The reaction CH_2 = CH - CH_3 + HBr rightarro | vedclass

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

Question

(B) In this reaction,$HBr$ undergoes heterolytic fission to produce an electrophile $H^{+}$:$HBr \rightarrow H^{+} + Br^{-}$The electrophile $H^{+}$ a

Vedclass · Accepted Answer

Electrophilic addition

Vedclass · Answer

(B) In this reaction,$HBr$ undergoes heterolytic fission to produce an electrophile $H^{+}$:$HBr \rightarrow H^{+} + Br^{-}$The electrophile $H^{+}$ attacks the double bond of propene to form a stable carbocation intermediate $(CH_3-\stackrel{+}{C}H - CH_3)$.Subsequently,the nucleophile $Br^{-}$ attacks the carbocation to form $2$-bromopropane.Since the initial step involves the attack of an electrophile on the alkene,this is an electrophilic addition reaction.

The reaction $CH_2 = CH - CH_3 + HBr \rightarrow CH_3CHBr - CH_3$ is

Similar Questions

Ethylene reacts with Baeyer's reagent to give

$But-1-ene \xrightarrow[(ii) NaBH_4]{(i) (CH_3COO)_2Hg/H_2O} ?$
The product in the above reaction is:

In the addition of $HOBr$ to $(I)$ $Cl_3C-CH=CH_2$ and $(II)$ $CH_3-CH=CH_2$:

Number of shared electrons between carbon-carbon atoms in an ethylene molecule is

The $IUPAC$ names of some alkenes are given below. Determine the correct stability order: $A$. $2$-Methylbut-$2$-ene,$B$. $cis$-But-$2$-ene,$C$. $2,3$-Dimethylbut-$2$-ene,$D$. Prop-$1$-ene.

Vedclass Test Series

Exam Paper Generator

Online Exam Module