How many isomers will form in the monochlorination reaction of 2,3−dimethylbutane under radical substitution conditions? what kind(s) of monochlorination products will form (choose all that apply)?
a. 1°
b. 2°
c. 3°
d. allylic

A and C

Explanation:

The monochlorination is a reaction in which we have to add only one Cl to the molecule. In this case we only have two types of carbons. Primary (the methyl groups) and the tertiary (the carbons in the middle).

Therefore we only have  2 type of isomers. The first one in which the Cl would bond to the primary carbons, the primary alkyl halide (1-chloro-2,3-dimethylbutane) and the second one in which the Cl would bond to the tertiary carbons, the tertiary alkyl halide (2-chloro-2,3 -dimethylbutane).

See the figure

hydrocarbons are organic compounds that contain only hydrogens and carbons. they can be saturated (those which contain only carbon-carbon single bonds) or unsaturated (those which contain carbon-carbon double or triple bonds) in nature. hydrocarbons can undergo several reactions like substitution, elimination etc.  

on substituting one or more hydrogen atoms in hydrocarbon it results in the formation of haloalkane. as the halogen atoms are large compared to the carbon and hydrogen atoms, due to which the molecular weight increases and the bond become polar because of presence electronegative halogen atom and results in the increase in boiling point of the haloalkane.

thus, the correct option is (b) that is the boiling point of the new compound increases on substituting a hydrogen atom with a halogen in a hydrocarbon.

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