In a set of the given reactions, acetic acid yields a product $C$ . Product $C$ would be:
$C H_{3} C O O H + P C l_{5} \to A C_{6} H_{6} - ----- \to A n h y . A l C l_{3} B C_{2} H_{5} M g B r / e t h e r - ---------- \to (i i) H^{+} C$ .

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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C H_{3} - C_{2} H_{5} | C - (O H) C_{6} H_{5}

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Solution

The correct option is C $C H_{3} - C_{2} H_{5} | C - (O H) C_{6} H_{5}$
When carboxylic acid reacts with $P C l_{5}$ , acid chloride formed and acid chloride on reaction with benzene (or its derivatives) forms ketones and this process is known as Freidal Craft acylation carried out in presence of anhydrous $A l C l_{3}$ . Ketones on reaction with Grignard reagent form tertiary alcohols.
$C H_{3} C O O H + P C l_{5} ⟶ C H_{3} C O C l C_{6} H_{6} - --------- \to A n h y d r o u s A l C l_{3} C H_{3} - C O C_{6} H_{5}$
$↓$ (i) $C_{2} H_{5} M g B r / e t h e r$ , (ii) $H^{+}$
$O H$
$H_{3} C - | C | - C_{6} H_{5}$
$C_{2} H_{5}$ $(2 - p h e n y l - b u t a n - 2 - o l)$

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