An alcohol ( A ) with molecular formula `C_(4)H_(10)O` on oxidation forms compound ( B ) `C_(4)H_(8)O`.The compound ( B ) gives iodoform test, but does not reduce ammonical silver nitrate, ( B ) on treatment with amalgamated zinc and HCl gives compound ( C ) `C_(4)H_(10)`. Identify A,B,C.

Options

A. `CH_(3)CH_(2)CH_(2)CH_(2)OH,CH_(3)CH_(2)CH_(2)CHO, CH_(3)CH_(2)CH_(2)CH_(3)`
B. `CH_(3)CHOHCH_(2)CH_(3),CH_(3)COC_(2)H_(5)CH_(3)CH_(2)CH_(2)CH_(3)`
C. `(CH_(3))_(3)COH,CH_(3)COCH_(3),CH_(3)CH_(2)CH_(2)CH_(3)`
D. `CH_(3)CH_(2)CH_(2)CH_(2)OH, CH_(3)CH_(2)CH_(2)CHO,(CH_(3))_(2)CHCH_(3)`

Answer

B

Explanation

a. -ve Silver mirror test `implies` B is not an aldehyde.
b. B undergoes Clemmensen's reaction `implies` B is a ketone. Therefore C is an alkane.
C. In oxidation,since number of C-atoms remains same. Therefore A is a `2^(0)` alcohol.
D. Since B gives iodoform test, it contains the group `:`
`CH_(3)-overset(O)overset(||)(C)-`
`:. B : CH_(3) -overset(O)overset(||)(C)-CH_(2)CH_(3)`
`:. A : CH_(3) -underset(OH)underset(|)(C)H-CH_(2)CH_(3)`
`:. C : CH_(3)CH_(2)CH_(2)CH_(3)`