Question:
What is the product P obtained in following reaction?
\(\text{CH}_3\text{CH}_2\text{Br}+\text{AgCN}_{(alc)}\rightarrow \text{P}\)
What is the product P obtained in following reaction?
\(\text{CH}_3\text{CH}_2\text{Br}+\text{AgCN}_{(alc)}\rightarrow \text{P}\)
\(\text{CH}_3\text{CH}_2\text{Br}+\text{AgCN}_{(alc)}\rightarrow \text{P}\)
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AgCN is covalent, so nitrogen attacks and an isocyanide forms.
Updated On: Oct 1, 2026
- \(\text{CH}_3\text{CH}_2\text{NC}\)
- \(\text{CH}_3\text{CH}_2\text{CN}\)
- \(\text{CH}_3\text{CH}_2\text{CONH}_2\)
- \(\text{CH}_3\text{CH}_2\text{NH}_2\)
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The Correct Option is A
Solution and Explanation
Step 1: Nature of AgCN
Silver cyanide is largely covalent. The cyanide group attacks through the nitrogen lone pair, since carbon is tied up in bonding with silver.
Step 2: Product
\(CH_3CH_2Br + AgCN \rightarrow CH_3CH_2NC + AgBr\). The product is ethyl isocyanide.
Step 3: Why others are wrong
\(CH_3CH_2CN\) forms with KCN, which is ionic and attacks through carbon. The amide and the amine are not formed under these conditions.
Final Answer:
The product is ethyl isocyanide, CH3CH2NC. \[ \boxed{\text{(A)}\ CH_3CH_2NC} \]
Silver cyanide is largely covalent. The cyanide group attacks through the nitrogen lone pair, since carbon is tied up in bonding with silver.
Step 2: Product
\(CH_3CH_2Br + AgCN \rightarrow CH_3CH_2NC + AgBr\). The product is ethyl isocyanide.
Step 3: Why others are wrong
\(CH_3CH_2CN\) forms with KCN, which is ionic and attacks through carbon. The amide and the amine are not formed under these conditions.
Final Answer:
The product is ethyl isocyanide, CH3CH2NC. \[ \boxed{\text{(A)}\ CH_3CH_2NC} \]
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