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Topic: Alcohol chemistry! Is my pushing mechanism drawing correct??  (Read 4020 times)

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Offline theanonymous

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Alcohol chemistry! Is my pushing mechanism drawing correct??
« on: October 30, 2012, 05:29:35 PM »
Problem 2. Alcohol chemistry.

Please provide the missing components of the following reactions. For missing products, provide only the major product.



I'm supposed to give the product of the formula at the top left.
Is that correct?

My teacher wrote it down on the board but I think he skipped a step and so I guessed the very last part. But My question is, how would I be able to tell that this is an E1 reaction and not an SN1 reaction???
The conditions for E1 reaction and SN1 reaction are almost the same!
Also, H+ isn't a base, so why did my teacher say this is supposed to be an E1 reaction?
« Last Edit: October 30, 2012, 05:39:49 PM by theanonymous »

Offline theanonymous

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Re: Alcohol chemistry! Is my pushing mechanism drawing correct??
« Reply #1 on: October 30, 2012, 05:38:29 PM »
Also, how would I solve for these two??



My teachers answer key says:

Top reaction:
1. SOCl2, NEt3
2. NaN3

Bottom reaction:
1. PBr3, pyridine
2. NaCN


How am I supposed to know what to treat the molecule with? i'm so confused - we have a test tomorrow at 8! :(

Offline Babcock_Hall

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Re: Alcohol chemistry! Is my pushing mechanism drawing correct??
« Reply #2 on: October 30, 2012, 05:39:51 PM »
One thing that I noticed were a couple of odd choices you made for the formal charges on oxygen in problem 2.  The oxygen atom of the hydroxy group is initially neutral but it becomes positively charged when it picks up a proton

For the other two problems, the hydroxyl group is not a good leaving group on its own; therefore, in each case you need to replace it with something better.  One way to do this sort of problem is to work backwards from the product, one step at a time.  If you do that, you can often identify the nucleophile, at the very least.

Offline theanonymous

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Re: Alcohol chemistry! Is my pushing mechanism drawing correct??
« Reply #3 on: October 30, 2012, 05:49:21 PM »
Quote
For the other two problems, the hydroxyl group is not a good leaving group on its own; therefore, in each case you need to replace it with something better.  One way to do this sort of problem is to work backwards from the product, one step at a time.  If you do that, you can often identify the nucleophile, at the very least.

Ok so for the top reaction. This is my reasoning...

1.It's not an Sn2 or an E2 reaction because there is an OH bond...so it must be an Sn1 or E1 reaction
2. It can't be Sn1 because there would be a racemic product (two products)... so it must be E1?

E1 reaction - you say to work backwards?
But how?

You mean:
I would basically make a double C-C bond near N3 first and think of a base that would come in to attack a hydrogen atom attached to the central carbon?


« Last Edit: October 30, 2012, 06:03:56 PM by theanonymous »

Offline Babcock_Hall

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Re: Alcohol chemistry! Is my pushing mechanism drawing correct??
« Reply #4 on: October 30, 2012, 06:28:33 PM »
The second and third problems you provided are substitutions, not eliminations.  They each need more than one step, and that is when working backwards sometimes helps.  For example, for the third problem, I would work as follows:  CH3(CH2)3CH2-X + a nucleophile  :rarrow: product.  X is understood to be a good leaving group.  Then I would identify the nucleophile and figure out how to convert my starting alcohol into some group X that can be displaced easily.

I am going to have to pass on the first question for now because I need to run.  But in general substitutions and eliminations are competing reactions, and the choice of conditions might favor one over the other.  Substitutions need decent nucleophiles.

Offline discodermolide

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Re: Alcohol chemistry! Is my pushing mechanism drawing correct??
« Reply #5 on: October 30, 2012, 08:47:43 PM »
In your two reactions with reagents.
The top product has an azide group in it therefore the reagents 1 belong to this one.
By default reagents 2 belong to the bottom one.
Now what happens, see if you can tell me.
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Offline theanonymous

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Re: Alcohol chemistry! Is my pushing mechanism drawing correct??
« Reply #6 on: October 30, 2012, 10:24:57 PM »
In your two reactions with reagents.
The top product has an azide group in it therefore the reagents 1 belong to this one.
By default reagents 2 belong to the bottom one.
Now what happens, see if you can tell me.

No wait, I'm asking how do I GET those answers... Because he's asking what would I treat each molecule with to get the product given. And I have no idea... Babcock_Hall said to "work backwards" by going from product to reactant and that would give me the nucleophile (?)

Offline discodermolide

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Re: Alcohol chemistry! Is my pushing mechanism drawing correct??
« Reply #7 on: October 30, 2012, 11:02:22 PM »
The first reaction (to give the azide). Firstly it is OH converted to Cl with thionyl chloride with inversion of configuration, followed by azide displacement of the chloride with inversion of configuration.
The second one is conversion of OH to Br with PBr3 then displacement of Br with CN.
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Offline Babcock_Hall

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Re: Alcohol chemistry! Is my pushing mechanism drawing correct??
« Reply #8 on: October 31, 2012, 09:30:59 AM »
In your two reactions with reagents.
The top product has an azide group in it therefore the reagents 1 belong to this one.
By default reagents 2 belong to the bottom one.
Now what happens, see if you can tell me.

No wait, I'm asking how do I GET those answers... Because he's asking what would I treat each molecule with to get the product given. And I have no idea... Babcock_Hall said to "work backwards" by going from product to reactant and that would give me the nucleophile (?)
The nucleophiles were actually given to you, but you have to recognize them as nucleophiles.  For example:  CH3(CH2)3CH2-X + CN1-  :rarrow:   product.  Can PBr3 react with your alcohol to produce a good leaving group?  Yes.

Offline Babcock_Hall

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Re: Alcohol chemistry! Is my pushing mechanism drawing correct??
« Reply #9 on: October 31, 2012, 07:09:39 PM »
I need to correct myself about one thing.  The nucleophiles were not given to you, but it is relatively easy to see what they must be.  Let me explain my approach for one problem, and you can try it for the other.  Every substitution needs a good nucleophile and a good leaving group.
1) Picture the reaction that leads to the product:
CH3(CH2)3CH2-X + CN1-  :rarrow: product + X-.
2) Picture a reaction that makes the intermediate:
CH3(CH2)3CH2-OH + PBr3  :rarrow: CH3(CH2)3CH2-X (the solvent is pyridine)
The bad leaving group (-OH) was turned into a good leaving group, -Br, and cyanide (a good nucleophile) displaced it.

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