Question

ОН (1) ethylene, heat (2) О, (3) Me,S (4) NaBH, (excess) (а) (5) Но* ОН Problem 13.49 Working backwards, deduce the starting material that led to the indicated product through the specified reactions.

Answer 1

a)

Excess NaBHA will reduce both aldehyde and ketone to sodium salt of alcohols. The treatment of aqueous acid will give the correspondig alcohols. So the structure may be OH ONa H OH ONa H30+ NaBHA Oz/MeS is ozonolysis reagent that breaks olefinic bond to aldehyde or ketone here we have mayo O3, Mens the previous step is Diels Alder reaction H ethylene, heat

buta-1,3-dien-2-ylbenzene is the starting material

b)

last step is olefin reduction OH OH Hz, Pd/C 3rd step is aldehyde oxydation OH H Jones oxydation 2nd step is Diels Alder H H heat 1st oxydation of alcohol OH Н. Н. Swern oxidation so the starting material is 1-propenol

c)

last step is methylation Meo. НО. Meo но" Nah, Mel 2nd step is epoxide ring breaking но. но" KOH, H20 1st step is epoxidation mCPBA so the starting material is cyclohexene