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Higher Alkynes from Alkynes

Accomplish th...

Question

Accomplish the following conversion. Assume general laboratory reagents are available at your disposal; Multiple steps may be needed:

B r_{2} / C H_{2} C l_{2}

; 2.

E t O^{-} / E t O H

B_{2} H_{6}

, followed by

H_{2} O_{2}, O H^{-}

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B r_{2} / C H_{2} C l_{2}

; 2.

E t O^{-} / E t O H

S i a_{2} B H

, followed by

H_{2} O_{2}, O H^{-}

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B r_{2} / C H_{2} C l_{2}

; 2.

E t O^{-} / E t O H

N a N H_{2} / N H_{3}

, followed by

C H_{3} I

L i, N H_{3}

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B r_{2} / C H_{2} C l_{2}

; 2.

E t O H

N a N H_{2} / N H_{3}

, followed by

C H_{3} I

L i, N H_{3}

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Solution

The correct option is C 1. $B r_{2} / C H_{2} C l_{2}$ ; 2. $E t O^{-} / E t O H$ 3. $N a N H_{2} / N H_{3}$ , followed by $C H_{3} I$ 4. $L i, N H_{3}$
The product has one extra carbon. So somehow, a methylation needs to happen:

After the methylation, dissolved metal reduction yields the required product:

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Similar questions

H_{2} O

E t O H

M e C O O^{⊝}

⊝ O H

E t O^{⊝}

C H_{3} C H_{2} I E t O^{-} - -- \to E 2 C H_{2} = C H_{2} + E t O H + I

D_{3} C - C H_{2} I E t O^{-} - -- \to E 2 D_{2} C = C H_{2} + E t O D + I

M e_{3} C - I E t O^{-} - ------ \to S N^{1} a n d E 1 M e_{3} C - O E t + (M e)_{2} C = C H_{2}

(C D_{3})_{3} C - I E t O^{-} - ------ \to S N^{1} a n d E 1 (C D_{3})_{3} C - O E t + (C D_{3})_{2} C = C D_{2}

B r_{2}

C H_{2} C l_{2}

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