3,3-Dimethylbut-1-ene, CH₂=CH-C(CH₃)₃, is treated with hydrogen chloride at room temperature, with…
Contributed by StemPy LabTry it on paper first. A hint, the key idea and the worked solution are here when you need them.
3,3-Dimethylbut-1-ene, , is treated with hydrogen chloride at room temperature, with no peroxide. A chemist might expect 2-chloro-3,3-dimethylbutane, because the first step forms a secondary carbocation on carbon 2. The product that forms in the larger amount is a different compound.
The secondary cation sits next to a carbon that has three methyl groups and no hydrogen atom. A methyl group, with its bonding pair of electrons, can move from that carbon to the positive carbon. Chloride ions are plentiful in the solution and attack any cation they meet. The two events compete, and the product mixture shows which one is faster and which cation is lower in energy. A student analyses the mixture. She finds that the chlorine atom is attached to a different carbon from the one that the double bond started on.
- Draw the cation after the shift, and say whether it is primary, secondary or tertiary.
- Give the name of the main product.
- Count the carbocations in the whole pathway to that product, and the alkyl groups on the positive carbon of the last one.
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
Have you tried it?
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.
Step 1. First cation. adds to carbon 1: . It is secondary.
Step 2. The shift. A methyl group moves from carbon 3 to carbon 2, and carbon 2 becomes a group with two methyl groups. Carbon 3 is left with two methyl groups and the positive charge: . It holds three alkyl groups: two methyl groups and an isopropyl group. The ion is tertiary.
Step 3. The product. Chloride attacks the tertiary cation: 2-chloro-2,3-dimethylbutane, .
Step 4. Count. There are two carbocations in the pathway, the secondary one and the tertiary one. The last has alkyl groups on the positive carbon.
Answer: tertiary cation, 2-chloro-2,3-dimethylbutane, cations and alkyl groups. The unrearranged product, 2-chloro-3,3-dimethylbutane, is the minor one.
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