# Alcoholic KOH mechanism with alkyl halides

Does alcoholic $\ce{KOH}$ only show $E_2$ mechanism with alkyhalides?

I have studied with two organic chemistry teachers, one told me that there is always $E_2$ mechanism, except in the case of tertiary alkylhalide and the other told me it can be any depending on the substrate with $\alpha$ hydrogens more than 4 or if it has resonance it will show $E_1$.

Which one is true?

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With a strong base $E2$ will always predominate over $E1$.

Both elimination mechanisms have a preference for tertiary substrates over secondary substrates over primary substrates. This behavior is on contrast with the substitution reactions, where steric factors are important for $S_N2$. Steric factors are less important for the proton transfer necessary for $E2$.

Since $E2$ and $E1$ have the same substrate preference order, you may feel that you cannot decide which is more likely. However, $E2$ is usually faster because $E2$ is bimolecular. Consider the rate laws:

• $\mathrm{rate}_{E2}=k\mathrm{[substrate][base]}$
• $\mathrm{rate}_{E1}=k\mathrm{[substrate]}$

If we make the assumption that the rate constants are of similar magnitude, then increasing the concentration of the base increases the rate of $E2$.

• E2 does have a stereochemical requirement though. The leaving group must be antiperiplanar to the hydrogen.
– bon
Sep 12 '15 at 15:20
• Means it will mainly show E2 mechanism.... Sep 12 '15 at 21:27