Q.A compound (A) with molecular formula C2H3N, on acid hydrolysis, gives (B), which reacts with thionyl chloride to give compound (C). Benzene reacts with compound (C) in the presence of anhydrous AlCl3 to give compound (D). Compound (D), on reduction with Zn/Hg and concentrated HCl, gives (E). Identify (A), (B), (C), (D) and (E). Write the equations.

Step 1. A, with molecular formula C2H3N, is methyl cyanide (acetonitrile), CH3-C#N -- the formula matches exactly (2 carbons, 3 hydrogens, 1 nitrogen).

Step 2. A, on acid hydrolysis (H3O+/H+, Section 12.10.2), gives B = acetic acid, CH3COOH, + NH3: CH3CN + 2 H2O gives CH3COOH + NH3.

Step 3. B, reacting with thionyl chloride, gives C = acetyl chloride (Section 12.12.B.1/12.14.2.1): CH3COOH + SOCl2 gives CH3COCl + SO2 + HCl.

Step 4. Benzene reacts with C in the presence of anhydrous AlCl3 by Friedel-Crafts acylation (Section 12.3.D.2) to give D = acetophenone: CH3COCl + C6H6 (AlCl3) gives C6H5COCH3 + HCl.

Step 5. D, reduced with Zn/Hg and concentrated HCl -- the Clemmensen reduction (Section 12.5.C.2) -- gives E = ethylbenzene, the ketone's C=O being fully reduced to CH2: C6H5COCH3 + 4[H] (Zn-Hg/conc. HCl) gives C6H5CH2CH3 + H2O.

Match the molecular formula C2H3N directly to methyl cyanide, then follow the standard nitrile-to-acid-to-acid-chloride-to-aryl-ketone-to-alkylbenzene chain, naming each named reaction (hydrolysis, SOCl2 chlorination, Friedel-Crafts acylation, Clemmensen reduction) as it is used.

• Confusing Clemmensen reduction (Zn-Hg/conc. HCl, full reduction of C=O to CH2, giving ethylbenzene) with a milder reduction that would stop at a secondary alcohol (1-phenylethanol) -- the harsh Zn-Hg/HCl conditions specifically drive the carbonyl all the way to the hydrocarbon.