CH_3-C equiv CH xrightarrow[dil. H_2SO_4]{HgS | vedclass

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

Question

(B) In reaction $1$,the hydration of propyne with $HgSO_4/dil. H_2SO_4$ follows Markovnikov's rule to form acetone $(CH_3-CO-CH_3)$. In reaction $2$,h

Vedclass · Accepted Answer

$NaOI$

Vedclass · Answer

(B) In reaction $1$,the hydration of propyne with $HgSO_4/dil. H_2SO_4$ follows Markovnikov's rule to form acetone $(CH_3-CO-CH_3)$. In reaction $2$,hydroboration-oxidation follows anti-Markovnikov's rule to form propanal $(CH_3-CH_2-CHO)$. Acetone,being a methyl ketone,gives a positive iodoform test with $NaOI$ (forming a yellow precipitate of $CHI_3$),while propanal does not. Therefore,they can be differentiated using $NaOI$.

$CH_3-C \equiv CH \xrightarrow[dil. H_2SO_4]{HgSO_4} (A)$
$CH_3-C \equiv CH \xrightarrow[(1) BH_3.THF]{(2) H_2O_2/HO^{-}} (B)$
Products $(A)$ and $(B)$ can be differentiated by:

Similar Questions

Ethene can be separated from acetylene by passing the mixture through :

What is $A$?

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.

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

Which of the following has acidic hydrogen?

Vedclass Test Series

Exam Paper Generator

Online Exam Module

$CH_3-C \equiv CH \xrightarrow[dil. H_2SO_4]{HgSO_4} (A)$ $CH_3-C \equiv CH \xrightarrow[(1) BH_3.THF]{(2) H_2O_2/HO^{-}} (B)$ Products $(A)$ and $(B)$ can be differentiated by: