hi,
if we look at the phenolates of 3-OH and 4-OH benzaldehydes, we see
that in the 4-OH case, the phenolate is in resonance with the formyl
group, whereas in the 3-OH case it isn't (if this is not clear, i
can make a quick sketch of what i mean).
1) does that mean that in the 4-OH case the negative charge is more
dislocated and thus the alkylation is _slower_ (maybe not noticeable)
than in the 3-OH case?
2) does that mean that in the 4-OH case, the formyl group can also
be alkylated? giving a strange compound (which i can't name)?
3) now suppose we have a MeO or other activating group in the 5 position.
this would speed up the 4-OH alkylation (two negative charges on the ring),
but would still leave the possibility of the formyl alkylation?
i hope i ain't annoying people with my questions! :-[
couch terrorist
As far as I know, aldehyde allylation require a catalyst like Sm/In/Mg etc...
Alkenylation of the formyl is done a la Perkin.
anisaldehyde + EtCOONa -> H2O + MeO-C6H4-CH=CMeCOONa
MeO-C6H4-CH=CMeCOOH - CO2 = anethole
Old trick from the perfume industry ;) . No Sm/In etc necessary, though Perkin forms propenyl compounds, and not allyl.
SWiM doesn't have the necessary articles handy to help you much further hypo. Till now, he mainly need DEmethylation reviews ;)
Doped(TM) since 19.... euhm... a long time :)
thanks.
rhodium: i see how normal aldehydes are pretty unreactive, but
shouldn't benzaldehyde and especially benzaldehydes with an -O
-in 4 position be pretty reactive?
GC_MS: alk(en)ylation on the formyl group is _not_ what i want. rather my
concern was if, when alkylating a 4-formyl-phenolate, etherification of the
formyl group would be a side reaction. if not, then why not?
this is the compound i'm talking about in the case of alkylation with
propyliodide:
Molecule: (https://www.the-hive.ws/forum/faq.pl?Cat=#applet)
watchamacallit ("O=C1C=CC(=COCCC)C=C1")
couch terrorist