Organic Chem mechanism

You could just post it up here (you can attach files to posts) so you can get a few peoples opinions :smile:

Reply 3

OP

Original post by jam12

Yeah sure. You should end up with Aspirin btw.

Original post by EierVonSatan

You could just post it up here (you can attach files to posts) so you can get a few peoples opinions :smile:

Thanks

Reply 4

Original post by _scott

Thanks

I'm afraid you got a couple of errors in there. Just use the phosphoric acid as an acid i.e. a H⁺ source that can protonate the anhydride. Then use the OH from the salicylic acid to attck it :smile:

Reply 5

jam12

1

You can't simply take OH's off a molecule and plonk them on another. Drawing OH groups fully e.g. O-H will help you understand that only H's are involved, hence these steps are called protonation/deprotonation.

Btw, remember mechanistic arrows can only come from a bond or lone pair - remember to do draw them on! :tongue:

(edited 13 years ago)

Reply 6

OP

Original post by EierVonSatan

I'm afraid you got a couple of errors in there. Just use the phosphoric acid as an acid i.e. a H⁺ source that can protonate the anhydride. Then use the OH from the salicylic acid to attck it :smile:

Original post by jam12

You can't simply take OH's off a molecule and plonk them on another. Drawing OH groups fully e.g. O-H will help you understand that only H's are involved, hence these steps are called protonation/deprotonation.

Btw, remember mechanistic arrows can only come from a bond or lone pair - remember to do draw them on! :tongue:

Ah yeah, I feel silly now :P obviously wasn't thinking properly when I said the acid loses the OH- !
So during the protonation of the anhydride, would the C=O be attacked or the C-O-C ?

Reply 7

Kyri

15

Original post by _scott

Ah yeah, I feel silly now :P obviously wasn't thinking properly when I said the acid loses the OH- !
So during the protonation of the anhydride, would the C=O be attacked or the C-O-C ?

Protonate a C=O oxygen and then have the OH lone pair from salicylic acid attack the other C=O carbon. This way you can draw arrows to spit out a molecule of acetic acid. I don't think you could do this by protonating the C-O-C oxygen. Well, you could but you'd end up having to transfer the proton onto the C=O oxygen at some point anyway.

Reply 8

OP

Original post by Kyri

Protonate a C=O oxygen and then have the OH lone pair from salicylic acid attack the other C=O carbon. This way you can draw arrows to spit out a molecule of acetic acid. I don't think you could do this by protonating the C-O-C oxygen. Well, you could but you'd end up having to transfer the proton onto the C=O oxygen at some point anyway.

Okay, thanks!
Would this be ok? step 2 mech.png

Reply 9

Hmm no. In the first step use the lone pair on the oxygen from the C=O bond to form C=O⁺-H then use the R-OH (on the salicylic acid's aromatic ring) to attack the carbon in the C=O⁺-H with a lone pair from the oxygen.

This might be simpler for you if you draw the mechanism without the acid catalyst?

Reply 10

jam12

1

Original post by Kyri

Protonate a C=O oxygen and then have the OH lone pair from salicylic acid attack the other C=O carbon. This way you can draw arrows to spit out a molecule of acetic acid. I don't think you could do this by protonating the C-O-C oxygen. Well, you could but you'd end up having to transfer the proton onto the C=O oxygen at some point anyway.

The OH attacks the protonated carbonyl group, not the other one.

@OP, draw without the full acid catalyst structure like EierVorSatan has suggested. The catalyst is only there to protonate one of the C=O groups (to C=O⁺-H) making it more electrophilic.

Reply 11

OP

Original post by EierVonSatan

...

Original post by jam12

...

Hi, sorry I haven't been on the computer for a few days. Sorry to bother you again, but would this be okay?

step 2 mechanism.png

Reply 12

This is quite tricky, so I think I'll just save some time and post it up - if you have any questions please ask :smile:

(edited 13 years ago)

untitled.bmp465.7KB

Reply 13

OP

Original post by EierVonSatan

This is quite tricky, so I think I'll just save some time and post it up - if you have any questions please ask :smile:

Thanks for your help!
Since it's quite tricky, do you think I should just leave this out my investigation report? It's for Advanced Higher chemistry

Reply 14