condensed formula for isopropyl alcohol can be written in four different ways:

4. Bond-line structural formula: For further simplification, only lines are used to represent the structures of organic molecules. In this bond-line structural formula of organic molecules, carbon, and hydrogen atoms are not shown. The lines denoting the carbon-carbon bonds are drawn in a zig-zag fashion. The only atoms especially written are the hetero atoms (oxygen, nitrogen etc). They are neither carbon nor hydrogen bonded to carbon.

The termini describe methyl (CH₃-) groups (unless denoted otherwise by a functional group). The line junctions indicate carbon atoms bonded to suitable number of hydrogens needed to satisfy the covalency of the carbon atoms: Thus

The bond-line representation is the quickest of all to write because it shows only the carbon skeleton.

Ring or cyclic compounds: An organic compound in which carbon atoms are not bonded in chain but are bonded in closed structures called rings are known as cyclic compounds. Such a compound containing one or more rings is represented by drawing the suitable ring (polygon) without indicating the carbon and hydrogen atoms. The corner of the polygon denotes a carbon atom and its sides represent a carbon-carbon bond. An atom or a group of atoms (other than hydrogen) bonded to the carbon is however shown in the structure.

Bond-line formulae of some cyclic compounds are:

Multiple bonds are also indicated in bond-line formulae.

Multiple bonds example:

Structural Representation Of Organic Compounds Examples

Example 1: Convert each of the following Lewis structures into complete structural formulae:

Solution: Use the dash to represent the shared electron pair.

Example 2: Convert each of the following complete structural formulae into condensed formulae.

Solution: Omitting some or all of the dashes and indicating the number of identical groups by a subscript we get

1. HO(CH₂)₂NH₂
2. CH₃C=C(CH₂)₃CH₃
3. H₃CCOCH₃
4. CH₃(CH₂)₃NO₂

Example 3: Write the condensed structural formulae for the compound that follows in different ways.

Solution:

Example 4: Write the bond-line formula for

Solution: First, outline the carbon skeleton, including the OH group as follows:

Thus, the bond-line formula is

Example 5: For each of the following condensed formulae write the corresponding bond-line formula.

1. (CH₃)₂CH(CH₂)₂CH₂OH
2. CH₃(CH2)₄CHICH₂CHO
3. (CH₃)₂CHCH₃CH(CH₂)₃NO₂
4. (CN)₂CHCH₂COCl

Solution:

In the bond-line formula carbon and hydrogen atoms except those that are part of the functional groups are not shown. We show only the carbon skeleton. The number of hydrogen atoms necessary to fulfill the carbon atoms’ valences are assumed to be present, but we do not write them in.

Other atoms (for example, Cl, O, N) are written in. Each intersection of two or more lines and the end of a line represent a carbon atom unless some other atom is written in.

Example 6: Draw all possible bond-line formulae for a cyclic compound, C₅H₁₀.

Solution: Start with the maximum number of carbon atoms in the ring and move towards the ring of 3 carbon atoms. Explore all sorts of possibilities on this route:

Three-dimensional representation of organic molecules

None of the formulae that we have described so far convey any information about how the atoms of a molecule are arranged in space. Shape or the three-dimensional (3-D) structure of organic molecules can be described on a paper (two-dimensional) by exploring certain conventions. For instance, by using solid and dashed wedge formula the 3-D image of an organic molecule can be perceived on a two-dimensional paper.

In solid-wedge and dashed-wedge description the solid-wedge denotes a bond projecting out of the plane of the paper towards the viewer. The dashed wedge depicts the bond projecting behind the plane of the paper and going away from the viewer. Both the wedges are drawn in such a way that the broad end of the wedge is near the viewer. The other two bonds lying in the plane of the paper are shown by using a normal line (________).

Let’s consider the wedge- and dashed-wedge representation of methane (CH₄) molecule: The two carbon-hydrogen bonds represented by normal lines are in the plane of paper, whereas the carbon-hydrogen bond represented with a solid wedge is aimed to be in front of the plane of paper.

The hydrogen bonded to carbon by dashed edge is intended to be behind the plane of paper. Note that the carbon atom is lying in the plane of paper. Wedge and dashed-wedge formulae are important tool for clearly showing three dimensions.