Compound a and compound b are constitutional isomers with molecular formula c3h7cl. when compound a is treated with sodium methoxide, a substitution reaction predominates. when compound b is treated with sodium methoxide, an elimination rection predominates. propose structures a and b.

Compound A is the primary alkyl halide, and B the secondary alkyl halide, which are represented below.

Explanation:

Constitutional isomers are compounds with the same molecular formula but with the difference in their flat structure. For C₃H₇Cl there are two constitutional compounds: a primary alkyl halide, and a secondary alkyl halide. The structures are represented below.

Sodium methodixe is both a stronger nucleophile and a strong base. When acts as a nucleophile, a substitution must occur, and for that, a stronger nucleophile (a compound with a negative partial charge that needs to bond with one with positive partial charge) must leave the molecule. So, chlorine must leave the molecule in substitution. For that, in the primary alkyl halide, it will be easy, because there's no steric hindrance.

When sodium methoxide acts as a base, an elimination occurs, for that, a strong electrophile (a compound with a positive partial charge that needs electrons) must leave the compound, and then a double bond will be formed. Hydrogen is a strong electrophile, so it must leave in the secondary alkyl halide because chlorine will be a steric hindrance to leave in a substitution reaction.

Then, compound A is the primary alkyl halide, and B the secondary alkyl halide, which are represented below.

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