I know when do you dehydrohalogenation of 2° alkyl halides, you can do either a strong base like KOCH2CH3 to get the most substituted product or a sterically hindered base like t-butoxide to get the less substitued product. However, I do not understand whether you would use the strong base or sterically hindered base for 1° alkyl halide because there are not two options here.


2 Answers 2


It's misleading to state your choice as between a strong base and a sterically hindered base, as t-butoxide is in fact even stronger than ethoxide. Rather, you choose between two strong bases, one sterically hindered and one not (or less so).

To answer your question: yes, there is only a single possible product for the elimination reaction (hydride/alkyl shift isn't generally an issue as primary carbocations aren't formed in the first place). However, all bases are potentially also nucleophiles, and can also engage in substitution, in this case giving an ether instead of the desired alkene. If you want to minimize this reaction it would pay to use the more hindered base, as substitution is more sterically demanding than elimination and would be slowed down more by bulky substituents on the nucleophile/base.


Simply put, by Saytzeff / Zaitsev rule, the favoured product in dehydrohalogenation reactions is the alkene with the greatest amount of alkyl groups connected to the doubly bound carbon atoms.

However in this case the base does not matter actually as both Saytzeff and Hoffmann alkenes will look same (except in the case carbocation's hydride shift / methyl shift takes place).

So you can use whatever base fits your budget!

Update: It seems that using a bulkier base will reduce the probability of getting the product of a nucleophilic displacement. It will then want to follow substitution and act as a nucleophile rather than as a base. Thus if you want ONLY elimination to take place the bulkier base is better as it offers higher steric hindrance. [inputs from comments]

Also you will have to note that even conditions make a difference in the products.

SOURCE: Haloalkanes, NCERT

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    $\begingroup$ Use of a sterically hindered base reduces the possibility of getting the product of nucleophilic displacement $\endgroup$
    – Waylander
    Commented Nov 29, 2023 at 7:30

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