An optically active carboxylic acid, X, has the molecular formula $C_{6} H_{10} O_{2}$ and can be shown by chemical and physical tests to contain the group $- C H = C H_{2}$ and at least one $- C H_{3}$ group. When X is reduced with hydrogen over platinum the produce formed is optically inactive. Which of the following best represents the structure of X?

C H_{2} = C H - C H (C H_{3}) - C H_{2} - C O O H

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C H_{2} = C H - C H (C H_{2} C H_{3}) - C O O H

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C H_{2} = C H - C H_{2} - C H (C H_{3}) - C O O H

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C H_{2} = C H - C (C H_{3})_{2} C O O H

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Solution

The correct option is B $C H_{2} = C H - C H (C H_{2} C H_{3}) - C O O H$
X will be $C H_{2} = C H - C H (C H_{2} C H_{3}) - C O O H$ .
As on reaction with $H_{2} / P t$ , it will form
$C H_{2} = C H - C H (C H_{2} C H_{3}) - C O O H H_{2} / P t - --- \to C H_{3} - C H_{2} - C H (C H_{2} C H_{3}) - C O O H$ , which is optically inactive compound due to absence of chiral carbon.
In other cases it will form optically active compound.

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Similar questions

(A) C H_{3} - C H_{2} - C | C O O H H - - C H - C H_{3}

(B) C H_{3} - C | C O O H H - C H_{2} - - C H - C H_{3}

(C) C | C O O H H_{2} - C H_{2} - C H_{2} - - C H - C H_{3}

C H_{3} - C O O H

C H_{2} = C H - C O O H

C_{6} H_{5} - C O O H

C_{6} H_{5} - C H_{2} - C O O H

C H_{3} - C H_{2} - C O O H

C H_{3} - C H | O H - C O O H

C H_{3} - C H_{2} - C H | C O O H - C H_{2} - C l | C H - C H_{2} - C H_{3}

C H_{3} - C H = C H - C H_{2} - O H | C H - C H_{3} ⟶ C H_{3} - C H = C H - C H_{2} - C O O H

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