Q.Suggest a suitable reagent to prepare a secondary alcohol with two identical alkyl groups using a Grignard reagent.

Step 1. Formaldehyde with one equivalent of Grignard gives a primary alcohol; a formate ester (H-COO-R') instead reacts with TWO equivalents of a Grignard reagent.

Step 2. The first equivalent displaces the alkoxide (R'O-) from the ester to give an aldehyde intermediate in situ (H-CO-R, the same alkyl group as the Grignard).

Step 3. The second equivalent of the SAME Grignard reagent then adds to that intermediate aldehyde's carbonyl carbon.

Step 4. Acid hydrolysis of the resulting alkoxide gives a SECONDARY alcohol, R-CH(OH)-R, bearing two IDENTICAL alkyl groups -- exactly as illustrated in the chapter for ethyl methanoate + 2 CH3MgBr giving propan-2-ol.

Recall that a formate ester reacting with two equivalents of one Grignard reagent is the standard route to a secondary alcohol with two identical alkyl groups.

• Trying to use an ordinary ketone for this purpose -- a ketone's two substituents are typically already fixed and different, whereas a formate ester lets BOTH final alkyl groups be introduced fresh, and identical, from the Grignard reagent itself.