Write equations to show how 2,3-dimethylbutane may be prepared from each of the following compounds. (i)an alkene (ii)A grignard reagent (iii)a haloalkane (iv)a sodium alkanoate?

i) Starting from any alkene

You can create this from 2-butene, or an alkene with a double bond on carbon-2. It theoretically doesn't matter whether it's cis or trans.

I would do this:

1. Basic bromination in dichloromethane
2. Add two equivalents of ##"LiCu"("CH"_3)_2##, a type of Gilman reagent, to essentially substitute both bromide groups with methyl groups like so

ii) Starting from a grignard reagent

I don't really see the point of starting from a Grignard reagent since it's usually a nucleophile... but:

1. Water gets rid of the magnesium bromide substituent and substitutes it with a hydrogen
2. Hydroboration adds anti-Markovnikov to give a hydroxide on the carbon where the magnesium bromide once was
3. ##"PBr"_3## substitutes the hydroxide with a bromide group
4. ##"MgSO"_4## acts as a drying agent to clear the reaction vessel of any water remaining from steps 1 and 2 (you may have used this in lab already); safe way of minimizing potential reactions with water
5. ##"LiCu"("CH"_3)_2## substitutes a methyl group in place of the bromide group

iii) Start from step 2 of part i) and do the same thing from that point on

iv) Starting from any sodium alkanoate

This'll take a while to do in real life...

1. Strong acid protonates the alkanoate to make a carboxylic acid
2. ##"MgSO"_4## dries out the reaction vessel to prevent overly violent reaction in step 3
3. ##"LiAlH"_4## acts as a strong reducing agent that is capable of reducing a carboxylic acid down to the corresponding alcohol
4. Dilute sulfuric acid terminates the reducing process
5. ##"PBr"_3## substitutes the hydroxide with a bromide group
6. ##"HBr"## with a peroxide causes a radical reaction; essentially, anti-Markovnikov addition of a proton to the dimethylated carbon (bottom left) and a bromide to the upper right carbon.
7. Two equivalents of ##"LiCu"("CH"_3)_2## substitutes with each bromide a methyl group (this is not likely to give that great of a yield due to the steric hindrance, but it's theoretical so it's OK)