Question

\(\text{RO}-\text{CH}_2-\text{C} \equiv \text{CH} \xrightarrow{\text{X}} \xrightarrow{\text{Y}} \text{RO}-\text{CH}_2-\text{C} \equiv \text{C}-\text{CH}_2\text{CH}_2-\text{Br}\)
To carry out the above conversion X and Y are respectively

• A. $\text{NaNH}_2$, $\text{Br}-\text{CH}_2-\text{CH}_2-\text{Cl}$
• B. $\text{NaNH}_2$, $\text{I}-\text{CH}_2-\text{CH}_2-\text{Br}$
• C. $\text{NaNH}_2$, $\text{F}-\text{CH}_2-\text{CH}_2-\text{Br}$
• D. $\text{NaNH}_2$, $\text{CH}_2=\text{CH}-\text{Br}$

Hint:

For selective nucleophilic substitution on unsymmetrical dihalides:
- The nucleophile will always attack the carbon attached to the better leaving group ($\text{I}^- > \text{Br}^- > \text{Cl}^-$).
- To retain a $\text{-Br}$ group in your product, the reacting end of the dihalide must have a superior leaving group like $\text{-I}$.

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
The starting material is a terminal alkyne, $\text{RO}-\text{CH}_2-\text{C} \equiv \text{CH}$. We need to identify reagents X and Y to convert it into the chain-extended terminal bromide product, $\text{RO}-\text{CH}_2-\text{C} \equiv \text{C}-\text{CH}_2\text{CH}_2-\text{Br}$.


Step 2: Detailed Explanation:
Let us break down the chemical transformation step-by-step:
1. Deprotonation of the terminal alkyne (Reagent X):
- The terminal hydrogen of an alkyne is weakly acidic.
- A strong base like sodium amide ($\text{NaNH}_2$) is used to deprotonate the terminal alkyne to form a highly nucleophilic acetylide carbanion:
\[ \text{RO}-\text{CH}_2-\text{C} \equiv \text{CH} \xrightarrow{\text{NaNH}_2} \text{RO}-\text{CH}_2-\text{C} \equiv \text{C}^- \text{Na}^+ \]
2. Nucleophilic Alkylation (Reagent Y):
- The nucleophilic acetylide anion must react with a dihaloalkane to introduce the $-\text{CH}_2\text{CH}_2-\text{Br}$ group.
- Crucially, the final product must retain the bromine atom ($\text{-Br}$) at the terminal end.
- To ensure that the nucleophile attacks selectively at one end and leaves the bromine intact on the other, the leaving group at the site of attack must be significantly better than bromine.
- The leaving group ability order is: $\text{I}^- > \text{Br}^- > \text{Cl}^- > \text{F}^-$.
- When $\text{I}-\text{CH}_2-\text{CH}_2-\text{Br}$ (option B) is used:
Iodide ($\text{I}^-$) is a much better leaving group than bromide ($\text{Br}^-$). Thus, the acetylide nucleophile attacks the carbon containing the iodine atom, selectively displacing iodide via an $\text{S}_\text{N}2$ mechanism:
\[ \text{RO}-\text{CH}_2-\text{C} \equiv \text{C}^- + \text{I}-\text{CH}_2\text{CH}_2-\text{Br} \rightarrow \text{RO}-\text{CH}_2-\text{C} \equiv \text{C}-\text{CH}_2\text{CH}_2-\text{Br} + \text{I}^- \]
- If $\text{Br}-\text{CH}_2-\text{CH}_2-\text{Cl}$ (option A) were used, bromine (being a better leaving group than chlorine) would be displaced, leaving the chlorine atom at the terminal position, which is not the desired product.
Thus, $\text{I}-\text{CH}_2-\text{CH}_2-\text{Br}$ is the ideal reagent Y.


Step 3: Final Answer:
The correct option is (B).