Q.Identify A, B and C: benzoic acid, treated with PCl5, gives A; A, treated with benzene/anhydrous AlCl3, gives B; B, treated with

Step 1. Benzoic acid + PCl5 replaces the acid's -OH with -Cl (Section 12.12.B.1) to give A = benzoyl chloride, C6H5COCl, + POCl3 + HCl.

Step 2. A + benzene, in the presence of anhydrous AlCl3, undergoes Friedel-Crafts acylation (Section 12.3.D.2) to give B = benzophenone, C6H5-CO-C6H5, + HCl.

Step 3. B (a ketone) + C6H5MgBr (i), then H3O+ (ii), undergoes nucleophilic addition of the phenyl Grignard's carbanion to the ketone's carbonyl carbon; since a ketone + RMgX always gives a TERTIARY alcohol (the Grignard concept, Section 12.5.A.0), the product carries three phenyl groups plus -OH on the same carbon. …

Follow the same acid-to-acid-chloride-to-aryl-ketone-to-tertiary-alcohol chain used elsewhere in this chapter (e.g. Question 6), substituting the phenyl analogues throughout (benzoyl chloride instead of acetyl chloride, benzop …

• Forgetting that a ketone (not an ester or acid) reacting with a Grignard reagent gives a TERTIARY alcohol carrying BOTH of the ketone's original substituents plus the new Grignard group …