A hydrocarbon C5H8 absorbs 2 mol of H2 over Pt to give pentane (CH3CH2CH2CH2CH3). Which of the following could it be?

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A hydrocarbon $C X_{5} H X_{8}$ absorbs 2 mol of $H X_{2}$ over Pt to give pentane ( $CH X_{3} CH X_{2} CH X_{2} CH X_{2} CH X_{3}$ ). Which of the following could it be?

Apenta-1,3-diene
Bpent-2-yne
Ccyclopentene
D2-methylbuta-1,3-diene

Answer: A, B

Pentane is the unbranched saturated product, so the starting material must be an unbranched, acyclic C5 chain with two pi bonds. Penta-1,3-diene and pent-2-yne qualify. Cyclopentene gives cyclopentane after one

H X_{2}

, and 2-methylbuta-1,3-diene gives 2-methylbutane.

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A hydrocarbon CX5​HX8​ absorbs 2 mol of HX2​ over Pt to give pentane (CHX3​CHX2​CHX2​CHX2​CHX3​). Which of the following could it be?

Answer: A, B

More questions on Degree of unsaturation and spectroscopic clues

A hydrocarbon $C X_{5} H X_{8}$ absorbs 2 mol of $H X_{2}$ over Pt to give pentane ( $CH X_{3} CH X_{2} CH X_{2} CH X_{2} CH X_{3}$ ). Which of the following could it be?