In th given reaction `R-I+NaCNoverset(DMF)rarrR-C-=N` R-I may not be

Options

A. Primary
B. Secondary
C. Tertiary
D. `CH_3-I`

Answer

To solve the question regarding the reaction of R-I with NaCN in the presence of DMF, we need to analyze the nature of the reactants and the mechanism involved.

### Step-by-Step Solution:

1. **Identify the Reactants**:
The reactants in the reaction are R-I (an alkyl iodide) and NaCN (sodium cyanide).

2. **Understand the Role of DMF**:
DMF (dimethylformamide) is a polar aprotic solvent that helps in stabilizing the nucleophile (CN⁻) and enhances its reactivity.

3. **Dissociation of NaCN**:
When NaCN is dissolved in DMF, it dissociates into Na⁺ and CN⁻. The CN⁻ ion is the nucleophile that will attack the alkyl iodide (R-I).

4. **Mechanism of Reaction**:
The reaction proceeds via an SN2 mechanism, where the nucleophile (CN⁻) attacks the electrophilic carbon of the alkyl iodide (R-I) and displaces the iodide ion (I⁻).

5. **Consider the Structure of R-I**:
The nature of R-I is crucial. If R is a primary alkyl group, the SN2 reaction will proceed efficiently. If R is secondary, the reaction may still occur, but with less efficiency. If R is tertiary, the SN2 reaction will not occur because tertiary alkyl halides favor the SN1 mechanism, which involves the formation of a carbocation.

6. **Conclusion**:
Since the question asks which R-I may not participate in the reaction, we conclude that if R is a tertiary alkyl group, the reaction will not proceed via the SN2 pathway. Therefore, the answer is that R-I cannot be a tertiary alkyl iodide.

### Final Answer:
R-I may not be a tertiary alkyl iodide.