Me - C - A C - C - Me Cs - liq- NH 3- EtOH B Compound B is-A-B- M tC- All of theseD-

The correct option is D Birch reduction converts (C≡ C) to (C=C) in anti addition of H_2. So, the stereochemistry around both the double bonds should be trans, ...

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Standard XII

Chemistry

Reaction with Metals

Me - C ≡ A C ...

Question

$M e - C \equiv C - \equiv C - M e (A) C s, l i q . N H_{3} - ------ \to E t O H (B)$

Compound (B) is :

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All of these

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Solution

The correct option is C
Birch reduction converts $(C \equiv C)$ to $(C = C)$ in anti-addition of $H_{2}$ . So, the stereochemistry around both the double bonds should be trans, so the answer is (c).

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Me−C≡C−≡C−Me(A)Cs, liq. NH3−−−−−−−→EtOH(B) Compound (B) is :

The correct option is C Birch reduction converts (C≡C) to (C=C) in anti-addition of H2. So, the stereochemistry around both the double bonds should be trans, so the answer is (c).

$M e - C \equiv C - \equiv C - M e (A) C s, l i q . N H_{3} - ------ \to E t O H (B)$

Compound (B) is :

The correct option is C
Birch reduction converts $(C \equiv C)$ to $(C = C)$ in anti-addition of $H_{2}$ . So, the stereochemistry around both the double bonds should be trans, so the answer is (c).