Which reaction among the following will yield | vedclass

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

Question

(A) The reaction of propyne $(CH_3-C \equiv CH)$ with two equivalents of hydrogen bromide $(HBr)$ follows Markovnikov's rule. In the first step,$HBr$

Vedclass · Accepted Answer

$CH_3-C \equiv CH + 2 HBr \longrightarrow$

Vedclass · Answer

(A) The reaction of propyne $(CH_3-C \equiv CH)$ with two equivalents of hydrogen bromide $(HBr)$ follows Markovnikov's rule. In the first step,$HBr$ adds to the triple bond to form $2-$bromopropene $(CH_3-C(Br)=CH_2)$. In the second step,another molecule of $HBr$ adds to the double bond,again following Markovnikov's rule,to form $2,2-$dibromopropane $(CH_3-C(Br)_2-CH_3)$. Therefore,the reaction $CH_3-C \equiv CH + 2 HBr \longrightarrow CH_3-C(Br)_2-CH_3$ is the correct one.

Which reaction among the following will yield $2,2-$dibromopropane?

Similar Questions

Identify $Z$ in the following sequence:
$C_3H_7OH$ $\xrightarrow{H_2SO_4} X$ $\xrightarrow{Br_2} Y$ $\xrightarrow{KOH} Z$

$CH_3-C \equiv C-CH_3$ $\xrightarrow[(1) \ O_3]{(2) \ H_2O} X$ $\xrightarrow[\Delta]{NaOH + CaO} Y$; $Y$ is

Which of the following is a homologue of ethyne?

The synthesis of $3-$octyne is achieved by adding a bromoalkane into a mixture of sodium amide and an alkyne. The bromoalkane and alkyne respectively are

$1, 2$-Dibromopropane on treatment with $X$ moles of $NaNH_2$ followed by reaction with ethyl bromide gives $2$-pentyne. The value of $X$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module