Compound H is optically active and has the molecular formula C6H10 and has a five carbon ring. On catalytic hydrogenation, H is converted to I (C6H12) and I is optically inactive. Propose structures for H and I. (Draw a three-dimensional formula for each using dashes and wedges around chiral centers.)

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gracemartin8203 gracemartin8203

10-06-2020
Chemistry

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Compound H is optically active and has the molecular formula C6H10 and has a five carbon ring. On catalytic hydrogenation, H is converted to I (C6H12) and I is optically inactive. Propose structures for H and I. (Draw a three-dimensional formula for each using dashes and wedges around chiral centers.)

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ajeigbeibraheem ajeigbeibraheem · Answer 1 · 2020-06-14T04:39:36+02:00

Answer:

Explanation:

Given that ;

Compound H is optically active and have a molecular formula of C6H10 and therefore undergo catalytic hydrogenation. Catalytic hydrogenation involves the use Platinum/Nickel to produce C6H12

i.e

[tex]C_6H_{10} +H_2 \to ^{Pt/Ni} \ \ \ C_6H_{12}[/tex]

The proposed H and I structures are shown in the diagrams attached below .

compound H represents 3- methyl cyclopentene

compound I represents methyl cyclopentane

However; 3- methyl cyclopentene posses just only one chiral carbon which is optically active at the third position and it R and S enantiomers are shown in the second diagram below.

pstnonsonjoku pstnonsonjoku · Answer 2 · 2021-11-21T07:24:28+01:00

The starting material is 3-methylcyclopentene while the optically inactive product is 1-methyl cyclopentane.

Hydrogenation refers to the addition of hydrogen across the double bond of an unsaturated compound. Hydrogenation results in the formation of a saturated compound having two more hydrogen atoms than the starting material.

The starting material is optically active 3-methylcyclopentene. The R and S enantiomers of the starting material is shown in image (1) attached. The optically inactive product is, 1-methyl cyclopentane is shown in image (2) attached.

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