To carry out the above conversion,(A) and (B) | vedclass

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

Question

(C) The reaction involves the alkylation of a terminal alkyne. Step $A$: The terminal alkyne is treated with a strong base like $NaNH_2$ to form an ac

Vedclass · Accepted Answer

$NaNH_2$,$I-CH_2-CH_2-CH_2-Br$

Vedclass · Answer

(C) The reaction involves the alkylation of a terminal alkyne. Step $A$: The terminal alkyne is treated with a strong base like $NaNH_2$ to form an acetylide ion. Step $B$: The acetylide ion acts as a nucleophile and undergoes an $S_N2$ reaction with an alkyl halide. To ensure the reaction occurs selectively at the $Br$ end,we use an alkyl halide with a better leaving group on the other side,such as $I-CH_2-CH_2-CH_2-Br$. Since $I^-$ is a better leaving group than $Br^-$,the nucleophilic attack occurs at the carbon attached to the iodine atom. Therefore,$(A) = NaNH_2$ and $(B) = I-CH_2-CH_2-CH_2-Br$.

To carry out the above conversion,$(A)$ and $(B)$ respectively are:

Similar Questions

The simplest alkyne is represented by which of the following?

The reaction of $1,1,2,2$-tetrabromoethane with alcoholic zinc gives $..........$.

Give chemical reactions for the following conversions:
$1$. $2-$bromobutane from $2-$butyne
$2$. $2-$propanol and $1-$propanol from propyne

The correct order of acidity of the following compounds is

$I. CH_3-C \equiv CH$

$II. F_3C-C \equiv CH$

$III. CH_3-C \equiv C-CH_3$

$IV. CH_3-CH=CH-CH_3$

The reaction of $2-$pentyne with dilute $H_2SO_4$ and $HgSO_4$ yields...............

Vedclass Test Series

Exam Paper Generator

Online Exam Module