Chemistry - ALCOHOL & ETHER Question with Solution | TestHub
Match each List-I transformation with its reagent sequence in List-II. Ph means C6H5.
List-I:
(P) PhCH(CH3)C(NH2)(CH3)2 -> PhC(OH)(CH3)C(OH)(CH3)2.
(Q) PhC(OH)(CH3)C(OH)(CH3)2 -> PhC(CH3)2C(=O)CH3.
(R) PhCH(CH3)C(Cl)(CH3)2 -> PhC(OH)(CH3)C(OH)(CH3)2.
(S) PhC(OH)(CH3)C(Cl)(CH3)2 -> PhC(CH3)2C(=O)CH3.
List-II: (1) Cold 1% KMnO4; (2) Alcoholic KOH, heat; (3) NaNO2 + HCl;
(4) Anhydrous SbCl5; (5) Concentrated H2SO4, heat.
Options:
Answer:
Solution:
Let Ph denote phenyl. Reagent numbers refer to List-II. P: (3) NaNO2/HCl generates nitrous acid and deaminates the primary aliphatic amine, releasing N2 and giving alcohol products through the carbocation pathway. (5) Heated concentrated H2SO4 dehydrates the alcohol to the most substituted central alkene PhC(CH3)=C(CH3)2. (1) Cold dilute KMnO4 adds two OH groups across that double bond to give the drawn vicinal diol. Thus P uses 3,5,1; reagent 3 alone is not the final two-OH conversion. Q: (5) Heated concentrated H2SO4 causes a pinacol rearrangement. Protonation and loss of the benzylic OH creates the resonance-stabilized carbocation on the Ph-bearing carbon. A methyl group migrates from the adjacent carbon to this cation while its remaining OH forms C=O. Deprotonation gives PhC(CH3)2C(=O)CH3, exactly the drawn ketone. Thus Q uses 5. R: (2) Hot alcoholic KOH removes HCl by beta-elimination to give PhC(CH3)=C(CH3)2. (1) Cold 1% KMnO4 hydroxylates the double bond, giving the same vicinal diol as in P. Thus R uses 2,1. Alcoholic KOH is used for elimination, not aqueous substitution. S: (4) Anhydrous SbCl5 is the Lewis acid shown in the image. It activates C-Cl cleavage in the vicinal chlorohydrin. A phenyl group migrates from the OH-bearing carbon to the adjacent carbon as the oxygen forms C=O: this is a semipinacol rearrangement. The ketone can be written CH3C(=O)C(Ph)(CH3)2, which is the same connectivity as PhC(CH3)2C(=O)CH3 in the diagram. Thus S uses 4. It is SbCl5, not SOCl2, and the carbon skeleton rearranges rather than simply replacing OH. Therefore P-3,5,1; Q-5; R-2,1; S-4, matching option C.
