[rozarria]

I am having trouble coming up with the mechanism for deacetylation of an amide to an amine. The image is below. Is anyone familiar with this? The reagents are given and so I'm constrained to this particular way of reducing the amide to the amine.

http://img238.imageshack.us/img238/7080/deacetylation.png




2.) Lastly, I have a question about the correct reagents to attach an amide to a phenol. The image is below:

http://img136.imageshack.us/img136/3241/acetylation2.png



Which of the following would I use as reagents to achieve this?
A) NH3 (1 eq.), heat; H2, Ni, anhydride, H2O
B) HNO3, H2SO4, Zn(Hg), HCl; acetyl chloride, H2O
C) Acetyl chloride, AlCl3; HNO3, H2SO4; Zn(Hg), HCl
D) HNO3, H2SO4; anhydride, H2O; H2, Ni

I think I've ruled out B and D ans nitration would result in the attachment of the strongest deactivating substituent. I am leaning toward A, since I'd add an amino substituent, followed by attachment of the anhydride, the protonation of which (using H2O) would cleave the anhydride at the central oxygen, leaving me with the product in the image above.

Help appreciated.

[macman104]

For problem 1, you need to look up hydrolysis of amides:

http://www.mhhe.com/physsci/chemistry/carey5e/Ch20/ch20-3-4-1.html

For 2, what was have you learned to put an NH2 group onto a ring?  It's not a one step process, you need to add something else, and then change that group to an NH2.

[rozarria]

Ah, yes, I have to nitrate first, and then use Clemmesen Reduction to reduce the nitrate to an amino substituent. Thanks for reminding me. I'm looking over the link, thanks.

[408]

While B is the most correct answer given, I would hassle whomever gave you that problem as getting phenol to stop at mononitro with sulfuric/nitric mix is pretty much impossible.  Either oxidation or higher levels of nitration will occur.  mononitration of phenols can usually be done with diluted nitric acid.

Extraneous for this assignment, yes, but something like this helps your professor get to know you or at least remember you so it can never hurt

[vRDang]

I have just finished question 1 and it seems that finishing step 5 ("Use the electrons of an adjacent oxygen to help "push out" the leaving group, a neutral ammonia molecule.") from the link it gives me the desired product.  So my question is, what does NaHCO3 + H2O do?  I must have did the reaction wrong, right? 

[Squirmy]

The reaction doesn't know you want it to stop at step 5

What do you get if it keeps going? How will that react with NaHCO3?

[vRDang]

I don't get it

[vRDang]

After step 5, does the NH2 get re-protonated to NH3? And the HCO3- takes away that proton and you get the product?

[vRDang]

nvm i got it--the thing that falls off reprotonates NH2 and the NaKO3 takes off that proton to give product.

[isup]

Hi. Actually I am working on a similar problem and I don't understand what happens after "step 5" After the NH2 comes off of the ring, does it then pick up an H off of the OH and get re-protonated to NH3? And then NaHCO3 removes that proton and then the NH2 reattaches to the ring? Why does that happen rather then just NH2 remaining on the ring and the rest getting pushed off. Also, why doesn't the lone pair from the N grab that H from the OH early on? THanks soo much for your help, its really appreciated. I'm really trying to understand this.

[isup]

Nevermind, I figured it out