Question

Match the Compounds (List - I) with the appropriate Catalyst/Reagents (List - II) for their reduction into corresponding amines.

• A. ( A ) − ( I I I ) , ( B ) − ( I V ) , ( C ) − ( I I ) , ( D ) − ( I )
• B. ( A ) − ( I I ) , ( B ) − ( I V ) , ( C ) − ( I I I ) , ( D ) − ( I )
• C. ( A ) − ( I I ) , ( B ) − ( I ) , ( C ) − ( I I I ) , ( D ) − ( I V )
• D. ( A ) − ( I I I ) , ( B ) − ( I I ) , ( C ) − ( I V ) , ( D ) − ( I )

Hint:

N/A

Answer 1

The Correct Option is A

Reaction (A): Reduction of Amides using LiAlH₄ 

\[ R - C(=O) - NH_2 \xrightarrow[\text{H}_2\text{O}]{\text{LiAlH}_4} R - CH_2 - NH_2 \]

Explanation:
In this reaction, the carbonyl group (\(C=O\)) of the amide is reduced by lithium aluminium hydride (LiAlH₄), a strong reducing agent. The oxygen is replaced by two hydrogen atoms, converting the amide into a primary amine.

Mechanism overview:
LiAlH₄ donates hydride ions (H⁻) that attack the electrophilic carbon of the carbonyl group. After protonation with water, the oxygen is completely removed, giving \( R-CH_2-NH_2 \).

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Reaction (B): Reduction of Nitrobenzene to Aniline

\[ \text{C}_6\text{H}_5\text{NO}_2 \xrightarrow[\text{HCl}]{\text{Sn}} \text{C}_6\text{H}_5\text{NH}_2 \]

Explanation:
Nitrobenzene (\( \text{C}_6\text{H}_5\text{NO}_2 \)) is reduced to aniline (\( \text{C}_6\text{H}_5\text{NH}_2 \)) using tin (Sn) and hydrochloric acid (HCl). Here, Sn acts as the reducing agent and HCl provides protons needed to complete the reduction process. This reduction proceeds via the intermediate formation of nitroso and hydroxylamine compounds.

Reaction path:
\[ \text{C}_6\text{H}_5\text{NO}_2 \rightarrow \text{C}_6\text{H}_5\text{NO} \rightarrow \text{C}_6\text{H}_5\text{NHOH} \rightarrow \text{C}_6\text{H}_5\text{NH}_2 \]

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Reaction (C): Hydrogenation of Nitriles

\[ R - C \equiv N \xrightarrow[\text{Ni}]{\text{H}_2} R - CH_2 - NH_2 \]

Explanation:
In this process, nitriles (R–C≡N) are hydrogenated using hydrogen gas in the presence of a nickel catalyst. Each of the triple bonds between carbon and nitrogen is broken, and hydrogen atoms are added to form a primary amine (\( R-CH_2-NH_2 \)).

Mechanism summary:
Hydrogen atoms are successively added across the C≡N triple bond. This is a catalytic reduction (hydrogenation), generally carried out under high temperature and pressure.

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Reaction (D): Hofmann Bromamide Degradation Reaction

\[ \text{Phthalimide derivative:} \quad \begin{aligned} &\text{Phthalimide N–R} + \text{Aq. NaOH} \longrightarrow RNH_2 + \text{Phthalamic salt} \end{aligned} \]

In the given reaction:

\[ \text{N–R phthalimide} \xrightarrow[\text{Aq. NaOH}]{} RNH_2 + \text{disodium phthalate} \]

Explanation:
This reaction is a typical example of the Hofmann degradation or ammonolysis of imides. Here, the nitrogen–carbon bond of the imide breaks, and the organic group \( R \) is converted into a primary amine. At the same time, the cyclic imide transforms into a salt of dicarboxylic acid.

Mechanism steps:
1. Hydroxide ion attacks the carbonyl carbon of the imide. 
2. The intermediate rearranges and liberates the corresponding amine. 
3. The residue forms a sodium salt of the corresponding acid.

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Summary Table

Reaction	Starting Compound	Reagent	Product
(A)	Amide (\(RCONH_2\))	LiAlH4 / H2O	Primary amine (\(RCH_2NH_2\))
(B)	Nitro compound	Sn / HCl	Aniline (Aromatic amine)
(C)	Nitrile (\(RCN\))	H2 / Ni	Primary amine (\(RCH_2NH_2\))
(D)	Phthalimide derivative	Aq. NaOH	Amine + Sodium phthalate

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Conclusion:

All the above reactions are examples of different methods for preparing primary amines. Among them, Reaction (A) and (C) yield aliphatic primary amines, Reaction (B) gives an aromatic amine, and Reaction (D) demonstrates deamination of imide derivatives to produce amines.