Organic chemistry problem?

Compare the effects of bases in substitution and elimination reactions (KOH, alkoxide, potassium tert-butoxide, etc)

please please help me, I am getting everything mixed up!!

Answer:
KOH and normal alkoxides are comparable as bases and as nucleophiles. Tert-butoxide is hindered and therefore is typically less reactive as a nucleophile but an excellent, strong base.
Firs, you have to be able to classify the bases (from the less strong) :
NaOH, KOH, R-O-, NH2- (so KOH is less strong than alkoxides)

The basic formula for alkoxides is R-O-.

To be able to say what alkoxide is stronger (in a comparison), you have to know the effect of the groups around the O-. Alkyl groups (R-) have an inductive effect that “pushes” the electrons to the O- (donor effet). This makes the base stronger (more negative = more reactive).
If around the O- you have groups with attractive effect (such as Br-) this makes the base less strong because it attracts the electrons so the O- will be less negative (that it could be) so the base is less strong = less reactive.

For example, if you compare potassium methoxide (CH3-O-K+) and potassium tert-butoxide, the last one is stronger. The alkyl (here CH3-) have an inductive effect that “pushes” the electrons to the O-. This makes the base stronger. So, because the tert-butoxide has more methyl groups, it is a stronger base than methoxide.

Usually, for eliminations, we used KOH (enough strong) but all the alkoxides work. For substations, use alkoxides and you can know which one is stronger.

Hope I helped you, if it wasn’t enough clear or if you have other questions, you can send me a mail...

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