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melamori03 [73]
3 years ago
7

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.)

Chemistry
1 answer:
Ksivusya [100]3 years ago
8 0

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.

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Answer:

(a) pH = 12.73

(b) pH = 10.52

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The net balanced reaction equation is:

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This amount of HBr will neutralize an equivalent amount of KOH (0.900 mmol), leaving the following amount of KOH:

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pOH = -log[OH⁻] = -log(0.0003344) = 3.476

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(c) The amount of HBr added in 38.00 mL of 0.1000 M HBr is:

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The pH of the solution can then be calculated:

pH = -log[H⁺] = -log(0.01176) = 1.93

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