Ok this is second year stuff anyways
I have a nitrogen double bonded to a carbon ring and attached to a hydroxide group and I am to treat it with an acid.
The product is a secondary amine where twO substituents are part of a ring. The ring has a -OH group on a carbon next to the nitrogen. From memory.
Can someone give me a hint for the mechanism?
The hydroxyl group can accept a proton thus making it a good leaving group.
When this leaves I get a positively charged nitrogen. Is this allowed?
And will this allow for a c-c bond to break and form a c-n bond?
And if water leaves how do I get a hydroxyl group on the carbon?
Any help welcome,
A positively charged nitrogen on the amine group is not as likely as a positively charged carbon on the ring. I believe its easier for a carbon to give up an electron than it is for a nitrogen, especially since the carbon is in a ring. This will leave the carbon two away from the amine group attachment site positive. The O in water can then attach itself to the carbon, leaving a proton as the leaving group.
Not totally sure since I took ochem a long time ago, but I believe the summary above is correct.
It does seem unlikely to have a positive nitrogen atom. Especially since secondary carbocations are relatively stable.
Can I ask why the positive charge would be on that particular carbon? I don't see it being any more nucleophilic than any other carbon.
The two immediately adjacent carbons to the C=N bond can be ruled out due for spatial reasons. Its harder to cram more atoms onto adjacent carbons in a carbon ring.
That leaves three carbons left on the ring. All three should have the same partial charge since the double bonds of the ring will try to compensate for the lack of an electron by forming partial bonds. This suggests that all three are equally possible. Since there are two carbons that can are two away from any attachment point compared to one carbon three away, it is the more likely nucleophillic attack target.
My description above assumes that you are not concerned about the chirality of the final product.
Thanks, that should be enough information to solve the problem.
Oh It was a C=N bonded to a ring of C-C bonds
Knowing my lecturer, chirality is probably important. But I should be ableto deal with those issues.
I know that C-O-C in a ring prefers to be in a 5 member ring, does nitrogen prefer to be in a 6 member ring?
Oh, I misunderstood your question. C=N is a good leaving group as is OH in acid. You're likely to have just benzene, hydrogen cyanide (HCN) and water.
I'm not sure about your nitrogen ring question. All I know is that benzene will definitely be more stable than a nitrogen and five carbons in a six member ring.
Ahhhh. There is no benzene in this question. However I made my explanation rather vague.
However I think I have a plausible mechanism now.
And your explanatiom or other principles were very useful. I aways forget my good leaving groups.
As for my question. My tutor walked us through a ring closing question where H-Br formed a bromonium ion and a ring formed in a 5member (4 C, 1 O) as opposed to the expected 6 member markovnikov product for the reason that oxygen likes to be in 5 membered rings.
|All times are GMT -7. The time now is 08:06 AM.|
Powered by vBulletin® Version 3.8.2
Copyright ©2000 - 2013, Jelsoft Enterprises Ltd.