Find out the major products from the following reaction sequence.

1. Step 1: Cyanohydrin Formation and Alcoholysis ($$\text{1) NaCN}$$, $$\text{2) EtOH, H}_2\text{O, H}^\oplus$$)

   The starting material is a ketone, 4-chloroacetophenone ($$\text{Cl--C}_6\text{H}_4\text{--COCH}_3$$).

   • First, nucleophilic addition of cyanide ($$\text{CN}^\ominus$$) to the carbonyl group yields a cyanohydrin intermediate: $$\text{Cl--C}_6\text{H}_4\text{--C(OH)(CN)(CH}_3)$$.
   • Second, treating the cyanohydrin with ethanol ($$\text{EtOH}$$) under acidic aqueous conditions hydrolyzes and esterifies the nitrile group ($$\text{--CN}$$) into an ethyl ester group ($$\text{--CO}_2\text{Et}$$).

   This gives the first intermediate shown in the scheme: ethyl 2-(4-chlorophenyl)-2-hydroxypropanoate.




2. Step 2: Grignard Reaction with Excess Methylmagnesium Bromide ($$\text{MeMgBr (excess)}$$ followed by $$\text{H}_3\text{O}^\oplus$$)

   The intermediate contains two key functional groups: a tertiary alcohol ($$\text{--OH}$$) and an ester ($$\text{--CO}_2\text{Et}$$).

   • The first equivalent of the Grignard reagent ($$\text{MeMgBr}$$) acts as a base and deprotonates the acidic hydroxyl proton ($$\text{--OH}$$), which is later regenerated during acidic workup.
   • Another equivalent of $$\text{MeMgBr}$$ undergoes nucleophilic acyl substitution at the ester carbonyl carbon, displacing the ethoxide leaving group ($$\text{--OEt}$$) to form a methyl ketone intermediate ($$\text{--COCH}_3$$).
   • Because the Grignard reagent is added in excess, a final equivalent immediately attacks this newly formed ketone group to yield a tertiary alkoxide.

   After acidic workup ($$\text{H}_3\text{O}^\oplus$$), the final product contains a highly branched tertiary alcohol unit attached to the original aromatic framework.

Structure of the Final Product:

The carbon atom alpha to the benzene ring preserves its original hydroxyl group ($$\text{--OH}$$) and methyl group ($$\text{--Me}$$). The ester group ($$\text{--CO}_2\text{Et}$$) is converted completely into a 2-hydroxypropan-2-yl group ($$\text{--C(OH)Me}_2$$).

Therefore, the structural formula of the final major product is:

$$\text{Cl--C}_6\text{H}_4\text{--C(OH)(Me)--C(OH)(Me)}_2$$

Conclusion:

The reaction sequence effectively introduces multiple methyl groups by transforming the ester functionality into a tertiary alcohol while maintaining the structural integrity of the chloro-substituted benzene ring.

Answer: The molecular structure labeled as 'C' in the reaction scheme, which corresponds to $$\text{Cl--C}_6\text{H}_4\text{--C(OH)(Me)--C(OH)(Me)}_2$$.