Two alkenes,
A and
B, have the same molecular formula C
5H
10, give the same alkane when treated with H
2/Pd and the same alcohol when treated with H
⊕/H
2O. However, they give different alcohols when treated with BH
3: THF followed by H
2O
2/
ΘOH. Treatment of alkene
A with O
3 followed by Zn/H
2O gave two carbonyl compounds
C (C
3H
6O) and
D (C
2H
4O) that showed the following NMR data.
C δ 2.07 singlet
D δ 9.79 quartet (1H) and δ 2.21 doublet
Give structures for A, B, C and D