One mole of 1,2-dibromopropane on treatment with X moles of `NaNH_(2)` followed by treatment with ethyl bromide gave a pentyne. The value of X is:

Options

A. One
B. Two
C. Three
D. Four

Explanation

To solve the problem, we need to determine how many moles of sodium amide (NaNH₂) are required to convert one mole of 1,2-dibromopropane into pentyne through a series of reactions. Let's break it down step by step.

### Step-by-Step Solution:

1. **Identify the Structure of 1,2-Dibromopropane:**
- The molecular structure of 1,2-dibromopropane is CH₃-CHBr-CH₂Br. This compound has two bromine atoms attached to the second and third carbon atoms of a three-carbon chain (propane).

2. **Reaction with Sodium Amide (NaNH₂):**
- Sodium amide acts as a strong base and can remove hydrogen atoms from carbon atoms adjacent to the bromine atoms.
- The first mole of NaNH₂ will remove one bromine atom (Br) from the 1,2-dibromopropane through an E2 elimination mechanism, leading to the formation of an alkene (propene).
- The structure after the first reaction will be CH₃-CH=CH₂ (propene).

3. **Second Reaction with Sodium Amide:**
- The second mole of NaNH₂ will remove the second bromine atom from the remaining brominated carbon, leading to the formation of an alkyne (specifically, propyne).
- The structure after this reaction will be CH₃-C≡C-H (propyne).

4. **Formation of Pentyne:**
- Now, to form pentyne, we need to introduce an ethyl group (C₂H₅) to the propyne. This is done by treating the propyne with ethyl bromide (C₂H₅Br).
- The propyne will act as a nucleophile and attack the ethyl bromide, leading to the formation of pentyne (C₅H₈), specifically 1-pentyne or 2-pentyne depending on the position of the ethyl group.

5. **Total Moles of NaNH₂ Used:**
- From the above steps, we used:
- 1 mole of NaNH₂ to remove the first bromine and form propene.
- 1 mole of NaNH₂ to remove the second bromine and form propyne.
- Therefore, we have used a total of 2 moles of NaNH₂.
- Additionally, we need 1 mole of NaNH₂ to deprotonate propyne before it can react with ethyl bromide.

6. **Final Calculation:**
- Total moles of NaNH₂ used = 2 (for bromine removal) + 1 (for deprotonation) = 3 moles.

### Conclusion:
The value of X, which represents the moles of NaNH₂ required, is **3**.