//M1//QN1//SUB//DL0

Write IUPAC names of the following compounds and classify them into primary, secondary and tertiary amines. (i) (CH3)2CHNH2 (ii) CH3(CH2)2NH2 (iii) CH3NHCH(CH3)2 (iv) (CH3)3CNH2 (v) C6H5NHCH3 (vi) (CH3CH2)2NCH3 (vii) m-BrC6H4NH2

//X

S.N.

Structure

IUPAC Name

Types

(i)

(CH3)2CHNH2

Propan 2-amine

1°

(ii)

CH3(CH2)2NH2

Propan 1-amine

1°

(iii)

CH3NHCH(CH3)2

N-Methyl propan-2-amine

2°

(iv)

(CH3)3CNH2

2-Methyl propan-2-amine

1°

(v)

C6H5NHCH3

N-Methyl benzenamine

2°

(vi)

(CH3CH2)2NCH3

N-Ethyl, N-Methyl Ethanamine

3°

(vii)

m-BrC6H4NH2

3-bromo benzenamine

1°

//M0//QN2//SUB//DL0//EQ

Give one chemical test to distinguish between the following pairs of compounds. (i) Methylamine and dimethylamine (ii) Secondary and tertiary amines (iii) Ethylamine and aniline (iv) Aniline and benzylamine (v) Aniline and N-methylaniline

//X

(i) Methylamine and dimethylamine: Methylamine and dimethylamine can be distinguished by carbyl amine test.
CH3NH2 + CHCl3 + 3 KOH
(CH3)2NH + CHCl3 + 3 KOH no reaction
(ii) Secondary and tertiary amines: Secondary amines give Libermann nitrosoamine test while 3° amines do not give 2° amines on treatment with HNO2.
It gives yellow coloured oily N-nitroso amine.
(iii) Ethylamine and aniline: Ethylamine and aniline can be distinguished by azo dye test.
CH3CH2NH2 + HO – N = O
CH3CH2N2+ Cl – CH3CH2OH
(unstable)
C6H5NH2 + NaNO2 + 2HCl C6H5N+2 Cl – + NaCl + 2H2O
(stable)
(iv) Aniline and Benzylamine: Aniline and benzylamine can be distinguished by nitrous acid test.
C6H5CH2NH2 C6H5CH2N2+ Cl –
C6H5CH2OH
(unstable)
C6H5NH2 + NaNO2 + 2HCl C6H5N+2 Cl – + NaCl + 2H2O
(stable)
(v) Aniline and N-methylaniline: Aniline and N-methyl aniline can be distinguished by carbyl amine test.
C6H5NH2 + CHCl3 + 3 KOH
C6H5NC + 3KCl + 3H2O
(foul smell)
C6H5 – NH – CH3 + CHCl3 + 3 KOH no reaction

//M0//QN3//SUB//DL0//EQ

Account for the following: (i) pKb of aniline is more than that of methylamine. (ii) Ethylamine is soluble in water, where as aniline is not. (iii) Methylamine in water reacts with ferric chloride to precipitate hydrated ferric oxide. (iv) Although amino group is o– and p-directing in aromatic electrophilic substitution reactions, aniline m-nitro aniline. (v) Aniline does not undergo friedel crafts reaction. (vi) Diazonium salts of aromatic amines are more stable than those of aliphatic amines. (viii) Gabriel phthalimide synthesis is preferred for synthesising primary amines.

//X

(i) pKb of aniline is more than that of methyl amine.
In aniline, the lone pair of electrons on N-atom is delocalized over the benzene ring. As a result, electron density on the nitrogen decreases. On the other hand in CH3NH2, +I effect of CH3 group increases the electron density on N-atom. Therefore, aniline is less basic than methylamine and hence, pKb of aniline is higher than that of methylamine.
(ii) Ethylamine is soluble in water where as aniline is not.
Ethylamine is soluble in water due to intermolecular hydrogen bonding, where as in aniline its hydrogen bonding with water is weak due to the large hydrophobic part. So aniline is insoluble in water.
(iii) Methyl amine in water react with ferric chloride to precipitate hydrated ferric oxide.
Methylamine is more basic than water and therefore, accepts a proton from water forming OH– ions.
(These OH – ions combine with Fe3+ ions to form brown ppt. of hydrate ferric oxide)
FeCl3 Fe3+ + 3Cl– 2Fe3+ + 6OH –
2Fe(OH)3 Or Fe2O3.3H2O
(iv) Although amino group is o- and p-directing in aromatic electrophilic substitution reactions, aniline on nitrogen gives a substitution reactions, aniline on nitrogen gives a substantial amount of m-nitro aniline.
Under strongly acidic conditions of nitration in the presence of mixture of conc. HNO3 + H2SO4, aniline. gets protonated and is converted into anilinium ion having –NH3+ group.
This group is deactivating group and is m-directing.
So, the nitrogen of aniline gives o, p nitro aniline (mainly p-product) while the nitration of anilinium ion gives m-nitro aniline.
(v) Aniline does not undergo Friedel crafts reaction.
Aniline being a Lewis base reacts with Lewis acid such as AlCl3 to form a salt.
C6H5NH2 + AlCl3 → C6H5NH2+ AlCl3–
As a result, N of aniline acquires +ve charge and hence, it acts as a strong deactivating group for electrophilic substitution reaction.
Hence, aniline does not undergo Friedel crafts reaction.
(vi) Diazonium salts of aromatic amines are more stable than those of aliphatic amines.
The diazonium salts of aromatic amines are more stable than those of aliphatic amines because of dispersal of positive charge on the benzene ring due to resonance.
This type of resonance stability is not possible in alkyl diazonium salts.
(vii) Gabriel phthalimide synthesis is preferred for synthesising primary amines.
Gabriel phthalimide reaction gives pure 1° amines without any impurity of 2° or 3° amines.
Therefore, it is preferred for the synthesis of
1° amines.

//M0//QN4//SUB//DL0

Arrange the following: (i) In decreasing order of the pKb values: C2H5NH2, C6H5NHCH3, (C2H5)2NH and C6H5NH2 (ii) In increasing order of basic strength: C6H5NH2, C6H5N(CH3)2, (C2H5)2NH and CH3NH2 (iii) In increasing order of basic strength: (a) Aniline, p-nitroaniline and p-toluidine (b) C6H5NH2, C6H5NHCH3, C6H5CH2NH2 (iv) In decreasing order of basic strength in gas phase: C2H5NH2, (C2H5)2NH, (C2H5)3N and NH3 (v) In increasing order of boiling point: C2H5OH, (CH3)2NH, C2H5NH2 (vi) In increasing order of solubility in water: C6H5NH2, (C2H5)2NH, C2H5NH2

//X

(i) In C6H5NH2 only one –C2H5 group is present while in (C2H5)2NH, two –C2H5 groups are present. Thus, the + I effect is more in
(C2H5)2NH than in C2H5NH2. Therefore, the electron density over the N-atom is more in (C2H5)2NH than in C2H5NH2. Hence, (C2H5)2NH is more basic than C2H5NH2. Also, both C6H5NHCH3 and C6H5NH2 are less basic than (C2H5)2NH and C2H5NH2 due to the delocalization of the lone pair in the former two. Further among C6H5NHCH3 and C6H5NH2, the former will be more basic due to +I effect of –CH3 group. Hence the order of increasing basicity of the given compounds is as follows:
C6H5NH2 < C6H5NHCH3 < C2H5NH2 < (C2H5)2NH we know that the higher the basic strength, the lower is the pkb values.
C6H5NH2 > C6H5NHCH3 > C2H5NH2 > (C2H5)2NH
(ii) C6H5N (CH3)2 is more basic than C6H5 NH2 due to the presence of the +I effect of two
–CH3 groups in C6H5N(CH3)2. Further, CH3NH2 contains one –CH3 group while (C2H5)2NH contains two –C2H5 groups. Thus (C2H5)2NH is more basic than CH3NH2. Now C6H5N(CH3)2 is less basic than CH3 NH2 because of the –R effect of –C6H5 group. Hence the increasing order of the basic strengths of the given compounds is as follows:
C6H5 NH2 < C6H5 N (CH3)2 < CH3 NH2 < (C2H5)2 NH
(iii)
(a) In p-toluidine, the presence of electron donating –CH3 group increases the electron density on the N atom. Thus, p-toluidine is more basic than aniline. On the other hand, the presence of electron withdrawing –NO2 group decreases the electron density over the N-atom in p-nitro aniline. Thus, p-nitroaniline is less basic than aniline. Hence, the increasing order of the basic strengths of the given compounds is as follows:
p-nitroaniline < aniline < p-toluidine
(b) C6H5NHCH3 is more basic than C6H5NH2 due to the presence of electron donating –CH3 group in C6H5NHCH3. Again in C6H5 NHCH3 the –R effect of –C6H5 group decreases the electron density over the N-atom. Therefore; C6H5CH2NH2 is more basic than C6H5NHCH3. Hence, the increasing order of the basic strengths of the given compounds is as follows:
C6H5NH2 < C6H5NHCH3 < C6H5CH2NH2.
(iv) In the gas phase, there is no solvation effect. As a result, the basic strength mainly depends upon the +I effect. The higher the +I effect, the stronger is the base. Also the greater the number of Alkyl groups, the higher is the +I effect. So, the given compounds can be arranged in the decreasing order of their basic strengths in the gas phase as follows:
(C2H5)3N > (C2H5)2 NH > C2H5NH2 > NH3
(v) The boiling points of compounds depend on the extent of H bonding present in that compound. The more extensive the H–bonding in the compound, the higher is the boiling point. (CH3)2NH contains only one H-atom where as C2H5NH2 contains two H-atoms. Then C2H5NH2 undergoes more extensive H-bonding than (CH3)2 NH. Hence the boiling point of C2H5NH2 is higher than that of (CH3)2 NH. Further O is more electronegative than N. Thus C2H5OH forms stronger H-bonds than C2H5NH2. As a result, the boiling point of C2H5OH is higher than that of C2H5NH2 and (CH3)2NH. Increasing order of boiling point:
(CH3)2NH < C2H5NH2 < C2H5OH
(vi) The solubility of amines decreases with increase in the molecular mass. This is because the molecular mass of amines increases with an increase in size of the hydrophobic part. The molecular mass of C6H5NH2 is greater than that of C2H5NH2 and (C2H5)2NH. Hence the increasing order of their solubility in water is as follows:
C6H5NH2 < (C2H5)2NH < C2H5NH2

//M0//QN5//SUB//DL0//EQ

How will you convert: (i) Ethanoic acid into methanamine (ii) Hexane nitrile into 1-amino pentane (iii) Methanol to ethanoic acid (iv) Ethanamine into methanamine (v) Ethanoic acid into propanoic acid (vi) Methanamine into ethanamine (vii) Nitro methane into dimethylamine (viii) Propanoic acid into ethanoic acid?

//X

(i) Ethanoic acid into methanamine
CH3COOH CH3COCl CH3CONH2 CH3NH2
Ethanoic acid Ethanoyl Chloride Ethanamide Methanamine
(ii) Hexane nitrile into 1-amino pentane
(iii) Methanol to ethanoic acid
CH3OH CH3Cl CH3CN CH3COOH
Methanol Chloromethane Ethane nitrile Ethanoic Acid
(iv) Ethanamine into methanamine
(v) Ethanoic acid into propanoic acid
(vi) Methanamine into ethanamine
(vii) Nitro methane into dimethylamine
CH3NO2 CH3NH2 CH3NC CH3NHCH3
Nitromethane Methanamine Methyl Dimethyl amine
Isocyanide
(viii) Propanoic acid into ethanoic acid

//M3//QN6//SUB//DL0

Describe a method for the identification of primary, secondary, and tertiary amines. Also write chemical equations of the reactions involved. OR Write a note on the reaction of amine compounds with aryl sulphonyl chloride and explain the usefulness of this process. OR Explain with equation how primary, secondary, and tertiary amine compounds can be separated using Hinsberg's reagent.

//X

Benzenesulphonyl chloride (C6H5SO2Cl) which is also known as Hinsberg's reagent, reacts with primary and secondary amines to form sulphonamides.
(a) The reaction of benzene sulphonylchloride with primary amine yields N-ethyl benzenesulphonamide.
The hydrogen attached to nitrogen in sulphonamide is strongly acidic due to the presence of strong electron withdrawing sulphonyl group. Hence it is soluble in alkali.
(b) In the reaction with secondary amine, N, N-diethyl benzene sulphonamide is formed.
Since, N, N-diethylbenzene sulphonamide does not contain any hydrogen atom attached to nitrogen atom, it is not acidic and hence, it insoluble in alkali.
(c) Tertiary amines do not react with benzenesulphonyl chloride. This property of amines reacting with benzenesulphonyl chloride in a different manner is used for the distinction of primary, secondary, and tertiary amines and also for the separation of a mixture of amines. However these days benzenesulphonyl chloride is replaced by p-toluenesulphonyl chloride. (PTS)

//M3//QN7//SN//DL0

Write short notes on the following: (i) Carbylamine reaction (ii) Diazotisation (iii) Hofmann’s bromamide reaction (iv) Coupling reaction (v) Ammonolysis (vi) Acetylation (vii) Gabriel phthalimide synthesis.

//X

(i) Refer Que. no. 23 Page no. 248
(ii) Refer Que. no. 31 Page no. 251
(iii) Refer Que. no. 11 Page no. 244
(iv) Refer Que. no. 38 Page no. 252
(v) Refer Que. no. 7 Page no. 242
(vi) Refer Que. no. 21 Page no. 247
(vii) Refer Que. no. 10 Page no. 244

//M0//QN8//SUB//DL0//EQ

Accomplish the following conversions: (i) Nitrobenzene to benzoic acid (ii) Benzene to m-bromophenol (iii) Benzoic acid to aniline (iv) Aniline to 2, 4, 6-tribromofluorobenzene (v) Benzyl chloride to 2-phenyl ethanamine (vi) Chlorobenzene to p-chloro aniline (vii) Aniline to p-bromoaniline (viii) Benzamide to toluene (ix) Aniline to benzyl alcohol

//X

(i) Nitrobenzene to benzoic acid
(ii) Benzene to m-bromophenol
(iii) Benzoic acid to aniline
(iv) Aniline to 2, 4, 6-tribromofluro benzene
(v) Benzyl chloride to 2-phenyl ethanamine
(vi) Chloro benzene to p-chloro aniline
(vii) Aniline to p-bromo aniline
(viii) Benzamide to toluene
(ix) Aniline to benzyl alcohol

//M0//QN9//SUB//DL0//EQ

Give the structures of A, B and C in the following reactions: (i) CH3CH2I A B C(ii) C6H5N2Cl A B C(iii) CH3CH2Br A B C(iv) C6H5NO2 A B C(v) CH3COOH A B C(vi) C6H5NO2 A B C

//X

(i) A = CH3CH2CN, B = CH3CH2 – – NH2 and C = CH3CH2 – NH2.
(ii) A = C6H5CN, B = C6H5COOH and
C = C6H5CONH2.
(iii) A = CH3CH2CN, B = CH3CH2CH2NH2 and
C = CH3CH2CH2OH.
(iv) A = C6H5NH2, B = C6H5N+=NCl – and
C = C6H5OH.
(v) A = CH3CONH2, B = CH3NH2 and C = CH3OH.
(vi) A = C6H5NH2, B = C6H5N+=NCl – and
C =

//M0//QN10//SUB//DL0//EQ

An aromatic compound 'A' on treatment with aqueous ammonia and heating forms compound 'B' which on heating with Br2 and KOH forms a compound 'C' of molecular formula C6H7N. Write the structures and IUPAC names of compounds A, B, C.

//X

Since the compound 'C' of molecular formula C6H7N formed from 'B' on treatment with Br2 and KOH, therefore the compound 'B' must be an amide and 'C' must be amine. The only aromatic amine having molecular formula is C6H5NH2 (Aniline). Thus, B is benzamine.
Since 'B' is formed from 'A' with aqueous ammonia and heating, therefore compound 'A' must be benzoic acid.
Compound A, B, C their structure and IUPAC names are as follows:

//M0//QN11//SUB//DL0//EQ

Complete the following reactions: (i) C6H5NH2 + CHCl3 + Alcoholic KOH → (ii) C6H5N2Cl + H3PO2 + H2O → (iii) C6H5NH2 + H2SO4 (Concentrated) → (iv) C6H5N2Cl + C2H5OH → (v) C6H5NH2 + Br2(aq) → (vi) C6H5NH2 + (CH3CO)2 O → (vii) C6H5N2Cl

//X

(i) C6H5NH2 + CHCl3 + 3KOH(alc) C6H5N+ ≡ C– + 3KCl + 3H2O
(ii) C6H5N2Cl + H3PO2 + H2O C6H6 + N2 + H3PO3 + HCl
(iii)
(iv) C6H5N2Cl + C2H5OH C6H6 + CH3CHO + N2 + HCl
(v)
(vi) C6H5NH2 + CH3CO – O – COCH3 C6H5NHCOCH3 + CH3COOH
(vii) C6H5N2Cl C6H5N2+BF4–
C6H5NO2 + N2 + NaBF4

//M2//QN12//SUB//DL0//EQ

Why can not aromatic primary amines be prepared by Gabriel phthalimide synthesis?

//X

The success of the Gabriel phthalimide synthesis rests on the nucleophilic attack of phthalimide ion on the organic halide compound.
Aryl halides can not be converted to aryl amines by Gabriel synthesis because they do not undergo nucleophilic substitution with potassium phthalimide. So, aromatic primary amines can not be prepared by this method.

//M2//QN13//SUB//DL0

Write the reactions of (i) aromatic and (ii) aliphatic primary amines with nitrous acid.

Or Explain the reaction of primary aliphatic amine and aromatic amine compounds with nitrous acid with necessary equations.

//X

Refer Que. no. 24 Page no. 248

//M0//QN14//SUB//DL0

Give plausible explanation for each of the following: (i) Why are amines less acidic than alcohols of comparable molecular masses? (ii) Why do primary amines have higher boiling point than tertiary amines? (iii) Why are aliphatic amines stronger bases than aromatic amines?

//X

(i) Amine compounds are less acidic than alcohol compounds of the same molecular masses because an N-H bond is less polar than an O-N bond, so amines liberate H+ more difficulty than alcohols.
(ii) Primary amines have two hydrogen atoms on the N–atom and so, form intermolecular hydrogen bonding. Tertiary amines do not have hydrogen atoms on the N–atom and so these do not form hydrogen bonds. As a result of hydrogen bonding in primary amines, they have higher boiling points than tertiary amines.
(iii) Because of the following reason aliphatic amines are stronger bases than aromatic amines.
(a) Both arylamines and alkyl amines are basic in nature due to the presence of pair on N-atom. But arylamines are less basic than alkyl amines. less basic character of aromatic amines are due to resonating structures.
(b) As a result of resonating structures the pair of electrons become less available for protonation. Hence aromatic amines is less basic than aliphatic amines.
Class 12 Chemistry (Part 2) 017