Question

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:

• A. Acetone
• B. 2-Butanol
• C. 1-Butanol
• D. 2-Pentanol

Hint:

Grignard reagents like EtMgBr are powerful nucleophiles and are commonly used in carbon-carbon bond formation reactions.

Answer 1

The Correct Option is B

Step 1: Hydrogenation of the alkene.
Hydrogenation of the alkene (C) using H\(_2\) and nickel (Ni) produces a saturated compound (A).

Step 2: Esterification.
The compound (A) is then treated with benzoic acid (\(PhCO_2H\)) and water, resulting in esterification to give (B).

Step 3: Nucleophilic Substitution.
In the next step, the ester (B) undergoes nucleophilic substitution with ethanol (\(EtO\)) to form (C).

Step 4: Grignard Reaction.
The compound (C) is then treated with ethylmagnesium bromide (EtMgBr) in anhydrous ether, followed by water work-up, resulting in the formation of 2-butanol as the final product (D).

Final Answer: \[ \boxed{\text{(2) 2-Butanol}} \]