[newbie!]

anyone help with these questions?


my answer is in the red box for the 1st, and i started the 2nd (red arrows).
for the 1st i think that would be the answer because the ring already has NO2 on it which is meta directing deactivator.  as for a mechanism though i have no idea hence i think my logic might be wrong.

for 2 pretty much no idea. the arrrows are what i instinctivly think, but i'm pretty sure i read in my notes that the MeO group - cant just attack the delta + carbon like that. i'm assuming there would be 2 lone paiirs on the OEt oxygen though... ?

[rucik5]

Hi.
i) I must say I'm not sure about the mech here. Will think about it.
ii) Ethers are electron releasing group, so you wouldn't really expect to stabilize the charge on that oxygen. However, you've got three nitro- groups in ortho- and para- positions which set up the ring perfectly for nucleophilic aromatic substitution via the addition-elimination.

You must bare in mind that the meta-directing and deactivating effect of electron withdrawing groups (ester,ketone,nitro etc.) occurs whilst considering electrophilic aromatic substitution (i.e. Friedel-Crafts), whereas in nucleophillic AS electron withdrawing groups in o- and p- positions help it.

Normally, you would have a halogen as the leaving group and O, C, N nucleophiles. Since OMe and OEt are similar I imagine, you'd need your methoxide in excess.
Hope it helps.

[Sepelio]

Try drawing a resonance structure for 1) to generate N+. This should create an electron poor position on the phenyl for the -OH to attack. Then you can probably push the electrons back down again to push out a hydride. Thats just off the top of my head, though.

[orgopete]

i) has an error.