Propyne on reaction with water in the presenc | vedclass

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

Question

(C) Step $1$: The reaction of an alkyne with water in the presence of dilute $H_2SO_4$ and $HgSO_4$ is known as Kucherov's reaction.Step $2$: Propyne

Vedclass · Accepted Answer

Propanone

Vedclass · Answer

(C) Step $1$: The reaction of an alkyne with water in the presence of dilute $H_2SO_4$ and $HgSO_4$ is known as Kucherov's reaction.Step $2$: Propyne $(CH_3-C \equiv CH)$ undergoes hydration following Markovnikov's rule.Step $3$: The addition of $H_2O$ across the triple bond forms an unstable enol intermediate: $CH_3-C(OH)=CH_2$.Step $4$: This enol undergoes tautomerization to form a stable ketone, which is Propanone $(CH_3-CO-CH_3)$.

Propyne on reaction with water in the presence of $40\% \ \text{H}_2\text{SO}_4$ and $1\% \ \text{HgSO}_4$ forms:

Similar Questions

When acetylene reacts with $HCl$ in the presence of $HgCl_2$,the product formed is.........

$CH_3-C \equiv C-CH_3 \, \xrightarrow[{(2) Br_2}]{{(1) H_2 /Pd/CaCO_3}} \, (X) \, , \, (X)$ is

Identify the alkyne in the following sequence of reactions:
Alkyne $\xrightarrow{\text{H}_2, \text{Lindlar's catalyst}} A$ $\xrightarrow{\text{Ozonolysis}} B \text{ (only)}$
Given that $B$ is obtained from $\text{CH}_2=\text{CH}_2$ via the Wacker process.

Which of the following reagents will be able to distinguish between $but-1-yne$ and $but-2-yne$?

The end product of the following sequence is
$CaC_2$ $\xrightarrow{H_2O} B$ $\xrightarrow[HgSO_4]{H_2O, H_2SO_4} C$

Vedclass Test Series

Exam Paper Generator

Online Exam Module