Question

An alcohol A (C4H10O) on oxidation with acidified potassium dichromate gives carboxylic acid B (C4H8O2). Compound A when dehydrated with conc.H2SO4 at 443K gives compound C. Treatment of C with aqueous H2SO4 gives compound D (C4H10O) which is an isomer of A. Compound D is resistant to oxidation but compound A can be easily oxidized. Identify the compounds and write their structures.

Answer 1

Hi,

Let us review the question once again;

An alcohol A (C4H10O) on oxidation with acidified potassium dichromate gives carboxylic acid B (C4H8O2). So here; A is 2-methylpropan-1-ol and B is 2-methylpropanoic acid.

Compound A when dehydrated with conc.H2SO4 at 443K gives compound C.

Compound C is 2-methylpropan-1-ene

Treatment of C with aqueous H2SO4 gives compound D (C4H10O) which is an isomer of A.

compound D is 2-methylpropan-2-ol

Answer 2

Hey dear,
The question is from JEE model question paper.

● Compounds -
(A) 2-methylpropanol
(B) 2-methylpropanoic acid
(C) 2-methylpropene
(D) 2-methylpropan-2-ol

● Structures -
(A) 2-methylpropan-1-ol
CH3-CH(CH3)-CH2-OH

(B) 2-methylpropan-1-oic acid
CH3-CH(CH3)-COOH

(C) 2-methylprop-1-ene
CH3-C(CH3)=CH2

(D) 2-methylpropan-2-ol
CH3-C(CH3)2-OH

Hope this helps you...