The synthesis of $3 - o c t y n e$ is achieved by adding a bromoalkane into a mixture of sodium amide and an alkyne. The bromoalkane and alkyne respectively are:

B r C H_{2} C H_{2} C H_{2} C H_{3}

and

C H_{3} C H_{2} C \equiv C H

No worries! We‘ve got your back. Try BYJU‘S free classes today!

B r C H_{2} C H_{2} C H_{3}

and

C H_{3} C H_{2} C H_{2} C \equiv C H

No worries! We‘ve got your back. Try BYJU‘S free classes today!

B r C H_{2} C H_{2} C H_{2} C H_{2} C H_{3}

and

C H_{3} C \equiv C H

No worries! We‘ve got your back. Try BYJU‘S free classes today!

B r C H_{2} C H_{2} C H_{2} C H_{3}

and

C H_{3} C H_{2} C \equiv C H

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct option is D $B r C H_{2} C H_{2} C H_{2} C H_{3}$ and $C H_{3} C H_{2} C \equiv C H$
$C H_{3} - C H_{2} - C \equiv C H N a N H_{2} \to C H_{3} - C H_{2} - C \equiv C^{-} N a^{+} + C H_{3} - C H_{2} - C H_{2} - C H_{2}$
$B r + C H_{3} - C H_{2} - C \equiv C^{-} N a^{+} \to C H_{3} - C H_{2} - C H_{2} - C H_{2} - C \equiv C - C H_{2} - C H_{3}$

Suggest Corrections

Similar questions

C H_{3} C H_{2} C \equiv C H (i) N a N H_{2}, N H_{3} - --------- \to (i i) C H_{3} C H_{2} B r H_{2} - ------------ \to L i n d l a r^{'} s c a t a l y s t, P d C H_{2} I_{2} - --- \to Z n (C u)

B (m i x) c o n c . H I \leftarrow ---- - (C H_{3})_{3} C - O - C H_{3} a n h y d H I - ----- \to A (m i x)

C h_{3} C = C - C H_{3}

C H_{3} C H_{2} - C H_{2} - C H_{3}

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

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

$B (m i x) C o n c . H I ⟵ ((C H_{3})_{3} C - O - C H_{3}) a n h y d . H I \to A (m i x)$

C H_{3} C H_{2} C \equiv C H

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

C H_{3} C H_{2} C H_{2} C H_{3}

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

Join BYJU'S Learning Program