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?




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](http://www.chemgapedia.de/vsengine/vlu/vsc/en/ch/12/oc/vlu_organik/substitution/sn_e_konkurrenz/sn2_e2_konkurrenz.vlu/Page/vsc/en/ch/12/oc/substitution/sn_e_konkurrenz/a2_3_sn2_e2_substrat/sn2_e2_substrat.vscml.html). 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 $E2$.


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