[Cooper]

Hi,

My book says that the Hofmann Elimination yields mainly non-Zaitsev alkenes, and I was a little confused by their explanation. It says that because in Hofmann Eliminations the LG is charged, the transition state resembles more of a carbanion than an alkene, like in an elimination with a neutral LG. Why is this so? Is it just fact?

Thanks

[orgopete]

You can look at it by rationalizing the products. The Hofmann product is the result of loss of the most acidic proton (that is a primary H). The Zaitsev product receives electrons from the most electron rich carbon.

The greater that one has E1-like conditions, the greater the Zaitsev product. The poorer the leaving group, the greater the Hofmann product.

Your book appears to be trying to capture how these products form in a transition state argument.

[Cooper]

The greater that one has E1-like conditions, the greater the Zaitsev product. The poorer the leaving group, the greater the Hofmann product.

But isn't the amine an excellent leaving group?

[orgopete]

But isn't the amine an excellent leaving group?

Let's think about it. Let's use the pKa of acids to judge leaving group ability. The stronger the acid, the better the leaving group.

HI, -10
HBr, -9
HCl, -7
HF, 3.7
Me3NH(+), 11

Iodide and bromide give the least amount of Hofmann products under all conditions. Chloride can give either in examples I have found. Ammonium salts give the least. I interpret this as an ammonium salt, although still a leaving group, has the least pull of neighboring electrons of the leaving groups. As a consequence, proton acidity plays a greater role.

[Cooper]

But isn't the amine an excellent leaving group?

Let's think about it. Let's use the pKa of acids to judge leaving group ability. The stronger the acid, the better the leaving group.

HI, -10
HBr, -9
HCl, -7
HF, 3.7
Me3NH(+), 11

Iodide and bromide give the least amount of Hofmann products under all conditions. Chloride can give either in examples I have found. Ammonium salts give the least. I interpret this as an ammonium salt, although still a leaving group, has the least pull of neighboring electrons of the leaving groups. As a consequence, proton acidity plays a greater role.

Oh, okay, makes sense. Thanks I was just thinking it would be a good leaving group because it would be neutral.