Preparation of Amines

ChemistryAminesFor JEE aspirants

The preparation of amines uses three families of reactions: nitrogen attacking a carbon electrophile (ammonolysis, Gabriel synthesis), reduction of N-containing groups (nitro, nitrile, amide, oxime, imine) and rearrangements that remove one carbon (Hofmann, Curtius, Schmidt, Lossen). Choosing a method means asking two questions: which class of amine is needed, and how many carbon atoms it must have. This page covers every preparation of amines in NCERT and the JEE syllabus, with mechanisms.

Key Formulas - Quick Reference

Ammonolysis: $RX + N H_{3} RN H_{3}^{+} X^{-}$ , then NaOH frees $RN H_{2}$ ; gives a mixture; reactivity RI > RBr > RCl
Nitro reduction: $RN O_{2} + 6 [H] Sn/HCl, Fe/HCl or H_{2} /Ni RN H_{2} + 2 H_{2} O$ (same carbon count)
Nitrile reduction: $RCN + 4 [H] Na/ C_{2} H_{5} OH or LiAl H_{4} RC H_{2} N H_{2}$ (one C more than R-X)
Amide reduction: $RCON H_{2} + 4 [H] LiAl H_{4} RC H_{2} N H_{2} + H_{2} O$ (same carbon count)
Oxime reduction: $R_{2} C = NOH + 4 [H] Na/ C_{2} H_{5} OH R_{2} CHN H_{2} + H_{2} O$
Reductive amination: $R_{2} C = O + N H_{3} + H_{2} Ni R_{2} CHN H_{2} + H_{2} O$
Gabriel: phthalimide, then KOH, R-X and hydrolysis, gives pure $RN H_{2}$ (1° aliphatic only)
Hofmann bromamide: $RCON H_{2} + B r_{2} + 4KOH RN H_{2} + K_{2} C O_{3} + 2KBr + 2 H_{2} O$ (one C fewer)
Curtius, Schmidt, Lossen: all pass through $R - N = C = O$ ; $RNCO + H_{2} O RN H_{2} + C O_{2}$
2° from isocyanides: $RNC + 4 [H] RNHC H_{3}$ ; 3° from $R_{4} N^{+} O H^{-} heat R_{3} N + alkene + H_{2} O$

1. Choosing a Preparation Method

Two questions decide the route. First, which class of amine is wanted: only some methods stop cleanly at a primary amine. Second, how many carbon atoms: some methods keep the carbon count of the starting compound, the nitrile route adds one carbon to an alkyl halide, and the Hofmann family removes one. The table summarises the methods on this page.

Method	Starting compound	Amine obtained	Carbon count
Ammonolysis	$R - X$ + $N H_{3}$	mixture of 1°, 2°, 3°, 4°	same as R-X
Ammonia + alcohol	$R - OH$ + $N H_{3}$	mixture	same as R-OH
Reduction of nitro compounds	$R - N O_{2}$ , $Ar - N O_{2}$	1° (including aniline)	same
Reduction of nitriles	$R - C \equiv N$	1° $RC H_{2} N H_{2}$	one more than R-X
Reduction of isocyanides	$R - N \equiv C$	2° $RNHC H_{3}$	adds N-methyl
Reduction of amides	$RCON H_{2}$ , $RCONH R^{'}$ , $RCON (R^{'})_{2}$	1°, 2° or 3°	same
Reduction of oximes	$R_{2} C = NOH$	1°	same
Reductive amination	aldehyde or ketone + $N H_{3}$ or amine	1°, 2° or 3°	same
Gabriel synthesis	potassium phthalimide + $R - X$	pure 1° aliphatic	same as R-X
Hofmann bromamide	$RCON H_{2}$	pure 1°	one fewer
Curtius, Schmidt, Lossen	$RCOCl$ , $RCOOH$ , $RCONHOH$	pure 1°	one fewer
Heating $R_{4} N^{+} O H^{-}$	quaternary ammonium hydroxide	3°	loses one R

Exam Trick

Nitrile: plus one. Hofmann: minus one. Reductions: no change. To go from $C H_{3} Br$ to ethylamine use $KCN$ then reduction (the CN carbon is added). To go from propanamide to ethylamine use $B r_{2}$ /KOH (the C=O carbon leaves as $C O_{2}$ ). Nitro, amide and oxime reductions keep every carbon.

Figure 1: Ask about the class first and the carbon count second; the answer points to one method.

2. Ammonolysis of Alkyl Halides

An alkyl halide heated with ethanolic ammonia in a sealed tube at about 373 K undergoes nucleophilic substitution: the nitrogen lone pair displaces the halide. This is called ammonolysis (Hofmann's method). The primary amine first forms as its ammonium salt; a strong base such as NaOH releases the free amine.

C_{2} H_{5} Br + N H_{3} ethanol, 373 K C_{2} H_{5} N H_{3}^{+} B r^{-}

C_{2} H_{5} N H_{3}^{+} B r^{-} + NaOH C_{2} H_{5} N H_{2} + NaBr + H_{2} O

The reaction does not stop there. The primary amine is itself a nucleophile, in fact a stronger one than ammonia, so it reacts with more alkyl halide to give the secondary amine, then the tertiary amine and finally the quaternary ammonium salt (Figure 2).

C H_{3} N H_{2} C H_{3} I (C H_{3})_{2} NH C H_{3} I (C H_{3})_{3} N C H_{3} I (C H_{3})_{4} N^{+} I^{-}

Figure 2: Ammonolysis does not stop at

RN H_{2}

: every product can attack

R - X

again, so a pure

1^{\circ}

amine needs Gabriel, Hofmann or a reduction method.

Order of reactivity of halides: RI > RBr > RCl, following the ease of breaking the C-X bond.
Excess ammonia makes the primary amine the main product; excess alkyl halide gives the quaternary salt (exhaustive alkylation).
Aryl halides hardly react, because the C-X bond has partial double-bond character and $S_{N} 2$ on a ring carbon is not possible.
The mixture of amines is separated by fractional distillation, the Hinsberg method or Hofmann's method with diethyl oxalate (see Analysis of Amines).

Figure 3: Ammonolysis is an

S_{N} 2

reaction on carbon; the product amine is itself a nucleophile, which is the root of over-alkylation.

3. Ammonia and Alcohols

Alcohol vapour and ammonia passed over heated alumina also give a mixture of 1°, 2° and 3° amines, which again must be separated. A large excess of ammonia over zinc chloride favours the primary amine.

C_{2} H_{5} OH + N H_{3} A l_{2} O_{3}, 623 K C_{2} H_{5} N H_{2} + H_{2} O

C_{2} H_{5} N H_{2} C_{2} H_{5} OH, - H_{2} O (C_{2} H_{5})_{2} NH C_{2} H_{5} OH, - H_{2} O (C_{2} H_{5})_{3} N

ROH + N H_{3} ZnC l_{2}, 573 K RN H_{2} + H_{2} O

4. Reduction of Nitro Compounds

Nitro compounds are reduced to primary amines by hydrogen over finely divided nickel, palladium or platinum, or by a metal and acid. This is the standard route to aniline and other aromatic amines.

C_{6} H_{5} N O_{2} + 3 H_{2} Ni, Pd or Pt C_{6} H_{5} N H_{2} + 2 H_{2} O

C H_{3} C H_{2} N O_{2} + 6 [H] Sn/HCl C H_{3} C H_{2} N H_{2} + 2 H_{2} O

With tin and hydrochloric acid the amine is obtained as its salt, which must be treated with alkali:

2 C_{6} H_{5} N O_{2} + 3Sn + 14HCl 2 C_{6} H_{5} N H_{3}^{+} C l^{-} + 3SnC l_{4} + 4 H_{2} O

C_{6} H_{5} N H_{3}^{+} C l^{-} + NaOH C_{6} H_{5} N H_{2} + NaCl + H_{2} O

Why iron scrap and HCl is preferred industrially. The iron(II) chloride formed is hydrolysed and releases hydrochloric acid again, so only a small amount of acid is needed to start the reaction:

FeC l_{2} + 2 H_{2} O \to Fe (OH)_{2} + 2HCl

Selective reduction. When two nitro groups are meta to each other, ammonium hydrogen sulphide, ammonium sulphide or $H_{2} S$ in ammonia reduce one group at a time (Zinin reduction). The first product dominates, so m-nitroaniline can be isolated; further reduction gives m-phenylenediamine.

C_{6} H_{4} (N O_{2})_{2} N H_{4} SH N O_{2} C_{6} H_{4} N H_{2} N H_{4} SH C_{6} H_{4} (N H_{2})_{2}

The products of nitrobenzene in neutral and alkaline media (phenylhydroxylamine, azoxy-, azo- and hydrazobenzene) are collected in Some Important Organic Compounds Containing Nitrogen.

5. Reduction of Nitriles and Isocyanides

Nitriles are reduced to primary amines by sodium and ethanol (Mendius reduction), by $LiAl H_{4}$ or by catalytic hydrogenation. The new amine has one more carbon than the alkyl halide the nitrile was made from, so this is a way of ascending a homologous series.

C H_{3} CN + 4 [H] Na/ C_{2} H_{5} OH or LiAl H_{4} C H_{3} C H_{2} N H_{2}

C H_{3} CN + 2 H_{2} Ni C H_{3} C H_{2} N H_{2}

C H_{3} Br KCN C H_{3} CN LiAl H_{4} C H_{3} C H_{2} N H_{2}

Isocyanides are reduced in the same way, but the carbon atom stays on nitrogen as a methyl group, so a secondary amine forms:

C H_{3} NC + 4 [H] Na/ C_{2} H_{5} OH or LiAl H_{4} N-methylmethanamine C H_{3} NHC H_{3}

6. Reduction of Amides and Oximes; Reductive Amination

6.1 Amides

Lithium aluminium hydride followed by water reduces the C=O of an amide to $C H_{2}$ . The amine keeps every carbon atom, and its class matches the amide: $RCON H_{2}$ gives a 1° amine, $RCONH R^{'}$ a 2° amine and $RCON (R^{'})_{2}$ a 3° amine.

RCON H_{2} + 4 [H] (i) LiAl H_{4} (ii) H_{2} O RC H_{2} N H_{2} + H_{2} O

RCONH R^{'} + 4 [H] LiAl H_{4} RC H_{2} NH R^{'} + H_{2} O

RCON (R^{'})_{2} + 4 [H] LiAl H_{4} RC H_{2} N (R^{'})_{2} + H_{2} O

6.2 Oximes

Aldehydes and ketones react with hydroxylamine to give oximes, which sodium and ethanol (or $LiAl H_{4}$ ) reduce to primary amines:

C_{6} H_{10} O + N H_{2} OH cyclohexanone oxime C_{6} H_{10} = NOH + H_{2} O

C_{6} H_{10} = NOH + 4 [H] Na/ C_{2} H_{5} OH cyclohexylamine C_{6} H_{11} N H_{2} + H_{2} O

Figure 4: Pick the starting compound by the amine you want: nitrile and amide give

RC H_{2} N H_{2}

, an isocyanide gives an N-methyl

2^{\circ}

amine.

6.3 Reductive Amination

An aldehyde or ketone and ammonia (or an amine) hydrogenated together over nickel give an amine directly. The carbonyl compound first forms an imine, which is reduced as it forms (Figure 5). Because each step adds exactly one carbon group to the nitrogen, over-alkylation does not occur.

C H_{3} CHO + N H_{3} + H_{2} Ni C H_{3} C H_{2} N H_{2} + H_{2} O

(C H_{3})_{2} CO + N H_{3} + H_{2} Ni propan-2-amine (C H_{3})_{2} CHN H_{2} + H_{2} O

C H_{3} C H_{2} CHO + C H_{3} C H_{2} N H_{2} + H_{2} Ni 2° amine C H_{3} C H_{2} C H_{2} NHC H_{2} C H_{3} + H_{2} O

Figure 5: Reductive amination adds one alkyl group to the nitrogen used:

N H_{3}

gives a

1^{\circ}

amine and

R^{'} N H_{2}

gives a

2^{\circ}

amine, with no over-alkylation.

7. Gabriel Phthalimide Synthesis

Phthalimide has one N-H flanked by two C=O groups, which makes it acidic. Ethanolic KOH converts it to potassium phthalimide, whose nitrogen anion attacks an alkyl halide by $S_{N} 2$ . Hydrolysis of the N-alkylphthalimide with aqueous NaOH (or 20% HCl under pressure) then releases the primary amine (Figure 6).

C_{6} H_{4} (CO)_{2} NH + KOH ethanol C_{6} H_{4} (CO)_{2} NK + H_{2} O

C_{6} H_{4} (CO)_{2} NK + RX C_{6} H_{4} (CO)_{2} NR + KX

C_{6} H_{4} (CO)_{2} NR + 2NaOH heat C_{6} H_{4} (COONa)_{2} + RN H_{2}

Figure 6: Phthalimide is a protected form of

N H_{3}

with only one replaceable H, so Gabriel synthesis gives pure

1^{\circ}

aliphatic amines.

Exam Trick

Gabriel = 1° and aliphatic only. Nitrogen in phthalimide has room for exactly one R group, so no 2° or 3° amines form. Aryl halides cannot be used because they do not undergo nucleophilic substitution with the phthalimide anion, so aniline and other aromatic amines are never made this way.

JEE Advanced

Hydrolysis of an N-alkylphthalimide is slow. In the Ing-Manske modification it is heated with hydrazine instead, which gives the amine and phthalhydrazide under much milder conditions. Because the key step is $S_{N} 2$ , Gabriel synthesis works best with methyl and primary halides; tertiary halides eliminate instead.

8. Hofmann Bromamide Degradation

A primary amide warmed with bromine and aqueous KOH (or NaOH) gives a primary amine with one carbon atom fewer than the amide. The carbonyl carbon ends up as carbonate.

C H_{3} CON H_{2} + B r_{2} + 4KOH C H_{3} N H_{2} + K_{2} C O_{3} + 2KBr + 2 H_{2} O

C_{6} H_{5} CON H_{2} + B r_{2} + 4KOH C_{6} H_{5} N H_{2} + K_{2} C O_{3} + 2KBr + 2 H_{2} O

Bromine and alkali first form hypobromite, $B r_{2} + 2KOH \to KOBr + KBr + H_{2} O$ , which brominates the nitrogen. The mechanism then runs in four steps (Figure 7):

N-Bromination: $RCON H_{2}$ becomes the N-bromoamide $RCONHBr$ .
Deprotonation: the Br atom makes the remaining N-H more acidic, and $O H^{-}$ removes it to give $RCO N^{-} Br$ .
Rearrangement: the R group moves from the carbonyl carbon to nitrogen at the same moment as $B r^{-}$ leaves, giving the isocyanate $R - N = C = O$ .
Hydrolysis: water adds to the isocyanate to give a carbamic acid $RNHCOOH$ , which loses $C O_{2}$ to leave $RN H_{2}$ .

Figure 7: The carbonyl carbon is lost as

C O_{2}

(as

K_{2} C O_{3}

), so

RCON H_{2}

gives

RN H_{2}

with one carbon fewer; R moves to N with retention.

JEE Advanced

Older books describe a free acyl nitrene $RCON :$ as the intermediate. Experiments show that migration of R and loss of $B r^{-}$ happen together, so no free nitrene forms. Because R never leaves the molecule, a chiral migrating group keeps its configuration: (S)-2-phenylpropanamide gives (S)-1-phenylethylamine. Aryl groups carrying electron-donating substituents migrate fastest.

9. Curtius, Schmidt and Lossen Rearrangements

These three reactions work exactly like the Hofmann degradation. In each, an alkyl or aryl group shifts from carbon to an electron-poor nitrogen (a 1,2-shift) while a leaving group departs, and the isocyanate formed is hydrolysed to the amine. Only the leaving group differs (Figure 8):

Reaction	Starting compound	Reagent	Leaving group
Hofmann	amide $RCON H_{2}$	$B r_{2}$ + KOH	$B r^{-}$
Curtius	acid chloride $RCOCl$ (via acyl azide $RCO N_{3}$ )	$Na N_{3}$ , then heat	$N_{2}$
Schmidt	carboxylic acid $RCOOH$	$H N_{3}$ , conc. $H_{2} S O_{4}$	$N_{2}$
Lossen	hydroxamic acid $RCONHOH$ (as its O-acyl derivative)	base, heat	carboxylate $R^{'} CO O^{-}$

Figure 8: Only the leaving group on nitrogen changes (

B r^{-}

N_{2}

R^{'} CO O^{-}

); every route passes through

R - N = C = O

and loses one carbon.

Curtius reaction. An acid chloride and sodium azide give an acyl azide, which on heating loses nitrogen to give the isocyanate:

RCOCl + Na N_{3} RCO N_{3} + NaCl

RCO N_{3} heat RNCO + N_{2}

RNCO + H_{2} O RN H_{2} + C O_{2}

Schmidt reaction. A carboxylic acid reacts with hydrazoic acid in the presence of conc. $H_{2} S O_{4}$ :

C H_{3} COOH + H N_{3} conc. H_{2} S O_{4} C H_{3} N H_{2} + C O_{2} + N_{2}

Lossen reaction. Hydroxylamine and an acid chloride give a hydroxamic acid $RCONHOH$ , which exists in equilibrium with its enol (hydroximic acid) form $RC (OH) = NOH$ . The O-acyl derivative of the hydroxamic acid, heated with base, loses a carboxylate ion as R migrates:

RCOCl + N H_{2} OH RCONHOH + HCl

RCONHOH + R^{'} COCl RCONHOCO R^{'} + HCl

RCONHOCO R^{'} + O H^{-} heat RNCO + R^{'} CO O^{-} + H_{2} O

JEE Advanced

Schmidt mechanism in brief. Conc. $H_{2} S O_{4}$ turns $RCOOH$ into the acylium ion $RC O^{+}$ (by protonation and loss of water). Hydrazoic acid attacks it to give the protonated acyl azide $RCONH - N_{2}^{+}$ . R migrates to nitrogen as $N_{2}$ leaves, and the protonated isocyanate is hydrolysed to $RN H_{2}$ and $C O_{2}$ . In all four rearrangements the migrating group keeps its configuration.

10. Other Methods for Particular Classes of Amines

10.1 Primary Amines

(i) Grignard reagent and chloramine give 1° amines, even with a tertiary alkyl group:

(C H_{3})_{3} CMgI + ClN H_{2} tert-butylamine (C H_{3})_{3} CN H_{2} + Mg (I) Cl

(ii) Decarboxylation of amino acids with barium hydroxide:

H_{2} NC H_{2} COOH Ba (OH)_{2}, heat glycine to methylamine C H_{3} N H_{2} + C O_{2}

(iii) Hydrolysis of isocyanides and isocyanates (the second is sometimes called Wurtz's method):

C H_{3} NC + 2 H_{2} O HCl C H_{3} N H_{2} + HCOOH

C H_{3} NCO + 2KOH C H_{3} N H_{2} + K_{2} C O_{3}

10.2 Secondary Amines

(i) Reduction of isocyanides (Section 5) gives N-methyl 2° amines.

(ii) A primary amine and one equivalent of alkyl halide:

C_{2} H_{5} N H_{2} + C_{2} H_{5} I diethylamine (C_{2} H_{5})_{2} NH + HI

(iii) Hydrolysis of p-nitroso-N,N-dialkylanilines with boiling alkali gives a pure dialkylamine. Aniline is methylated twice, then nitrosated at the para position:

C_{6} H_{5} N H_{2} 2 C H_{3} I C_{6} H_{5} N (C H_{3})_{2} HN O_{2} ON C_{6} H_{4} N (C H_{3})_{2}

ON C_{6} H_{4} N (C H_{3})_{2} + NaOH boil ON C_{6} H_{4} ONa + (C H_{3})_{2} NH

10.3 Tertiary Amines

(i) Ammonia with excess alkyl halide (in practice the reaction tends to run on to the quaternary salt):

N H_{3} + 3C H_{3} I (C H_{3})_{3} N + 3HI

(ii) Heating a tetraalkylammonium hydroxide. Moist silver oxide converts the quaternary iodide to the hydroxide, which on heating loses an alkene (Hofmann elimination, explained in Properties of Amines). Tetramethylammonium hydroxide, which has no $β$ -hydrogen, gives methanol instead.

(C_{2} H_{5})_{4} N^{+} I^{-} + AgOH (C_{2} H_{5})_{4} N^{+} O H^{-} + AgI

(C_{2} H_{5})_{4} N^{+} O H^{-} heat (C_{2} H_{5})_{3} N + C H_{2} = C H_{2} + H_{2} O

(C H_{3})_{4} N^{+} O H^{-} heat (C H_{3})_{3} N + C H_{3} OH

Figure 9: Every method either makes a new C-N bond (substitution, amination) or keeps one while changing its oxidation level (reduction, rearrangement).

11. Solved Examples

Solved Example 1

The amide that gives 1-phenylethylamine on Hofmann bromamide reaction is
(A) 2-phenylpropanamide
(B) 3-phenylpropanamide
(C) 2-phenylethanamide
(D) N-phenylethanamide

Solution:

Answer: (A). The Hofmann reaction replaces $- CON H_{2}$ by $- N H_{2}$ on the same carbon. 1-Phenylethylamine is $C_{6} H_{5} CH (C H_{3}) N H_{2}$ , so the amide must be $C_{6} H_{5} CH (C H_{3}) CON H_{2}$ , 2-phenylpropanamide.

C_{6} H_{5} CH (C H_{3}) CON H_{2} + B r_{2} + 4KOH C_{6} H_{5} CH (C H_{3}) N H_{2} + K_{2} C O_{3} + 2KBr + 2 H_{2} O

Solved Example 2

Phthalimide is treated with (i) KOH in ethanol and (ii)

C_{6} H_{5} C H_{2} Cl

to give X, and X is hydrolysed to give Y. Y is
(A) 2-carbamoylbenzoic acid
(B)

C_{6} H_{5} C H_{2} N H_{2}

Solution:

Answer: (B). This is Gabriel synthesis. X is N-benzylphthalimide (option D is the intermediate, not the final product), and hydrolysis releases benzylamine.

C_{6} H_{4} (CO)_{2} NH KOH C_{6} H_{4} (CO)_{2} NK C_{6} H_{5} C H_{2} Cl C_{6} H_{4} (CO)_{2} NC H_{2} C_{6} H_{5} NaOH(aq) C_{6} H_{5} C H_{2} N H_{2}

Solved Example 3

Which of these amines cannot be prepared by Gabriel phthalimide synthesis: (i)

C H_{3} C H_{2} N H_{2}

, (ii)

(C_{2} H_{5})_{2} NH

, (iii)

C H_{3} C H_{2} C H_{2} N H_{2}

Solution:

(ii) diethylamine. It is a secondary amine, and Gabriel synthesis gives only primary amines because phthalimide nitrogen can carry just one alkyl group. (i) and (iii) are 1° aliphatic amines and are made easily from bromoethane and 1-bromopropane.

Solved Example 4

Bicyclo[2.2.1]heptane-2-carboxamide (norbornane-2-carboxamide) is treated with

B r_{2}

/KOH to give A; A with excess

C H_{3} I

gives B; B heated with moist AgOH gives C. Which statement is correct?
(A) B is a tertiary amine
(B) C is bicyclo[2.2.1]hept-2-ene
(C) C is bicyclo[2.2.1]hept-1-ene, with the double bond at the bridgehead
(D) None of these

Solution:

Answer: (B).

A: Hofmann degradation removes the C=O carbon, giving bicyclo[2.2.1]heptan-2-amine.
B: excess $C H_{3} I$ converts it to the quaternary salt $R - N (C H_{3})_{3}^{+} I^{-}$ , so (A) is wrong.
C: Hofmann elimination of the hydroxide removes a $β$ -hydrogen from C-3, giving norbornene. A double bond at the bridgehead carbon (C) is ruled out by Bredt's rule.

Solved Example 5

Which compound on hydrolysis gives a carboxylic acid and a secondary amine?
(A)

C H_{3} CON (C H_{3}) C_{2} H_{5}

(B)

C_{6} H_{5} CON H_{2}

(C)

C H_{3} C H_{2} NC

(D)

C H_{3} CONHC H_{3}

Solution:

Answer: (A). An N,N-disubstituted amide carries two alkyl groups on nitrogen, so hydrolysis frees a 2° amine. (B) gives ammonia, (C) gives the 1° amine ethylamine and formic acid, and (D) gives the 1° amine methylamine.

C H_{3} CON (C H_{3}) C_{2} H_{5} + H_{2} O H^{+}, heat C H_{3} COOH + C H_{3} NH C_{2} H_{5}

Solved Example 6

The reaction that does NOT produce an amine is
(A)

RCON H_{2}

with

B r_{2}

and NaOH
(B)

RCON H_{2}

heated with

P_{2} O_{5}

C H_{3} N H_{2}

(D) None of these

Solution:

Answer: (B). $P_{2} O_{5}$ dehydrates an amide to a nitrile, $RCON H_{2} \to RCN + H_{2} O$ . (A) is the Hofmann reaction, which gives $RN H_{2}$ . In (C) the nitro group activates the ring, so methylamine displaces chloride to give the amine N-methyl-4-nitroaniline.

Solved Example 7

Complete the sequence and name X, Y and Z:

(C H_{3})_{2} CHN H_{2}

is treated with (i) excess

C H_{3} I

and (ii) moist

A g_{2} O

to give X; X on heating gives Y + Z.

Solution:

X is the quaternary hydroxide; heating removes a $β$ -hydrogen (Hofmann elimination):

(C H_{3})_{2} CHN H_{2} + 3C H_{3} I (C H_{3})_{2} CHN (C H_{3})_{3}^{+} I^{-} + 2HI

(C H_{3})_{2} CHN (C H_{3})_{3}^{+} O H^{-} heat C H_{3} CH = C H_{2} + (C H_{3})_{3} N + H_{2} O

X = isopropyltrimethylammonium hydroxide; Y = propene; Z = trimethylamine.

Solved Example 8

A ketone (A) reacts with

C H_{3} MgBr

followed by

H_{3} O^{+}

to give 2-methylbutan-2-ol. (A) with hydroxylamine gives (B), and (B) with

LiAl H_{4}

gives (C). Identify (A), (B) and (C).

Solution:

Adding $C H_{3}$ to the carbonyl carbon of (A) gives $(C H_{3})_{2} C (OH) C H_{2} C H_{3}$ , so (A) must carry one $C H_{3}$ and one $C_{2} H_{5}$ : butanone.

C H_{3} COC H_{2} C H_{3} N H_{2} OH C H_{3} C (= NOH) C H_{2} C H_{3} LiAl H_{4} C H_{3} CH (N H_{2}) C H_{2} C H_{3}

(A) = butanone; (B) = butanone oxime; (C) = butan-2-amine.

Solved Example 9

How will you carry out these conversions? (a) cyclohexanecarboxamide into N-methylcyclohexanamine; (b) cyclohexylamine into cyclopentylamine.

Solution:

(a) Hofmann degradation gives cyclohexylamine; the carbylamine reaction converts it to the isocyanide; reduction turns the isocyanide carbon into an N-methyl group.

C_{6} H_{11} CON H_{2} B r_{2} /KOH C_{6} H_{11} N H_{2} CHC l_{3} /KOH C_{6} H_{11} NC LiAl H_{4} C_{6} H_{11} NHC H_{3}

(b) Convert the amine to cyclohexene, cleave the ring to adipic acid, close it to cyclopentanone and finish by reductive amination.

C_{6} H_{11} N H_{2} NaN O_{2} /HCl C_{6} H_{11} OH conc. H_{2} S O_{4}, heat C_{6} H_{10}

C_{6} H_{10} KMn O_{4} / H^{+} HOOC (C H_{2})_{4} COOH BaO, heat C_{5} H_{8} O N H_{3}, H_{2} /Pt C_{5} H_{9} N H_{2}

Solved Example 10

Phenyl cyanide on reduction with Na/

C_{2} H_{5} OH

gives
(A)

C_{6} H_{5} C H_{2} N H_{2}

(B)

C_{6} H_{5} NHC H_{3}

C_{6} H_{5} N H_{2}

Solution:

Answer: (A). Mendius reduction turns $- C \equiv N$ into $- C H_{2} N H_{2}$ , keeping the C-C bond to the ring, so benzylamine forms.

C_{6} H_{5} CN + 4 [H] Na/ C_{2} H_{5} OH C_{6} H_{5} C H_{2} N H_{2}

Solved Example 11

Cyclohexanone reacts with

C H_{3} C H_{2} N H_{2}

in the presence of

H_{2}

/Ni to give X. X is
(A)

C_{6} H_{11} OC H_{2} N H_{2}

(B) 1-(aminomethyl)-1-methoxycyclohexane
(C)

C_{6} H_{10} = NC H_{2} C H_{3}

(D)

C_{6} H_{11} NHC H_{2} C H_{3}

Solution:

Answer: (D). Reductive amination: the ketone and the primary amine form the imine (C), which is hydrogenated at once to the secondary amine N-ethylcyclohexanamine.

C_{6} H_{10} O + C H_{3} C H_{2} N H_{2} + H_{2} Ni C_{6} H_{11} NHC H_{2} C H_{3} + H_{2} O

Solved Example 12

Isopropylamine can be obtained by
(A) acetone +

N H_{2} OH

, then

LiAl H_{4}

(B) acetone +

N H_{3}

, then

H_{2}

/Ni
(C) propan-2-ol +

N H_{3}

at room temperature
(D) both A and B

Solution:

Answer: (D). Route A reduces acetone oxime and route B is reductive amination; both give $(C H_{3})_{2} CHN H_{2}$ . An alcohol does not react with ammonia at room temperature; it needs a catalyst such as $A l_{2} O_{3}$ at high temperature and even then gives a mixture.

Solved Example 13

In the reaction of

RCON H_{2}

with

B r_{2}

and KOH to give

RN H_{2}

, the intermediate involved is
(A)

RCONHBr

(B)

RNHBr

(C)

RCONB r_{2}

(D)

R - C = N = O

Solution:

Answer: (A). The N-bromoamide $RCONHBr$ is the first intermediate (Figure 7). The isocyanate is also an intermediate, but it is $R - N = C = O$ with R on nitrogen, not the structure written in (D).

Solved Example 14

Which of the following does NOT give ethylamine on reduction?
(A) methyl cyanide
(B) ethyl cyanide
(C) nitroethane
(D) acetamide

Solution:

Answer: (B). Ethyl cyanide is $C H_{3} C H_{2} CN$ , a three-carbon nitrile, so it gives propylamine $C H_{3} C H_{2} C H_{2} N H_{2}$ . Methyl cyanide $C H_{3} CN$ , nitroethane $C H_{3} C H_{2} N O_{2}$ and acetamide $C H_{3} CON H_{2}$ all reduce to $C H_{3} C H_{2} N H_{2}$ .

Solved Example 15

A sequence of reactions is carried out as shown. What is the reagent for step 3?

RC H_{2} COOH (1) SOC l_{2} RC H_{2} COCl (2) N H_{3} RC H_{2} CON H_{2}

RC H_{2} CON H_{2} (3) RC H_{2} N H_{2} (4) RC H_{2} OH (5) RCOOH

(A) NaBr
(B) bromine and alkali
(C) HBr
(D)

P_{2} O_{5}

Solution:

Answer: (B). Step 3 converts an amide into an amine with one carbon fewer, the Hofmann bromamide reaction. Step 4 is nitrous acid (1° amine to alcohol) and step 5 is oxidation.

Solved Example 16

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

Solution:

The key step is nucleophilic substitution of the halide by the phthalimide anion. In aryl halides the C-X bond has partial double-bond character from resonance, the carbon is $sp^{2}$ and the ring blocks backside attack, so the phthalimide anion cannot displace the halogen. Aniline is instead made by reducing nitrobenzene.

Solved Example 17

Name the amine and count its carbon atoms when (a) propanamide is treated with

B r_{2}

/KOH, (b) propanenitrile is reduced with

LiAl H_{4}

, (c) 1-nitropropane is reduced with Sn/HCl.

Solution:

(a) Hofmann removes one carbon: ethylamine $C H_{3} C H_{2} N H_{2}$ , 2 C.
(b) The CN carbon becomes $C H_{2} N H_{2}$ : propan-1-amine $C H_{3} C H_{2} C H_{2} N H_{2}$ , 3 C.
(c) Nitro reduction keeps every carbon: propan-1-amine, 3 C.

Practice Questions

Cyclohexylamine is treated with 3 $C H_{3} I$ to give A, A with $A g_{2} O$ / $H_{2} O$ to give B, and B is heated to give C + D + $H_{2} O$ . Identify A to D.Answer: A = N,N,N-trimethylcyclohexanaminium iodide; B = the corresponding hydroxide; C = cyclohexene; D = trimethylamine
Cyclohexene oxide is opened with $Na N_{3}$ in dioxane-water to give A, and A is hydrogenated ( $H_{2}$ /Pt) to give B. Identify A and B.Answer: A = trans-2-azidocyclohexan-1-ol; B = trans-2-aminocyclohexan-1-ol
Cyclopropanecarboxylic acid is heated with $N H_{3}$ to give C; C with $P_{2} O_{5}$ gives D; D with $LiAl H_{4}$ then water gives E. Identify C, D and E.Answer: C = cyclopropanecarboxamide; D = cyclopropanecarbonitrile; E = cyclopropylmethanamine
C (from Q3) is heated with KOH/ $B r_{2}$ to give F; F with $CHC l_{3}$ /alc. KOH gives G; G with $H_{3} O^{+}$ gives F + H. Identify F, G and H.Answer: F = cyclopropylamine; G = cyclopropyl isocyanide; H = formic acid
Methyl cyanide is reduced with $H_{2}$ /Pt or with $LiAl H_{4}$ . What is the product?Answer: ethylamine, $C H_{3} C H_{2} N H_{2}$ , in both cases
An alcohol A (C, H, O; gives a colour with ceric ammonium nitrate) with $PC l_{5}$ gives B; B with KCN gives C; C with Na/ $C_{2} H_{5} OH$ gives D; D on heating gives E and $N H_{3}$ ; E with nitrobenzene gives pyridine. Identify A to E.Answer: A = propane-1,3-diol; B = 1,3-dichloropropane; C = pentanedinitrile; D = pentane-1,5-diamine; E = piperidine
How can the formation of 2° and 3° amines be avoided when a 1° amine is made by alkylation?Answer: use a large excess of ammonia; better, choose Gabriel synthesis, Hofmann degradation or a reduction method
Acetophenone (A) with $N H_{2} OH$ ·HCl gives two oximes B and C, which rearrange in acid to D and E ( $C_{8} H_{9} NO$ ). D boiled with alc. KOH gives an oil F ( $C_{6} H_{7} N$ ) that reacts with $C H_{3} COCl$ to give back D; E with alkali gives G ( $C_{7} H_{6} O_{2}$ ). Identify A to G.Answer: B = (E)-oxime and C = (Z)-oxime; D = acetanilide; E = N-methylbenzamide; F = aniline; G = benzoic acid (Beckmann rearrangement: the group anti to OH migrates)
A (M = 135) boiled with NaOH gives $N H_{3}$ and, after acidification, B (M = 136); A with $HN O_{2}$ also gives B, and with $B r_{2}$ /KOH gives C, which with cold $HN O_{2}$ gives an alcohol D. An isomer E of A gives, with dilute HCl, an acid F (M = 136) that is oxidised and heated to an anhydride G used to make anthraquinone. Identify A to G.Answer: A = phenylacetamide; B = phenylacetic acid; C = benzylamine; D = benzyl alcohol; E = 2-methylbenzamide; F = 2-methylbenzoic acid; G = phthalic anhydride
An optically inactive acid A ( $C_{5} H_{8} O_{5}$ ) loses $C O_{2}$ on heating to give a resolvable acid B ( $C_{4} H_{8} O_{3}$ ). B with $H_{2} S O_{4}$ gives C; its ethyl ester with $H_{2}$ /Pt gives D; D with conc. $N H_{3}$ gives E ( $C_{4} H_{9} NO$ ); E with $B r_{2}$ /KOH gives F ( $C_{3} H_{9} N$ ); F with $HN O_{2}$ gives G, and G is oxidised to H. G and H both give the iodoform test. Identify A to H.Answer: A = $HOC H_{2} C (C H_{3}) (COOH)_{2}$ ; B = 3-hydroxy-2-methylpropanoic acid; C = methacrylic acid; D = ethyl 2-methylpropanoate; E = 2-methylpropanamide; F = propan-2-amine; G = propan-2-ol; H = acetone
A neutral compound A ( $C_{5} H_{11} N O_{2}$ ) is reduced to a base B ( $C_{5} H_{13} N$ ). B with excess $C H_{3} I$ and moist $A g_{2} O$ gives C ( $C_{8} H_{21} NO$ ), which on heating gives trimethylamine and 2-methylbut-1-ene. Identify A, B and C.Answer: A = 2-methyl-1-nitrobutane; B = 2-methylbutan-1-amine; C = (2-methylbutyl)trimethylammonium hydroxide
A chlorine compound X with $N H_{3}$ gives a solid Y (C 49.31%, H 9.59%, N 19.18%), which with $B r_{2}$ and NaOH gives a base Z; Z with $HN O_{2}$ gives ethanol. Identify X, Y and Z.Answer: Y = propanamide $C_{3} H_{7} NO$ (empirical formula from the analysis); X = propanoyl chloride; Z = ethylamine

Common Mistakes to Avoid

Watch out

Choosing direct ammonolysis to make a pure primary amine. It always gives a mixture; use Gabriel, Hofmann or a reduction.
Proposing Gabriel synthesis for aniline or for a secondary amine. It gives only 1° aliphatic amines.
Forgetting the carbon count: Hofmann, Curtius, Schmidt and Lossen lose one carbon, nitrile reduction keeps the CN carbon (one more than R-X), and nitro, amide and oxime reductions keep all carbons.
Writing a 1° amine as the product of isocyanide reduction. $RNC$ gives the 2° amine $RNHC H_{3}$ .
Balancing the Hofmann bromamide reaction with 2 KOH. It needs 4 KOH: two for the bromination and deprotonations, two to trap $C O_{2}$ as $K_{2} C O_{3}$ .
Expecting $LiAl H_{4}$ to reduce nitrobenzene to aniline. It gives azobenzene; use Sn/HCl, Fe/HCl or $H_{2}$ /Ni.
Forgetting that Sn/HCl gives the anilinium salt. Alkali is needed to free the amine.
Confusing Hofmann bromamide degradation (amide to amine), Hofmann elimination (quaternary hydroxide to alkene) and Hofmann ammonolysis (alkyl halide + ammonia).
Assuming the migrating group racemises in the Hofmann or Curtius reaction. It migrates with retention of configuration.

Frequently Asked Questions

Which methods give pure primary amines?

Gabriel phthalimide synthesis, Hofmann bromamide degradation and the Curtius, Schmidt and Lossen rearrangements give only primary amines, as do the reductions of nitro compounds, nitriles, primary amides and oximes. Direct ammonolysis of alkyl halides does not, because the amine formed reacts further to give secondary, tertiary and quaternary products.

Why can aromatic primary amines not be made by Gabriel synthesis?

Gabriel synthesis depends on the phthalimide anion displacing a halide by nucleophilic substitution. Aryl halides do not undergo this reaction, because their carbon-halogen bond has partial double-bond character and the ring blocks backside attack. Aromatic primary amines such as aniline are therefore made by reducing nitro compounds instead.

Why does the Hofmann bromamide reaction give an amine with one carbon less?

In the key step the alkyl group moves from the carbonyl carbon to nitrogen, giving an isocyanate. Hydrolysis of the isocyanate turns that carbonyl carbon into carbon dioxide, which is removed as potassium carbonate. The amine therefore contains every carbon of the amide except the carbonyl carbon.

Why is iron and hydrochloric acid preferred for reducing nitrobenzene?

Iron scrap is cheap, and the iron(II) chloride formed during the reaction is hydrolysed, releasing hydrochloric acid again. Only a small amount of acid is needed to start the reduction, which makes the process economical on an industrial scale. Tin and hydrochloric acid work too but use much more acid.

How are the Hofmann, Curtius, Schmidt and Lossen reactions related?

All four convert a carboxylic acid derivative into a primary amine with one carbon fewer. In each, the alkyl or aryl group shifts from carbon to nitrogen while a leaving group departs, giving an isocyanate that is hydrolysed. The leaving group is bromide in Hofmann, nitrogen gas in Curtius and Schmidt and a carboxylate ion in Lossen.

How can an amine with one more carbon than an alkyl halide be prepared?

Convert the alkyl halide into a nitrile with potassium cyanide, then reduce the nitrile with lithium aluminium hydride, sodium and ethanol or hydrogen over nickel. The cyanide carbon becomes the CH2 attached to nitrogen, so methyl bromide gives ethylamine. This is called ascent of the homologous series.

Which preparation methods of amines are asked in NEET?

NEET follows NCERT closely: reduction of nitro compounds with tin or iron and hydrochloric acid, ammonolysis of alkyl halides, reduction of nitriles and amides with lithium aluminium hydride, Gabriel phthalimide synthesis and the Hofmann bromamide degradation. Questions often ask for the product, the reagent or why Gabriel synthesis cannot give aniline.

What does JEE Main ask about the preparation of amines?

JEE Main favours reagent-product questions and short conversion sequences. Typical items are the carbon count after Hofmann degradation or nitrile reduction, the intermediate in the Hofmann reaction, reductive amination products, identifying amines from multi-step clues, and the Curtius, Schmidt and Lossen rearrangements that also pass through isocyanates.

Previous year questions on Preparation of Amines

18 questions from past papers, each with a step-by-step solution.

Show all 18 questions

Ready to master Amines?

Take a full mock test, practice concept-by-concept, and get an AI-powered rank prediction — all on Fundamenthol.

Practice questions →Take a mock test