alkylation of 3-OH vs 4-OH benzaldehydes?

[hypo]

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

[Rhodium]

As far as I know, aldehyde allylation require a catalyst like Sm/In/Mg etc...

[GC_MS]

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  

[hypo]

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:

watchamacallit ("O=C1C=CC(=COCCC)C=C1")


couch terrorist