Destroyer Ochem #208 help plz-

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whawha

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The reaction molecule is too big to draw...

How do we know which alpha-hydrogen to remove from? I was going to remove the one on the left because I figured that it was closer to the additional ethyl group attached, and it will cause some inductive effect. But, the solution is removing the one on the right to reach the solution it provides...

Can anyone explain why this is so? Thank you!
 
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I'm not sure which question you're referring to, but the general rule of thumb is that the proton that you're removing must be trans (opposite) to your leaving group!

Hope this helps!
 
Does anyone know the correct answer to this poster's question?

nze82's answer doesnt answer the question.

I dont have time to scan this question up, but anyone has Detroyer 2009 version who can answer why Dr. Romano chose to remove the acidic hydrogen from the right side of the diester as oppose to the left side (Note: the molecule is not symmetric so product's would be different based on which hydrogen you pick off)
 
Depends on the type of reaction. Are you doing an internal reaction forming a ring? If so, you remove the alpha hydrogen that allows the ring formation of a 5-7 member ring. 6 member ring is most stericly stable so if that is an option choose that alpha hydrogen. Removal of the alpha hydrogen that will allow the nucleophile to attack your electrophilic site on of the carbonyl group
 
This is just my reasoning (sont know if its absolutely correct) but if you place the ethyl group in the second position with the ester right next to it, it would be unstable due to steric hindrance..i.e bulky groups too close to each other. Thats how I rationalized it..
 
Is there an alpha hydrogen between 2 carbonyls? If so, that one. Not sure what the molecule is. Can you explain any more?
 
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