alpha hydrogen

[PAyankee]

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if you have carbon between two carbonyl groups and methyl groups attached to the other side of each carbonyl, and you carry out a reaction using 1.NaOCH3/2. CH3CH2Br why is the alpha hydrogen between the two carbonyl groups deprotonated instead of one of the methyl groups?

 

[MowgliR]

if you are referring to a beta-keto ester...

its the most acidic H

 

[sonofastunna]

Take off that hydrogen, and draw the resonance structures. you will see that you can draw 2 structures, whereas by deprotonating the methyl you only get 1 structure. More resonance structures = more stable

 

[UndergradGuy7]

if you have carbon between two carbonyl groups and methyl groups attached to the other side of each carbonyl, and you carry out a reaction using 1.NaOCH3/2. CH3CH2Br why is the alpha hydrogen between the two carbonyl groups deprotonated instead of one of the methyl groups?

Because then the negative charge on that carbon when you take off the hydrogen can be put in resonance between 2 C=O groups.

If it was on the end, it would be able to do resonance with only one C=O group.

Resonance between two C=O groups makes the conjugate base more stable, and its acid stronger.

 

[Troyvdg]

Take off that hydrogen, and draw the resonance structures. you will see that you can draw 2 structures, whereas by deprotonating the methyl you only get 1 structure. More resonance structures = more stable

^exactly.

it sounds like your talking about an aldol type rxn. Because the two carbonyl groups are electron withdrawing, the alpha carbon between the two carbonyl groups is left with a slight positive charge, also making it more acidic--easiest to deprotinate 👍

 

[PAyankee]

Thanks for the great responses, another question I have in a similar vein is why 2-iodopropane reacted with NaOCH3 in a solution of ethanol would cause an E2 and SN2 product. I thought that reactions went Sn1/E1 with polar protic solvents like ethanol.

 

[UndergradGuy7]

Thanks for the great responses, another question I have in a similar vein is why 2-iodopropane reacted with NaOCH3 in a solution of ethanol would cause an E2 and SN2 product. I thought that reactions went Sn1/E1 with polar protic solvents like ethanol.

The solvent is usually the least important thing. The type or nucleophile strong or weak, and the type of carbon primary, secondary, tertiary are more important.

Also the example you gave would give more E2 than Sn2 I think?

 

[loveoforganic]

I'd think it's pretty up in the air. Hard to tell with a secondary substrate and unhindered strong base.

 

[Troyvdg]

The solvent is usually the least important thing. The type or nucleophile strong or weak, and the type of carbon primary, secondary, tertiary are more important.

Also the example you gave would give more E2 than Sn2 I think?

yes, the example he gave would give an SN2 or E2 with a little bit of heat

the nucleophile is the most important thing.
if strong-can do E2 or SN2
if weak- E1 or SN1