Five isomeric alkenes A–E each undergo catalytic hydrogenation to give 2-methylpentane. The IR spectra of these five alkenes have the following key absorptions (in cm–1): Compound A: 912 (s), 994 (s), 1643 (s), 3077 (m) Compound B: 833 (s), 1667 (w), 3050 (weak shoulder on C–H absorption) Compound C: 714 (s), 1665 (w), 3010 (m) Compound D: 885 (s), 1650 (m), 3086 (m) Compound E: 967 (s), no absorption 1600–1700, 3040 (m) The alkene structures are given below. Identify each compound.

Solution and Explanation:

Medium peak at 3077cm−1 corresponds =C-H stretching.

Strong peaks at 912cm−1 and 994cm−1 corresponds to alkene =C-H bending. This tells the alkene is monosubstituted.

2. Weak peak at  3077cm−1 corresponds =C-H stretching.

The compound shows strong peak at 833cm−1 corresponds to alkene =C-H bending and weak peak at 1667cm−1 corresponds to alkene C=C stretching. This tells the alkene is trisubstituted.

3. Weak peak at 1665cm−1 corresponds to alkene C=C stretching.

Medium peak at 3086cm−1 corresponds =C-H stretching.

Strong peaks at 714cm−1 corresponds to alkene =C-H bending. This tells the alkene is cis-disubstituted.

4. Medium peak at 1643cm−1 corresponds to alkene C=C stretching.

Medium peak at 3010cm−1 corresponds =C-H stretching.

Strong peaks at 885cm−1 corresponds to alkene =C-H bending. This tells the alkene is gem-dialkylsubstituted.

5. Medium peak at 1650cm−1 corresponds to alkene C=C stretching.

Medium peak at 3040cm−1 corresponds =C-H stretching.

Strong peaks at 967cm−1 corresponds to alkene =C-H bending. This tells the alkene is trans-disubstituted.

a electronegativity decreases beacuse the size of nucleus is bigger.