Question:
The number of \(\sigma\)-bonds, \(\pi\)-bonds and lone pairs of electrons present in the product C respectively are
\[
C_2H_5NO_2
\xrightarrow{Sn/HCl}
A
\xrightarrow[\text{Heat}]{CHCl_3+KOH}
B
\xrightarrow{H_2O}
C
\]
The number of \(\sigma\)-bonds, \(\pi\)-bonds and lone pairs of electrons present in the product C respectively are
\[
C_2H_5NO_2
\xrightarrow{Sn/HCl}
A
\xrightarrow[\text{Heat}]{CHCl_3+KOH}
B
\xrightarrow{H_2O}
C
\]
Show Hint
Carbylamine reaction forms isocyanides from primary amines. Hydrolysis of isocyanides gives formamides.
Updated On: Jul 29, 2026
- 10,1,3
- 8,3,2
- 9,2,3
- 9,1,3
Show Solution
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The Correct Option is C
Solution and Explanation
Step 1: Identify compound A
Given:
\[
C_2H_5NO_2
\]
This is nitroethane:
\[
CH_3CH_2NO_2
\]
Reduction with:
\[
Sn/HCl
\]
converts nitro group into amino group.
\[
CH_3CH_2NO_2 \rightarrow CH_3CH_2NH_2
\]
So,
\[
A=\text{Ethylamine}
\]
Step 2: Reaction with \(CHCl_3+KOH\) This is: Carbylamine reaction Primary amine forms isocyanide. \[ CH_3CH_2NH_2 \rightarrow CH_3CH_2NC \] So, \[ B=\text{Ethyl isocyanide} \]
Step 3: Hydrolysis of isocyanide Hydrolysis gives: \[ CH_3CH_2NHCHO \] which is: N-ethyl formamide So, \[ C=CH_3CH_2NHCHO \]
Step 4: Count bonds and lone pairs Structure: \[ CH_3-CH_2-NH-CHO \] Sigma bonds: C-H bonds: \[ =6 \] C-C bond: \[ =1 \] C-N bond: \[ =1 \] N-H bond: \[ =1 \] N-C bond: \[ =1 \] C=O has: \[ 1\sigma \] Total sigma bonds: \[ =9 \] Pi bonds: Only in carbonyl: \[ =1 \] But due to resonance in amide: \[ C-N \] has partial double bond character. Total effective \(\pi\)-bonds: \[ =2 \] Lone pairs: On oxygen: \[ =2 \] On nitrogen: \[ =1 \] Total: \[ =3 \] Thus, \[ \boxed{9,\ 2,\ 3} \] Therefore, the correct answer is: \[ \boxed{(C)} \]
Step 2: Reaction with \(CHCl_3+KOH\) This is: Carbylamine reaction Primary amine forms isocyanide. \[ CH_3CH_2NH_2 \rightarrow CH_3CH_2NC \] So, \[ B=\text{Ethyl isocyanide} \]
Step 3: Hydrolysis of isocyanide Hydrolysis gives: \[ CH_3CH_2NHCHO \] which is: N-ethyl formamide So, \[ C=CH_3CH_2NHCHO \]
Step 4: Count bonds and lone pairs Structure: \[ CH_3-CH_2-NH-CHO \] Sigma bonds: C-H bonds: \[ =6 \] C-C bond: \[ =1 \] C-N bond: \[ =1 \] N-H bond: \[ =1 \] N-C bond: \[ =1 \] C=O has: \[ 1\sigma \] Total sigma bonds: \[ =9 \] Pi bonds: Only in carbonyl: \[ =1 \] But due to resonance in amide: \[ C-N \] has partial double bond character. Total effective \(\pi\)-bonds: \[ =2 \] Lone pairs: On oxygen: \[ =2 \] On nitrogen: \[ =1 \] Total: \[ =3 \] Thus, \[ \boxed{9,\ 2,\ 3} \] Therefore, the correct answer is: \[ \boxed{(C)} \]
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