In the following reaction sequence, identify the major product R:

Step 1: p-Chlorobenzaldehyde + NaCN/HOAc → P (Cyanohydrin formation)

$\ce{p-ClC6H4-CHO + HCN -> p-ClC6H4-CH(OH)-CN}$

NaCN provides $\ce{CN-}$ (nucleophile) which attacks the aldehyde carbonyl. HOAc (acetic acid) provides H+ to protonate the alkoxide. P = 4-chloromandelonitrile ($\ce{p-ClC6H4-CH(OH)-CN}$, a cyanohydrin)

Step 2: P + EtOH/H+ → Q (Acid-catalysed esterification of CN... actually Fischer esterification of OH)

Actually, EtOH/H+ in the presence of CN → the CN can be converted to $\ce{-COOH}$ and then esterified, OR the CN itself reacts with EtOH under acid to give an imidate/amide/ester.

More precisely: CN + EtOH/H+ → imino ester → hydrolysis → ethyl ester: $\ce{p-ClC6H4-CH(OH)-COOEt}$ (ethyl 4-chloromandelate) Q = ethyl 4-chloromandelate ($\ce{p-ClC6H4-CH(OH)-COOEt}$)

Step 3: Q + 2 MeMgBr, then H3O+ → R

Grignard addition to an ester (2 equivalents):

• First equivalent of MeMgBr attacks the ester carbonyl → tetrahedral intermediate → collapse → ketone intermediate.
• Second equivalent of MeMgBr attacks the ketone intermediate → alkoxide → after H3O+ → tertiary alcohol.

The -OH on the alpha carbon may also participate, but the ester carbon (C of COOEt) reacts with 2 equivalents of MeMgBr to give: $\ce{p-ClC6H4-CH(OH)-C(CH3)2-OH}$ (diol with tertiary alcohol).

R = a tertiary diol: p-Cl-phenyl group bearing both a secondary alcohol from original cyanohydrin OH and a tertiary -C(CH3)2OH from ester reaction.

Key Points to Remember:

• Cyanohydrin = nucleophilic addition of HCN to aldehyde/ketone.
• Ester + 2 Grignard → tertiary alcohol (overall).
• The first Grignard addition to ester gives ketone intermediate; second gives tertiary alcohol.