bulky base and E2 reaction

[yueru]

Is the product always the less substituted alkene?
or only with bulky halide?


Thanks!

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[invictusx]

Is the product always the less substituted alkene?
or only with bulky halide?


Thanks!

Whenever you have a bulky base, for example t-butoxide, then you do an elimination. You cannot do an SN2 with a bulky base simply because of steric hindrance. You should form the MOST stable (more substituted) alkene.

I'm not sure what you mean by bulky halide. Do you mean the reactant? When the reactant is tertiary, no SN2 can occur because, again, of steric hindrance. Therefore, you resort to E2.

Is this clear?

 

[wall1two]

the most common way to get hoffman (spelling? and least subbed) alkene is with a "bulky base, ie LDA or tert-but oxide. but another method of giving the less substituted product is with a bulky leaving group!

(CH3)2C(NR2+)CH3 with KOH and heat will give

(CH3)2C=CH2 so bulky base or a bulky l.g. will give the less substituted product

edit- i got hoffman and Zaitsev's confused. these 12 hour days are killing me. Zaitsev's is the more sub"zait"tuded alkene. but still bulky base or bulky leaving group gives you hoffman

 

[yueru]

thanks!
Is the product Hoffmann product? I got confused as I watched chad's vedio and he mentioned that only when bulky base and bulky halide ( bulky LG) will produce Hoffmann product.

the most common way to get zetzev (spelling? and least subbed) alkene is with a "bulky base, ie LDA or tert-but oxide. but another method of giving the less substituted product is with a bulky leaving group!

(CH3)2C(NR2+)CH3 with KOH and heat will give

(CH3)2C=CH2 so bulky base or a bulky l.g. will give the less substituted product

 

[yueru]

Not really Thanks anyway.
bulky halide = bulky LG, yes, the reactant

Whenever you have a bulky base, for example t-butoxide, then you do an elimination. You cannot do an SN2 with a bulky base simply because of steric hindrance. You should form the MOST stable (more substituted) alkene.

I'm not sure what you mean by bulky halide. Do you mean the reactant? When the reactant is tertiary, no SN2 can occur because, again, of steric hindrance. Therefore, you resort to E2.

Is this clear?

 

[invictusx]

Not really Thanks anyway.
bulky halide = bulky LG, yes, the reactant

Ah, if you have a bulky LG, then you go hoffman, depending on the LG.

Okay, let's try again. If you have a bulky reactant (tertiary molecule), you cannot do a SN2 because what does a SN2 reaction entail? An inverted attack; however, tertiary molecules prevent any SN2 attack because of the steric hindrance of the molecule (see it in 3D). That's why you resort to E2. I wasn't quite sure what you meant by bulky halide because.... hoffman LG isn't a halide.