Propyne reacts with excess HBr to form the co | vedclass

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

Question

(D) The reaction of propyne $(CH_3-C \equiv CH)$ with excess $HBr$ follows Markovnikov's rule.In the first step,$HBr$ adds across the triple bond to f

Vedclass · Accepted Answer

$CH_3CBr_2CH_3$

Vedclass · Answer

(D) The reaction of propyne $(CH_3-C \equiv CH)$ with excess $HBr$ follows Markovnikov's rule.In the first step,$HBr$ adds across the triple bond to form $2-$bromopropene $(CH_3-C(Br)=CH_2)$.In the second step,another molecule of $HBr$ adds to the alkene following Markovnikov's rule,where the hydrogen atom attaches to the carbon with more hydrogens and the bromine atom attaches to the more substituted carbon.This results in the formation of $2,2-$dibromopropane $(CH_3-CBr_2-CH_3)$.

Propyne reacts with excess $HBr$ to form the compound $Z$. The structure of $Z$ is:

Similar Questions

The reagent needed for converting $Ph-C \equiv C-Ph$ to trans-stilbene is:

$C_6H_5 - C \equiv C - CH_3 \xrightarrow{HgSO_4, H_2SO_4} A$

$(A)$ $\xrightarrow[dil. H_2SO_4]{HgSO_4} (B)$ $\xrightarrow{LiAlH_4} \underset{\text{racemic mixture}}{(C)}$
$\therefore$ reactant $(A)$ is

$Mg_2C_3 + H_2O \xrightarrow{X}$ (organic compound).
Compound $X$ is

Ozonolysis of acetylene $(HC \equiv CH)$ gives:

Vedclass Test Series

Exam Paper Generator

Online Exam Module