Butane is organic compound of an alkane row. It is colourless gas which is formed in the course of processing (cracking) of oil. At high concentrations butane is poisonous, also this hydrocarbon goryuch and is explosive. Receive it in laboratory and in the industry in various ways. However on a practical training in chemistry use two of them.
Instruction
1. In both cases initial substance is ethane. The thing is that in school or institute receivinglaboratory butane it is possible with the minimum expenses of energy resources. Since cracking of oil products, process of dehydration of butanol and hydration of other various substances are feasible only under big pressure with a temperature in several hundreds and even thousands of degrees Celsius.
2. In these conditions of the most rational Vyurts's reaction is represented. On initial ethane (CH₃-CH ₃) work solution of iodine, bromine or chlorine. Since all saturated hydrocarbons are perfectly halogenated, as a result of reaction at you halogen-ethane will turn out: CH₃-CH ₃ + Br ₂ → CH₃-CH -Br + HBr.
3. Then add equivalent amount of metal sodium to the received brometan: 2C₂H₅Br (brometan) + 2Na (sodium) → C₄H ₁₀ (butane) + 2NaBr (sodium bromide). Two molecules of sodium CH₃-CH -Br + Na interact with each two molecules of a brometan. Metal takes away halogen, and two ethyl radicals of CH₃-CH ₂-+ - CH₂-CH ₃ unite with formation of new connection – butane (CH₃-CH -CH -CH ₃).
4. Other way is less rational, but if you are faced by a task to receive butane from ethane with an intermediate stage at which propane is formed such option quite will approach. As well as in the first case, subject ethane to halogenation: CH₃-CH ₃ + Br ₂ → CH₃-CH -Br +HBr.
5. At the second stage add to the formed brometan brommetan and sodium, from each of connections metal sodium takes away bromine. So you receive propane: CH₃-CH -Br (brometan) + CH₃Br (brommetan) + 2Na → CH₃-CH -CH ₃ (propane) + 2NaBr (sodium bromide). Further you subject to halogenation already propane: C₃H ₈ + Br ₂ → C₃H₇Br (brompropan) + HBr.
6. To the received brompropan add brommetan and metal sodium. As a result of an otshchepleniye bromide ions and connections of propyl and methyl radicals you receive butane: CH₃-CH -CH -Br (brompropan) + CH₃Br (brommetan) + 2Na → CH₃-CH -CH ₂-CH ₃ (butane) + 2NaBr.