Draw the product formed when the compound shown below undergoes a reaction with hbr in ch2cl2. interactive 3d display mode

Answer : The complete question is attached in the answer.

Explanation : In the given question an alkene is treated with alkylhalide in presence of methylene dichloride it gives a corresponding alkylbromide species.

The mechanism follows markonikov's addition rule, and the major product thus formed is bromine which is attached to least substituted carbon atom ad minor is bromine attached to highly substituted carbon atom.

Products formed follow markovnikov's rules: 2.2 dimethyl 3 bromo butane

Alkenes are alkane-derived compounds

Alkene is an unsaturated hydrocarbon compound that has a double bond

Alkene general formula: CnH2n

The nomenclature of the alkenes is generally the same as the alkanes

Alkenes including hydrocarbons are more reactive compared to alkanes because of the double bonds they have

One of the reactions that occur in alkenes is the addition reaction, which is generally the removal of the double bond due to the addition of another compound

In the addition reaction of HX gas (X = Cl, Br, or I) which is a halogen gas, Markovnikov rules apply,

1. If a carbon bonded double carbon atom binds to a different number of hydrogen atoms, then the halogen atom (X) will be bound to carbon atoms that hold fewer hydrogen atoms

2. If the number of carbon atoms in the double bond binds the same number of hydrogen atoms, then atom X will be bound to atom C which has the longest carbon chain

In the addition reaction of HBr to 3,3-dimethylbutene-1 in a solution of CH₂Cl₂ (CH₂Cl₂ does not react, is solvent) then the alkene compound 3,3-dimethylbutena-1 which has a double bond is found in atoms C no 1 and 2

In C-1 atoms, carbon atoms bind to 2 H atoms

In C-2 atoms, carbon atoms bind to 1 H atom

Then the Br Atom on HBr will bind to the C-2 atom that hold fewer hydrogen atoms, while the H atom on the HBr will bind to the C-2 atom

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