Suppose the following reaction sequence: CH e | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) $1$. $CH \equiv CH + NaNH_2 \rightarrow CH \equiv C^- Na^+ (A)$.$2$. $CH \equiv C^- Na^+ + 2 CH_3 - I \rightarrow CH_3 - C \equiv C - CH_3 (B)$ (B

Vedclass · Accepted Answer

Geometrical isomer

Vedclass · Answer

(A) $1$. $CH \equiv CH + NaNH_2 ightarrow CH \equiv C^- Na^+ (A)$.$2$. $CH \equiv C^- Na^+ + 2 CH_3 - I ightarrow CH_3 - C \equiv C - CH_3 (B)$ (But$-2-$yne).$3$. $CH_3 - C \equiv C - CH_3 + H_2 / Pd - BaSO_4 ightarrow 	ext{cis-but-2-ene} (C)$.$4$. The reaction $C \xrightarrow{H_2 / Pd - BaSO_4} \dots$ is redundant or implies further reduction,but the final step $NaNH_2 / 	ext{liq. } NH_3$ is typically used for dehydrohalogenation. Assuming the sequence leads to different isomers of butene or related structures,$C$ (cis-but$-2-$ene) and $D$ (trans-but$-2-$ene) are geometrical isomers.

Similar Questions

Explain the addition of $H_2O$ to an alkene.

What will be the product $B$ formed in the following reaction sequence?
$CH_3CH_2CH_2OH$ $\xrightarrow{PCl_5} A$ $\xrightarrow{alc. KOH} B$

On reaction with hot conc. $H_2SO_4$,which one of the following compounds loses a molecule of water?

In methyl alcohol solution,bromine reacts with ethylene to yield $BrCH_2CH_2OCH_3$ in addition to $1, 2-dibromoethane$ because:

When a tertiary alcohol $A$ $(C_4H_{10}O)$ reacts with $20\% \ H_3PO_4$ at $358 \ K$,it gives a compound $B$ $(C_4H_8)$ as a major product. The $IUPAC$ name of the compound $B$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module