Give the main organic product of each reaction, and name the rule or effect that decides it…
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Give the main organic product of each reaction, and name the rule or effect that decides it.
- 2-Methylbut-2-ene with hydrogen bromide, no peroxide.
- 2-Methylbut-2-ene with hydrogen bromide in the presence of benzoyl peroxide.
- But-1-ene with hydrogen iodide.
- 3-Methylbut-1-ene with hydrogen bromide, no peroxide.
Need a reminder? The key ideaShow
An alkene adds HX by Markovnikov's rule: the more stable carbocation forms, so the halide goes to the carbon with more alkyl groups. A carbocation can rearrange to a more stable one by a hydride or methyl shift. With a peroxide, HBr adds by a radical chain to give the anti-Markovnikov product, because the more stable radical forms.
The worked solution
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Work it out on paper first. It is fine to get stuck or make a mistake: that is how we learn. When you are ready, go through the solution one step at a time, and check each step against your own.
(a) The hydrogen goes to the carbon with one hydrogen, which leaves the tertiary cation. Bromide adds there: 2-bromo-2-methylbutane, . Markovnikov's rule.
(b) With a peroxide the bromine atom adds first, to the carbon with one hydrogen atom, so the more stable tertiary radical forms on the other carbon. The product is 2-bromo-3-methylbutane, . Peroxide effect, anti-Markovnikov.
(c) The cation forms on carbon 2: 2-iodobutane, . Markovnikov's rule. Hydrogen iodide shows no peroxide effect.
(d) The first cation, secondary, forms on carbon 2. A hydride shift from carbon 3 makes a tertiary cation. Bromide adds to it: 2-bromo-2-methylbutane is the main product. A little 2-bromo-3-methylbutane forms without the shift. Rearrangement.
Answer: (a) and (d) give 2-bromo-2-methylbutane, (b) gives 2-bromo-3-methylbutane, and (c) gives 2-iodobutane.
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