[L3ul]

A and B are two aliphatic and acyclic hydrocarbons, which are not isomers, have the same empirical formula, but belong to different classes of substances. If A has 10% H, and hydrocarbon C, obtained by partial hydrogenation of B, has three geometrical isomers and the same degree of unsaturation as A.

a) Find the structural formulas of A, B and C.
...
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I tried working out some relations so that I'll find their degree of unsaturation or something else to help me find what I'm dealing with.
The only thing that I managed to discover is that B has 3 more carbons than A.
Could you, please, guide me on the right track?

[mjc123]

How did you discover that B has 3 more carbons than A? Telling us might help us see whether you are on the right track.
Can you work out the empirical formula from the information that A has 10% H?
Do you know how to work out the degree of unsaturation from the molecular formula?

[L3ul]

I wrote the compounds in this manner:

A: C_xC_(y+2-z)

B: C_aC_(b-z)

C: C_aC_(b+2-z)

z/2 represents the degree of unsaturation of A and C.

Since we know that A and B have the same empirical formula, we find the empirical formula CH_1.33 and retrieve the following relations:
4x = 3(y+2-z)
4a = 3(b-z)

Which brings us to: 4x-4a-6 = 3y-3b

Knowing that A and C have the same degree of unsaturation, we find:
2x+2-y-2+z = 2a+2-b-2+z (=)
2x-y = 2a-b

We write the underline relations like this:
2x = 2a+y-b
4x = 3y+4a+6-3b

From which we get:
4a+2y-2b = 3y+4a+6-3b (=)
b = y+6 =>
C has 6 more hydrogen atoms than A
B has 4 more hydrogens than A

We have:
2x = 2a+y-b
b = y+6
=> 2x = 2a-6 (=) x = a-3 => A has 3 less carbon atoms than B

[mjc123]

OK, now think of what possible molecular formulas you could have (with whole numbers of C and H atoms) consistent with the empirical formula. What structures could they have? What structural units could be susceptible to partial hydrogenation? What are the possibilities for geometrical isomerism?

[L3ul]

We have the following possibilities:

I.

A: C₃H₄
1.

2.


B: C₆H₈
1.

2.

3.

4.

5.


C: C₆H₁₀
1.

2.

3.

4.

5.



For C, #3 is the single structure that has three geometrical isomers.
Therefore:
A can be #1 and #2 (which one is it? how do I find that?)
B is #3
C is #3

Here's the problem: I can go further and assume A is C₆H₈ and still find possible structures that follow the problem. How do I know which one is the right one?

[mjc123]

I think either 1 or 2 would do for A, but since the question comes from the chapter on alkynes, I think they are thinking of 1. (Have you done allenes yet? Are you supposed to know that 2 exists?)
Possibly there is not only one solution to the problem; if you can find more, credit to you. Can you draw the structures you're thinking of? (not all the possible ones, but those that satisfy the question.)

[L3ul]

Yup, alkenes and allenes were the last chapter I worked out.

Here's another solution for B and C:

B: C₉H₁₂


C: C₉H₁₄


Now that I think about it, it's pretty absurd to search for other solutions with a higher number of carbons since there will be a ton of possibilities for A, and there's no way you can choose which one is correct.

Also, there's a way you can find what A: by looking at the last sub-point of the problem. But I'm pretty unsure about using that as an argument.
May I ask you: when it comes to competitions, am I allowed to use an incoming sub-point as an argument?

[mjc123]

I'm afraid that doesn't quite work. C has 4 geometrical isomers - trans-trans, cis-cis, cis-trans and trans-cis. Because the molecule is not symmetrical, the last two are different compounds, not the same as with a symmetrical diene. I've actually found it very difficult to think of higher solutions that satisfy all the criteria, apart from your original one.