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OCR C&R - Halogenoalkanes

I have 4 halogenoalkanes:

1) 1-bromobutane
2) 2-bromobutane
3) 2-bromo-2-methylpropane
4) 1-bromo-2-methylpropane

How do I know which could react with:

Hot aqueous sodium hydroxide to produce butan-2-ol?
Hot aqueous sodium hydroxide to produce a tertiary alcohol?

Which two could react with hot ethanolic sodium hydroxide to produce but-1-ene?
Reaction with sodium hydroxide to produce an alcohol: the substitution must take place where the bromine is attached.

In other words, to produce propan-2-ol you must start off with 2-bromopropane.

but-1-ene can only be formed if there is a hydrogen atom attached to the carbon atom adjacent to the carbon atom holding the bromine. (I hope that doesn't confuse you)

CH2Br-CH(OH)- etc

or CH2(OH)-CHBr- etc
Yeah as charco said.

Remember aqueous NaOH= substitution (nucleophilic) where OH substitutes the Br. Its a straight swap.

But ethanolic NaOH= elmination where Br removed forms NaBr with sodium from NaOH, and the H on the adjacent carbon atom to the Br forms water with OH from NaOH. You're left with an alkene.

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