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?

1 Answer
Truong-Son N.
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(CH3)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. PBr3 substitutes the hydroxide with a bromide group
  4. MgSO4 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(CH3)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. MgSO4 dries out the reaction vessel to prevent overly violent reaction in step 3
  3. LiAlH4 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. PBr3 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(CH3)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)
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