Question:
In which of the following sets, reagents required to convert propene (X) to an amine (Y) with one carbon atom more than X, are correctly arranged?
I. \(HBr/(C_6H_5CO)_2O_2;\ NH_3\)
II. \(HBr;\ AgCN;\ H_2/Ni\)
III. \(HBr/(C_6H_5CO)_2O_2;\ NaCN/C_2H_5OH;\ LiAlH_4,H_2O\)
IV. \(HBr/(C_6H_5CO)_2O_2;\ AgNO_2;\ Sn,HCl\)
In which of the following sets, reagents required to convert propene (X) to an amine (Y) with one carbon atom more than X, are correctly arranged?
I. \(HBr/(C_6H_5CO)_2O_2;\ NH_3\)
II. \(HBr;\ AgCN;\ H_2/Ni\)
III. \(HBr/(C_6H_5CO)_2O_2;\ NaCN/C_2H_5OH;\ LiAlH_4,H_2O\)
IV. \(HBr/(C_6H_5CO)_2O_2;\ AgNO_2;\ Sn,HCl\)
Show Hint
Whenever a question asks for an amine with one extra carbon atom, think of the sequence:
\[
R-X \rightarrow R-CN \rightarrow R-CH_2NH_2
\]
using KCN/NaCN followed by reduction.
Updated On: Jun 17, 2026
- I, II only
- I, IV only
- II, III only
- III, IV only
Show Solution
Verified By Collegedunia
The Correct Option is C
Solution and Explanation
Concept:
The product amine must contain one extra carbon atom compared with propene.
Therefore a cyanide intermediate must be formed because:
\[
R-X \rightarrow R-CN
\]
increases the carbon chain by one carbon atom.
Step 1: Examine Set-II.
\[ CH_3CH=CH_2 \xrightarrow{HBr} CH_3CHBrCH_3 \] \[ \xrightarrow{AgCN} CH_3CH(CH_3)NC \] \[ \xrightarrow{H_2/Ni} \text{amine} \] Carbon number increases by one. Hence Set-II is correct.
Step 2: Examine Set-III.
Anti-Markovnikov addition: \[ CH_3CH=CH_2 \xrightarrow{HBr/peroxide} CH_3CH_2CH_2Br \] \[ \xrightarrow{NaCN} CH_3CH_2CH_2CN \] \[ \xrightarrow{LiAlH_4} CH_3CH_2CH_2CH_2NH_2 \] A primary amine having one extra carbon atom is formed. Hence Set-III is correct.
Step 3: Reject remaining sets.
Set-I does not increase carbon number. Set-IV forms nitro compounds through \(AgNO_2\), not the required homologated amine route. Therefore, \[ \boxed{\text{II and III}} \] \[ \boxed{\text{Correct Option (3)}} \]
Step 1: Examine Set-II.
\[ CH_3CH=CH_2 \xrightarrow{HBr} CH_3CHBrCH_3 \] \[ \xrightarrow{AgCN} CH_3CH(CH_3)NC \] \[ \xrightarrow{H_2/Ni} \text{amine} \] Carbon number increases by one. Hence Set-II is correct.
Step 2: Examine Set-III.
Anti-Markovnikov addition: \[ CH_3CH=CH_2 \xrightarrow{HBr/peroxide} CH_3CH_2CH_2Br \] \[ \xrightarrow{NaCN} CH_3CH_2CH_2CN \] \[ \xrightarrow{LiAlH_4} CH_3CH_2CH_2CH_2NH_2 \] A primary amine having one extra carbon atom is formed. Hence Set-III is correct.
Step 3: Reject remaining sets.
Set-I does not increase carbon number. Set-IV forms nitro compounds through \(AgNO_2\), not the required homologated amine route. Therefore, \[ \boxed{\text{II and III}} \] \[ \boxed{\text{Correct Option (3)}} \]
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