pretty advanced (i think) organic chemistry question...?

Question

I'm shown a molecule, which I have drawn and uploaded here:

http://ul.hamzahkhan.com/files/3/Lignin.bmp

I am asked what the product's structure will be if I react it with
(i) Aqueous NaOH
(ii) Hot, acidified aqueous potassium dichromate(vi)
(iii) Aqueous bromine (Br2 (aq))

Is anybody able to help?

Answer 1

i). NaOH ; No reaction

ii) The hydroxy functional group( at the end of the saturated chain) will be oxidised to the corresponding carboxylic acid.
3-[4-hydroxyphenyl]prop-2-enoic Acid

iii). The alkene functional group in the chain will be brominated. A bromine atom will attach to each of the Carbon atoms on either end of the alkene.
3-[4-hydroxyphenyl]-2,3-dibromopropan-1-ol

Answer 2

Don't be too hard on yourself if you get a low grade on an exam. Organic Chemistry is one of THE most difficult courses you can take in school. We did a little bit in my chemistry for emergency responders class and it near blew up my mind some of the stuff he was going over. Keep at it my friend

Answer 3

i) I think you'll get the disodium salt. the phenolic and the aliphatic OH-groups will be deprotonatet -> R-O^- Na^+

ii) oxidation to the aldehyde

iii) bromination of the double bond

Edit: ii) Lenpol7 is right. The aldehyde is only the intermediate