Ch3-chcl-ch2+koh(alchol) saytzeff rule for alpha beta carbon
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we’ve only looked at some very simple elimination reactions. 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 treat the alcohol below with a strong acid (like sulfuric acid, H2SO4) and heat, you obtain one major product (an alkene) and a minor product (also an alkene). As we talked about last time, the fact that we’re forming new C-C π bonds here is a sign that these are elimination reactions.
What’s interesting about this? Well, if you look closely you should see that actually twoelimination products are possible here, but only one is formed as the major product. Note that the alkene which is “tetrasubstituted” – that is, attached to four carbon atoms – is the major product, and not the “disubstituted” alkene.

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”:

So what’s going on here? Note that in both of the elimination reactions we’re seeing that the major product is the one where the more substituted alkene is being formed (that is, the alkene attached to the most carbons)? Why might this be?
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]
This agrees with the trend that’s observed for elimination reactions. The major product of an elimination reaction is 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.
For example, if you treat the alcohol below with a strong acid (like sulfuric acid, H2SO4) and heat, you obtain one major product (an alkene) and a minor product (also an alkene). As we talked about last time, the fact that we’re forming new C-C π bonds here is a sign that these are elimination reactions.
What’s interesting about this? Well, if you look closely you should see that actually twoelimination products are possible here, but only one is formed as the major product. Note that the alkene which is “tetrasubstituted” – that is, attached to four carbon atoms – is the major product, and not the “disubstituted” alkene.

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”:

So what’s going on here? Note that in both of the elimination reactions we’re seeing that the major product is the one where the more substituted alkene is being formed (that is, the alkene attached to the most carbons)? Why might this be?
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]
This agrees with the trend that’s observed for elimination reactions. The major product of an elimination reaction is 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.
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setzaff rule mean markonikoff rule.
markonikovs rule state that if aqueous koh is added then oh attaches to double bonded carbon which has
less no.of hydrogen
markonikovs rule state that if aqueous koh is added then oh attaches to double bonded carbon which has
less no.of hydrogen
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