Question

the product $ Z $ of the following reaction is:

 $ A){H_3}CC{H_2}CHB{r_2} $
 $ B){H_3}CCHBrC{H_2}Br $
 $ C){H_3}CCB{r_2}C{H_3} $
 $ D)BrC{H_2}C{H_2}Br $

Answer 1

Hint: This is an electrophilic substitution reaction. In this alkenes react with hydrogen bromide in the cold. The double bond breaks and a hydrogen atom ends up attached to one of the carbons and a bromine atom to the other. Symmetrical alkenes are dealt with first. These are alkenes where identical groups are attached to each end of the carbon-carbon double bond.

Complete answer:
Hydrogen bromide adds to alkenes to create alkyl halides. In this the reaction is that the pi bond of the alkene acts as a weak nucleophile and reacts with the electrophilic proton of $ HBr $ . In the first step of the reaction as the protonation of the pi bond. A carbocation intermediate is formed along with the bromide anion during the initial step of the reaction.
In unsymmetrical alkenes, the more stable of the two possible carbocations will form predominantly. Carbocations are stabilized by either the presence of alkyl groups, adjacent pi bonds, or atoms with a lone pair attached to the cationic carbon atom.
In this reaction, the propyne reacts with hydrogen bromide and forms $ 2,2 - $ dibromopropane.

So, the correct answer is $ (C){H_3}CCB{r_2}C{H_3} $ .

Note:
Hydrogen bromide is highly corrosive, anhydrous appears as a colorless gas with a pungent irritating odor. Heavier than air. Prolonged exposure to fire or intense heat may result in the violent rupture and rocketing of the container and also irritating to inhalation severe irritant to the eyes, skin, and nasal passages high concentrations may penetrate to the lungs resulting in edema and hemorrhage