l1 ch-12 sol set-1 aldehydes ketones and carboxylic acids

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Level - I

SECTION - A

School/Board Exam. Type Questions

Very Short Answer Type Questions :

1. Write the IUPAC name of following compound?

CH CH CH3 2 2

N

CH3

C H2 5

Sol. N-Ethyl-N-methylpropan-1-amine

2. What does Kb value for a base stand for?

Sol. Kb stands for dissociation constant of a base. It determines basic strength of a base. Higher the value of K

b,

stronger base is the base.

3. How is phenol obtained from aniline?

Sol.

NH2

N Cl2

OH

NaNO + dil. HCl2

Boiling H O / H2

+

Aniline Benzenediazonium

chloride

Phenol

273 - 278 K

4. Give a chemical test to distinguish between primary and secondary amines.

Sol. Carbylamine reaction : Primary amine will form offensive smelling isocyanide whereas secondary amine does

not give this reaction.

5. Why do amines behave as nucleophiles?

Sol. They act as nucleophiles because there is a lone pair of electrons on N which it can donate.

6. Why primary amines have higher boiling point than tertiary amines?

Sol. Primary amines are associated due to H-bonding. Tertiary amine have no hydrogen bond.

Amines

Solutions (Set-1)

Chapter 13

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68 Amines Solutions of Assignment (Set-1) (Level-I)

7. State the reaction taking place when : Bromine water is added to the aqueous solution of aniline.

Sol.

NH2

NH2

+ 3Br (aq)2

+ 3HBrBrBr

Br

8. How is the basic strength of aromatic amines affected by the presence of an electron-releasing group on the

benzene ring?

Sol. Basic strength of aromatic amines is increased by the presence of electron releasing group, due to increase

in electron density on the amino group.

9. Arrange the following in the order of their increasing basic character in solution :

NH3, C

2H5NH

2, (C

2H5)2NH, (C

2H5)3N

Sol. NH3 < C

2H5NH

2 < (C

2H5)3N < (C

2H5)2NH

10. Account for the following : Ammonolysis of alkyl halide does not give a corresponding amine in pure state.

Sol. This is because the amine formed reacts further with more of the alkyl halide to give 2°, 3° amines and finally

quaternary salt. Thus, a mixture of amines is obtained.

Short Answer Type Questions :

11. (i) Arrange the following in an increasing order of basic strength in water :

C6H5NH

2, (C

2H5)2NH, (C

2H5)3N and NH

3

(ii) Arrange the following in increasing order of basic strength in gas phase :

C2H5NH

2, (C

2H5)2NH, (C

2H5)3N and CH

3NH

2

Sol. (i) Out of the amines, aromatic amine C6H5NH

2 is the weakest base followed by NH

3. For aliphatic amines,

the basic strength is a function of inductive effect, solvation effect and steric effect of the alkyl group.

Increasing order of basic strength, keeping in view all these factors is :

C6H5NH

2 < NH

3 < (C

2H5)3N < (C

2H5)2NH (in aqueous solution)

(ii) In the gas phase, only the inductive effect of the alkyl group determines the basic strength of the base.

Greater the number of alkyl groups attached to the nitrogen, greater is the electron density and hence

greater is the basic strength. Methyl group being smaller than ethyl group exerts smaller +I effect,

therefore methylamine is weakest base. The increasing order of basic strength of the given compounds

in the gas phase is :

CH3NH

2 < C

2H5NH

2 < (C

2H5)2NH < (C

2H5)3N

12. Account for the following :

(i) Amines are basic substances while amides are neutral.

(ii) Aromatic amines are weaker bases than aliphatic amines.


69Solutions of Assignment (Set-1) (Level-I) Amines

Sol. (i) In amines, alkyl group is electron releasing which increases electron density on nitrogen making them

basic whereas in amides R C

O

group is electron withdrawing, therefore, they are neutral.

(ii) It is because aryl group is electron withdrawing which decreases electron density on nitrogen making

them less basic whereas alkyl group is electron releasing which makes alkylamines more basic.

Moreover, the lone pair of electrons on N in aromatic amine is delocalised due to resonance with benzene

ring.

13. (a) How will you convert an alkyl halide to a primary amine having one more carbon atom than the parent

alkyl halide?

(b) Why are amines more basic than corresponding alcohols?

Sol. (a) CH Cl + KCN3

CH C N3

CH CH NH3 2 2

Na / C H OH2 5

Chloromethane Ethanenitrile Ethanamine

4[H]

(b) Oxygen is more electronegative than nitrogen. Therefore, oxygen has less tendency to donate the

electrons compared to nitrogen. That is why amines containing N are more basic than alcohols containing

oxygen.

14. Give reasons for the following observations :

(i) It is difficult to prepare pure amines by ammonolysis of alkyl halides.

(ii) Electrophilic substitution in case of aromatic amines takes place more readily than in benzene.

Sol. (i) It is because primary amines formed react with more alkyl halide to form 2° and 3° amines. That is why

we get mixture of amines, which need to be separated.

(ii) It is because –NH2 group is electron releasing and increases electron density on benzene ring.

15. Identify A and B in the following reactions :

(i) C H C N2 5

ALiAlH

4HNO

2

B

(ii)

NH2

ABr

2NaNO / HCl

2

BH O

2 273 K

Sol. (i) C H C N2 5

LiAlH4

C H CH NH2 5 2 2

HNO2

C H CH OH2 5 2

Propanenitrile (A)

Propanamine

(B)

Propanol

(ii)

NH2

NH2

N Cl2

Br2

NaNO / HCl2

H O2

273 K

Br BrBr Br

Br Br

(A) (B)

2, 4, 6-Tribromoaniline 2, 4, 6-Tribromobenzenediazonium chloride



16. State the reactions for obtaining benzoic acid from aniline.

Sol.

NH2

CN COOHN Cl2

NaNO / HCl2

H O / H2

+

CuCN

273 - 278 K KCN

Aniline Benzene

diazonium chloride

Benzonitrile Benzoic acid

17. Before reacting aniline with HNO3 for nitration, it is converted to acetanilide. Why is this done and how is

nitroaniline obtained subsequently?

Sol. It is because aniline gets oxidised, protonated and gives 47% m-nitroaniline on direct nitration. Therefore, it

is converted into acetanilide and then nitrated. Anilide group activates the ring at o- and p-positions. We get

o- and p-nitroacetanilide which on hydrolysis gives o- and p-nitroaniline.

18. For the following conversion reactions write the chemical equations :

(i) Ethylisocyanide to ethylamine

(ii) Aniline to N-phenylethanamide

Sol. (i)C H NC + 2H O

2 5 2C H NH + HCOOH

2 5 2

H+

Ethylisocyanide Ethylamine Methanoic acid

(ii)

NH2

NHCOCH3

NaOH

Aniline N-Phenylethanamide

+ CH COCl3

+ HCl

19. Give reason :

(i) C H N (CH ) OH6 5 3 3

is stronger base than NH4OH.

(ii) Aniline is acetylated to prepare monobromo derivative.

Sol. (i) C H N (CH )6 5 3 3

is more stable than 4

NH

because methyl groups are electron releasing whereas hydrogen

is not.

(ii) Aniline is acetylated to deactivate –NH2 group so as to prepare monobromo derivative. Otherwise, we

would get tribromo derivative.


(i) Aniline is weaker base than methylamine.

(ii) Aryl cyanides can not be formed by the reaction of aryl halides and sodium cyanide.



Sol. (i) Aniline is weaker base than methylamine because C6H5 group is electron withdrawing in aniline whereas

CH3 group is electron releasing in methylamine. Moreover, lone pair of electrons on N in aniline is

delocalised due to resonance with benzene ring. This reduces electron density on amino group in aniline.

(ii) It is due to double bond character between carbon and halides in aryl halides which cannot be broken

easily. Therefore, Cl cannot be easily replaced by –CN to get aryl cyanide.

21. How will you convert :

(i) Nitrobenzene to phenol

(ii) Aniline to chlorobenzene

Sol. (i)

NO2

NH2

N Cl2

OH

Nitrobenzene Aniline Phenol

+ 6[H]Sn / HCl NaNO / HCl

2H O

2

273 - 278 K boil

(ii)

NH2

ClN Cl2

Aniline Chlorobenzene

+ NaNO + HCl2

+ N2

273 - 278 K HCl

CuCl

22. What is the role of HNO3 in the nitrating mixture used for nitration of benzene?

Sol. HNO3 acts as a base in the nitrating mixture (Conc. HNO

3 + Conc. H

2SO

4). H

2SO

4 acts on HNO

3 to generate

the electrophile, 2

NO

(nitronium ion)

H SO2 4

H + HSO+ –

4

HONO2

O NO+

2NO

2 2

+

+ H OH

+ H

H

23. Predict the product of reaction of aniline with bromine in non-polar solvent such as CS2.

Sol. In non-polar solvents, the resonance structures involving separation of +ve and –ve charges are not stabilized

by dipole-dipole interactions with the non polar solvent. As a result, activating effect of the NH2 is reduced

and hence monosubstitution occurs only at o- and p-positions giving a mixture of 2-bromoaniline (minor) and

4-bromoaniline (major).

24. A compound Z with molecular formula C3H9N reacts with C

6H5SO

2Cl to give a solid, insoluble in alkali.

Identify Z.

Sol. Since the amine Z on treatment with C6H5SO

2Cl gives a product which is insoluble in alkali, therefore, the

product does not have a replaceable hydrogen on the N-atom. In other words, the amine Z is a 2° amine.

The only 2° amine having the molecular formula C3H9N is ethylmethylamine, i.e., the amine (Z) is

ethylmethylamine, CH3CH

2NHCH

3.



25. A primary amine, RNH2 can be reacted with CH

3 – X to get secondary amine, R – NHCH

3 but the only

disadvantage is that 3° amine and quaternary ammonium salts are also obtained as side products. Can you

suggest a method where RNH2 forms only 2° amine.

Sol. 1° amines react with CHCl3 in presence of alcoholic KOH to form isocyanides which upon catalytic reduction

give 2° amines.

R NH

2R N C R NH CH

3

KOH / CHCl3

(Carbylamine reaction)

H / Pt2

(Catalytic reduction)1° amine 2° amine

26. Suggest a route by which the following conversion can be accomplished :

C NH2

NH CH3

O

Sol.

O

Br / KOH2

CHCl / KOH3C NH

2NH

2N C

NHCH3

(Hofmann bromamide reaction)

Cyclohexylamine Cyclohexyl carbylamine

N-Methylcyclohexylamine

H / PCl2

27. How is classification of amines different from that of alcohols?

Sol. Classification of amines is made on the basis of the nature of nitrogen atom (whether it has one, two or no

hydrogen atom) to which the alkyl groups are attached. The classification of alcohols, on the other hand, is

made on the basis of the nature of carbon atom to which the –OH group is attached.

–NH2

NH –CH OH2

1° amine 2° amine 3° amine 1° alcohol 2° alcohol 3° alcohol

N C OHCHOH

28. Although trimethylamine and n-propylamine have the same molecular weight, the former boils at a lower

temperature (276 K) than the later (332 K). Explain.

Sol. n-Propylamine has two H–atoms on the N-atom and hence undergoes intermolecular H-bonding thereby raising

its boiling point.

Trimethylamine, (CH3)3N being a 3° amine does not have a H-atom on the N-atom. As a result, it does not

undergo H-bonding and hence its boiling point is low.

29. Why is it necessary to add excess of mineral acid to diazotisation of amines?

Sol. During diazotisation of arylamines, excess of mineral acid is used. In general, one mole of amine is treated

with approximate three moles of acid : (i) one mole to dissolve the amine, (ii) one mole to liberate HNO2 from

NaNO2 and (iii) one mole to maintain the proper acidity of the reaction mixture to prevent the coupling of

diazonium salt.



30. How will you carry out the following conversion?

NO2

NH2

Sol.Conc. HNO + Conc. H SO

3 2 4

(Nitration)

Benzene

NO2

Sn / HCl

H / H O+

2

Nitrobenzene

NH2

NHCOCH3

(CH CO) O / Pyridine3 2

NHCOCH3

NO2

p-Nitroacetaniline

NH2

NO2

p-Nitroaniline

Conc. HNO3

Conc. H SO2 4

Long Answer Type Questions :

31. Account for the following observations :

(i) pKb of aniline is more than that of methylamine.

(ii) Aniline does not undergo Friedel-Crafts reaction.

Sol. (i) Lone pair of electrons on nitrogen in aniline is delocalised due to resonance with benzene ring. Thus,

electron density on the amino group in aniline decreases. Consequently, its basic strength also

decreases. On the other hand, electron density on amino group in methylamine increases due to +I effect

of CH3 group. Thus, it has a greater basic strength (K

b). pK

b and K

b are inversely related. Therefore,

pKb for aniline is more than that for methylamine.

(ii) AlCl3 is used as a catalyst in Friedel Crafts reaction. AlCl

3 is a Lewis acid (electron deficient). AlCl

3

gets attached to the lone pair of electron in aniline. Amino group is thus not able to activate the benzene

ring for electrophilic substitution. Moreover electrophiles like CH3 and CH CO3

are not formed as the

catalyst is consumed in association with amino group. Therefore, aniline does not undergo Friedel Crafts

reaction.

32. State the reactions and reaction conditions for the following conversions :

(i) Benzene diazonium chloride to nitrobenzene

(ii) Aniline to benzene diazonium chloride

(iii) Ethanamide to methylamine

Sol. (i)NaNO

2

Cu

Benzene diazonium

tetrafluoroborate

N Cl2

N BF2 4

+ –

NO2

+ HBF4

Fluoroboric

acid

+ N + NaBF2 4



(ii)

NH2

N Cl2

+ –

+ NaNO + HCl2

+ NaCl + 2H O2

Aniline

273 - 278 K

(iii) CH CONH + Br + 4KOH3 2 2

CH NH + K CO + 2KBr + 2H O3 2 2 3 2

Ethanamide Methylamine

33. How would you achieve the following conversions :

(i) Nitrobenzene to aniline

(ii) An alkyl halide to a quaternary ammonium salt

(iii) Aniline to benzonitrile

Write the chemical equation with reaction conditions in each case.

Sol. (i)

NO2

+ 6[H]

Nitrobenzene

Sn / HCl

NH2

+ H O2

Aniline

(ii) C H Cl + NH2 5 3

C H NH2 5 2

Chloroethane Ethanamine

(C H ) NH2 5 2

N-Ethylethanamine

(C H ) N2 5 3

N, N-Diethylethanamine

(C H ) N Cl+ –

2 5 4

Tetraethylammonium chloride

C H Cl2 5

C H Cl2 5

C H Cl2 5

(iii)

NH2

Aniline

NaNO + HCl2

N Cl2

CuCN

C N

Benzonitrile

273 - 278 K

34. (i) Stating the necessary reaction conditions write chemical reaction equations to obtain the following :

Chlorobenzene from aniline

(ii) Identify A and B in the following :

(a)

CH Br2

CN–

ALiAlH

4

B

(b)NH

3

ANi / H

2

BR CO2



Sol. (i)

NH2

NaNO + HCl2

N Cl2

+ –

CuCl

Cl

Chlorobenzene

273 - 278 K HCl

(ii) (a)

CH Br2

CN–

Bromomethylbenzene

CH CN2

(A)

Benzyl cyanide

LiAlH4

CH CH NH2 2 2

(B)

2-Phenylethanamine

(b) R CO2

Ketone

NH3

R C NH

(A)

R

Ni / H2

R CH NH2

(B)

Dialkylethanamine

R


(i) CH CH CONH3 2

CH3

Br / NaOH2

AHNO

2

B

(ii)

NO2

ABr

2 / Fe

3+

Sn / HClB

(iii) AH , H O

+

2LiAlH

4

BCH CN3

Sol. Identification of A and B can be made with the help of the following reactions :

(i) CH CHCONH3 2

CH3

Br / NaOH2

CH CH NH3 2

CH3

HNO2

(A)

2-Methylethanamine

CH CH OH3

CH3

(B)

Propan-2-ol

(ii)

NO2

Fe / Br3+

2

Nitrobenzene

NO2

(A)

3-Bromonitrobenzene

Br

Sn / HCl

NH2

(B)

-Bromoanilinem

Br

(iii) CH CN3

H O / H2

+

CH COOH3

LiAlH4

(A)

Ethanoic acid

Ethanenitrile

CH CH OH3 2

(B)

Ethanol



36. Why is NH2 group of aniline acetylated before carrying out nitration?

Sol. Due to strong activating effect of NH2 group, aromatic amines readily undergo electrophilic substitution reactions

and it is difficult to stop the reaction at the mono substitution stage. Usually the reaction occurs to give 2,

4, 6-trisubstituted amines. For example, aniline on treatment with bromine water gives 2, 4, 6-tribromoaniline.

If, however, we wish to stop the reaction at the monosubstitution stage, the activating effect of the –NH2 group

is reduced by acetylation. The acetyl group being electron withdrawing attracts the lone pair of electrons on

the N-atom towards itself.

H N C CH3

O

H N C CH3

O–

As a result, the lone pair of electrons on the N-atom is not exclusively available for donation to the benzene

ring and hence the activating effect of the –NH2 group is reduced. This method is known as protection of the

amino group by acetylation and can be used to stop the reaction at the monosubstitution stage by preventing

the formation of di- and trisubstitution products. For example, acetylation of aniline gives acetanilide.

Bromination of acetanilide with Br2/CH

3COOH followed by acid hydrolysis gives p-bromoaniline as the major

product.

37. tert-Butylamine can not be prepared by the action of NH3 on tert-butyl bromide. Explain why?

Sol. Tert-butyl bromide a 3° alkyl halide on treatment with a base prefers to undergo elimination rather than

substitution. Therefore, the product is isobutylene rather than tert butylamine.

CH C Br3

CH3

CH3

(Substitution)X

tert-Butyl bromide

NH3

CH C Br3

CH H2

CH3

Elimination+ NH

3

CH C CH

3 2

CH3

Isobutylene

38. Why does bromination of aniline, even under very mild conditions, gives 2,4, 6-tribromoaniline instantaneously?

Sol. Due to strong electron donating effect of the –NH2 group, the electron density increases at the o-, p-positions.

Further, when aniline is treated with Br2, the Br+ attacks the benzene ring at o- and p-positions to form

carbocation intermediates which are stabilized not only by the usual resonance of the benzene ring but also

by the –NH2 group as shown below :

o-Bromination :

NH2

Br+

NH2

(I)

H

Br

NH

2

(II)

H

Br

NH2

(III)

H

Br

NH2

(IV)

H

Br



p-Bromination :

NH2

+ Br+

NH2

H Br

NH2

H Br

(V) (VI)

NH2

H Br

NH2

H Br

(VII) (VIII)

In case of o-bromination, the carbocation intermediate is stabilized not only by usual resonating structure

(I, III and IV) but is also stabilized by resonance structure (II) in which the lone pair of electrons on the

N-atom interacts with the positively charged carbon of the ring. Similarly, in case of p-bromination, the

carbocation is stabilized not only by the usual resonating structures (V, VI and VIII) but is also stabilized

by the resonance structure (VII) in which the lone pair of electrons on the N-atom interacts with the positively

charged carbon of the ring. These additional resonating structures (II and VII) increase the stability of

the carbocation to such an extent that bromination occurs instantaneously at p- and two o-positions giving

2, 4, 6-tribromoaniline.

39. Arrange the following amines in order of decreasing basicity :

N N N N

H H H

(I) (II) (III) (IV)

O

Sol. (i) In compound II (pyridine), the lone pair of electrons on N is present in a sp2 orbital while in compounds

I (piperidine) and III (morpholine), the lone pair of electrons on N is present in a sp3 orbital. Since a sp2-

orbital has more s-character (33.33%) than a sp3 orbital (25%), therefore, the lone pair of electrons on

N is more readily available for protonation in I and III than in II. In other words, compound (II) is less

basic than I and III. Among I and III, III contains an oxygen atom which has –I effect. As a result, it will

attract the lone pair of electrons on N towards itself. Consequently, the lone pair of electrons on N in

III is less readily available for protonation than in I. In other words, compound I is more basic than III.

(ii) Compound IV (pyrrole) is aromatic in character. Therefore, in accordance with Huckel’s rule it has a cyclic

cloud of six- electrons. Out of these, four are contributed by two double bonds while the remaining two

are the lone pair of electron on the N-atom.

In other words, the lone pair of electrons on N-atom is contributed towards the aromatic sextet formation

and hence is not at all available for protonation. Therefore, compound IV is the least basic. In fact, it is

such a weak base that it is weakly acidic in character and thus reacts with K metal when heated to

form the corresponding potassium salt.

Thus, the basicity of the above four compounds decreases in order.

I > III > II > IV




(i)

NO2

ANaNO , HCl

2CuCl

B273 KNH

2

(ii) ANaCN LiAlH

4

BCH CH CH Br3 2 2

(iii)

NO2

AH SO / SO

2 4 3 Fe / HClB

373 K

Sol. (i)

NO2

NaNO , HCl2

NH2

273 K

NO2

CuCl

N Cl2

+ –

NO2

Cl

(A)

3-Nitrobenzene

diazonium chloride

(B)

3-Chloronitrobenzene

(ii) CH CH CH Br3 2 2

NaCNCH CH CH C N

3 2 2

LiAlH4

(A)

Butanenitrile

CH CH CH CH NH3 2 2 2 2

(B)

Butanamine

(iii)

NO2

H SO / SO2 4 3

333 K

NO2

SO H3

(A)

-Nitrobenzene

sulphonic acid

m

Nitrobenzene

Fe / HCl

NH2

SO H3

(B)

-Aminobenzene

sulphonic acid

m

SECTION - B

Model Test Paper

Very Short Answer Type Questions :

1. Direct nitration of aniline is not carried out. Explain why.

Sol. In presence of HNO3, benzene ring contains +ve charge which does not allow nitronium

2(NO ) ion to attack.

2. Write the IUPAC names for :

(a) CH CH CH CONH3 2

NH2

Cl

(b)

N(CH )3 2

NO

Sol. (a) 3-Amino-2-chlorobutanamide

(b) p-Nitroso-N, N-dimethylaniline



3. Write a chemical reaction in which the iodide ion replaces the diazonium group in a diazonium salt.

Sol.

N Cl2

+ –

+ KI

I

+ N + KCl2

4. Why are aqueous solution of amines basic in nature?

Sol. R NH + H O2 2

RNH + OH3

–

Formation of OH– indicates that it is basic in nature.

5. Give chemical tests to distinguish between the following pair of compounds : CH3 – CH

2 – NH

2 and (CH

3)2NH

Sol. Add CHCl3 and KOH solution to the compound. Ethylamine will give offensive smell of ethyl isocyanide.

(CH3)2NH does not give this reaction.

6. How is phenylamino methane obtained from phenylnitrile?

Sol.Na / C H OH

2 5

C N

Phenylnitrile

+ 4[H]

CH NH2 2

Phenylaminomethane

7. Complete the following reaction :

NH2

Aniline

NaNO / HBF2 4

273 K

?

Sol.

NH2

NaNO / HBF2 4

273 K

N BF2 4

Cu /

F

+ BF + N3 2

Benzene diazonium

tetrafluoroborate

Fluorobenzene

8. For the following conversion reactions write the chemical equation :

(a) Ethyl isocyanide to ethylamine

(b) Aniline to N-phenylethanamide

Sol. (a) C H N C + 2H O2 5 2

H+

C H NH + HCOOH2 5 2

EthylamineEthylisocyanide

(b)

NH2

Base

Aniline

+ CH COCl3

Acetylchloride

NH C CH3

N-phenylethanamide

+ HCl

O




9. Give one chemical test to distinguish between methylamine and dimethylamine.

Sol. Methylamine and dimethylamine can be distinguished by carbylamine test. When heated with alcoholic solution

of KOH and chloroform, methylamine gives foul smell of methyl isocyanide.

CH NH + CHCl + 3KOH (alc.)3 2 3

CH NC + 3KCl + 3H O3 2

Methylamine Methylisocyanide

(foul smell)

Dimethylamine does not give carbylamine test.

(CH ) NH3 2

Dimethylamine

No reaction.CHCl , KOH

3

10. Assign reason for the following : Amines are less acidic than alcohols of comparable molecular masses.

Sol. Loss of proton from amines gives amide ion whereas loss of a proton from alcohol gives an alkoxide ion.

RNH2

RNH + H– +

Amine Amide ion

ROH RO + H– +

Alcohol Alkoxide ion

Since O is more electronegative than N, therefore, RO– can accommodate the –ve charge more easily than

RNH–. Thus RO– is more stable than RH–. Hence amines are less acidic than alcohol.

11. Account for the following observation : pKb for aniline is more than that for methylamine.

Sol. In aniline, the lone pair of electrons on the N-atom are delocalised over the benzene ring. As a result, electron

density on the nitrogen decreases. In contrast, in CH3NH

2 + I effect of CH

3 increases the electron density

on the N-atom. Therefore, aniline is a weaker base than methylamine and hence its pKb value is higher than

that of methylamine.

12. Account for the following observation : Aniline does not undergo Friedel-Crafts reaction.

Sol. Aniline being a Lewis base reacts with Lewis acid, AlCl3 to form a salt.

C H NH + AlCl6 5 2 3

C H NH AlCl6 5 2 3

Lewis base Lewis acid

As a result, N of aniline acquires +ve charge and hence it acts as a strong deactivating group for electrophilic

substitution reaction.

Consequently, aniline does not undergo substitution reaction.



13. Write chemical tests to distinguish between aniline and N-methyl aniline.

Sol. Warm each compound separately with few drops of chloroform and alcoholic KOH. Aniline (Primary amine)

gives offensive smell of isocyanide while N-methyl aniline (Secondary amine) does not give carbylamine test.

C H NH + CH Cl + 3KOH (alc.)6 5 2 3

C H NC + 3KCl + 3H O6 5 2

Carbylamine

Warm

14. Account for the following : like ammonia, amines are good nucleophiles.

Sol. Amines are good nucleophiles. It is because alkyl group in an amine shows electron releasing effect. This

increases the electron density of N of amino group. This makes the amines very good nucleophiles.

R NH2

(+I effect)

15. Account for the following observations :

(i) Tertiary amines do not undergo acylation reaction.

(ii) Aniline readily reacts with bromine to give 2,4,6-tribromoaniline.

Sol. (i) Tertiary amines do not undergo acylation reaction because they do not contain -H-atom on the

N-atom.

(ii) Aniline readily reacts with bromine to give 2, 4, 6-tribromoaniline. This is due to strong activating effect

of the amino group, halogenation of amines occur very fast and the halogen enters the p and both o-

positions even in the absence of a catalyst.


16. Explain giving reasons :

(i) The boiling points of amines are lower than those of corresponding alcohols.

(ii) Silver chloride dissolves in methylamine solution.

Sol. (i) As oxygen is more electronegative than nitrogen therefore hydrogen bonding among alcohol is stronger

than among amine molecules, so alcohols have high boiling point than amines.

(ii) Due to formation of a soluble complex : [CH H N3 2

Ag NH CH ]2 3

17. How are the following conversions accomplished?

(i) Aniline to chlorobenzene

(ii) Nitrobenzene to phenol

(iii) Aniline to benzoic acid

Sol. (i)NaNO / HCl

2

NH2

273 K

N Cl2

+ –

CuCl

HCl

Cl



(ii)H / Pt

2

NO2

NaNO / HCl2

NH2

273 K

H O2

N Cl2

Boil

OH

(iii)NaNO / HCl

2

NH2

273 K

N Cl2

+ –

CuCN

HCN

CN

H O / H2

+

COOH

18. (a) How can you convert an amide into an amine having one carbon less than the starting compound?

(b) Name the reaction.

(c) Give the IUPAC name and structure of the amine obtained by the above method if amide is

3-chlorobutanamide.

Sol. (a) Hoffmann developed a method for preparing of primary amines by treating an amide with bromine in an

aqueous medium of sodium hydroxide.

2 2 2 2 3 2RCONH Br 4NaOH RNH Na CO 2NaBr 2H O

(b) Hoffmann Bromamide degradation reaction.

(c) CH CH CH CONH3 2 2

3-Chlorobutanamide

Cl

CH CH CH NH3 2 2

2-Chlorobutanamine

Cl

Br / KOH2

19. Predict, giving reasons, the order of basicity of the following compounds (CH3)3N, (CH

3)2NH, CH

3NH

2 and NH

3

in (i) gaseous phase and (ii) in aqueous solutions.

Sol. (i) The order of basicity in gaseous phase is (CH3)3N > (CH

3)2NH > CH

3NH

2 > NH

3.

Due to +I effect of alkyl group, there is more density at N at tertiary amine.

(ii) The order of basicity in aqueous state is (CH3)2NH > CH

3NH

2 > (CH

3)3N > NH

3.

The inductive effect, solvation effect, H-bonding and steric hinderance of the alkyl groups decide the basic

strength of alkyl amines in the aqueous state.

20. Complete the following reactions :

(i) C6H5NH

2 + CH

3COCl

(ii) C2H5NH

2 + C

6H5SO

2Cl

(iii) C2H5NH

2 + HNO

2

Sol. (i)6 5 2 3 6 5 3

C H NH CH COCl C H NHCOCH

(ii)2 5 2 6 5 2 6 5 2 2 5

C H NH C H SO Cl C H SO NHC H

(iii) 2H O

2 5 2 2 2 5 2 2 5 2

Unstable

C H NH HNO [C H N Cl ] C H OH N HCl



21. (a) How is aniline obtained from benzene?

(b) Why are secondary amines more basic than primary amine? Explain.

(c) Write the complete chemical reactions for the conversion of aniline to sulphanilic acid.

(d) Mention the two important uses of sulphanilic acid.

(e) Write a chemical reaction of aniline which may distinguish it from ethylamine.

Sol. (a)Conc. H SO

2 4

333 K+ HNO

3

NO2

Fe / HCl

NH2

AnilineNitrobenzene

(b) In a primary amine, only one alkyl group is attached to N-atom while in a secondary amine two alkyl

groups are attached to N atom. So electron density on N atom is more in a secondary amine than in a

primary amine due to +I effect of alkyl groups. Secondary amines can donate electron pair more readily

and hence are more basic than primary amines.

R N R N

H R

1° amine 2° amine

H H

(c)

NH2

+ H SO2 4

Aniline

NH HSO3 4

+ –

Anilium hydrogensulphate

455-475 K

–H O2

NH2

SO H3

Sulphanilic acid

Rearrangement

NH3

+

SO3

–

Zwitter ion

(d) It is used to manufacture (i) dyes (ii) drugs.

(e) Aniline gives white or brown precipitate with bromine water.

NH2

+ 3Br2

Aniline

NH2

+ 3Br2

BrBr

Br

Ethyl amine does not react with bromine water.

Long Answer Type Questions :


(i) Electrophilic substitution in case of aromatic amines takes place more readily than benzene.

(ii) CH3CONH

2 is a weaker base than CH

3CH

2NH

2.

(iii) Nitro compounds have higher boiling points than hydrocarbons having comparable molecular masses.



Sol. (i) Benzene ring in aromatic amines is highly activated. This is due to the displacement of lone pair

electrons of nitrogen towards the ring which results in the increase in the electron density in the ring.

This facilitates the electrophilic attack on the ring.

(ii) In CH3CONH

2, the lone pair of electrons on nitrogen atom is involved in resonance with the carbonyl group.

So, the electron pair of nitrogen is not easily available for protonation. Hence CH3CONH

2 is a weaker base

than CH3CH

2NH

2.

(iii) Nitro compounds are polar compounds whereas hydrocarbons are non polar. Due to their polarity, nitro

compounds have higher boiling points than the hydrocarbons having almost same molecular mass.

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Objective Type Questions

(Classification and Nomenclature of Amines)

1. CH C N3

H

H

CH3

CH3

is a

(1) Primary amine (2) Secondary amine

(3) Tertiary amine (4) Ammonium salt

Sol. Answer (1)

NH2

This is primary amine.

2. The correct IUPAC name for

CH CH CH NH CH2 2 3

is

(1) Allyl methylamine (2) 2-Amino-4-pentene

(3) 4-Aminopent-I -ene (4) N-Methylprop-2-en-1-amine

Sol. Answer (4)

NH

CH3

N-methyl prop-2-ene 1-amine

3. The compound having the molecular formula C3H

9N can represent

(1) Trimethylamine (2) n-propylamine

(3) Isopropylamine (4) All of these

Sol. Answer (4)

C3H

9N may be trimethylamine or n-propylamine or may be isopropylamine.

4. A secondary amine is

(1) A compound with two –NH2 group

(2) A compound with two carbon atoms and a –NH2 group

(3) A compound with a –NH2 group on the carbon atom in number two position

(4) A compound in which two of the hydrogens of NH3 have been replaced by alkyl or aryl groups

Sol. Answer (4)

CH – N – CH3 3

H

is secondary amine.

Solutions (Set-2)



(Preparation of Amines)

5. Hoffmann bromamide degradation reaction is shown by

(1) Ar – NH2

(2) Ar – CONH2

(3) Ar – NO2

(4) Ar – CH2NH

2

Sol. Answer (2)

C – NH2

O

will show Hoffmann degradation reaction.

6. Best method for preparing primary amines from alkyl halides without changing the number of carbon atoms

in the chain is

(1) Hoffmann bromamide reaction (2) Gabriel phthalimide synthesis

(3) Sandmeyer reaction (4) Reaction with NH3

Sol. Answer (2)

1° amine can be prepared by Gabriel phthalimide reaction

7. Gabriel phthalimide reaction is used for the preparation of

(1) Primary aromatic amines (2) Secondary amines

(3) Aliphatic primary amines (4) Tertiary amines

Sol. Answer (3)

Gabriel phthalimide reaction 1° primary amine

8. When acetamide is treated with Br2 and caustic soda, then we get

(1) Bromoacetic acid (2) Acetic acid (3) Methylamine (4) Ethane

Sol. Answer (3)

CH – C – NH3 2

O

Br OH21

–

CH – NH3 2

9.2 2 2

ReductionCONH CH NH

In above reaction hybridisation state of carbon changes from

(1) sp sp2

(2) sp sp3 (3) sp

2

sp3 (4) sp

2

sp

Sol. Answer (3)

– C – NH 2

Reduction– CH NH

2 2

sp2

sp2

O

10. A compound ‘A’ when treated with HNO3 (in presence of H

2SO

4) gives compound B, which is then reduced

with Sn and HCl to aniline. The compound is

(1) Toluene (2) Benzene (3) Ethane (4) Acetamide

Sol. Answer (2)

NitrationNO

2

Sn/4HClNH

2

(Benzene) (Aniline)



(Physical and Chemical Properties)

11. Amines are basic in character because they have

(1) A lone pair of electrons on the nitrogen atom

(2) A hydroxyl group in the molecule

(3) Replaceable hydrogen atom

(4) Saturated C atom

Sol. Answer (1)

R – NH2

:

because of low pair of electron amines are basic

12. Which is most basic?

(1) Benzylamine (2) Aniline

(3) Acetamide (4) o-methyl aniline

Sol. Answer (1)

NH2

: most basic, no conjugation (lone pair not in conjugation)

13. Aliphatic amines are ______ basic than NH3

but aromatic amines are ______ basic than NH3.

(1) More, less (2) Less, more

(3) More, more (4) None of these

Sol. Answer (1)

Because in aromatic amine lone pair is in conjugation with benzene ring.

14. Amongst the following, the strongest base in aqueous medium is

(1) CH3NH

2(2) (CH

3)3N (3) (CH

3)2NH (4) C

6H

5NHCH

3

Sol. Answer (3)

(Me)2NH > MeNH

3 > (Me)

3N > Ph(Me)

2N (Me = methyl)

15. Which of the following is the weakest Bronsted base?

(1)

NH2

(2)

H

N

(3)

NH2

(4) CH3NH

2

Sol. Answer (1)

NH2

:

NH2

+NH

2

+NH

2

+NH

2

one pair on nitrogen in conjugation.



16. The correct increasing order of basic strength for the following compounds is

NH2

NH2

NH2

NO2

CH3

(I) (II) (III)

(1) II < III < I (2) III < I < II (3) III < II < I (4) II < I < III

Sol. Answer (4)

CH3

NH2

NH2

NH2

NO2

(+H) (–M)Basic strength

17. Which of the following is the least basic?

(1)

NH2

NO2

(2)

NH2

OCH3

(3)

NH2

C H6 5

(4) All are equally basic

Sol. Answer (1)

NH2

NO2

is least basic because of presence of –I & –R (–NO2) group.

18. The decreasing order of the basic character of NH3, CH

3NH

2, C

2H

5NH

2 and C

6H

5NH

2 is

(1) NH3 > CH

3NH

2 > C

2H

5NH

2 > C

6H

5NH

2(2) C

2H

5NH

2 > CH

3NH

2 > NH

3 > C

6H

5NH

2

(3) C6H

5NH

2 > C

2H

5NH

2 > CH

3NH

2 > NH

3(4) CH

3NH

2 > C

2H

5NH

2 > C

6H

5NH

2 > NH

3

Sol. Answer (2)

NH2

CH NH3 2

NH3

NH2

19. The one which is the least basic

(1) NH3

(2) C6H

5NH

2(3) (C

6H

5)3N (4) (C

6H

5)2NH



Sol. Answer (3)

N

:

is least basic

20. Strongest base is

(1) CH2 = CH – CH

2 – NH

2(2) CH C – CH

2 – NH

2

(3) CH3CH

2CH

2NH

2(4)

NH2

Sol. Answer (3)

NH2 is the strongest base among the other shown bases because of max e– density over N-atom.

21. CH3CH

2NH

2 is more basic than CH

3CONH

2 because

(1) Acetamide is amphoteric in character

(2) In CH3CH

2NH

2 the electron pair on N-atom is delocalised by resonance

(3) In CH3CH

2NH

2 there is no resonance, while in acetamide the lone pair of electrons on N-atom is

delocalised and therefore less available for protonation

(4) Acetamide is derivative of acid

Sol. Answer (3)

Lone pair of nitrogen in conjugation with bond i.e., ( –LP conjugation)

NH2

O

NH2

:

resonance

O

NH2

+

22. Which one is the most volatile?

(1) CH3CH

2CH

2NH

2(2) (CH

3)3N (3)

CH2

CH3

NH (4) CH3OH

Sol. Answer (2)

N

CH3

CH3

CH3

is most volatile in above choices

23. Aniline on reaction with acetyl chloride gives

(1) Phenol (2) Acetamide (3) Acetanilide (4) Benzene

Sol. Answer (3)

NH +2

Cl

O

NH–C–CH3

O

acetanilide



24. The compound obtained by heating a mixture of a primary amine and chloroform with ethanolic potassium

hydroxide is

(1) An alkyl isocyanide (2) An alkyl halide

(3) An amide (4) An amide and nitro compound

Sol. Answer (1)

OH

2 3R – NH + CHCl R N C

−

⎯⎯⎯→ − =

��

alkyl isocyanide

25. A compound X has the molecular formula C7H

7NO. On treatment with Br

2 and KOH, X gives an amine Y.

The latter gives carbylamine test. Y upon diazotisation and coupling with phenol gives an azo dye. Thus X

is

(1) C6H

5CONH

2(2) C

6H

5CH

2NO (3) C

6H

5COONH

4(4) All of these

Sol. Answer (1)

C – NH –2

O

Br , OH 2

–

NH2

N= C

Y

CHCl , OH3

–

26. An organic compound (A) on reduction gave a compound (B). Upon treatment with HNO2, (B) gave ethyl

alcohol and on warming with CHCl3 and alcoholic KOH, (B) gave an offensive smell. The compound (A) is

(1) CH3NH

2(2) CH

3NC (3) CH

3CN (4) C

2H

5CN

Sol. Answer (3)

CH CN3

[H]NH

2

HONOOH

CHCl , OH3

–

CH –CH –N =C3 2

27. An amine reacts with C6H

5SO

2Cl and the product is soluble in alkali, amine is

(1) 1° amine (2) 2° amine (3) 3° amine (4) All of these

Sol. Answer (1)

R–NH + Cl–S–2

R–N–S–

O

OH

BaseSoluble

(1°)

O

O

28. Which would not react with benzene sulphonyl chloride in aqueous NaOH?

(1) Aniline (2) N, N-dimethylaniline (3) p-toluidine (4) N-ethyl aniline

Sol. Answer (2)

N

Me

Me

+ Cl–S

O

O

No reaction

Because of lack of H on nitrogen.



29. A mixture of 1°, 2° and 3° amines can be separated by Hinsberg’s reagent which is

(1) Benzoyl chloride (2) Acetyl chloride

(3) Benzene sulphonyl chloride (4) Benzyl chloride

Sol. Answer (3)

Hinsberg's reagent

O

O

–S–Cl

Benzene sulphonyl chloride

30. Bromine water reacts with aniline to give

(1) o-Bromoaniline (2) m-Bromoaniline

(3) p-Bromoaniline (4) 2, 4, 6-Tribromoaniline

Sol. Answer (4)

–NH + Br2 2

H O2 Br –

Br

NH2

Br

(2, 4,6 Tribromo anilne)

31. Identify X in the series.

NHCOCH3

HNO3

H SO2 4

IntermediateH O

2

X

(1)

NH2

NH2

(2)

NH2

NO2

(3)

NHCOCH3

NO2

(4)

NH2

NO2

NO2

Sol. Answer (2)

NH–C–CH3

O Nitration

NH–C–CH3

O

H O2

NH2

NO2

32. 2, 4, 6-tribromo aniline is a product of

(1) Electrophilic addition on C6H

5NH

2(2) Electrophilic substitution on C

6H

5NH

2

(3) Nucleophilic addition on C6H

5NH

2(4) Nucleophilic substitution on C

6H

5NH

2

Sol. Answer (2)

–NH + Br2 2

H O2 Br –

Br

NH2

Br

(2, 4,6 Tribromo aniline)



33. Amines are highly soluble in

(1) CCl4

(2) Diethyl ether (3) Benzene (4) Water

Sol. Answer (4)

Amines are highly soluble in H2O because of hydrogen bonding.

34. Which one of the following compounds gives dye test?

(1) Aniline (2) Methylamine (3) Diphenylamine (4) Ethylamine

Sol. Answer (1)

Aniline gives a dye test.

35. The strongest ortho-para and strongest meta directing groups are respectively

(1) –NO2 and –NH

2(2) –CONH

2 and –NH

2(3) –NH

2 and –NO

2(4) –NH

2 and CONH

2

Sol. Answer (3)

– NH2 Strongest o/p directing because of + R nature

– NO2 Strongest m directing because of – R nature

(Diazonium Salts)

36. Which of the following forms a stable diazonium salt at 273 - 278 K?

(1) Ethylamine (2) Aniline (3) Dimethylamine (4) Benzylamine

Sol. Answer (2)

NH 2

NaNO , dil HCl2

O–5°CN

2

+

Cl–

37. Reaction of nitrous acid with aliphatic primary amine in the cold gives

(1) A diazonium salt (2) An alcohol (3) A nitrite (4) A dye

Sol. Answer (2)

R–NH2 + HONO R–OH

(Alcohol)

38. The gas evolved when methylamine reacts with nitrous acid is

(1) NH3

(2) N2

(3) H2

(4) C2H

6

Sol. Answer (2)

CH3NH

2 + 2HNO

2 C-H

3–O–N =O + N

2 + H

2O

Methyl nitrite

2CH3NH

2 + 2HNO

2 CH

3–O–CH

3 + 2N

2 + 3 H

2O

Methoxy methane

1° Amine reacts with nitrous acid to form alcohol except methyl amine at ordinary temperature.

39. Benzenediazonium chloride on reaction with water gives

(1) Phenol (2) Aniline (3) Benzylamine (4) Benzaldehyde

Sol. Answer (1)

N Cl + H O2 2

+ –

OH



40. Benzenediazonium chloride on reaction with phenol in weakly basic medium gives

(1) Diphenyl ether (2) p-Hydroxyazobenzene (3) Chlorobenzene (4) Benzene

Sol. Answer (2)

N Cl + 2

+ –

OHOH

–

N

N

OH

41. When benzenediazonium chloride is heated with fluoro boric acid, fluorobenzene is formed. This reaction is

called

(1) Balz Schiemann reaction (2) Gattermann reaction

(3) Sandmeyer reaction (4) Hofmann bromamide reaction

Sol. Answer (1)

N Cl + H BF2 4

+ – + – F

This is Balz Schiemann reaction.

42. Benzenediazonium chloride is reduced to benzene by

(1) Water (2) Hypophosphorous acid (3) Hypophosphoric acid (4) Phosphine

Sol. Answer (2)

N Cl + H PO2 3 2

+–

(Benzene)

43. Diazotisation can be carried out by the action of NaNO2 and dilute HCI at ice cold temperature on

(1) Aromatic secondary amine

(2) Aromatic primary amine

(3) Aromatic nitro compound

(4) Aliphatic amine

Sol. Answer (2)

NH + HONO2

O–5°CN

2

44. When benzene diazonium chloride is treated with cuprous chloride and HCI the product formed is

(1) Chlorobenzene (2) Benzene

(3) Phenol (4) Chloroazobenzene



Sol. Answer (1)

N Cl + Cu Cl2 2 2

+ – HClCl (Sandmeyr's reaction)

45. The end product (Z) of the following reaction is

N Cl2

+ –

CuCN(X)

Complete hydrolysis(Y)

NaOH

CaO, (Z)

(1) A cyanide (2) A carboxylic acid (3) An amine (4) An arene

Sol. Answer (4)

CuCN

N Cl2

+ – CN

H+

H O2

COOH

(X) (Y)

NaOH

Cao,

(An arene)

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l1 ch-12 sol set-1 aldehydes ketones and carboxylic acids

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