Question

Suggest structure for C4H10O : 3.92(singlet, 1H), 3.38(doublet,2H), 1.76(multiplet, 1H), 0.90(doublet,6H)

Answer 1

First calculate Double Bond Equivakents (DBE)

DBE = C-(H/2)+1

C- no. Of carbons

H- no. Of hydrogens

DBE = 4-(10/2)+1 =0

DBE = 0, means in given compound has no any double bonds.

At 0.90 (doublet,6H) corresponds to two methyl group and two methyl groups are giving doublets means, these two methyl groups are attached one carbon, which has only one hydrogen.

3.38(doublet,2H)

At 3.38 range only those hydrogens come, next to that carbon oxygen is present. It is doublet these hydrogens are attached one carbon, which has only one hydrogen.

3.92(singlet, 1H), at this range only OH proton will come.

So structure is