Question:medium
The final product (D) in the above conversion is:
\[
\text{Me} - \text{CH}_3 \xrightarrow{\text{H}_2, \text{(mol)}, \text{Ni}} \text{(A)} \xrightarrow{\text{PhCO}_2\text{H}, \text{H}_2\text{O}} \text{(B)} \xrightarrow{\text{NaOEt, EtOH}} \text{(C)} \xrightarrow{\text{EtMgBr, Et}_2\text{O}} \xrightarrow{\text{H}_2\text{O}} \text{(D)}
\]
The options are:
The final product (D) in the above conversion is:
\[
\text{Me} - \text{CH}_3 \xrightarrow{\text{H}_2, \text{(mol)}, \text{Ni}} \text{(A)} \xrightarrow{\text{PhCO}_2\text{H}, \text{H}_2\text{O}} \text{(B)} \xrightarrow{\text{NaOEt, EtOH}} \text{(C)} \xrightarrow{\text{EtMgBr, Et}_2\text{O}} \xrightarrow{\text{H}_2\text{O}} \text{(D)}
\]
The options are:
Show Hint
Grignard reagents like EtMgBr are powerful nucleophiles and are commonly used in carbon-carbon bond formation reactions.
Updated On: Feb 23, 2026
- Acetone
- 2-Butanol
- 1-Butanol
- 2-Pentanol
Show Solution
The Correct Option is B
Solution and Explanation
Step 1: Alkene Hydrogenation.
Alkene (C) is hydrogenated with H\(_2\) and Ni to yield saturated compound (A).
Step 2: Esterification.
Compound (A) reacts with benzoic acid (\(PhCO_2H\)) and water via esterification to form (B).
Step 3: Nucleophilic Substitution.
Ester (B) undergoes nucleophilic substitution with ethanol (\(EtO\)) to produce (C).
Step 4: Grignard Reaction.
Compound (C) reacts with ethylmagnesium bromide (EtMgBr) in anhydrous ether, followed by aqueous work-up, to yield 2-butanol (D).
Final Answer: \[\boxed{\text{(2) 2-Butanol}}\]
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