SN2 Substitution Products: (2R, 3S)-3-methyl-2-methoxypentane

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The discussion centers on the SN2 reaction of (2S, 3S)-2-chloro-3-methylpentane with the nucleophile CH3O-. Participants clarify that the nucleophile's attack leads to the inversion of configuration at the carbon attached to chlorine, resulting in the formation of (2R, 3S)-3-methyl-2-methoxypentane. There is confusion regarding whether the product is an alcohol or an ether, but it is confirmed that the product is indeed an ether due to the nature of the nucleophile. The Williamson ether synthesis is referenced to explain how the ether is formed. The final consensus is that the correct substitution product is (2R, 3S)-3-methyl-2-methoxypentane.
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Homework Statement



Find the substitution products of ( 2S, 3S)-2-chloro-3-methylpentane + CH3O- in SN2 reaction.





The Attempt at a Solution



Will the anti-attack of the nucleophile result in the change of S to R for both the chiral carbons or just for the one with Cl connected to it? How can I know this?
 
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assaftolko said:

Homework Statement



Find the substitution products of ( 2S, 3S)-2-chloro-3-methylpentane + CH3O- in SN2 reaction.





The Attempt at a Solution



Will the anti-attack of the nucleophile result in the change of S to R for both the chiral carbons or just for the one with Cl connected to it? How can I know this?

The configuration of the carbon attached with Cl changes. The configuration changes only when the bonds to the chiral carbon break or change.
 
So I'll get (2R,3S)-3-methyl-2-pentanol?
 
assaftolko said:
So I'll get (2R,3S)-3-methyl-2-pentanol?

How do you get an alcohol? Shouldn't that be an ether?
 
How can it be ether? Cl is the leaving group and you get OH that is attached to the Cl's carbon from the other side. Now This carbon is directly attached to O, H, R and R', I don't see how the oxygen can be attached to 2 carbons to form ether...
 
assaftolko said:
How can it be ether? Cl is the leaving group and you get OH that is attached to the Cl's carbon from the other side. Now This carbon is directly attached to O, H, R and R', I don't see how the oxygen can be attached to 2 carbons to form ether...

Williamson ether synthesis.
 
assaftolko said:
How can it be ether? Cl is the leaving group and you get OH that is attached to the Cl's carbon from the other side. Now This carbon is directly attached to O, H, R and R', I don't see how the oxygen can be attached to 2 carbons to form ether...

You said in your first post that the nucleophile was CH3O-, not OH-. This would account for a carbon on one side of the oxygen. As this nucleophile attacks, it kicks away a Cl-, causing inversion of configuration and a new bond to carbon. That makes the oxygen bridge two carbons, forming an ether. I believe the correct answer would be (2R, 3S)-3-methyl-2-methoxypentane.
 
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