Ok, now with the question, here we go.
R is an ether (R-O-R).
The OCH2CH2CH3 substituent is electron donating by both
resonance and induction, hence is ortho and para directing.
arigatoast suggested
:
>>> 2nd step : For both NO2 to get reduce, u need 2
mols of H2 . so to get T, u need only 1 mol of H2 for 1 mol
of S. i think need to state the mol ratio. Thats what i think.
<<<
That's smart thinking; although in practice, instead of
uniformly obtaining only one product
of 5-nitro-2-propoxyaniline, you would end up with a mixture
of several different products (eg. in some molecules, both nitro
groups would be reduced; in some other molecules, neither of the
nitro groups would be reduced; in yet other molecules, the para
instead of ortho nitro group would be reduced) in other words, a
horribly low yield).
Which is why, while in theory it seems very easy : just add
aqueous HNO3, then reduce one of the nitro groups to amine group
(by using Sn and conc HCl).
But in practice, here are two serious problems :
1) When using the aqueous HNO3, how can the chemist ensure only
di-nitration occurs, instead of mono or tri nitration? (the propoxy
group directs the incoming nitro electrophile to the ortho and para
positions).
2) When carrying out reduction, how can the chemist ensure only
one of the two nitro groups (and furthemore, to be precise, the
nitro group ortho to the ethoxy group) gets reduced to
amine?
I would think the yield would be too low for such a synthesis
pathway to be viable.
However, since this compound is a commercially synthesized
artificial sweetener (albeit a toxic one), there should be a better
synthesis strategy (which will require a lot more than 2 steps) for
this compound, and hence, it is possible that this 2008
H2 SEAB-Cambridge exam question is not a fair one (ie. it
probably isn't a fair question if the only 2-step synthesis
possible, gives too low a yield to be viable).
Ah well.
