Question

An alcohol [A] with molecules formula `(C_(4)H_(10)O)` o oxidation with aciddified potassium dichromate gives acid [B] `(C_(4)H_(8)O_(2))`. Compound [A] when dehydrated with conc. `H_(2)SO_(4)` at `443 K` gives compound [C]. Treatment of [C] with aqueous `H_(2)SO_(4)` gives compound [D] `(C_(4)H_(10)O)` 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

`A :CH_(3)-overset(underset(|)(CH_(3)))(CH)-CH_(2)-OH " " B : CH_(3)-overset(underset(|)(CH_(3)))(CH)-COOH`
`C :CH_(3)-overset(underset(|)(CH_(3)))(CH)-CH_(2)=CH_(2) " " D : CH_(3)-overset(underset(|)(CH_(3)))underset(overset(|)(CH_(3)))(C)-CH_(3)`
A and D are position isomers.