View Full Version : polar aprotic solvents and Sn1
talkalot24 04-21-2008, 08:03 PM Why/how do polar aprotic solvents destabalize nucleophiles? Basically, why do they favor Sn2 rxns?
I understand why protic solvents favor Sn1, but not why aprotics favor Sn2.
Ok, one other question: Why does E2 require a strong base?
Kaustikos 04-22-2008, 11:23 AM Why/how do polar aprotic solvents destabalize nucleophiles? Basically, why do they favor Sn2 rxns?
I understand why protic solvents favor Sn1, but not why aprotics favor Sn2.
Ok, one other question: Why does E2 require a strong base?
I tend to look at it in terms of what I would need to do to draw a reaction to completion. If I wanted to react a bulky base with a halogenated/substituted alkyl group, I would use a protic solvent because the bulky base can "shed", as kaplan says, the protic solvent and bind the positively charged alkyl group.Not the mention the H-bonds formed between the carbo cation and the protic solvents tends to favor the release of the halogen from the alkyl group so that it can be attacked by Sn2.
For Sn2, I would need a solvent that caters in that way as well. ie; drives the reaction to completion. The aprotic solvent is used because if you used a protic solvent, it would make it less reactive by "bonding" with the nucleophile you're wanting to react with teh alkyl group. You want to the nucleophile to be as reactive as possible, and H-bonding would definitely NOT help.
edit
E2 requires a strong base to attack the hydrogen instead of a strong nucleophile to attack the carbonyl group in Sn2. If you used a strong nucleophile, you run the risk of Sn2 instead of E2.
Phlame217 04-23-2008, 08:29 PM I agree with above but to keep it shorter:
A protic solvent will ionize and form charged particles which stabilizes the formation of a carbocation in SN1, but the negative charge will repell the nucleophile in an SN2 due to like charges and slow the reaction.
A non protic solvent (aka non polar) will not interfere due to lack of a charge to interfere.
bluemonkey 04-23-2008, 11:42 PM I agree with above but to keep it shorter:
A protic solvent will ionize and form charged particles which stabilizes the formation of a carbocation in SN1, but the negative charge will repell the nucleophile in an SN2 due to like charges and slow the reaction.
A non protic solvent (aka non polar) will not interfere due to lack of a charge to interfere.
Not quite-
A polar protic solvent will actually hydrogen bond to the leaving group thus helping to distribute the negative charge that the leaving group carries. This, in turn, stabilizes the carbocation by preventing the leaving group from reforming a bond with the carbocation.
With an SN2 reaction, the nucleophile must approach the substrate from the backside of the leaving group. Since steric factors, rather than electronic factors as in SN1, are of paramount importance with SN2, anything that interferes with the nucleophile attacking will slow the reaction rate. Polar protic solvents, by definition, are capable of hydrogen bonding. If there is a polar protic solvent with a strong nucleophile, the solvent molecules hydrogen bond to the nucleophile forming a "solvation shell" which greatly increases the effective size of the nucleophile. This in turn prevents the nucleophile from attacking from the backside of the leaving group and greatly slows the reaction rate.
Kaustikos 04-24-2008, 05:31 AM Not quite-
A polar protic solvent will actually hydrogen bond to the leaving group thus helping to distribute the negative charge that the leaving group carries. This, in turn, stabilizes the carbocation by preventing the leaving group from reforming a bond with the carbocation.
Nice:thumbup:
Vihsadas 04-24-2008, 07:49 AM Not quite-
A polar protic solvent will actually hydrogen bond to the leaving group thus helping to distribute the negative charge that the leaving group carries. This, in turn, stabilizes the carbocation by preventing the leaving group from reforming a bond with the carbocation.
With an SN2 reaction, the nucleophile must approach the substrate from the backside of the leaving group. Since steric factors, rather than electronic factors as in SN1, are of paramount importance with SN2, anything that interferes with the nucleophile attacking will slow the reaction rate. Polar protic solvents, by definition, are capable of hydrogen bonding. If there is a polar protic solvent with a strong nucleophile, the solvent molecules hydrogen bond to the nucleophile forming a "solvation shell" which greatly increases the effective size of the nucleophile. This in turn prevents the nucleophile from attacking from the backside of the leaving group and greatly slows the reaction rate.
Yeah, this is how I think of it as well. Just as a word of advice for your orgo studying. The main question you should almost *always* be asking yourself when doing orgo problems is:
"What will stabilize the carbocation?"
And as monkey is explaining, you should also be able to explain logically why the carbocation is stabilized.
sleepy425 04-24-2008, 10:23 PM "What will stabilize the carbocation?"
or the carbanion :D
SaintJude 04-13-2012, 03:09 PM Not quite-
A polar protic solvent will actually hydrogen bond to the leaving group thus helping to distribute the negative charge that the leaving group carries. This, in turn, stabilizes the carbocation by preventing the leaving group from reforming a bond with the carbocation.
With an SN2 reaction, the nucleophile must approach the substrate from the backside of the leaving group. Since steric factors, rather than electronic factors as in SN1, are of paramount importance with SN2, anything that interferes with the nucleophile attacking will slow the reaction rate. Polar protic solvents, by definition, are capable of hydrogen bonding. If there is a polar protic solvent with a strong nucleophile, the solvent molecules hydrogen bond to the nucleophile forming a "solvation shell" which greatly increases the effective size of the nucleophile. This in turn prevents the nucleophile from attacking from the backside of the leaving group and greatly slows the reaction rate.
I like this answer---need to refer to it later.
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