Compounds A and B (both C10H14) show prominent peaks in their mass spectrum at m/z 134 and 119. Compound B also shows a less prominent peak at m/z 91. On vigorous oxidation with chromic acid, compound A is nonreactive while compound B yielded terephthalic acid. From this information, deduce the structures of both compounds, and then draw the structure of B.
You do not have to consider stereochemistry
You do not have to explicitly draw H atoms

1-methyl-4-propylbenzene

Explanation:

In this case, we have several clues that can help.

1) We have the production of terephthalic acid for compound B

2) The mass spectrometry info.

With the first clue, we can deduce that the initial molecule has a benzene ring and that in this ring we have alky groups bonded in para position. Additionally, we have to add 4 more carbons if we want to obtain the mass of the compound (134).

So, the question is ¿How we have to add these carbons? To answer this we have to check the mass spectrometry info. In this case, we have a mass in 119, therefore we have a loss of 15 m/z, this indicates that we have a methyl group. Also, we have a mass on 91, this can be explain as a loss of a propyl group. If we take into account that we have to have these groups on para position (due to the first clue) we will have a 1-methyl-4-propylbenzene. (See figure 1)

For molecule A we only know that we dont have any benzene ring because it doesn't react with the chromic acid. So, probably we will have a linear structure on which we have several double or triples bonds. Or we can have several rings that fit with the molecular mass. We will need more information for structure A.

I hope it helps!

i believe the answer is three water molecules