Deprotonation of alpha-protons

[MedPR]

What is the final product after acetone is treated first with NaH, followed by iodoethane, and subsequently followed by workup?

A. 2-methyl-2-butanol
B. 3-methyl-2-butanone
C. 2-pentanone
D. 2-butanone

A few questions.

What purpose does the NaH serve? Is it thrown in to distract you and try to make you forget that treating a ketone with alkyl halide lengthens the ketone?

Also what does "followed by workup" imply?

I understand the mechanism and product formation, but only because I decided to ignore whatever NaH does.

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[Ssina]

wow totally didn't see that one coming! totally got a diff thing
I found this... but if we follow the top reaction with acetone and ethane... then shouldn't we get C?!

I know it's a strong base, so it has to be involved, otherwise I would've had the carbonyl attach the iodoethane and would've gotten like a tertiary alcohol...maybe?!

thanks for sharing btw!

 

[MedPR]

Oh I see, the NaH is a strong base and it deprotonates one of the CH3's on Acetone, then the electrophile (ethyliodine) gets attacked by the newly formed carbanion.

I'm not sure about your link though, sorry 🙁

 

[Ssina]

no that works,I get it, it's C!

 

[chiddler]

wow totally didn't see that one coming! totally got a diff thing
I found this... but if we follow the top reaction with acetone and ethane... then shouldn't we get C?!

I know it's a strong base, so it has to be involved, otherwise I would've had the carbonyl attach the iodoethane and would've gotten like a tertiary alcohol...maybe?!

thanks for sharing btw!

why doesn't the hydride ion attack the carbonyl carbon?

workup usually is acid/base addition to remove or add excess/surplus protons. i'm not sure how it fits in this context since it doesn't really require workup because the product is a ketone.

 

[MedPR]

why doesn't the hydride ion attack the carbonyl carbon?

workup usually is acid/base addition to remove or add excess/surplus protons. i'm not sure how it fits in this context since it doesn't really require workup because the product is a ketone.

It probably does, but it would just reform the original compound, then get deprotonated again.

 

[chiddler]

It probably does, but it would just reform the original compound, then get deprotonated again.

teehee that's what i did initially and thought "what a wasteful reaction!"

 

[MedPR]

teehee that's what i did initially and thought "what a wasteful reaction!"

🙂

 

[maxx52188]

why doesn't the hydride ion attack the carbonyl carbon?
.

I asked my Ochem professor this almost exact question when we started on alpha acidity. Essentially, many of the carbonyl reactions ARE occurring due to individual equilibriums. Since this isnt in the lab, we dont really have to worry about all the different reactions. We should be able to know what reaction it is based on the context that the question is presented to us in IE the chapters were being tested on, the possible answers, any reactants that are primarily used on SPECIFIC reactions (like iodoethane), etc.

 

[Jay2910]

What is the final product after acetone is treated first with NaH, followed by iodoethane, and subsequently followed by workup?

A. 2-methyl-2-butanol
B. 3-methyl-2-butanone
C. 2-pentanone
D. 2-butanone

A few questions.

What purpose does the NaH serve? Is it thrown in to distract you and try to make you forget that treating a ketone with alkyl halide lengthens the ketone?

Also what does "followed by workup" imply?

I understand the mechanism and product formation, but only because I decided to ignore whatever NaH does.

Sodium Hydride, is a really strong base and I think it attacks one of the CH3 groups so that it becomes CH2Na+ . ..then if you put in iodoethane, you will get a carbon carbon bond to make 2 pentanone.

 

[Jay2910]

wow . .it looks like you already got it 🙂 Thanks for sharing though.