Hydrocarbon and stereoisomerism?

Question

A hydrocarbon C14H12 contains 93.33% carbon and 6.67% hydrogen by mass.
The hydrocarbon has 2 structural isomers, W and X that contains 2 benzene rings respectively. W exhibits geometrical isomerism but X does not. In their reaction with HBr, W forms Y that is optically active, while X forms Z that is optically non-active. Determine the structure of W, X, Y, and Z.

Can you guys please teach me how to elaborate on ways to determine W, X, Y, and Z?

Answer 1

If you have two benzene rings, thats 2 x C6H5 = C12H10. That leaves only >C=C< + 2H to make C14H12. I write >C=C<, because one isomer shows further geometrical isomerism. So X is (C6H5)2C=CH2. X + HBr ===> (C6H5)2CBr*CH3, which is optically inactive, because two C6H5- groups are the same. Then W is C6H5CH=CHC6H5, cis and trans. Because HBr addition is trans, cis-W gives a threo-bromide, which can be resolved into two enantiomers, while trans-W gives an erythro-bromide, which is resolvable into two more enantiomers.

Answer 2

I think it looks like this

In the first structural isomer, you have 2 carbons connected with a double bond as the backbone. Each carbon has a benzene ring atttached. So the geometrical isomer is cis and trans around the double bond.

The second isomer also has 2 carbons attached by a double bond. But both benzene rings are on the same carbon, so no geometrical isomer.

When you react HBr, a Br will add to the double bond in both cases, creating chiral (Optical) isomer in the first one only. You must have 4 different substitutes on the carbon to create a chiral isomer.

Answer 3

It has a naphthalene skeleton. The substituent is butene.