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Offline noodlesoup

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SN2 reaction
« on: January 30, 2009, 07:22:46 PM »
Hi guys, nice to meeting you all here. I have a question about SN2 reaction that I want to ask you guys for insight and suggestion. I'm still wondering how the reaction proceeds. please see my attachment to the quesiton below. For question 1, is that how you guys think the reaction goes, i'm not so sure; and for question 2, i'm kind of lost, can you guys please walk me through it by giving your insight or suggestion. I really appreciate if you guys can help me out. Thank you in advance.

Offline macman104

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Re: SN2 reaction
« Reply #1 on: January 31, 2009, 03:03:05 AM »
It seems pretty reasonable, otherwise, you could also have the chloride attack the carbon and kick out the protonated oxygen in one concerted step.  Also, becareful with drawing the positive charge, because right now your picture technically shows a 4-carbon chain, not 3 (you've got that extra little tail).

For number 2, you know it is sn2, so look for a leaving group and a nucleophile (Note, the nucleophile may not be depicted like that, but reaction conditions may change that).  Can you see two possible nucleophiles and only one possible leaving group?

Offline english

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Re: SN2 reaction
« Reply #2 on: January 31, 2009, 03:17:34 AM »
For #1 a 1,3 H-shift may be more appropriate before the substitution.  The cation you have drawn is very short lived etc.

A better pathway would place your pushing arrow on the other side of the O, breaking the other C-O bond.  This does not rule out the first pathway, but to me it looks more reasonable.

also, this is more like an SN1 pathway, not SN2.  The reaction takes as long as the C-O bond heterolysis.

Offline macman104

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Re: SN2 reaction
« Reply #3 on: January 31, 2009, 03:23:56 AM »
Actually, now that I think about it, they are asking for an SN2 reaction.  So there should be no intermediate cation formed, even though it is possible to put this through an SN1 mechanism.  So, you have the nucleophile attack the most nucleophilic site, which would be the carbon to the right of the oxygen (because it has the least electron donation coming to it).

However, if you were to consider a SN1 pathway, spanglish's suggestions would probably be accurate (I would break to form the more stable carbocation like you suggested).

Offline noodlesoup

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Re: SN2 reaction
« Reply #4 on: January 31, 2009, 10:40:14 AM »
It seems pretty reasonable, otherwise, you could also have the chloride attack the carbon and kick out the protonated oxygen in one concerted step.  Also, becareful with drawing the positive charge, because right now your picture technically shows a 4-carbon chain, not 3 (you've got that extra little tail).

For number 2, you know it is sn2, so look for a leaving group and a nucleophile (Note, the nucleophile may not be depicted like that, but reaction conditions may change that).  Can you see two possible nucleophiles and only one possible leaving group?

thank your for your insight. By the way, what do you mean when you mention that my picture shows a 4-carbon chain not 3? and oh for question 1, the final product has been given to me, it is the one on the right. So i was trying to draw the mechanism that looks reasonable for that pathway. can you be a little bit specific? sorry i'm kind of slow.
and for question 2, since it is asked for SN2 reaction, i was thinking that the Br is the Nu and it would leave when the OH from NaOH does the backside attack, but then i'm kind of unsure about how the rxn continues.

Offline noodlesoup

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Re: SN2 reaction
« Reply #5 on: January 31, 2009, 10:44:55 AM »
For #1 a 1,3 H-shift may be more appropriate before the substitution.  The cation you have drawn is very short lived etc.

A better pathway would place your pushing arrow on the other side of the O, breaking the other C-O bond.  This does not rule out the first pathway, but to me it looks more reasonable.

also, this is more like an SN1 pathway, not SN2.  The reaction takes as long as the C-O bond heterolysis.

thank you for your opinion. yeah for the 1st question, it didn't ask for SN2 reaction, it just asks for draw a reasonable mechanism for the reaction using pushing arrows. anyway, when you mention about 1,3 H-shift, i get your poiint there but how would you come up to the final structure of the product that been given on the right of the reaction if you do a 1,3 H-shift? can you explain abit more about how it goes.

Offline macman104

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Re: SN2 reaction
« Reply #6 on: January 31, 2009, 10:51:12 AM »
thank your for your insight. By the way, what do you mean when you mention that my picture shows a 4-carbon chain not 3?
In your mechanism, when you show the positive charge.  If you count the carbons, there are four (you go down, up, down, up, and then the positive charge).  It should be down, up, down, then draw the positive charge.
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and oh for question 1, the final product has been given to me, it is the one on the right. So i was trying to draw the mechanism that looks reasonable for that pathway. can you be a little bit specific? sorry i'm kind of slow.
Right, so you need to draw an SN2 mechanism to arrive at that compound.  So your first step is correct, protonating oxygen.  Then, since it's SN2 you need to perform a concerted mechanism, and have the nucleophile attack and then kick out the protonated OH.  Also, you'll want to attack from the most nucleophilic site, which is the site on the right (which would then give you the correct compound).
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and for question 2, since it is asked for SN2 reaction, i was thinking that the Br is the Nu and it would leave when the OH from NaOH does the backside attack, but then i'm kind of unsure about how the rxn continues.
Right, so you have two possible OH groups, correct?  Can you draw out the possible intramolecular and the intermolecular reaction/mechanism and post it here?

Offline noodlesoup

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Re: SN2 reaction
« Reply #7 on: January 31, 2009, 07:04:00 PM »
In your mechanism, when you show the positive charge.  If you count the carbons, there are four (you go down, up, down, up, and then the positive charge).  It should be down, up, down, then draw the positive chargeRight, so you need to draw an SN2 mechanism to arrive at that compound.  So your first step is correct, protonating oxygen.  Then, since it's SN2 you need to perform a concerted mechanism, and have the nucleophile attack and then kick out the protonated OH.  Also, you'll want to attack from the most nucleophilic site, which is the site on the right (which would then give you the correct compound).

Right, so you have two possible OH groups, correct?  Can you draw out the possible intramolecular and the intermolecular reaction/mechanism and post it here?
thank you for your thought. ok so i i've been thinking about the questions closesly again and here is the drawing that i came up with, for question 2, i get stuck at the end, i was thinking the O then will the backside attack the OH group leave, but then the OH is a bad leaving group so i was kind of stuck right there, don't know if my pathway is correct. If you guys can give me your thought and suggestion. it would be very helpful. Thank you.


Offline macman104

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Re: SN2 reaction
« Reply #8 on: January 31, 2009, 08:35:56 PM »
For question 1, you are still drawing an SN1-type mechanism.  After you perform the initial protonation, you need to have the nucleophile attack, do not form a carbocation!  Look at the attachment below.

For question 2, the NaOH is not going to attack anything, it is only there to act as a base.  It is there because OH is not a good nucleophile, but O- (if a base removes the H *hint hint*).  So, you have two possible places to remove a H, and then the O- can attack the leaving group (do you know what the best leaving group is on the molecule?)

Offline noodlesoup

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Re: SN2 reaction
« Reply #9 on: January 31, 2009, 10:45:13 PM »
For question 1, you are still drawing an SN1-type mechanism.  After you perform the initial protonation, you need to have the nucleophile attack, do not form a carbocation!  Look at the attachment below.

For question 2, the NaOH is not going to attack anything, it is only there to act as a base.  It is there because OH is not a good nucleophile, but O- (if a base removes the H *hint hint*).  So, you have two possible places to remove a H, and then the O- can attack the leaving group (do you know what the best leaving group is on the molecule?)

Thanks alot macman for your explanantion. your mechanism picture makes it easier to see and understand.
for question 2, i still don't get why NaOH is not going to attack anything; and i think the best LG on the molecule is Br, since OH can't be because it is a bad LG. i was thinking about react the OH group with tosylate to make it better LG but don't think it is fit in this situation beccause the question provided all the reactants and it asks for SN2.

Offline macman104

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Re: SN2 reaction
« Reply #10 on: January 31, 2009, 10:49:10 PM »
Sorry, I meant OH is not going to do any "nucleophile" attacking.  It will still act as a base in this case (and there are 2 possible sources for it to do so!) So, you've correctly identified the leaving group.  Now, we agree that OH is not that great of a nucleophile, but remember, O- is good!  Like I mentioned, do you see two possible sources for a O- attacking group?  Can you see how one attack of bromine is going to be intramolecular and one will be intermolecular?

EDIT:  Also, understand that my mechanism is correct, but only because you asked for SN2.  Your idea for the SN1 is still somewhat correct (we would have to make the modifications suggested by "g english" earlier) and it's likely this reaction will undergo SN1 as well, but it's not an SN2 reaction.

Offline noodlesoup

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Re: SN2 reaction
« Reply #11 on: February 01, 2009, 11:46:32 AM »
Sorry, I meant OH is not going to do any "nucleophile" attacking.  It will still act as a base in this case (and there are 2 possible sources for it to do so!) So, you've correctly identified the leaving group.  Now, we agree that OH is not that great of a nucleophile, but remember, O- is good!  Like I mentioned, do you see two possible sources for a O- attacking group?  Can you see how one attack of bromine is going to be intramolecular and one will be intermolecular?


thanks macman. ok, i see what you meant somehows,  so i look at it again and here is what i came up with. i m not sure if it is correct, especially the pathway near the end. but i hope you can give me your insight. Thanks


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