this is an ozonolysis problem and addition of hydrogen?

Question

Analysis of compound "X" (molar mass 72) containing C, H, and O gave and analysis of 66.7% C and 11.1% H. the compund neither reacted with pheny hydrozine nor gave a positive iodoform test (that's the HALOFORM test). reaction with OZONE and subsequent hydolysis gave METHANAL as on of the products. A mole of "X" absorbed a mole of hydrogen when catalytically hydrogenated to give a product "Y" which could be oxidzed by KMNO4 (pottasium permanganate) to a product "Z" (molar mass 72) which did not react with Fehling's solution. Identify the compounds X, Y and Z.

so far i have found the molecular formula of the compound to be C8H8O. and the degree of unsaturation to be 1. i wanted to really know what is the structure of this compound.

i thought it would be .......CH2=CHCH2CH3
.....................................................|.....................................................
....................................................OH.............................................

An OH group is attached to the second carbon

oops my bad. i meant i got C4H8O.

Answer 1

First you need to know the molecular formula.
Divide each element's percentage by it's atomic weight (e.g. 66.7/12.011 = 5.5). Divide each of this numbers by the smallest (which turns out to be oxygen), to get the formula of C4H8O. This corresponds to one degree of saturation (fully saturated is C4H10O).

Iodoform test: X is not a methyl ketone
Phenylhydrazine: X is not any other ketone or aldehyde.
Ozone: The molecule ends with a --CH=CH2.
Hydrogenation: This, along with the ozonolysis reaction, confirms the claim above that there is one degree of unsaturation.
KMnO4: Oxidation indicates conversion of an alcohol into a ketone or alcohol.
Fehling's solution: Reacts only with aldehydes. You oxidized a secondary alcohol into a ketone.

Summary:
C4H10
one double bond, one secondary OH group.

Compound X is H2C=CH-CH(OH)-CH3, or 3-buten-2-ol.
Compound Y is 2-butanol
Compound Z is 2-butanone

Answer 2

First of all you need to find the empirical formula:
You have
Element .. % .. .. .. %/AW .. Divide by smaller
O .. .. .. .. 22.2 .. .. 1.3875 .. .. .. 1
C.. .. .. .. .66.7 .. .. 5.5583 .. .. .. 4
H.. .. .. .. 11.1 .. .. .. 11.1 .. .. .. . 8

So the empirical formula is C4H8O
The molecular weight is 72=(16+4*`2+8)n => n=1.
Thus the molecular formula is the same as the empirical
From the molecular formula we have a degree of unsaturation =1 (since 2 H are needed to make it saturated)

You know that X can be reduced with hydrogen. That means that the degree of unsaturation corresponds to a double bond and not to a ring structure.
Since it gave methanal upon ozonolysis it should have the group
CH2=
It doesn’t react with phenyl hydrazine so there is no carbonyl group. (Thus you can’t have CH2=CH(OH)CH2CH3 as you say since that would be in equilibrium with its tautomer, 2-butanone)
It doesn’t give the haloform reaction, so it can’t be
an a-methyl ketone. It also can't be CH2=CH-CH(OH)-CH3, since this alcohol can give the haloform reaction.
When you do the reduction with H2, the double bond becomes single. Thus the MW of Y is 74 (2 H atoms are added). Then your compound is oxidized by KMnO4 and loses 2 H (the MW becomes again 72) which means that you should have the O atom in a –OH group and not as an ether.
Compound Z doesn’t give a reaction fehling’s solution so it is not an aldehyde, but a ketone. this means that -OH cannot be attached to a primary C atom.

Taking all these into account, there is no solution to your problem. The closest solution would be to assume that the alcohol
CH2=CH-CH(OH)-CH3 cannot give the haloform test, and then all the rest data fall in place with Y being
CH3-CH2-CH(OH)-CH3 and Z CH3-CH2-CO-CH3