Organic chemistry- Some Basic principles and techniques

1) What are the hybridisation states of each carbon atom in the following compounds?

CH2=C=O, CH3CH=CH2, (CH3)2CO, CH2=CHCN, C6H6

Solution

Carbon in all its compound undergoes any of three hybridisations –sp3, sp

2 or sp,

depending upon the number of σ bonds formed.

• If the C atom forms 4σ- bonds, then it is sp3 hybridised.

• If the C atom forms 3σ- bonds, then it is sp2 hybridised.

• If the C atom forms 2σ- bonds, then it is sp hybridised.

By writing the structural formula, the type of hybridisation of the C atom can be

known

a)

c)

d)

e)

2) Indicate the σ and π bonds in the following molecules:

C6H6, C6H12, CH2C12, CH=C=CH2, CH3NO2 , HCONHCH3

Solution

All the single covalent bonds are sigma(σ) bonds . A double covalent bond has one σ and

one π bond. A triple covalent bond has σ and two π bonds.

Then,

a)

b)

c)

d)

e)

f)

3) Write bond line formulas for : Isopropyl alcohol, 2,3 –Dimehtyl butanal, Heptan -4-one.

Solution

4) Give the IUPAC names of the following compounds:

Solution

a) Propyl benzene

b) 3- Mthyl pentanenitrile

c) 2, 5 –Dimehtylpentane

d) 3-Bromo -3-chloroheptane

e) 3- Chloropropanal

f) 2,2 –Dichloro ethanol

5) Which of the following represents the correct IUPAC name for the compounds

concerned?

a) 2,2 –Dimethylpentane or 2-Dimethylpentane

b)2,4,7 –trimethylloctane or 2,5,7-Trimethyloctane

c) 2-Chloro-4-methylpentane or 4-Chloro-2 –methylpentane

d) But-3 –yn-1-ol or But-4-ol-1-yne

Solution

a) 2,2 –Dimethylpentane

b) 2,4,7 –trimethylloctane

c) 2-Chloro-4-methylpentane

d) But-3 –yn-1-ol

6) Draw formulas fro the first five members of each homologous series beginning with the

following compounds .a)H-COOH b) CH3COCH3 c) H-CH=CH2

Solution

In a homologous series, adjacent members differ in their molecular formula by one-CH2

unit.

So, the members are

7) Give condensed and bond line strucutural formulas and identify the functional group(s)

present , if any , for:

a) 2,2,4-Trimethylpentane

b) 2-Hydroxy -1,2,3-propanetricarboxylic acid

c)Hexanedial

Solution

8) Identify the functional groups in the following compounds:

Solution

9) Which of the two: O2NCH2CH2O- or CH3CH2O

- is expected to be more stable and why?

Solution

O2NCH2CH2O- is more stable than CH3CH2O

- as the –NO2 group is an electron-withdrawing

group (-I effect group), due to which the negative charge on the O- ion is dispersed. On the

other hand, in case of CH3CH2O- , the CH3CH2 group is an electron –releasing group (+I

effect group), due to which the negative charge on the O-atom is intensified . The dispersal

of charge in O2NCH2CH2O- makes it more stable than CH3CH2O

- .

10) Explain why alkyl groups act as electron donors when attached to a π system.

Solution

Alkyl groups act as electron donors when attached to a π system due to hyperonjugation.

Due to the availability of an electron pair on the C atom, an alkyl group behaves as an

electron donor.

11) Draw the resonance structures for the following compounds. Show the electron shift

using curved-arrow notation.

a) C6H5OH b) C6H5NO2 c) CH3CH=CH-CHO

Solution

d) C6H5CHO e) C6H5-CH2 f) CH3CH=CHCH2

Solution

12) What are electrophiles and nucleophiles ? Explain with examples.

Solution

Electrophiles

Nucleophiles

i) Reagents are attracted towards the negatively

charged centre of a molecule are called

electrophiles.

ii)They attack the high electron density area of the

substrate.

iii) They accept an electro pair during reaction, and

are also called electron pair acceptors.

E.g. X-, NO2

+ , HSO3

+, RCH2

+, H

+

i) Reagents that are attracted towards

the positively charged centre of a

molecule are called nucleophiles

ii) They attack the low electron

density area of the substrate.

iii) They donate an electron pair

during reaction, and are called

electron pair donors.

E.g. OH- , CN

-, NO

-2, RO

-, NH3 , H2O

13) Identify the reagents shown in bold in the following equation as nucleophiles or electrophiles:

a) CH3COOH + HO- →CH3COO

- + H2O

b) CH3COCH3 + CN → (CH3)2C(CN)(OH)

c) C6H5 + CH3CO → C6H5 CO CH3

Solution

OH- is the nucleophile (-vely charged)

CN- is the nucleophile (-vely charged)

CH3CO+ is the electrophile (+vely charged)

14) Classify the following reactions in one of the reaction type studied in this unit.

a) CH3CH2Br + HS- → CH3CH2SH+ Br

-

b) (CH3)2C= CH2+ HCl → (CH3)2ClC-CH3

c) CH3CH2Br + HO- → CH2= CH2+ H2O + Br

-

d) (CH3)3 C-CH2OH +HBr →)(CH3)2CBrCH2CH3 + H2O

Solution

a) Nucleophilic substitution

b) Electrophilic addition

c) Elimination reaction

d) Nucleophilic substitution with rearrangement

15) What is the relationship between the members of following pairs of structures?

Are they structural or geometrical isomers or resonance contributors?

Solution

a)Position isomers and metamers

b)Geometrical isomers

c) Resonating structures

16) For the following bond cleavages , use curved arrows to show the electron flow and

classify each as hololysis or heterolysis . Identify reactive intermediate produced as free

radical, carbocation and carbanion.

Solution

16)c-d

Solution

17) Explain the terms Inductive and Electromeric effects. Which electron displacement effect

explains the following correct orders of acidity of the carboxylic acids?

a) Cl3CCOOH> Cl2CHCOOH > ClCH2COOH

b) CH3CH2COOH > (CH3)2CHCOOH > (CH3)3C.COOH

Solution

Inductive effect:

The delocalisation of the σ-bond electrons towards the more electronegative group is called the

inductive effect.

Electromeric effect

The shifting of the π electrons of the C-C double or triple bond under the influence of an attacking

reagent is called electromeric effect.

a) –I effect: Cl has –I effect, which increases the extent of polarisation in the O-H bond of the –

COOH group, and hence , acidic strength increases with an increase in the number of Cl atoms.

b) +I effect: The alkyl group has +I effect due to which there is a decrease in the extent of

polarisation in the O-H bond of the –COOK group , and hence acidic strength decreases with

increase in number of alkyl groups.

18) a) Give a brief description of the principles of the crystallisation taking an example.

Solution

Crystallisation:

The method of purification of an impure sample on the basis of difference in solubility of the

compound and impurities is called crystallisation.

The impure compound is dissolved in a suitable solvent and heated to concentrate it to a saturated

solution. On cooling this solution, the pyre compound crystallises out, leaving the impurities

behind in the filtrate.

b) Give a brief description of the principles of the distillation taking an example.

Solution

Distillation

The separation of two or more liquids , and even a liquid from soluble non-volatile impurities, on

the basis of the difference in their boiling points, is called distillation. Distillation may be simple

distillation, fractional distillation, steam distillation, etc.

a) Simple distillation is carried out when the difference in the boiling points of the liquids is wid,

i.e about 30-50 K.

For example , the distillation of a mixture of ether(bp=308 K) and toluene (bp=384 K)is done by

simple distillation.

b) Fractional distillation is carried out when the difference in the boiling points of the component

liquids is not so large(10-20K). In this process, vapours of the liquid mixture are passed through a

fractionating column before condensation.

For example, the separation of methanol (bp=337.8 K) and acetone (bp=329 K ) by fractional

distillation.

c) Steam distillation is used to separate organic compounds that are steam volatile, immiscible in

water and contain non-volatile impurities . Steam distillation makes the substance with the higher

boiling point to distill at a lower temperature withour decomposition.

For example, aniline and bromobenzene are separated by steam distillation.

c) Give a brief description of the principles of the chromatography taking an example.

Solution

Chromatography:

The principle of chromatography is based on the difference in the rates at which the components of

a mixture are adsorbed on a suitable solvent or adsorbent. The substance used for the adsorption of

different components of a mixture is called the stationary phase. The solvent in which the mixture

is dissolved is called the moving phase. The moving phase is allowed to move over stationary

phase, and the components of the mixture get separated .

Example:

Amino acids are detected by this technique, using ninhydrin solution.

19) Describe the method, which can be used to separate two compounds with different solubilities

in a solvent S.

Solution

Two compounds that have different solubility in a solvent S can be separated by fractional

crystallisation .A hot saturated solution of the mixture is prepared and is filtered to remove the

insoluble impurities and then allowed to cool. The component with respect to which the solution

becomes saturated separates out first, while the component with respect to which the solution in

unsaturated remains in the solution. Crystals of the first component separate out and are separated

by filtration. The mother solution is then allowed to cool, after which crystals of the second

component separate out.

20)What is the difference between distillation, distillation under reduced pressure and steam

distillation?

Solution

Distillation Distillation under reduced

pressure

Steam distillation

This method is used to

separate volatile liquids from

non-volatile impurities having

sufficient difference in their

boiling points.

Example: Mixture of

chloroform (bp 334 K) and

aniline (bp 457 K) is separated

by this method.

This method is employed to

purify liquids that have very

high boiling points and

decompose near their boiling

point. Due to reduced pressure,

the liquids start boiling at lower

temperature without getting

decomposed.

Example: Glycerol is separated

from spent-lye in the soap

industry by this method.

This method is employed

to purify liquids that are

steam volatile, immiscible

in water and contain non-

volatile impurities. Steam

distillation makes the

liquid with the higher

boiling point to boil at a

lower temperature without

decomposition .

Example: A mixture of

aniline and water is

separated by steam

distillation.

21) Discuss the chemistry of Lassaigne’s test.

Solution

Lassaigne’s test is used to detect the presence of extra elements like nitrogen, sulphur , halogens

and phosphorus in organic compounds . The elements present in the compound are converted inot

their ionic from from covalent form by fusing with sodium metal.

C,N,S,X and P are from the carbon compound. The compounds formed are separated from the

fused mass by boiling it with water. The extract thus obtained i called sodium fusion extract, which

is used to detect different elements.

a) Detection of nitrogen

The Prussian blue colour is due to the formation of iron(III) hexacyano ferrate (II) complex.

Detection of sulphur

Reactions

2Na+S →Na2 S

Na2 S + Pb (CH3COO)2 → PbS + 2CH3COONa

The black precipitate is PbS . This confirms the presence of S in the given carbon compound.

b) Detection of halogens(Cl, Br, I)

c) Detection of phosphorus

22) Differentiate between the principle of estimation of nitrogen in an organic compound

by Dumas method and Kjeldahl’s method.

Solution

Dumas method Kjeldahl’s method

1. The nitrogen –containing organic

compounds is heated with copper oxide

in an atmosphere of carbon dioxide.

2. Nitrogen in the organic compound is

converted into nitrogen gas.

3. The volume of nitrogen obtained is

measured at room temperature.

1. The nitrogen-containing

organic compound is heated with

concentrated sulphuric acid.

2. Nitrogen in the organic

compound is converted into

ammonium sulphate.

3. The ammonium sulphate

mixture is heated with sodium

hydroxide. The ammonia obtained

is absorbed in an excess of

standard solution of HCl or H2SO4

.The amount of acid neutralised

by ammonia is determined.

23) Discuss the principle of estimation of halogens, sulphur and phosphorus present in an

organic compound.

Solution

a) Estimation of halogen by the Carius method

A known mass of the organic compound is treated with fuming HNO3 and some crystals of AgNO3

in a furnance. The halogen present in the compound is quantitatively converted into silver

halide(AgX). By knowing the weight of AgX, the estimation of halogen is given as:

b) Estimation of sulphur by the Carius method

A known mass of organic compound is heated with fuming HNO3 acid. As a result , sulphur

is quantitatively oxidised to sulphuric acid, which is than precipitated as barium sulphate by

adding excess of barium chloride solution. By knowing the mass of BaSO4 precipitated ,

sulphur is estimated as

c) Estimation of phosphorus by the Carius method

A known mass of organic compound is heated with fuming HNO3 acid, due to which the

phosphorus present in the compound is quantitatively oxidised to phosphoric acid (H3PO4) .

By adding ammonia and ammonium molybdate to the solution , it is converted into

ammonium phosphomolybdate. By knowing the mass of ammonium phosphomolybdate ,

phosphorus s estimated as

24) Explain the principle of paper chromatography.

Solution

Paper chromatography is based on the principle of partition of a s solid between two liquids

. It is based upon continuous differential distribution of various components of the mixture

beteen the mobile phase and the stationary phase.

The mixture to be separated is dissolved in a suitable solvent. A spot of the solution is

applied on the starting line. Chromatographic paper is then suspended in a suitable solvent

or mixture of solvents , which acts as a mobile phase. The different components of the

mixture travel different distances depending upon their distribution or adsorption between

the stationary phase and the mobile phase. When the solvent reaches the top of the

chromatographic paper, the paper is removed and then dried . The different components of

the mixture can be identified by spraying a suitable reagent on the paper strip.

25) Why is nitric acid added to sodium extract before adding silver nitrate for testing

halogens?

Solution

Nitric acid is added to sodium extract before adding AgNO3 for testing halogens to

decompose the cyanide or sulphide during Lassaigne’s test, to avoid their interference with

silver nitrate during the test for halogens.

NaCN + HNO3 → NaNO3 + HCN

26) Explains the reason for the fusion of an organic compound with metallic sodium for

testing nitrogen, sulphur and halogens.

Solution

Organic compounds do not give ions in solution. Hetero atoms in the organic compound

can be easily tested when it is fused with sodium.

N, Cl, Br, I, S etc. present in the compound are converted into ionic compounds like NaCN,

NaCl, NaBr, Na2S, etc. now, the solution is tested for CN-,Cl

-, Br

-, S

- and PO4

-3 ions to

detect presence of N, Cl, Br, S and P in the organic compounds.

27) Name a suitable technique of separation of the compounds from a mixture of calcium

sulphate and camphor.

Solution

Calcium sulphate and camphor can be separated from any of the two methods:

a) Sublimation : Camphor being sublimable can be separated from non-volatile CaSO4 by

heating.

b) By solvent extraction: Camphor being an organic compound is soluble in organic

solvent, while CaSO4 is insoluble in organic solvents. By dissolving camphor in

chloroform, it goes into the solution, leaving behind CaSO4 as solid residue. The solutions

is filtered and then evaporated to get solid camphor.

28) Explain , why an organic liquid vaporises at a temperature below its boiling point in its

steam distillation?

Solution

In steam distillation , the liquid to be purified and water is heated together. Each liquid

exerts its own vapour pressure. So, the vapour pressure of the mixture will be the sum of the

parial vapour pressure of the two liquids in the mixture. When it is in equal to atmospheric

pressure, the mixture boils.

P= p1 + p2

P= Atmospheric pressure

p1 =partial pressure of liquid

p2 = partial pressure of water

p1 = P - p2

Since, p1 is lower than P, the organic liquid vaporises at a temperature lower than its boiling

point.

29) Will CCl4 give white precipitate of AgCl on heating it with silver nitrate? Give reason

for your answer.

Solution

CCl4 will not give precipitate of AgCl on heating with silver nitrate, as it does not ionise to

form Cl- ions. Only ionic chlorides that furnish chloride ions in their aqueous solution can

give AgCl precipitate with silver nitrate.

30) Why is a solution of potassium hydroxide used to absorb carbon dioxide evolved during

the estimation of carbon present in an organic compound?

Solution

Potassium hydroxide (KOH) being a strong base readily absorbs CO2 , which is acidic in

nature.

2KOH + CO2 →K2CO3 +H2O

Since, only CO2 can be absorbed by KOH , to estimate the mass of carbon in the organic

compound, KOH is used. An increase in the mass of KOH in the U-tube corresponds to the

mass of CO2 . From the mass of CO2 , carbon present in the organic compound can be

estimated.

31) Why is it necessary to use acetic acid and not sulphuric acid for acidification of sodium

extract for testing sulphur by lead acetate test?

Solution

Lead acetate is soluble in acetic acid and remains dissolved. In place of acetic acid if H2SO4

is used, then lead sulphate will get precipitated , which will interfere with the reaction.

(CH2COO)2Pb+ H2SO4 →PbSO4 + 2CH3COOH

32) An organic compound contains 69% carbon and 48% hydrogen, the remainder being

oxygen. Calculate the masses of carbon dioxide and water produced when 0.20 g of this

substance is subjected to complete combustion.

Solution

33) A sample of 0.50 g of an organic compound was treated according to Kjeldahl’s method

. The ammonia evolved was absorbed in 50 mL of 0.5 M H2SO4 . The residual acid required

60 mL of 0.5 M solution of NaOH for neutralisation . Find the percentage composition of

nitrogen in the compound.

Solution

0.5 M H2SO4 = 2 x 0.5 NH2SO4

= 1.0 NH2SO4

Number of milliequivalent of H2SO4 taken

= N x V

= 1 x50=50

Number of milliequivalents of NaOH

= Nx V

=0.5 x 60

= 30

Number of milliequivalent of NaOH used= Number of milliequivalents of acid left = 30

So, number of milliequivalents of acid neutralised by ammonia = 50-30 =20

Hence,

= 1.4 x 20/0.5

=56

Percentage of nitrogen = 56

34) 0.3780 g of an organic chloro compound gave 0.5740 g of silver chloride in Carius

estimation. Calculate the percentage of chlorine present in the compound.

Solution

Mass of AgCl = 0.5740 g

Mass of organic compound=0.3780 g

Molar mass of AgCl=143.5 g

= 35.5 x 0.5740 x 100/143.5 x 0.3780

= 37.566

% of Cl =37.57

35) In the estimation of sulphur by Carius method, 0.468 g of an organic sulphur compound

afforded 0.668 g of barium sulphate. Find out the percentage of sulphur in the given

compound.

Solution

Mass of BaSO4 = 0.668 g

Mass of organic compound=0.468 g

Molar mass of S= 32 g

Molar mass of BaSO4 =233 g

Now,

= 32/233 x 0.668/0.468 x 100

= 19.60

= % of S= 19.60

36) In the organic compound CH=CH-CH2-CH2-C≡ CH, the pair of hybridised orbitals

involved in the formation of : C2-C3 bond is:

a) sp-sp2 b) sp-sp

3 c) sp

2 – sp

3 d) sp

3-sp

3

Solution

An unsaturated hydrocarbon with C=C and C≡C is numbered in such a way that the carbon

with C=C gets the least number, or the numbering is done from C=C.

So, the pair of hybridised orbitals involved in the formation of C2-C3 bond is sp2-sp

3

orbitals.

Hence , option (c) is correct.

37) In the Lassaigne’s test for nitrogen in an organic compound, the Prussian blue colour

is obtained due to the formation of :

a) Na4[Fe(CN)6] b) Fe4[Fe(CN)6]3 c)Fe2 [Fe(CN)6] d)Fe3 [Fe(CN)6]4

Solution

The Prussian blue colour obtained in Lassaigne’s test for detection of nitrogen is due to the

formation of Fe4[Fe(CN)6]3

Hence, option (b) is the answer.

38) Which of the following carbocation is most stable?

Solution

The order of stability of the carbocation ion is

1°<2°<3°

Among the four options, option (b) is a tertiary carbocation, so it is the most stable among all

carbocations given.

Hence, answer is (b).

39) The best and latest technique for isolation, purification and separation of organic compounds

is:

a) Crystallisation b) Distillation c) Sublimation d) Chromatography

Solution

The best and latest technique for the isolation , purification and separation of organic compound is

chromatography.

Hence, option (d) is correct answer.

40)The reaction CH3CH2I +KOH(aq) →CH3CH2OH +KI

Is classified as :

a) electrophilic substitution b) nucleophilic substitution

c)elimination d)addition

Solution

The given reaction is a nucleophilic substitution as the attacking reagent OH- ion is a nucleophile,

which is substituting I- ion, another nucleophile.

Hence, option (b) is correct.