An organic compound $X (C_{5} H_{8} O_{2})$ liberated $C O_{2}$ on treatment with $N a H C O_{3}$ . X exist in two stereoisomeric forms but none of them is optically active. X on selective olefinic hydrogenation yields $Y (C_{5} H_{10} O_{2})$ which can be resolved into enantiomers. The correct deduction is /are :

X

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

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Y

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

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X

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

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Y

2 - m e t h y l

b u t a n o i c a c i d .

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Solution

The correct option is A $X$ is $C H_{3} - C H = C (C H_{3}) C O O H$
The organic compound (X) is 2-methyl but-2-enoic acid. $C H_{3} - C H = C (C H_{3}) C O O H$ .

It reacts with $N a H C O_{3}$ to liberate $C O_{2}$ gas.

$C H_{3} - C H = C (C H_{3}) C O O H + N a H C O_{3} \to C H_{3} - C H = C (C H_{3}) C O O N a + H_{2} O + C O_{2} ↑$

X exists in two stereoisomeric forms (geometrical isomers Z and E) but none of them is optically active.
X on selective olefinic hydrogenation yields Y $(C_{5} H_{10} O_{2})$ which can be resolved into enantiomers. The chiral C atom is marked with asterik in the product below.

$C H_{3} - C H = C (C H_{3}) C O O H selective olefinic hydrogenation - ------------------------- \to C H_{3} - C H_{2} C^{*} H (C H_{3}) C O O H$

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Similar questions

C H_{3} - C H = C = C H - C H = C = C = C H - C H = C = C H - C H_{3}

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

C H_{3} - C H = C = C H - C H_{3}

⎛ ⎜ ⎜ ⎝ C H_{3} - C H | C l - C H = C H - C H | C l - C H_{3} ⎞ ⎟ ⎟ ⎠

(X)

(X) \Rightarrow P h - C H = C H - C | H C H_{3} - C H = C H - P h

C H_{3} - C H_{2} - C H = C H - C H_{3} + H C l \to X + Y

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