How do you synthesize 2-phenylpropanal from benzene?

Answer 1

I can't say for sure, this isn't one I've thought much about but lets look at it logically. To a stable molecule like a benzene or an aryl group, there isn't much you can do in the absene of a catalyst. YOu can halogenate, formylate, acylate plus numerous more complex stuff.

I think there would be a route via styrene oxide, at least there is in the literature. so for that youd acylate to acetophenone then reduce with borohydride then dehydrate to the alkene followed by performic acid epoxidation to styrene oxide. a grignard methyl magnesium bromide reagent will open the oxide to an alcohol in some percentage the direction you want and maybe a bit the opposite. Now you need to oxidize a terminal alcohol without over doing it and making the carboxylic acid and manganese dioxide is out because you need an adjacent double bond or benzene ring alfa to it to work Permanganate would be like taking a tank to a knife fight, dchromate would be ok in conjunction with pyridine, just get a text on oxidations or look at one online and you[ll find your answer. nitric acid is my go to choice for a terminal alchohol but it may be too much here, maybe very cold and slow addition of acid but it will be too much. good luck, that is a pretty significant sythesis problem for a student that is doing anything but advanced organic chemistry.

alternatives would be a bromination of benzene to bromobenzene followed by a grignard with bromoethene to get styrene.

start with benzaldehyde , do grignard with methylmagnesium bromide and dehydrate the alcohol

start with benzonitrile and grignard with methylbromide to acetophenone

reductions with borohydride can maybe use zinc instead or aluminum amalgam, certainly ataolysis with palladium

solvents used would be thf or ether for grignards, the acylation of benzene could likely self solvate, the reduction with borohydride in methanol or methanol/dcm or water will work too. the best system here would be probably

methanol/dcm, look at intermediates in sn1, sn2 reactions and you'll figure out why. remember you need an intermediate to form often when you want a reaction to go, if the intermediate is not compatible with the solvent it won't form.