Question:
The total number of bromine atoms in the final product ‘X’ of the following reactions is:
The total number of bromine atoms in the final product ‘X’ of the following reactions is:
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One mole of Br\(_2\) adds across triple bond giving tetra-substituted dibromo intermediate (total 4 Br atoms).
Updated On: Jun 20, 2026
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The Correct Option is C
Solution and Explanation
Step 1: Understand the starting compound.
The starting compound is propyne (CH\(_3\)-C≡CH), which is a terminal alkyne. It can undergo deprotonation, alkylation, and halogen addition reactions stepwise.
Step 2: First step - NaNH\(_2\) liq NH\(_3\).
NaNH\(_2\) is a strong base and removes acidic hydrogen from terminal alkyne forming an acetylide ion. This ion is highly nucleophilic and ready for alkylation.
Step 3: Second step - CH\(_3\)I.
The acetylide ion attacks CH\(_3\)I via SN2 mechanism, forming an internal alkyne (CH\(_3\)-C≡C-CH\(_3\)). No bromine introduced yet.
Step 4: Third step - Br\(_2\)/CCl\(_4\).
Bromine adds across the triple bond. One equivalent of Br\(_2\) adds twice to form a tetrabromo alkane derivative. Each carbon of the triple bond gets two bromine atoms after addition.
Step 5: Count bromine atoms in product.
After addition of Br\(_2\), total bromine atoms introduced = 4 (two on each carbon of former triple bond).
Step 6: Final conclusion.
Thus, total bromine atoms in product X = 4.
Final Answer: \[ \boxed{4} \]
The starting compound is propyne (CH\(_3\)-C≡CH), which is a terminal alkyne. It can undergo deprotonation, alkylation, and halogen addition reactions stepwise.
Step 2: First step - NaNH\(_2\) liq NH\(_3\).
NaNH\(_2\) is a strong base and removes acidic hydrogen from terminal alkyne forming an acetylide ion. This ion is highly nucleophilic and ready for alkylation.
Step 3: Second step - CH\(_3\)I.
The acetylide ion attacks CH\(_3\)I via SN2 mechanism, forming an internal alkyne (CH\(_3\)-C≡C-CH\(_3\)). No bromine introduced yet.
Step 4: Third step - Br\(_2\)/CCl\(_4\).
Bromine adds across the triple bond. One equivalent of Br\(_2\) adds twice to form a tetrabromo alkane derivative. Each carbon of the triple bond gets two bromine atoms after addition.
Step 5: Count bromine atoms in product.
After addition of Br\(_2\), total bromine atoms introduced = 4 (two on each carbon of former triple bond).
Step 6: Final conclusion.
Thus, total bromine atoms in product X = 4.
Final Answer: \[ \boxed{4} \]
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