See image — Alcohols Phenols and Ethers Chemistry Question

💡 Solution & Explanation

Step 1 - Identify the substrate and reaction type: The substrate is a tertiary alcohol with a chiral center bearing four different substituents: CH3, CD3, Ph, and OH. It reacts with HBr, which is a substitution reaction (OH replaced by Br). Step 2 - Determine the mechanism: Tertiary alcohols react with HBr via an SN1 mechanism. In SN1, the OH is first protonated by HBr to form a good leaving group (water), which then departs to generate a planar, sp2-hybridized carbocation intermediate. Step 3 - Analyze the carbocation intermediate: The tertiary carbocation formed is planar (trigonal planar). The three substituents (CH3, CD3, Ph) lie in a plane, and the empty p-orbital is perpendicular to this plane. This means the carbocation has no defined stereochemistry at that center. Step 4 - Nucleophilic attack by Br-: The bromide nucleophile can attack the planar carbocation from either face (top or bottom) with approximately equal probability. This gives a roughly equal mixture of both enantiomers (R and S configurations) at the carbon bearing Br. Step 5 - Conclusion: Because attack occurs from both faces with equal likelihood, racemization (loss of optical activity, production of a racemic mixture) occurs at the carbon undergoing substitution. Step 6 - Why other options fail: - (a) Inversion requires SN2 mechanism with backside attack only; SN1 does not proceed with exclusive inversion. - (b) Retention would require a neighboring group participation or front-side attack preference, which is not the case here. - (d) The carbon IS stereogenic (it bears four different groups: CH3, CD3, Ph, and Br in the product), so this option is incorrect. Therefore, the correct answer is C.