Elimination reactions usually occur such that they are removing a hydrogen from the carbon attached to the fewest hydrogens.
So when you form an alkene in an elimination reaction, make sure you form the most substituted alkene (i.e. the one with the most carbon atoms directly attached).
Another way of stating Zaitsev’s rule is that “the poor get poorer”. In other words, the carbon with the fewest hydrogens loses the hydrogen.
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In Elimination Reactions, The “More Substituted” Alkene Tends To Be The Major Product
So far, we’ve only looked at some simple elimination reactions where only one product is possible. In this post we’ll look at some examples where we start to see some of the extra “wrinkles” that can be present in elimination reactions.
For example, if you heat the alcohol below with a strong acid (like sulfuric acid, H2SO4) you obtain one major product (an alkene) and a minor product (also an alkene).
What’s interesting about this? Well, if you look closely you should see that actually two elimination products are possible here, but only one is formed as the major product.
The alkene which is “tetrasubstituted” – that is, attached to four carbon atoms – is the major product, and not the “disubstituted” alkene, which is attached to two carbon atoms and two hydrogen atoms.
(The fact that we’re forming a new C-C π bond at the expense of sigma bonds on adjacent carbons is characteristic of elimination reactions.)
Similarly, look at the product of this next reaction. Taking an alkyl bromide and adding a strong base, we again get a “major” product and a “minor product”.
Again, the major product is “more substituted” than the minor product. Of the 4 atoms directly attached to the alkene in the major product, 3 of them are carbon and 1 is hydrogen. In the minor product, 2 carbon atoms and 2 hydrogen atoms are directly attached to the alkene.
So what’s responsible for this preference for the “more substituted” alkene in elimination reactions?
The Stability Of Alkenes Increases As C-H Bonds Are Replaced With C–C Bonds
Well, this correlates nicely with an observation that’s been made regarding the heats of formation of various alkenes. As an alkene becomes more substituted (i.e. more carbons attached, fewer hydrogens attached) it becomes more thermodynamically stable. [This observation comes from measuring the enthalpy of hydrogenation for various alkenes – click here for data] . [See post: Alkene Stability]
This agrees nicely with the trend that’s observed for elimination reactions. The major product of an elimination reaction tends to be the more substituted alkene. This is because the transition state leading to the more substituted alkene is lower in energy and therefore will proceed at a higher rate.
FAQ
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