See image — Aromatic Hydrocarbons Chemistry Question

💡 Solution & Explanation

Step 1: Identify the starting material. Ph2C(OH)CH2NH2 is 2-amino-1,1-diphenylethanol. It has a diphenyl carbinol carbon bearing an OH group, and a CH2NH2 group. Step 2: Deamination with NaNO2/HCl converts the primary amine (-NH2) to a diazonium salt in situ, which then loses N2. The carbocation or rearrangement intermediate formed at the CH2 position adjacent to the Ph2C(OH) group can undergo a 1,2-hydride or 1,2-phenyl shift. Step 3: The primary carbocation at CH2+ is adjacent to Ph2C(OH)-. A 1,2-phenyl shift from the adjacent carbon (Ph2C(OH)) migrates one Ph group to the carbocation center, giving a more stable tertiary or secondary carbocation. After phenyl migration: the carbocation moves to the carbon that originally bore two phenyl groups and an OH, i.e., Ph(OH)C+ with one Ph migrated. This intermediate then loses a proton or the OH acts accordingly. Step 4: After the 1,2-phenyl shift, the rearranged carbocation is Ph-C+(OH)-CH2-Ph or upon loss of proton from OH gives Ph-C(=O)-CH2-Ph (desyl / phenacyl phenyl, i.e., deoxybenzoin). Step 5: Verify with oxidation condition. Oxidation of Ph-C(=O)-CH2-Ph (deoxybenzoin): the CH2 between the carbonyl and Ph can be oxidized. Vigorous oxidation (e.g., KMnO4) cleaves both Ph-CO- and Ph-CH2- groups to give benzoic acid from both the PhCO- (PhCOOH) and the PhCH2- (PhCOOH) portions. Thus oxidation gives benzoic acid only, consistent with the problem statement. Step 6: Check other options. (a) Ph-CH2-CH2-Ph (bibenzyl): oxidation gives benzoic acid only as well, but this product would not arise from deamination with rearrangement of Ph2C(OH)CH2NH2 in a straightforward manner; also no oxygen functionality retained. (c) Ph-CO-CO-Ph (benzil): oxidation would give benzoic acid but its formation mechanism doesn't fit. (d) Ph-CH2-NH-Ph: this is an amine, not a deamination product consistent with rearrangement. Step 7: The product P is Ph-C(=O)-CH2-Ph, deoxybenzoin, which is option (b). Therefore, the correct answer is B.