Question

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.)

Answer 1

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

$C_6H_{10} +H_2 \to ^{Pt/Ni} \ \ \ C_6H_{12}$

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.