JEE Main 2019

LEVELJEE Advanced

Animated Solution for Chemistry - Organic Chemistry: Which of the following is not a correct method of the preparation of benzylamine from cyanobenzene?

Select Answer:

Visualized Solution

The Objective: $P h - C N \to P h - C H_{2} - N H_{2}$

$Objective: Convert Cyanobenzene to Benzylamine$
$P h - C N ⟶ P h - C H_{2} - N H_{2}$

Option (a): $H_{2} / N i$

$Option (a): Catalytic Hydrogenation$
$P h - C \equiv N H_{2} / N i, Δ P h - C H_{2} - N H_{2}$
$Result: Benzylamine is formed.$

Option (c): $L i A l H_{4}$

$Option (c): Reduction with L i A l H_{4}$
$P h - C \equiv N (i) L i A l H_{4}, (ii) H_{3} O^{+} P h - C H_{2} - N H_{2}$
$Result: Benzylamine is formed.$

Option (d): $S n C l_{2} / H C l$ and $N a B H_{4}$

$Option (d): Modified Stephen Reduction$
$P h - C \equiv N (i) S n C l_{2}, H C l (g) [P h - C H = N H_{2}]_{2} S n C l_{6}^{2 -}$
$[P h - C H = N H_{2}]_{2} S n C l_{6}^{2 -} (ii) N a B H_{4} P h - C H_{2} - N H_{2}$
$Result: Benzylamine is formed.$

Option (b): $H C l / H_{2} O$ and $N a B H_{4}$

$Option (b): Hydrolysis and Reduction$
$P h - C N H C l, H_{2} O P h - C O O H$
$P h - C O O H N a B H_{4} P h - C H_{2} - O H$
$Result: Benzyl alcohol is formed, not Benzylamine.$

Conclusion

$Conclusion: Option (b) is the incorrect method for preparing Benzylamine.$

The Way Forward

$To convert P h - C O O H to P h - C H_{2} - N H_{2} :$
$P h - C O O H N H_{3}, Δ P h - C O N H_{2} L i A l H_{4} P h - C H_{2} - N H_{2}$

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The Sigma Insight: Amines

Solution Diagram

The Objective

Reducing the Nitrile

We are tasked with finding the incorrect method for converting cyanobenzene (

P h - C N

) to benzylamine (

P h - C H_{2} - N H_{2}

).

This transformation requires the complete reduction of the carbon-nitrogen triple bond (

C \equiv N

) into a single bond (

C H_{2} - N H_{2}

). Let us evaluate each option systematically to see which one fails to achieve this.

Option (a)

Catalytic Hydrogenation

The first option uses hydrogen gas (

H_{2}

) in the presence of a Nickel (Ni) catalyst.

This is a classic and highly effective method for reducing nitriles. The transition metal catalyst facilitates the addition of hydrogen across both

π

-bonds of the nitrile group.

P h - C \equiv N H_{2} / N i, Δ P h - C H_{2} - N H_{2}

This brute-force reduction successfully yields benzylamine. Therefore, this method is correct.

Option (c)

The Power of Lithium Aluminium Hydride

Option (c) employs Lithium Aluminium Hydride (

L i A l H_{4}

) followed by an acidic workup (

H_{3} O^{+}

).

L i A l H_{4}

is one of the most powerful hydride donors in organic chemistry. The nucleophilic hydride ion (

H^{-}

) attacks the electrophilic carbon of the nitrile, eventually reducing it completely to a primary amine.

P h - C \equiv N (i) L i A l H_{4}, (ii) H_{3} O^{+} P h - C H_{2} - N H_{2}

This is a standard laboratory procedure for synthesizing primary amines from nitriles. Thus, this method is also correct.

Option (d)

A Modified Stephen Reduction

Option (d) presents a fascinating two-step sequence: first, Tin(II) chloride (

S n C l_{2}

) with

H C l

gas, followed by Sodium Borohydride (

N a B H_{4}

).

The first step is identical to the famous Stephen reduction. The

S n C l_{2} / H C l

mixture reduces the nitrile to an intermediate imine salt (specifically, a hexachlorostannate complex).

P h - C \equiv N S n C l_{2}, H C l [P h - C H = N H_{2}]_{2} S n C l_{6}^{2 -}

Normally, we would hydrolyze this imine to get an aldehyde (benzaldehyde). However, the second step here introduces

N a B H_{4}

. While

N a B H_{4}

is a mild reducing agent, it is perfectly capable of reducing an imine (

C = N

) to an amine (

C H - N H

).

[P h - C H = N H_{2}]_{2} S n C l_{6}^{2 -} N a B H_{4} P h - C H_{2} - N H_{2}

This clever modification successfully produces benzylamine. Hence, this method is correct.

Option (b)

The Trap of Hydrolysis

Finally, let us examine option (b), which involves acidic hydrolysis (

H C l / H_{2} O

) followed by

N a B H_{4}

.

When a nitrile is subjected to acidic hydrolysis, it is converted into a carboxylic acid. In this case, cyanobenzene hydrolyzes to form benzoic acid.

P h - C N H C l, H_{2} O P h - C O O H

Now, we must ask: can

N a B H_{4}

reduce benzoic acid? The answer is a resounding no.

N a B H_{4}

is a chemoselective, mild reducing agent that can reduce aldehydes and ketones, but it is entirely unreactive towards the resonance-stabilized carboxylic acid group.

Even if we hypothetically assumed a stronger reducing agent was used (like

L i A l H_{4}

), the reduction of a carboxylic acid yields a primary alcohol (

P h - C H_{2} - O H

), not an amine.

Therefore, this sequence fails to produce benzylamine, making it the incorrect method.

Final Conclusion

By systematically analyzing the reagents, we have determined that option (b) leads to benzoic acid (and potentially benzyl alcohol if a stronger reductant were used), completely missing the target of benzylamine. Thus, option (b) is the correct answer to the question.

Common Traps & FAQs

Why doesn't

N a B H_{4}

reduce the carboxylic acid in option (b)?

N a B H_{4}

is a mild reducing agent. It is chemoselective and can only reduce highly electrophilic carbonyls like aldehydes and ketones. Carboxylic acids are resonance stabilized and less electrophilic, requiring a stronger agent like

L i A l H_{4}

or

B_{2} H_{6}

.

What exactly is the intermediate formed in the Stephen reduction?

The intermediate is an iminium salt, specifically a hexachlorostannate complex

[R - C H = N H_{2}]_{2} S n C l_{6}^{2 -}

. It's formed when

S n C l_{2}

reduces the nitrile in the presence of

H C l

.

Can we use

H_{2} / P d

instead of

H_{2} / N i

for this reaction?

Absolutely! Both Palladium (Pd) and Nickel (Ni) are excellent transition metal catalysts for the hydrogenation of nitriles to primary amines.

Why does hydrolysis of a nitrile give a carboxylic acid?

Acidic hydrolysis adds water across the

C \equiv N

bond to first form an amide (

R - C O N H_{2}

). Further hydrolysis cleaves the amide bond, releasing ammonia (or ammonium ion in acid) and yielding the carboxylic acid.

Is it possible to stop the reduction at the imine stage using

L i A l H_{4}

?

No,

L i A l H_{4}

is too reactive and will completely reduce the nitrile to the primary amine. To stop at the aldehyde (via imine hydrolysis), you need a bulkier, milder reducing agent like DIBAL-H.

Similar Questions

The most appropriate reagent for conversion of $C_{2} H_{5} CN$ into $CH_{3} CH_{2} CH_{2} NH_{2}$ is

Na(CN)BH_{3}

LiAlH_{4}

NaBH_{4}

CaH_{2}

Ethylamine ( $C_{2} H_{5} NH_{2}$ ) can be obtained from N-ethylphthalimide on treatment with

NaBH_{4}

NH_{2} NH_{2}

H_{2} O

CaH_{2}

LEVELJEE Advanced

Which of the following reaction(s) will not give $p$ -aminoazobenzene?

Which one of the following methods is neither meant for the synthesis nor for separation of amines ?

Curtius reaction

Hinsberg method

The total number of reagents from those given below, that can convert nitrobenzene into aniline is ........ (Integer answer) I. $Sn - HCl$ II. $Sn - NH_{4} OH$ III. $Fe - HCl$ IV. $Zn - HCl$ V. $H_{2} - Pd$ VI. $H_{2} -Raney nickel$

Benzene diazonium chloride on reaction with aniline in the presence of dilute hydrochloric acid gives

JEE Advanced 2021

LEVELJEE Advanced

Correct option(s) for the following sequence of reactions is(are)

* Multiple Correct Options

_{2}

_{4}

R = benzenamine, V = KCN

_{2}

, R = phenylmethanamine

_{4}

An amine on reaction with benzene sulphonyl chloride produces a compound insoluble in alkaline solution. This amine can be prepared by ammonolysis of ethyl chloride. The correct structure of amine is

CH_{3} CH_{2} NH_{2}

CH_{3} CH_{2} CH_{2} NHCH_{3}

CH_{3} CH_{2} CH_{2} NHCH_{2} CH_{3}

Fluorobenzene ( $C_{6} H_{5} F$ ) can be synthesized in the laboratory

by heating phenol with HF and KF

from aniline by diazotisation followed by heating the diazonium salt with

HBF_{4}

by direct fluorination of benzene with

F_{2}

by reacting bromobenzene with NaF solution

LEVELJEE Advanced

The major products obtained from the reactions in List-II are the reactants for the named reactions mentioned in List-I. Match each entry in List-I with the appropriate entry in List-II and choose the correct options.

List-I

Stephen reaction

Sandmeyer reaction

Hoffmann bromamide degradation reaction

Cannizzaro reaction

List-II

(i) CrO_{2} Cl_{2} / CS_{2}, (ii) H_{3} O^{+}

(i) PCl_{5}, (ii) NH_{3}, (iii) P_{4} O_{10}, Δ

(i) Fe, HCl, (ii) HCl, NaNO_{2} (273 - 278 K), H_{2} O

(i) Cl_{2} / h ν, H_{2} O, (ii) Tollen’s reagent, (iii) SO_{2} Cl_{2}, (iv) NH_{3}

(i) (CH_{3} CO)_{2} O, Pyridine, (ii) HNO_{3}, H_{2} SO_{4}, 288 K, (iii) aq. NaOH