In a set of the given reactions, acetic acid yielded a product $C$ .

$C H_{3} C O O H + P {C l}_{5} ⟶ A C_{6} H_{6} ⟶ A l {C l}_{3} B (i) C_{2} H_{5} M g B r / e t h e r ⟶ (i i) H_{3} O^{\oplus} C$ ; product $C$ would be

C H_{3} C H (O H) C_{2} H_{5}

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C H_{3} C O C_{6} H_{5}

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C H_{3} C H (O H) C_{6} H_{5}

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\begin{matrix} C_{2} H_{5} | C H_{3} - & C (O H) & C_{6} H_{5} \end{matrix}

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Solution

The correct option is D $\begin{matrix} C_{2} H_{5} | C H_{3} - & C (O H) & C_{6} H_{5} \end{matrix}$

The compound (C) is 2-phenylbutan-2-ol.
Acetic acid reacts with phosphorous pentachloride to form acetyl chloride. Friedel Craft's acylation of benzene with acetyl chloride in presence of anhydrous aluminium chloride gives acetophenone. Acetophenone reacts with Grignard reagent (ethyl magnesium bromide) to form 2-phenylbutan-2-ol which is a tertiary alcohol.

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Similar questions

\begin{matrix} O H | C H_{3} - & C - & C_{6} H_{5} | C_{2} H_{5} \end{matrix}

O H^{⊖}

C H_{3} - C H | C_{2} H_{5} - B r

C H_{3} - B r | C | C_{2} H_{5} - C H_{3}

C H_{3} - C H | C_{2} H_{5} - C H_{2} B r

{C H}_{3} - O - C_{2} H_{5} H_{2} {S O}_{4}, H_{2} O - ------ \to c o n c

C H_{3} C O C H_{2} C O O C_{2} H_{5}

C H_{3} C O C H_{2} C O C H_{3}

C_{2} H_{5} O O C C H_{2} C O O C_{2} H_{5}

I . C_{2} H_{5} - I + {N H}_{3} ⟶ C_{2} H_{5}_{3} + I^{⊖}

I I . C_{2} H_{5} - I + \cdot \cdot P H_{3} ⟶ C_{2} H_{5}_{3} + I^{⊖}

I I I . C_{2} H_{5} - I + E t O^{⊖} ⟶ C_{2} H_{5} - O E t + I^{⊖}

I V . M e - \oplus S - {M e}_{2} + ⊖ O H ⟶ M e - O H + {M e}_{2} S

S_{N} 2

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