Halogen Alkanes

In aqueous alkali solution, you will get substitution (OH- ions will act as the nucleophile), i.e. the C-Br bond is broken and replaced by the OH- nucleophile, giving you the C-OH bond, i.e. the alcohol. This is the only possible substitution. This will then give you 2-methylbutan(e)-2-ol, or 2-hydroxy-2-methylbutane (but the former is preferred).

For aqueous ethanolic solution, you will then get the elimination reaction. This is where you form your alkenes. You can only form alkenes on the adjacent carbons to the carbon with the halogen, i.e. the carbons next to the carbon with the halogen. The adjacent carbon MUST have a hydrogen that can be removed. This leaves three options, either the carbon to the left (the methyl group) or to the right (the ethyl group) or the bottom (i.e. the methyl group) of the carbon connected to the halogen. Two of them will give you the same product however. I'll attach the mechanisms as a pic in a second. Then you just need to name all of them with all the mechanisms.

I hope this all makes sense. If I have made a mistake, please anyone feel free to correct me.

Thank you so much!!

Is there a mechanism for the production of 2-methylbutan-2-ol?

Yesss - nucleophilic substitution. I'll explain it but if you do not understand it, then I'll draw it .

Basicaly the :OH- nuclephile will attack the carbon that's bonded to the halogen (MAKE SURE TO DRAW CURLY ARROWS AND NOT STRAIGHT ONES and also make sure it goes to the carbon and NOT the bond). Then the second curly arrow will be from the C-Br bond to the bromine, if that makes sense? If you want to draw it, post a pic and I'll tell you if it's right (and also make sure to draw the product because that's always worth 1 mark! I made the important notes capital because a lot of people lose marks there)

So kinda like that?

perfecto. just one thing though, make sure the curly line starts on the actual bond, i.e. there is no gap. some examiners are really particular and will deduct a mark if it's not connected to the bond. otherwise, good stuff!

Okay thank you so much!!

Also does anyone know anything about Maxwell–Boltzmann curves?

Draw, with labelled axes, a curve to represent the Maxwell–Boltzmann distribution of molecular energies in a gas. Label this curve T1. On the same axes, draw a second curve to represent the same sample of gas at a lower temperature. Label this curve T2. Use these curves to explain why a small decrease in temperature can lead to a large decrease in the rate of a reaction. (7 marks)