Question:
What are X and Y in the following reactions?
\[
(CH_3)_3C-ONa + C_2H_5Br \rightarrow X
\]
\[
(CH_3)_3C-Br + C_2H_5ONa \rightarrow Y
\]
What are X and Y in the following reactions?
\[
(CH_3)_3C-ONa + C_2H_5Br \rightarrow X
\]
\[
(CH_3)_3C-Br + C_2H_5ONa \rightarrow Y
\]
Show Hint
Primary halides favor Williamson ether synthesis, while tertiary halides generally undergo elimination with strong bases.
Updated On: Jun 18, 2026
- X = Ether, Y = Ether
- X = Ether, Y = Alkene
- X = Alkene, Y = Ether
- X = Ether, Y = Alkane
Show Solution
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The Correct Option is B
Solution and Explanation
Concept:
Williamson synthesis works best with primary alkyl halides whereas tertiary alkyl halides generally undergo elimination.
Step 1: First reaction.
\[ (CH_3)_3CONa + C_2H_5Br \] The alkoxide ion attacks the primary alkyl halide through \(S_N2\). \[ (CH_3)_3COC_2H_5 \] is formed. Thus \[ X=\text{Ether} \]
Step 2: Second reaction.
\[ (CH_3)_3CBr + C_2H_5ONa \] Since tert-butyl bromide is tertiary, elimination predominates. \[ (CH_3)_3CBr \rightarrow (CH_3)_2C=CH_2 \] Thus \[ Y=\text{Alkene} \] Therefore, \[ \boxed{X=\text{Ether},\quad Y=\text{Alkene}} \]
Step 1: First reaction.
\[ (CH_3)_3CONa + C_2H_5Br \] The alkoxide ion attacks the primary alkyl halide through \(S_N2\). \[ (CH_3)_3COC_2H_5 \] is formed. Thus \[ X=\text{Ether} \]
Step 2: Second reaction.
\[ (CH_3)_3CBr + C_2H_5ONa \] Since tert-butyl bromide is tertiary, elimination predominates. \[ (CH_3)_3CBr \rightarrow (CH_3)_2C=CH_2 \] Thus \[ Y=\text{Alkene} \] Therefore, \[ \boxed{X=\text{Ether},\quad Y=\text{Alkene}} \]
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