See image — Alcohols Phenols and Ethers Chemistry Question

💡 Solution & Explanation

Step 1 - Identify the starting material: The compound shown has two stereocenters. Reading the wedge-dash notation, it is an amino alcohol: one carbon bears Ph, NH2, and H; the adjacent carbon bears OH, CH3, Ph, and H. The arrangement shown is (1R,2S) or (1S,2R) - specifically the threo or erythro isomer consistent with the drawn configuration. The compound is 2-amino-1,2-diphenyl-1-propanol (with the amino group on C2 and OH on C1, or vice versa), i.e., it is a beta-amino alcohol. Step 2 - Reaction with HNO2: HNO2 converts primary amines to diazonium ions (ArNH2 → ArN2+, or aliphatic RNH2 → RN2+ which rapidly loses N2). For an aliphatic beta-amino alcohol, the reaction with HNO2 proceeds via a diazonium intermediate that undergoes neighboring group participation by the adjacent OH. This is a semipinacol-type or Whitmore rearrangement. Step 3 - Mechanism: The NH2 is converted to N2+ (diazonium). The adjacent C–OH carbon's substituents can migrate to the carbocation center with departure of N2. Specifically, the phenyl group (or hydride) on the carbon bearing OH migrates to the carbon that lost N2, with inversion at that carbon. This is a 1,2-phenyl shift (or 1,2-hydride shift). Given the stereochemistry of the starting beta-amino alcohol, the migration occurs with anti periplanar geometry (the migrating group and the leaving N2 must be anti). The result is an aldehyde or ketone product. Step 4 - Product identity: The migration of Ph from the carbon bearing OH to the diazonium carbon gives an oxocarbenium that collapses to an aldehyde. The product is PhCH(CH3)-CHO rearranged, or more precisely Ph-CO-CH(CH3)Ph. Actually: starting from Ph(NH2)(H)C-C(OH)(CH3)(Ph)H, the NH2 carbon becomes the electrophilic center; Ph migrates from the adjacent carbon (bearing OH, CH3, Ph) to give Ph-CHO with retention/inversion considerations, and the other carbon becomes Ph-C(=O)- with the CH3 and remaining Ph. The product is Ph-C(=O)-CH(Ph)(CH3), i.e., an alpha-phenyl propiophenone derivative: Ph-CO-CH(CH3)(Ph). Step 5 - Stereochemistry of product: Because the 1,2-phenyl shift occurs with inversion at the migrating terminus, and the starting material has a defined configuration, the product retains a single stereocenter at the alpha-carbon. The reaction is stereospecific (anti migration), so a single diastereomeric starting material gives a single enantiomeric product - not a racemate, not a mixture of diastereomers. Step 6 - Matching to options: Option (a) shows Ph-C(=O)(H) which would be an aldehyde with an extra H on the carbonyl carbon - this represents PhCHO attached to CH(CH3)(Ph), i.e., the product where H migrates. Option (b) shows Ph-C(=O)-CH(CH3)(Ph) with defined stereochemistry (wedge-dash), representing the product of Ph migration giving a single stereoisomeric ketone. Since phenyl migrates preferentially over H (phenyl has better migratory aptitude), and the stereochemistry is defined by the anti migration from a single starting stereoisomer, option (b) - a single stereoisomeric ketone - is the major product. Step 7 - Why other options fail: (c) Racemic would imply loss of stereochemical information, but the anti migration is stereospecific. (d) Diastereomers would imply two stereocenters in the product or a mixture, but the product has only one stereocenter and the reaction is stereospecific giving one isomer. Therefore, the correct answer is A.