Amines

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NCERT Textual Exercise (Solved)

1. Write IUPAC names of the following compounds and classify them into primary, secondary and tertiary amines.

(a) (CH3)2CHNH2 (b) CH3(CH2)2NH2

(c) CH3NHCH(CH3)2 (d) (CH3)3CNH2

(e) C6H5NHCH3 (f) (CH3CH2)2 NCH3

(g) m-Br-C6H4NH2

Sol. (a) Propan-2-amine (1º) (b) Propan-1-amine (1º) (c) N-Methylpropan-2-amine (2º) (d) 2-methylpropan-2-amine (3º) (e) N-methylbenzenamine or N-methylaniline (2º) (f ) N-ethyl-N-methylethanamine (3º) (g) 3-Bromobenzenamine or 3-bromoaniline (1º) 2. Give one chemical test to distinguish between the following pairs of

compounds: (a) Methylamine and dimethylamine (b) Secondary and tertiary amines (c) Ethylamine and aniline (d) Aniline and benzylamine (e) Aniline and methylaniline. Sol. (a) Methylamine and dimethylamine – by Carbylamine test.

CH NH CHCl KOH CH NCM

3 21

3 33º Amine ethyl isocyanide

(offensiv

+ + →∆

ee smell)

+ +3 3 2KCl H O

( )º

/ ( )CH CH NH CHCl KOH alc3 2 22

3

Amine∆

→ No reaction/ or Hinsberg’s test.

(b) Secondary and tertiary amine by Liebermann nitroamine test. Secondary amines gives Liebermann nitroamine test while tertiary amines do not.

( ) (CH CH NH HO N O CH CHHCl NaNO3 2 2

23

Diethylamine2º Amine

22 2 2) N N O H O

Yellow colour

or Hinsberg’s test.

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(c) Ethylamine and aniline by azo-dye test:

C2H5NH2 + HONO 273 278− →K C2H5OH + ↑+N H OBriskeffervesence

2 2

(d) Aniline and benzylamine by nitrous acid test:

5 C6H5CH2OH+N2↑+HCl

(e) Aniline and N-methylaniline by carbylamine test:

C H NH

Aniline6 5 2 + CHCl3 + 3KOH ∆ → C H NC

Offensivesmell

6 5(

)

+ 3KCl + 3H2O

C6H5 — NH — CH3CHCl KOH3 3+ → No reaction

3. Account for the following (a) pKb of aniline is more than that of methylamine (b) Ethylamine is soluble in water whereas aniline is not. (c) Methylamine in water reacts with ferric chloride to precipitate

hydrated ferric oxide. (d) Although amino group is o, p-directing in aromatic substitution

reactions, aniline on nitration gives a substantial amount of m-nitroaniline.

(e) Aniline does not undergo Friedel Crafts reaction. (f) Diazonium salts of aromatic amines are more stable than those of

aliphatic amines. (g) Gabriel phthalimide synthesis is preferred for synthesising primary

amines. Sol. (a) In aniline, the lone pair of electrons on the N-atoms are delocalised

over the benzene ring. As a result, electron density on the nitrogen decreases. In contrast, in CH3NH2, + I-effect of CH3 increases the electron density on the N-atom. Therefore, aniline is a weaker base

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NCERT Textual Exercise (Solved)

than methylamine and hence its pKb value is higher than that of methylamine.

(b) Ethylamine dissolves in water due to intermolecular H-bonding. However, in aniline, due to the large hydrophobic part, i.e., hydrocarbon part, the extent of H-bonding decreases considerably and hence aniline is insoluble in water.

(c) Methylamine being more basic than water, accepts a proton from water liberating OH– ions.

These OH– ions combine with Fe3+ ions present in H2O to form

brown precipitate of hydrated ferric oxide. FeCl3→Fe3+ + 3Cl–

2Fe3+ + 6OH– → 2 33 2 3 2Fe OH or Fe O H OBrown ppt

( ) .( .)

(d) Nitration is usually carried out with a mixture of conc HNO3 + conc H2SO4. In presence of these acids, most of aniline gets protonated to form anilinium ion. Therefore, in presence of acids, the reaction mixture consists of aniline and anilinium ion. Now, –NH2 group in aniline is o, p-directing and activating while the +NH3 group in anilinium ion is m-directing and deactivating. Whereas nitration of aniline (due to steric hindrance at o-position) mainly gives p-nitroaniline, the nitration of anilinium ion gives m-nitroaniline. In actual practice, approx a 1:1 mixture of o-nitroaniline and m-nitroaniline is obtained. Thus, nitration of aniline gives a substantial amount of m-nitroaniline due to protonation of the amino group.

Amines

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NCERT Textual Exercise (Solved)

(e) Aniline being a Lewis base reacts with Lewis acid AlCl3 to form a salt.

C6H5NH2 + AlCl3→C6H5NH2+AlCl3

–

As a result, N of aniline acquires positive charge and hence it acts as a strong deactivating group for electrophilic substitution reactions. Consequently, aniline does not undergo Fridel Craft reaction.

(f) The diazonium salts of aromatic amines are more stable than those of aliphatic amines due to dispersal of the positive charge on benzene ring as:

(g) Gabriel phthalimide reaction gives pure primary amines without

any contamination of secondary and tertiary amines. Therefore, it is preferred for synthesising primary amines.

4. Arrange the following: (a) In decreasing order of pKb values: C2H5NH2, C6H5NHCH3, (C2H5)2 NH and C6H5NH2

(b) In decreasing order of basic strength: C6H5NH2, C6H5N(CH3)2, (C2H5)2 NH and CH3NH2. (c) In increasing order of basic strength: (i) Aniline, p-nitroanitine and p-toluidine (ii) C6H5NH2, C6H5NHCH3, C6H5CH2NH2

(d) Decreasing order of basic strength in the gas phase: C2H5NH2, (C2H5)2NH, C2H5NH2

(e) Increasing order of boiling point: C2H5OH, (CH3)2NH, C2H5NH2

(f) Increasing order of solubility in water: C6H5NH2, (C2H5)2NH, C2H5NH2

Sol. (a) Due to delocalisation of lone pairs of electrons of the N-atom over the benzene ring, C6H5NH2 and C6H5NHCH3 are far less basic than C2H5NH2 and (C2H5)2NH. Due to +I-effect of the CH3

– group, C6H5NHCH3 is little more basic than C6H5NH2. Among C2H5NH2 and (C2H5)2NH, (C2H5)2NH is more basic than C2H5NH2 due to greater

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NCERT Textual Exercise (Solved)

+ I-effect of two C2H5 – groups. Therefore.

(C H ) NH C H NH C H NHCH C H NH2 5 2 2 5 2 6 5 3 6 5 2Decreasi

> > > → nng basic strength

Decreasing pk valueb←

(b) Among CH3NH2 and (C2H5)2NH, primarily due to the greater +I-effect of the two C2H5-groups over one CH3 group, (C2H5)2NH is more basic than CH3NH2.

∴

(C H ) NH CH NH C H NHCH C H NH2 5 2 3 2 6 5 3 6 5 2

Decreasin> > >

→ gg basic strength

(c) (i) The presence of electron donating group increases while the presence of electron withdrawing group decreases the basic strength of amines.

∴

p-nitroaniline < aniline < p-toluidine

Increasin → gg basic strength

(ii) In C6H5NH2 and C6H5NHCH3, N is directly attached to the benzene ring. As a result, the lone pair of electrons on the N-atom is delocalised over the benzene ring. Therefore, both C6H5NH2 and C6H5NHCH3 is a stronger base than C6H5NH2.

∴

C H NH C H NHCH C H CH NH6 5 2 6 5 3 6 5 2 2

Increasing basic< <

→ sstrength

(d) In the gas phase, solvent effects, i.e., stabilization of the conjugate acids due to H-bonding are absent. Therefore, basic strength mainly depends upon the +I-effect of the alkyl groups. Also the +I-effect increase with number of alkyl groups.

∴

(e) Since the electronegativity of O is higher than that of N, therefore, alcohols form stronger H-bonds than amines. Also, the extent of H-bonding depends upon the number of H-atoms on the N-atom.

∴

(f ) Solubility decreases with increase in molecular mass of amines due to increase in the size of the hydrophobic hydrocarbon part and with decrease in the number of H-atoms on the N-atom which undergo H-bonding.

∴

Amines

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NCERT Textual Exercise (Solved)

5. How will you convert the following: (a) Ethanoic acid into methanamine (b) Hexanenitrile into 1-amino pentane (c) Methanol to ethanoic acid. (d) Ethanamine into methanamine (e) Ethanoic acid into propanoic acid (f ) Methanamine into ethanamine (g) Nitromethane into dimethylamine (h) Propanoic acid into ethanoic acid

Sol. (a) NH3

�CH CONH3 2

(b) � NH3

(c) CH3OH PClPOCl

53−

→ CH3Cl KCN alc( ) → CH3CN H O3+

→ CH3COOH

(d) �

(e)

(f )

(g) CH NO CH NH CH NHFe HCl CH BrHBr3 2 3 2 3 23+

− → → ( )

(h)

3 2

3

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NCERT Textual Exercise (Solved)

6. Describe the method for the identification of primary, secondary and tertiary amines. Also write chemical equations of the reactions involved.

Sol. The three types of amines can be distinguished by Hinsberg test. In this test, the amine is shaken with benzene sulphonyl chloride (C6H5SO2Cl) is the presence of excess of NaOH or KOH. A primary amine reacts to give a clear solution, which on acidification yields an insoluble compound.

RNH2 + C6H5SO2Cl → C6H5SO2NHR

KOH

6 5 2(Clear solution)

H6 5 2

(InsolubleC H SO NRK C H SO NHR

+ → →+

))

A secondary amine forms an insoluble compound, which remains insoluble even on acidification.

R NH C H SO Cl C H SO NR6 5 2

OH6 5 2 2

(Insoluble)

H- +

2 + → → No reaction

A tertiary amine does not react with the reagent, but dissolves in acid.

R N H [R NH]3

+

(Clear solution)2 + →+

7. Write a short notes on the followings: (a) Carbylamine reaction (b) Diazotization (c) Hofmann Bromamide (d) Coupling reaction (e) Ammonolysis (f) Acetylation (g) Gabriel phthalimide synthesis Sol. (a) Carbylamine reactions: Both aliphatic and aromatic primary amines

when warmed with chloroform and an alcoholic solution of KOH, produces isocyanides or carbylamines which have very unpleasant odours. This reaction is called carbylamine reaction.

R — NH2 + CHCl3 + 3KOH (alc) → R — N ≡ C + 3KCl + 3H2O (b) Diazotisation: The process of conversion of a primary aromatic

amino compound into a diazonium salt, is known as diazotisation. This process is carried out by adding an aqueous solution of sodium nitrite to a solution of primary aromatic amine (e.g., aniline) in excess of HCl at a temperature below 5 ºC.

(c) Hoffmann Bromamide reaction: When an amide is treated with

bromine in alkali solution, it is converted to a primary amine that has one carbon atom less than the starting amide. This reaction is known as Hoffmann’s bromamide degradation reaction.

Amines

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NCERT Textual Exercise (Solved)

C H CONH C H NHBr NaOHNaOBr6 5 2 6 5 22+ →( )

(d) Coupling reaction: In this reaction, a molecule of diazonium salt reacts with a molecule of an aromatic amino compound (in acidic medium) or a phenol (in alkaline medium). The reaction generally takes place at para position to the hydroxy or amino group. If ortho and para positions are occupied, then no coupling takes place.

+ –

(e) Ammonolysis: It is a process of replacement of either halogen atom in alkyl halides (or aryl halides) or hydroxyl group in alcohols (or phenols) by amino group. The reagent used for ammonolysis is alcoholic ammonia. Generally, a mixture of primary, secondary and tertiary amine is formed.

ROH + NH RNH + H OAl O ,”High temp, Pressure

2 33 2 2 →

(f ) Acetylation: The process of introducing an acetyl (CH3CO–) group into molecule using acetyl chloride or acetic anhydride is called acetylation.

+

NH – CO – CH3

(g) Gabriel phthalimide synthesis: It is a method of preparation of pure aliphatic primary amines. Phthalimide on treatment with ethanolic KOH gives potassium phathalimide which on heating with a suitable alkyl halides gives N-substituted phthalimides, which on hydrolysis with dil HCl gives primary amines.

C H NH

2 5 2

8. Accomplish the following conversions: (a) Nitrobenzene to benzoic acid (b) Benzene to m-bromophenol

Amines

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NCERT Textual Exercise (Solved)

(c) Benzoic acid to aniline (d) Aniline to 2, 4, 6-tribromofluorobenzene (e) Benzylchloride to 2-phenylethanamine (f ) Chlorobenzene to p-chloroaniline (g) Aniline to p-bromoaniline (h) Benzamide to toluene (i) Aniline to benzylacohol. Sol. (a)

(b)

(c)

(d)

Heat

(e)

Amines

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NCERT Textual Exercise (Solved)

(f )

(g)

(h)

(i)

9. Give the structures of A, B and C in the following reactions:

(a) CH CH I A B2 2NaCN OH

Partial hydrolysisNaOH Br-

2 → → →+ CC

(b) C H N Cl A B C5 2CuCN H O H NH2

+3

6 → → →/

(c) CH CH Br A B C2KCN LiAlH HNO4 2

3 → → →

(d) C H NO A B C2Fe HCl NaNO HCl

KH O H2 2

+

6 5 273/ / → → →+

(e) CH COOH A B CNH NaOBr NaNO HCl3 23 ∆

→ → →/

Amines

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NCERT Textual Exercise (Solved)

(f ) C H NO A B CFe HCl HNOK

C H OH2 6 56 5 2 273

/ → → →

Sol. (a) A = CH3CH2CN,

O

B C H C NH||

2 5 2 and C = C2H5 — NH2.

(b) A = C6H5CN, B = C6H5COOH and C = C6H5COONH4. (c) A = CH3CH2CN, B = CH3CH2CH2NH2 and C = CH3CH2 CH2 OH. (d) A = C6H5NH2, B = C6H5N

+ ≡ NCl– and C = C6H5OH. (e) A = CH3CONH2, B = CH3NH2 and C = CH3OH.

(f ) A = C6H5NH2, B = C6H5N+ ≡ NCl– and C =

10. 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 compound A, B and C.

Sol. Since the compound ‘C’ with molecular formula C6H7N is formed from compound ‘B’ on treatment with Br2 / KOH, therefore, compound ‘B’ must be an amide and ‘C’ must be an amine.

The only amine having the molecular formula C6H7N, i.e., C6H5NH2 is aniline. Since ‘C’ is aniline, therefore, the amide from which it is formed must be

benzamide (C6H5CONH2). Thus, compound ‘B’ is benzamide. Since compound ‘B’ is formed from compound ‘A’ with aqueous ammonia

and heating, therefore, compound ‘A’ must be benzoic acid. 11. Complete the following reactions: (a) C6H5NH2 + CHCl3 + Alc. KOH (b) C6H5N2Cl + H3PO2 + H2O→ (c) C6H5NH2 + H2SO4 (conc)→ (d) C6H5N2Cl + C2H5OH→ (e) C6H5NH2 + Br2 (aq)→ (f ) C6H5NH2 + (CH3 CO)2 O→

(g) C H N Cl (i) HBF(ii) NaNO Cu

426 5 2 / ,∆ →

Sol. (a) C6H5NH2 + CHCl3 + alc KOH ∆ → C6H5NC + 3KCl + 3H2O

(b) C6H5N2Cl + H3PO2 + H2OCu+ → C6H6 + N2 + H3PO3 + HCl

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NCERT Textual Exercise (Solved)

(c)

Zwitterion

(d) C6H5N2Cl + C2H5OH Reduction → C6H6 + CH3CHO + N2 + HCl

(e)

(f) C6H5NH2 + CH3CO — O — COCH3 CH COOH3 → C6H5NHCOCH3

+ CH3COOH

(g) C6H5N2Cl HBF

HCl4

− → C6H5N2

+BF4–

NaNO Cu2 /∆

→ C6H5NO2 + BF3

+ NaF 12. Why aromatic primary amines cannot be prepared by Gabriel phthalimide

synthesis? Sol. The success of Gabriel phthalimide reaction depends upon the nucleophilic

attack by the phthalimide anion on the organic halogen compound. Since aryl halides do not undergo nucleophilic substitution reactions easily,

therefore, arylamines, i.e., aromatic, primary amines cannot be prepared by Gabriel phthalimide reaction.

13. How do aromatic and aliphatic primary amines react with nitrous acid? Sol. Both aromatic and aliphatic primary amines react with HNO2 at 273-278 K

to form aromatic and aliphatic diazonium salts respectively. But aliphatic

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diazonium salts are unstable even at this low temperature and thus decompose readily to form a mixture of compounds.

14. Give plausible explanation for each of the following: (a) Why are amines less acidic than alcohols? (b) Why are primary amines have higher boiling point than tertiary

amines? (c) Why are aliphatic amines stronger bases than aromatic amines? Sol. (a) Loss of proton from an amine gives an amide ion while loss of a

proton from alcohol give an alkoxide ion. R — NH2→R — NH– + H+

R — O — H→R — O– + H+

Since O is more electronegative that N, therefore, RO– can accommodate the negative charge more easily than the can accommodate. Thus, RO– is more stable than . Thus, alcohols are more acidic than amines. Conversely, amines are less acidic than alcohols.

(b) Due to the presence of two H-atoms on N-atom of primary amines, they undergo extensive intermolecular H-bonding while tertiary amines due to the absence of a H-atom on the N-atom do not undergo H-bonding. As a result, primary amines have higher boiling points than tertiary amines of comparable molecular mass.

(c) Aromatic amines are far less basic than ammonia and aliphatic amines because of following reasons:

(i) Due to resonance in aniline, the lone pair of electrons on the nitrogen atom gets delocalized over the benzene ring and thus is less easily available for protonation. Therefore, aromatic amines are weaker bases than ammonia.

(ii) Also aniline is more stable than anilinium ion, Hence, aniline has very less tendency to combine with a proton to form anilinium ion, and is thus less basic.