Alkyl halide reacts with AgCN to from -

Options

A. Alcohol
B. Cyanide
C. Isocyanide
D. Both B & C

Answer

To solve the question of what product is formed when an alkyl halide reacts with AgCN, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Reactants**:
- We start with an alkyl halide, which can be represented as R-X, where R is an alkyl group and X is a halogen (like Cl, Br, or I).
- The other reactant is silver cyanide (AgCN).

2. **Dissociation of AgCN**:
- When AgCN is added to the alkyl halide, it dissociates into silver (Ag⁺) and cyanide (CN⁻) ions.
- However, in this case, we consider the formation of free radicals: Ag free radical and CN free radical.

3. **Nucleophilic Attack**:
- The cyanide radical (CN⁻) acts as a nucleophile and attacks the alkyl halide (R-X).
- The bond between R and X breaks, and the CN group attaches to the alkyl group.

4. **Formation of Isocyanide**:
- The product formed from this reaction is isocyanide (also known as isocyanide or carbylamine), which can be represented as R-N≡C (where R is the alkyl group and N≡C indicates a triple bond between nitrogen and carbon).
- The silver ion (Ag⁺) will combine with the halogen (X) to form AgX.

5. **Final Products**:
- The overall reaction can be summarized as:
\[ R-X + AgCN \rightarrow R-N≡C + AgX \]
- Thus, the main product is isocyanide.

### Conclusion:
The product formed when an alkyl halide reacts with AgCN is **isocyanide**.