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I read an amazing PDF from nptel on elimination reactions...but I was really confused when it used used $\ce{^+NMe3}$ as a leaving group in the reaction pathway via:
My organic chemistry teacher also did syn-elimination but in the problem solving book and on various online sites it took the $\mathrm{E2}$ elimination pathway.
There is a range of elimination reactions with E1cb at one end, E1 at the other end and E2 in between. It is not uncommon for these different reaction pathways to compete with one another. For example, in some elimination reactions, the E1 and E2 pathways can operate in competition with one another. Activation energy is associated with each of these 3 reaction pathways. Whichever pathway has the lowest activation energy will be the major pathway followed. By changing solvent, reaction temperature, the relative strength of the nucleophile, the relative strength of the base, leaving group stability, etc., we can raise or lower the activation energy for each of these 3 pathways and shift a reaction towards one side of this mechanistic range or the other.
(Taken from Ron's answer, in order to start an explanation)(1)
From Wikipedia:
The first step of an E1cB mechanism is the deprotonation of the of β-carbon, resulting in the formation of an anionic transition state, such as a carbanion. The greater the stability of this transition state, the more the mechanism will favour an E1cB mechanism.
So on further thinking about the Hofmann elimination, you notice the following
$\ce{NR_3^+}$ is a poor leaving group (Smith, March. Advanced Organic Chemistry 6th ed. (501-502))
Due to this, the rate of cleavage of the $\ce{C-H}$ bond cleavage exceeds the rate of the $\ce{C-N}$ bond cleavage
In fact, the three well known poor leaving groups namely $\ce{-F}$, $\ce{-SMe^2+}$, $\ce{NR^3+}$ all have the same observation
What's the conclusion?
Of course, these undergo E1cB when heated to a suitable temperature
There are sources which infer the same -
Organic mechanisms by Xiaoping Sun (3)
Some examples of compounds that contain poor leaving groups and can undergo the E1cB mechanism are alcohols and fluoroalkanes.
The $\ce{-NMe3}$ group is electronegative, it can activate beta hydrogen to make it acidic, vulnerable to attack by a base. This gives rise to carbanion character (e1cb) for the transition state.
Hence, E1cB is the correct answer
Honestly, leaving groups usually do not determine whether it is syn or ante elimination. Hence, although the leaving group is an amine, it does not change how you view the problem. What determines syn or ante elimination is the type of reaction occurring, whether it is E1, E2 or E1cB.
E2 would always go through an anti elimination due to the requirement of the antiperiplanar transition state.
Whereas for E1 and E1cB, there is no fixed answer to whether it is syn or ante. The product that ultimately forms is dependent on the thermodynamic stability of the product.
Now, let us go back to your "E1" example. That, my friend, is not an E1 elimination! That is in fact an E2 elimination. Your E2 elimination has the requirement of having your hydrogen and leaving group be coplanar. Since the D and ammonium quaternary cation are coplanar, though it is fixed in a syn position, they are forced to go through a syn coplanar E2 elimination. The reason it is not E1 is that your leaving group is on a secondary carbon and therefore E1 is unfavourable.
