See image — Aldehydes Ketones and Carboxylic Acids Chemistry Question

💡 Solution & Explanation

Concept: Esterification of an alcohol with an acid chloride (acyl chloride) in the presence of a base proceeds via a nucleophilic acyl substitution mechanism. The key point is that the reaction occurs at the carbonyl carbon of the acid chloride, NOT at the chiral carbon of the alcohol. Step 1: Identify the chiral center. In CH3CH2CH(OH)CH(CH3)2, the chiral center is the carbon bearing the -OH group (C3), which has four different substituents: -OH, -CH2CH3, -CH(CH3)2, and -H. Step 2: Understand the mechanism. When the alcohol reacts with CH3COCl (acetyl chloride) in the presence of a base, the oxygen of the alcohol acts as the nucleophile and attacks the carbonyl carbon of CH3COCl. The C-O bond of the alcohol is retained; no bond at the chiral carbon is broken or formed. Step 3: Since the C-O bond at the chiral center is NOT broken during the reaction (the oxygen remains bonded to the same carbon), the spatial arrangement (configuration) at the chiral carbon is completely retained. Step 4: The product CH3CH2CH(OCOCH3)CH(CH3)2 retains the R-configuration at the chiral center because the reaction does not involve any bond-breaking at that carbon. Why other options fail: - (a) Inversion would require an SN2-type attack at the chiral carbon, which does not occur here. - (b) A racemic mixture would result if the chiral center were planarized (e.g., via carbocation or radical intermediate), which does not happen here. - (d) Optically inactive implies either racemic or meso, neither of which applies since configuration is fully retained. Therefore, the correct answer is C.