[orgostudent]

this is way above me...any input would be greatly appreciated!

Compound H (C12 H16 O) gives I (C5 H9 BR) and J (C7 H8 O) upon treatment with 48% aqueous HBr. Compound I gives compound K (C5 H8) when I is treated wiht potassium tert-butoxide in tert-butanol. When K is treated with ozone, followed by zinc metal, the only product isolated is compound L (C5 H8 O2), which is a dialdehyde. Compound J gives a negative reaction to Jones reagent and gives a gas when treated with a sodium metal. Compound J is soluble in dilute aqueous KOH but insoluble in dilute aqueous sodium bicarbonate. The proton NMR of J shows the following: an aromatic "quartet" at 7.21 ppm, a singlet at 2.30 ppm, and a broad singlet (1 H; exchanges wiht D2 O) at 5.50 ppm. Provide structures for H, I, J, K, and L which are consistent with this data.

any help would really be great

[kryptoniitti]

I'd start working from the J part of the problem. Because the NMR gives you aromaticy, it narrows the problem significantly since J is only C7H8O. "broad singlet (1 H; exchanges wiht D2O)", ok here you have another nugget of info, exchanging protons, broad singlet at 5.50 ppm--> what functional group? With the knowledge of aromaticy and a functional group, you are able to draw the few isomers for C7H8O and you'll see why Jones reagent doesn't give any reaction. Fine structure, "an aromatic "quartet" at 7.21 ppm" should nail the right isomer...

After this start adding stuff to the probable connection/cleavage point until you get the original compound H (bearing in mind the reaction conditions at starting point and molecular formula of I). From your estimation what H is, you should just work it forward and tune the structure until everything you do to I will give the desired products. Hope this helps a bit.

[g-bones]

I'd start working from the J part of the problem. Because the NMR gives you aromaticy, it narrows the problem significantly since J is only C7H8O. "broad singlet (1 H; exchanges wiht D2O)", ok here you have another nugget of info, exchanging protons, broad singlet at 5.50 ppm--> what functional group? With the knowledge of aromaticy and a functional group, you are able to draw the few isomers for C7H8O and you'll see why Jones reagent doesn't give any reaction. Fine structure, "an aromatic "quartet" at 7.21 ppm" should nail the right isomer...

After this start adding stuff to the probable connection/cleavage point until you get the original compound H (bearing in mind the reaction conditions at starting point and molecular formula of I). From your estimation what H is, you should just work it forward and tune the structure until everything you do to I will give the desired products. Hope this helps a bit.

Just going off of what this kryponiitti said, I would say that J is toluene with a phenolic alcohol in the ortho position since you have a quartet and not a doublet of doublets.  having a methyl group one carbon over from the alcohol would make all of the protons unequivelent where as having them in the para position would make 2 protons equivelent to each other as well as the other two on the ring.  and since phenol cannot be further oxidized, jones wouldn't due anything. the broad singlet that kryptoniitti mentioned is because phenol is fairly acidic because the negative charge can be deloaclized into the ring and this can exchange with a dueterium. 

hope this helps

[venkatjohnny]

Hi,
your starting material H is a ether (cyclopentyltolyl ether)
J is p-hydroxy toluene
I is cyclopentyl bromide
K is cyclopentene
L is pentandial ( dialdehyde)
am i right??? if not post the answer, have a nice time

[sjb]

the broad singlet that kryptoniitti mentioned is because phenol is fairly acidic because the negative charge can be deloaclized into the ring and this can exchange with a dueterium. 

More just to do with the exchange ArOH <-> ArOD. To my knowledge D₂O does not exchange aromatic protons in benzene derivatives, and indeed the fact that the NMR still shows 4H in that region is proof to that effect.

J is p-hydroxy toluene

Without knowledge of the fine structure of the "aromatic quartet", it's difficult to say, if the question is as written (i.e. no actual spectra) then perhaps it's ambiguous at best. O-disubsitution, as in my experience would perhpas lead to a multiplet, but not really describable as a quartet. P-disubsitution would probably lead to two clear doublets (which is probably the closest visually to what's described), and m-disubsitution perhaps a singlet and three other multiplets. But in each case you could get coincident signals or at least partial overlapping, so all bets are off, as far as I'm concerned.