Question

An alcohol A (C₄H₁₀O) on oxidation with acidified potassium dichromate gives acid B (C₄H₈O₂). Compound A when dehydrated with conc. H₂SO₄ at 443 K 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 oxidised. Identify A, B, C and D. Name the type of isomerism exhibited by A and D

Answer 1

An alcohol Compound A is 2-methylpropanol , when 2- methylpropanol acidified potassium dichromate gives 2-methylpropanioic acid , hence, compound B is 2-methylpropanoic acid .

       Now when 2-methylpropanol is dehydrated with concentrated sulphuric acid (e.g., H₂SO₄ ) gives compound 2-methylpropene. Hence, compound C is 2-methylpropene.

And when 2-methylpropanol is dehydrated with aqueous sulphuric acid gives methylpropan-2-ol . Hence, compound D is methylpropan-2-ol.

compound A and compound D are position isomers.

Let’s see attached  figure , where chemical equations are mentioned to understand better.

                                 

Answer 2

An alcohol Compound A having molecular formula C₄H₁₀O  is 2-methylpropanol. On oxidation with acidified potassium dichromate, 2- methylpropanol gives 2-methylpropanioic acid. Hence, compound B is 2-methylpropanoic acid .

                               Now when 2-methylpropanol is dehydrated with concentrated sulphuric acid (e.g., H₂SO₄ ) at 443 K, A gives compound 2-methylpropene. Hence, compound C is 2-methylpropene.

                                Now when 2-methylpropene is treated with aqueous sulphuric acid, C gives methylpropan-2-ol . Hence, compound D is methylpropan-2-ol.

The type of isomerism exhibited by A and D is positional isomer.