Question

The major product formed is:
\[{\text{C}}{{\text{H}}_{\text{3}}}{\text{ - CH = C}}{{\text{H}}_{\text{2}}}{\text{ + HBr}} \to {\text{?}}\]
A.\[{\text{C}}{{\text{H}}_{\text{3}}}{\text{ - C}}{{\text{H}}_{\text{2}}}{\text{ - C}}{{\text{H}}_{\text{2}}}{\text{ - Br}}\]
B.\[{\text{C}}{{\text{H}}_{\text{3}}}{\text{ - CH}}\left( {{\text{Br}}} \right){\text{ - C}}{{\text{H}}_{\text{3}}}\]
C.\[{\text{C}}{{\text{H}}_{\text{2}}}{\text{Br - CH = C}}{{\text{H}}_{\text{2}}}\]
D.\[{\text{C}}{{\text{H}}_{\text{2}}}{\text{ = C = C}}{{\text{H}}_{\text{2}}}\]

Answer 1

Hint: We also remember that in organic chemistry, hydrocarbons are an important topic. The hydrocarbons are majorly classified as three groups. There are alkane, alkene and alkyne. The alkane means carbon-carbon single bond. The alkene has a carbon-carbon double bond. The alkyne means carbon-carbon having triple bonds in the molecule.

Complete answer:
We also need to know that the Markovnikoff’s rule is one of the important rule for hydrogen bromide addition. In this law some rules for addition of hydrogen and halogen in alkene.
According to this halogen that means most electronegative atom in the reagent going to addition in the carbon having least hydrogen atom in alkene or alkyne molecule. The hydrogen in the reagent is going to attack in the carbon having more number of hydrogen atoms in the alkene or alkyne.
According to the above discussion the hydrogen bromide addition in propene is given below,
\[{\text{C}}{{\text{H}}_{\text{3}}}{\text{ - CH = C}}{{\text{H}}_{\text{2}}}{\text{ + HBr}} \to {\text{C}}{{\text{H}}_{\text{3}}}{\text{ - CH}}\left( {{\text{Br}}} \right){\text{ - C}}{{\text{H}}_{\text{3}}}\]
According to Markovnikoff’s rule, hydrogen in hydrogen bromide to add in carbon atoms having more number of hydrogen accounts in alkene. Bromide is going to add in carbon having less number of hydrogen accounts in alkene.
Hydrogen bromide addition in propene to give 2-Bromopropane.

Hence, option B is the correct answer.

Note:
We must know the conversion of one type of hydrocarbon to another hydrocarbon by oxidation and reduction. The oxidation of alkane to give alkene. The oxidation of alkene to give alkyne. The reduction of alkyne to give alkene. The reduction of alkene to give alkane. It has some general formulas. The general formula of alkane is \[{{\text{C}}_{\text{n}}}{{\text{H}}_{{\text{2n + 2}}}}\]. The general formula of alkene is \[{{\text{C}}_{\text{n}}}{{\text{H}}_{{\text{2n}}}}\]. The general formula of alkyne is \[{{\text{C}}_{\text{n}}}{{\text{H}}_{{\text{2n - 2}}}}\].