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 reaction predominates. Required:
Propose structures A and B.

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 reaction predominates.

Explanation:

Constitutional isomers are the one which differs in the structural formula.

When compound A is treated with sodium methoxide, a substitution reaction predominates.

That means sodium methoxide is a strong base and a strong nucleophile.

But when it reacts with primary alkyl halides it forms a substitution product and when it reacts with secondary alkyl halide it forms mostly elimination product.

The reaction and the structures of A and B are shown below:

k

explanation:

k

given:

mass of na = 115 g

excess cl2

to determine:

mass of nacl produced

explanation:

given reaction is-

2na(s) + cl2(g) → 2nacl(s)

since cl2 is in excess, na will be the limiting reagent

as per the reaction stoichiometry na: nacl = 1: 1

i.e. moles of na reacted = moles of nacl formed

now, # moles of na = mass of na/atomic mass

                                  = 115 g/23 g.mol-1 = 5 moles

therefore, moles of nacl = 5

molar mass of nacl = 58 g/mol

mass of nacl = 5 moles * 58 g.mol-1 = 290 g

ans: amount of nacl produced = 290 g