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?

Aug 26, 2015

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. ${\text{PBr}}_{3}$ substitutes the hydroxide with a bromide group
4. ${\text{MgSO}}_{4}$ 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. ${\text{MgSO}}_{4}$ dries out the reaction vessel to prevent overly violent reaction in step 3
3. ${\text{LiAlH}}_{4}$ 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. ${\text{PBr}}_{3}$ substitutes the hydroxide with a bromide group
6. $\text{HBr}$ 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)