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Topic: Sn1, Sn2, E1, E2 reactions  (Read 31541 times)

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nemzy

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Sn1, Sn2, E1, E2 reactions
« on: December 07, 2004, 01:40:12 AM »
okay, i understand how these reactions works and in which circumstances, but there are some general concepts that i am confused about.

first of all, i know that in Sn2 reactions, generally weak bases are best as nucleophiles..but how can you know what a weak base is? for example, why is CH3CO2- a weak base (sn2 will predominate) and CH3CH2O- a strong base (E2 will predominate)?

And..how can u tell what is an aprotic or protic solvent?

for example, CH3OH is protic, but CH3CN is aprotic?

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Re:Sn1, Sn2, E1, E2 reactions
« Reply #1 on: December 07, 2004, 02:07:25 AM »
Don't know why for a strong base E2 should dominant. you want to differentiate sn2 and E2? for a strong bases but weak nucleophile, like t-butO-, E2 will dominant. The substrate is usually secondary or tertiary. For a good nucleophile, whether it's a strong base like CH3CH2O-, or a weak base like I-, SN2 will donimant, providing that the substrate is primary or secondary. Otherwise, both can occur.

nemzy

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Re:Sn1, Sn2, E1, E2 reactions
« Reply #2 on: December 07, 2004, 04:30:33 AM »
yeah, i know that..

but my questoin was..how can u tell what  strong or weak acid is?

and how can u tell what is a protic or aprotic solvent?

thx

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Re:Sn1, Sn2, E1, E2 reactions
« Reply #3 on: December 07, 2004, 12:33:12 PM »
A weak base is generally the conjugate base of a strong acid, for example SO42- is a very weak base, as it is the conjugate base of sulfuric acid.  Similar reasoning will suggest that acetate is a relatively weak base because it is the conjugate base of an acid.

To put it in more qualitative terms, a weak base will be a base will be an ion with a stabilized negative charge, such as a carboxylate group where the charge is delocalized onto the two oxygens by resonance.  Another class of weak bases is those that don't have a negative charge, but do have an available lone pair of electrons; an example would be an amine base such as triethylamine.

Protic solvents are defined by possesing a highly, or slightly acidic proton which is available for hydrogen bonding.  Alcohol solvent meet this criteria, as does water.  Aprotic solvent do not have the ability to hydrogen bond and therefor their solubility properties are very different.  An example of an aprotic solvent is DMSO (or acetonitrile, as you pointed out), since they have no acidic protons under standard conditions and are incapable of hydrogen bonding.

nemzy

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Re:Sn1, Sn2, E1, E2 reactions
« Reply #4 on: December 07, 2004, 10:38:39 PM »
hmm, then why is CH3CH2O- a strong base then? it has a negative ion?

Also, how do u know if a reaction will go under an E1 or E2 reaction in a tertiary substrate?  I know that in E2, only ANTI configuration of the leaving group and the hydrogen has to be present to work, and E1 configuration can take place in any circumstances , whether it be anti or syn....But , how would u know , for example, if a tertiary substarte is already in the anti configuration, if an E1 or E2 reaction will take place?  FOr E2, does the nucleophile have to be basic, and for E1, only under neutral conditions will work?  Like heating in pure ethanol??

also, in Sn1 reaction, i know that the best leaving groups are the one that are most stable, in other words, the one that fullfills the octet rule??  But isn't tertiary alchohols the best leaving group for Sn1 since they give the most stable carbocation intermediates?  But my book says that a tertiary alcohol is one of the weakest leaving groups...

thanks

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Re:Sn1, Sn2, E1, E2 reactions
« Reply #5 on: December 08, 2004, 12:06:34 AM »
Ethoxide, CH3CH2O-, is a strong base because it is essentially a non-stabilized anion.  There are no resonance forms like there are in the case of the acetate ion.

The E2 mechanism requires that one of the reactants be at least somewhat basic.  E1 mechanisms will predominate when you have an exceptionally good leaving group.  For example, t-butanol under acidic conditions will protonate and then lose water to make a tertiary carbocation.  Then a base can remove one of the protons to make a double bond.  The key is that the leaving group on the substrate molecule leaves spontaneously.  So, no E1 does not have to be under neutral conditions.

Alcohols are poor leaving groups in general because you form an alkoxide ion which is essentially non-stabilized (see above).  As I described, the leaving group in the case of a tertiary alcohol is water (assuming the alcohol has been protonated).  The leaving group ability depends on how stable that leaving group will be away from the molecule it is attached to, and how stable the remainder of the molecule will be once the leaving group has left (given that the reaction is reversible, as it probably is).  For example, supose you treated ethanol and t-butanol with a solution of HCl and methanol (no water).  In both cases you would expect to see a substitution reaction, but the mechanism would probably be different.  In each case the first step is to protonate the alcohol so that it becomes a good leaving group.  In the ethanol case, the water molecule can either leave spontaneously or be displaced by a molecule of methanol via an SN2 mechanism (to give methyl ethyl ether as product).  The latter is more likely since spontaneous departure of water would form a primary carbocation.  In the t-butanol case, there is no problem with carbocation stability, but there is a problem with the approach of the nucleophile.  The SN1 mechanism predominates and t-butyl methyl ether would be the product.

I hope that all makes sense.

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