Question

Compound A, C7H13Br, is a tertiary alkyl bromide. On treatment with CH3CH2ONa, A is converted into B, C7H12. Ozonolysis of B with O3 gives C as the only product. What are the structures of compounds A and B

Answer 1

Answer:

Compound A:    1-bromo-1-methylcyclohexane

Compound B:     1-methylcyclohex-1-ene

Explanation:

In this question, we can start with the I.D.H (hydrogen deficiency index):

$I.D.H~=~\frac{(2C)+2+(N)-(H)-(X)}{2}$

In the formula we have 7 carbons, 13 hydrogens, and 1 Br, so:

$I.D.H~=~\frac{(2*7)+2+(0)-(13)-(1)}{2}=1$

We have an I.D.H value of one. This indicates that we can have a cyclic structure or a double bond.

We have to keep in mind that the Br atom must be bonded to a tertiary carbon. We can not have a double bond because in the ozonolysis reaction we have only 1 product, therefore, we can not have a double bond in the initial molecule (if we have a double bond in the initial molecule we will have more than 1 product in the ozonolysis reaction).

With this in mind, we will have a cyclic structure. If we have 7 carbons and we need a tertiary alkyl halide. We can have a cyclic structure of 6 members and a methyl group bonded to a carbon that also is bonded to a Br atom (1-bromo-1-methylcyclohexane).

In the reaction with $CH_3CH_2ONa$ we will have an elimination reaction. In other words, we have the production of a double bond inside of the cyclic structure (1-methylcyclohex-1-ene).

See figure 1 for further explanations.

I hope it helps!