Question

`CH_(3)-C-= CH underset("dil." H_(2)SO_(4))overset(HgSO_(4))(to) (A)`
`CH_(3) - C-= CHunderset((2) H_(2)O_(2)//HO^(-))overset((1) BH_(3).THF)(to) (B)`
Product (A) and (B) is differentiated by :
A. 2-4 DNP
B. NaOI
C. Na-metal
D. `NaHSO_(3)`

Answer 1

Correct Answer - B
(b) `CH_(3)-C-=CH underset(H_(2)O)overset(Hg^(2+)//H^(+))(to) CH_(3)-overset(OH)overset(|)(CH)=CH_(2)hArr underset("Iodoform test with" NaOI)underset(" A ketone"("gives + ve"))(CH_(3)-overset(O)overset(||)(C)-CH_(3))`
`CH_(3)-C-=CHunderset((2) H_(2)O_(2)//OH^(+))overset((1) BH_(3). THF)(to) CH_(3)- CH = CH_(2)-OH hArr underset((-ve "Iodoform test because" - underset(O)underset(||)(C)-CH_(3)"group is absent")) underset(("An aldehyde"))(CH_(3)-CH_(2)-CH=O`