CH_3-C equiv C-O-CH_3 xrightarrow{dil. H_2SO_ | vedclass

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

Question

(A) The reaction of an alkyne with water in the presence of $dil. H_2SO_4$ and $HgSO_4$ is a hydration reaction. For the given substrate $CH_3-C \equi

Vedclass · Accepted Answer

Methyl propionate $(CH_3CH_2COOCH_3)$

Vedclass · Answer

(A) The reaction of an alkyne with water in the presence of $dil. H_2SO_4$ and $HgSO_4$ is a hydration reaction. For the given substrate $CH_3-C \equiv C-O-CH_3$,the hydration follows Markovnikov's rule. The $OH^-$ group attacks the more substituted carbon (or the carbon attached to the electron-donating group) to form an enol intermediate. However,in the presence of an alkoxy group $(-OCH_3)$,the reaction proceeds to form an ester. The hydration of the alkyne $CH_3-C \equiv C-O-CH_3$ leads to the formation of methyl propionate $(CH_3CH_2COOCH_3)$ as the major product.

$CH_3-C \equiv C-O-CH_3 \xrightarrow{dil. H_2SO_4, HgSO_4} \text{major product}$

Similar Questions

The product $(B)$ of the reaction is:
$CH(CO_2H)=CH(CO_2H)$ $\xrightarrow[2 \ mole]{NaOH} (A)$ $\xrightarrow{\text{electrolysis}} (B)$

$CH_3-CH_2-CH_2-C\equiv CH + LiNH_2 \to (A) \xrightarrow{(CH_3)_2SO_4} (B)$
Give the structural formula of compound $(B)$.

Which of the following will be the final product when $C_2H_2$ reacts with $HCl$?

One mole of $1,2$-dibromopropane on treatment with $X$ moles of $NaNH_2$ followed by treatment with ethyl bromide gave a pentyne. The value of $X$ is

Give the main product of the following reactions:
$(i)$ $1,1,2,2-$tetrabromopropane + zinc powder in methanol.
$(ii)$ Reaction occurring on passing hydrogen gas through carbon electrodes at high temperature in an electric arc.

Vedclass Test Series

Exam Paper Generator

Online Exam Module