Chemical Forums
Chemistry Forums for Students => Organic Chemistry Forum => Topic started by: Twickel on April 04, 2012, 09:46:13 AM
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Hi all.
Am having trouble with figuring out the Sn2 mechanism with acetonirtrile as a nucleophile.
Its stucture is CH3Cn, but the carbon bonded to the nitrogen has four bonds ( 3 to N 1 to the adjacent C), so how does it react as the nucleophile. Note it is also the solvent.
Thank you
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Hi all.
Am having trouble with figuring out the Sn2 mechanism with acetonirtrile as a nucleophile.
Its stucture is CH3Cn, but the carbon bonded to the nitrogen has four bonds ( 3 to N 1 to the adjacent C), so how does it react as the nucleophile. Note it is also the solvent.
Thank you
Then CH3 hydrogen atoms can be abstracted by a strong base, e.g. LDA to give the anion which is stabilized through the nitrile group. This anion can act as a carbon nucleophile and displace bromides, tosylates etc. in the usual fashion.
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Hmm, I would have eventually figured that out. But my question has CH3CN as the reagent, it does not mention any base.
What do I do in a situation like this?
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Did you draw a structure of acetonitrile? Think about the Lewis octet rule.
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Hi all.
Am having trouble with figuring out the Sn2 mechanism with acetonirtrile as a nucleophile.
Its stucture is CH3Cn, but the carbon bonded to the nitrogen has four bonds ( 3 to N 1 to the adjacent C), so how does it react as the nucleophile. Note it is also the solvent.
Thank you
How about telling us the reaction. Disco made a guess and you said, "Wrong". Just tell us.
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Ok, sorry for any trouble.
I have attached a document with the reactions I am stuck with.
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Ok, sorry for any trouble.
I have attached a document with the reactions I am stuck with.
In the first case the SH will probably attach the bromine to give the spire compound with inversion.
In the second case the I of NaI will displace the tosylate with inversion to give the corresponding iodide.
I would say that CH3CN plays no part in the reaction except to act as a solvent.
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Ok, so why does the SH react. I am confused. I have never seen an Sn2 reaction where the nuc is already attached to the substrate.
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Ok, so why does the SH react. I am confused. I have never seen an Sn2 reaction where the nuc is already attached to the substrate.
S is a particularly good nucleophile.
Such Sn2 reactions are well documented and work the same as any other.
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Hypothetically, then Oxygen groups with oxygen having a singal bond or amines can act as nucleophiles aswell.
So what about, when the solvent is ethanol, what I get confused about is the basicity of the ethanol. What reacts is acutally Et-O- which is a strong base,so why does it proceed as SN2 instead of E2?
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Hypothetically, then Oxygen groups with oxygen having a singal bond or amines can act as nucleophiles aswell.
So what about, when the solvent is ethanol, what I get confused about is the basicity of the ethanol. What reacts is acutally Et-O- which is a strong base,so why does it proceed as SN2 instead of E2?
To generate ethoxide you usually require a base, e.g. NaH.
Ethanol itself rarely reacts.
Depending on the leaving group and other factors ethoxide can cause elimination or substitution.
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(slight tangent) When trimethylsilylchloride (TMSCl) and NaI are used to deprotect esters dissolved in acetonitrile, it may be that the acetonitrile temporarily picks up the TMS group, with production of NaCl (Olah, et al., J. Org Chem. 44, 1979, 1247). So, I have to wonder whether acetonitrile can sometimes be a nucleophilic catalyst when it is the solvent. However, in the present case, I see no reason to believe that acetonitrile plays an analogous role.
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Can someone please do the second example with the tosylate group.
I can not do it, I am cunfused as to what the Tos actaully looks like, how does it leave?
Also for leaving groups, we want it to be as least basic as possible right? The more basic it is the worst the leaving group is.
Thats why we like I over say Fl?
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Is this it?
The carbon of the alkyl chain is bonded to an oxygen which is bonded to Ts. The oxygen receives two electrons ( bond breaks). Then the I simply takes over its spot
To draw the inversion so I make the Me group be going into the plane of the page.
I remember my lecturer saying for inversion of stereochemistry, the group replacing the leaving group must not have a higher prioirty?
So confused with that
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Can someone please do the second example with the tosylate group.
I can not do it, I am cunfused as to what the Tos actaully looks like, how does it leave?
Also for leaving groups, we want it to be as least basic as possible right? The more basic it is the worst the leaving group is.
Thats why we like I over say Fl?
Here is a drawing.
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Thank you. So the TsO is actually bonded to the chain like so?
The benzene ring is attached to the carbon, the S is bonded to the ring, and the 3 oxygens are bonded to the sulfur. How does it leave?
If I draw a curly arrow going to the ring it has a carbon bonded to 5 things. Sorry if I am not making any sense
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I drew a structure, is that what TsO looks like?
Ignore the H bonded to the O, pretends its just an O
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Thank you. So the TsO is actually bonded to the chain like so?
The benzene ring is attached to the carbon, the S is bonded to the ring, and the 3 oxygens are bonded to the sulfur. How does it leave?
If I draw a curly arrow going to the ring it has a carbon bonded to 5 things. Sorry if I am not making any sense
Here is another picture
This is what tosyl looks like
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Ok, so now get it, weird I clicked Tos in my chem sketch and it added it like so. Simply forms an anion.
There is no TsO option.
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One more questions.
This is a radical reaction.
So I did the reaction and mechanisms and in the end, I simply replaced the Br with a hydrogen.
How do they get the rings as the products?
I was thinking anti markovnikov addition, but it doesnt add to the double bond, the radical reaction is on the other side.
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One more questions.
This is a radical reaction.
So I did the reaction and mechanisms and in the end, I simply replaced the Br with a hydrogen.
How do they get the rings as the products?
I was thinking anti markovnikov addition, but it doesnt add to the double bond, the radical reaction is on the other side.
Sorry I don't follow, what is a radical reaction?
Your document shows the same two reactions
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Hi
Sorry fixed it up. Now attached.
I am talking about a radical chain reaction. With initiation and propagation.
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Hi
Sorry fixed it up. Now attached.
I am talking about a radical chain reaction. With initiation and propagation.
It may be better to start a new thread so it is differentiated from the previous one.
I'm sure one of the admins can arrange this
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Will now make a new thread. Thanks for helping me out.