Compound A has the formula C9H10. It reacts rapidly with acidic KMnO4 but reacts with only 1 equivalent of H2 over a palladium catalyst to produce compound B, C9H12. On hydrogenation under conditions that reduce aromatic rings, A reacts with 4 equivalents of H2, and hydrocarbon C, C9H18, is produced. The reaction of A with KMnO4 gives CH3COOH and a carboxylic acid D, C7H6O2. Draw the structure of compund D.

In the picture you have the reactions involved, and the products formed. Now, in order to give you an idea of which compound can be, we have two hints here.

First, we can calculate the number of insaturations in the original compound A

this value (i will call it "n") is calculated with the following expression:

n = C+1 - 1/2(H + N - X)

Where X is halogens.

Now, if we calculate the number of insaturations for compound A we have:

n = 9+1 - 1/2(10) = 5

We have 5 insaturations in this compound.

Reading the exercise we can see that you reduce compound A, but it says aromatic rings. This is the second hint, so, if A is an aromatic ring, we have 4 insaturations there, one of the ring, and the other 3 from double bonds. Which would be the remaining insaturation? there has to be a double bond outside the aromatic ring.

How can we know this? we know that C9H10 reacts in acidic conditions, with KMnO4, This is oxydation of alkenes, and that gives as products, carboxilic acid always, and depending on the location of the double bond, you can generate a ketone and acid, two acids, or an acid and CO2.

According to the formula of D, we have two oxygens there, so it can't be CO2 + acid, and a Ketone only has one oxygen, os it has to be two acids.

See the picture below.