See image — Alcohols Phenols and Ethers Chemistry Question

💡 Solution & Explanation

Step 1 - Identify the starting material and reagents: The starting material is 1-phenyl-2-propanol (Ph-CH2-CH(OH)-CH3). The first reagent is K (potassium metal), and the second reagent is C2H5Br (ethyl bromide). Step 2 - Reaction with K: Potassium metal reacts with the alcohol to form the corresponding alkoxide (Ph-CH2-CH(O^- K^+)-CH3). This is a simple acid-base reaction where K deprotonates the OH group. The stereochemistry at the chiral center is fully retained during alkoxide formation because no bond to the chiral center is broken. Step 3 - Reaction of alkoxide with C2H5Br: The alkoxide acts as a nucleophile and reacts with ethyl bromide (C2H5Br) via an SN2 mechanism. However, in this step, the nucleophile (alkoxide oxygen) attacks the ethyl bromide carbon (a primary carbon), NOT the chiral center of the original molecule. The C-O bond at the chiral center is not broken during this etherification step. Step 4 - Stereochemical outcome: Since the bond at the chiral carbon is never broken throughout the entire reaction sequence (K deprotonates O-H, not C-H; and SN2 occurs at the ethyl carbon of EtBr, not at the chiral carbon), the configuration at the chiral center is fully retained. Step 5 - Why option (a) fails: Option (a) claims inversion of configuration. Inversion would require an SN2 attack at the chiral carbon itself, which does not occur here. The nucleophilic attack is on the primary ethyl carbon of C2H5Br, so no inversion at the chiral center takes place. Step 6 - Product: The product is Ph-CH2-CH(OEt)-CH3 with retention of configuration at the chiral center. Therefore, the correct answer is B.