When an alkyl halide is dehydrohalogenated by sodium hydroxide, which mechanisms does it undergo?

If it's "dehydrohalogenated", then it loses a hydrohalide, i.e. #"HBr"#, #"HCl"#, etc.

So, if you lose both, the simplest case you can have for an alkyl halide is turning it into an alkene. It's similar in principle to the dehydration of an alcohol, and is an elimination reaction.

With sodium hydroxide...

• For primary alkyl halides, a one-step #\mathbf(E2)# (second-order elimination) process tends to occur because the primary carbocation intermediate is unstable.
• For secondary and tertiary alkyl halides, a two-step #\mathbf(E1)# (first-order elimination) process might also occur, due to the greater thermodynamic stability of the carbocation intermediate.

In general, the most effective way to accomplish "dehydrohalogenations", is to use a strong base that is a poor nucleophile.

In other words, a thermodynamically stable negatively-charged compound (readily lost a proton) that is bulky and slow (lots of stuff weighing it down).

tert-butoxide fits the bill and is something you should remember if you want to easily perform an elimination reaction on an alkyl halide:

(requires antiperiplanar conformation for #E2#)