Starting with butane, explain how you could prepare: 2-bromobutane 1-bromobutane

Question

Starting with butane, explain how you could prepare: 2-bromobutane 1-bromobutane Note: For the following targets, you may also choose to begin with either 2-bromobutane or the 1-bromobutane as prepared above without repeatedly showing how you made these two compounds. Be sure to choose the best starting material for each job! 1-butene 3-methylheptane 5-methyl-3-heptyne

Explanation / Answer

Leaving groups: HOH >> OH- > Br- Nucleophiles: OH- > Br- (>> HOH) in polar protic solvent. 1)2-bromobutane via mixed SN1/SN2 if you drive it with a little heat, 2-butene via E1 if you drive it with a lot of heat. OH- is a much better leaving group after it's protonated, so that's always the first step if there is H+ available. There's no reason to prefer either substitution mechanism; probably both paths are active. Br- though is a very poor base, so E2 isn't very likely. Substituting hydroxy with halogen is kinetically unfavorable, so you must drive the rxn with heat. Elimination is favored entropically (more pieces are in the pot after it occurs) though it's energetically unfavorable at room temperature, so if you drive it very hard with heat you'll favor elimination over substitution. However, you'll only change the mix - you'll still get both substitution and elimination products. 2)if you drive it with heat. The hydroxide abstracts a proton from HBr, making it a good HOH leaving group, which Br- then displaces. However, the rxn, R-OH + H+ + Br- --> R-Br + HOH, is entropically unfavorable, particularly in weak (as in concentration) aqueous acid, since HOH becomes part of the almost-pure solvent. 3)Butene via E1, if you drive it hard with heat, but not butanol. H2O is still a much better leaving group than Br-, and no nucleophile whatsoever. (HOH is not the same beast as OH-.) 4)t-butylbromide via Sn1 with a little heat, 2-methylpropene via E1 with a little more heat. Both rxns are kinetically unfavorable, elimination more than substitution, but elimination is entropically favored while substitution is not, so the mix of products depends on rxn temperature.

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