In the given reactions identify A and B . $\mathrm{CH}_{3}-\mathrm{C} \equiv \mathrm{C}-\mathrm{CH}_{3}+\mathrm{H}_{2} \xrightarrow{\mathrm{Na} / \text { Liquid } \mathrm{NH}_{3}}$ "B"

(A) A : 2-Pentyne B : trans - $2-$ butene

(B) A : n - Pentane B : trans - $2-$ butene

(C) A : 2 - Pentyne B : Cis - 2 - butene

(D) A : n - Pentane B : Cis - 2 - butene

JEE Main · 2024 · 01 Feb 2024 · Shift 2 · Hydrocarbons

In the given reactions identify A and B .

CH_{3} - C \equiv C - CH_{3} + H_{2} Na / Liquid NH_{3}

"B"

(A)A : 2-Pentyne
B : trans - $2 -$ butene✓ Correct
(B)A : n - Pentane
B : trans - $2 -$ butene
(C)A : 2 - Pentyne
B : Cis - 2 - butene
(D)A : n - Pentane
B : Cis - 2 - butene

Answer: (A)

Solution

2-Pentyne,

C H_{3} - C \equiv C - C_{2} H_{5}

, hydrogenated over Pd/C undergoes syn addition of hydrogen across the triple bond, placing both alkyl groups on the same side of the resulting double bond and giving cis-2-pentene; the starting alkyne here is A, 2-pentyne.

2-Butyne,

C H_{3} - C \equiv C - C H_{3}

, reduced with sodium in liquid ammonia goes through a radical-anion mechanism that adds hydrogen anti across the triple bond, placing the two methyl groups on opposite sides of the double bond and giving trans-2-butene as B.

JEE Main · 2024 · 01 Feb 2024 · Shift 2 · Single correct option

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JEE Main · 2024 · 01 Feb 2024 · Shift 2 · Hydrocarbons

Solution

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