Step 1: Understanding the Question:
The goal is to identify which reaction conditions produce benzyl isocyanide (\(C_6H_5CH_2NC\)). We evaluate each given reaction sequence based on the reagents provided. Step 2: Detailed Explanation:
- Reaction A: Benzyl bromide (\(C_6H_5CH_2Br\)) reacting with \(AgCN\). Since silver cyanide is covalent, the lone pair on nitrogen is more available for nucleophilic attack, leading to the formation of the isocyanide (\(C_6H_5CH_2NC\)). This is correct.
- Reaction B: Benzylamine (\(C_6H_5CH_2NH_2\)) reacting with chloroform (\(CHCl_3\)) and aqueous \(NaOH\). This is the Carbylamine reaction, which converts primary amines into isocyanides. It produces benzyl isocyanide. This is correct.
- Reaction C: Aryl halide (Bromobenzene) reacting with \(AgCN\). Aryl halides generally do not undergo nucleophilic substitution under standard conditions due to partial double bond character of the \(C-Br\) bond and steric hindrance.
- Reaction D: Aniline reacting with \(CHCl_3/NaOH\). This produces phenyl isocyanide (\(C_6H_5NC\)), not benzyl isocyanide.
- Reaction E: Benzyl bromide reacting with \(KCN\). \(KCN\) is ionic and provides cyanide ions (\(CN^-\)), where the carbon is more nucleophilic. This produces benzyl cyanide (\(C_6H_5CH_2CN\)), not isocyanide. Step 3: Final Answer:
Reactions A and B produce the desired product.