In a set reaction, acetic acid yields a product S. The compound S would be:
$C H_{3} C O O H S O C l_{2} - --- \to P Benzene - ------ \to A n h y . A l C l_{3} Q H C N - -- \to R H_{3} O^{+} - -- \to$

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Solution

The correct option is A
Step 1: Carboxylic acids react with phosphorous trichloride $(P C l_{3})$ , phosphorous pentachloride $(P C l_{5})$ , thionyl chloride $(S O C l_{2})$ , and phosphorous tribromide $(P B r_{3})$ to form acyl halides.
Step 2: Friedel craft's acylation reaction.
Step 3: Nucleophilic addition of hydrogen cyanide to form cyanohydrin.
Step 4: cyanohydrin on hydrolysis produces carboxylic acid.

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Similar questions

C H_{3} C H_{2} C O O H S O C l_{2} - --- \to A C_{6} H_{6} - ----- \to A n h y . A l C l_{3} B H C N - -- \to C H O H - -- \to D

C

C

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 O O H S O C l_{2} \to A b e n z e n e \to a n h y d A l C l_{2} B H C N \to C H_{2} O^{+} \to D

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

C

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