Give the full mechanism for the reaction of 3-bromo-3-methylpentane with methanol.
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To provide the full mechanism for the reaction of 3-bromo-3-methylpentane with methanol, we need to understand the process of nucleophilic substitution. Nucleophilic substitution is a reaction in which a nucleophile replaces a leaving group in a molecule.
Here's the step-by-step mechanism for the reaction of 3-bromo-3-methylpentane (also known as tert-butyl bromide) with methanol:
1. Step one: Formation of the carbocation intermediate
- Initially, the lone pair of electrons on the oxygen atom in methanol attacks the carbon atom of the bromine in 3-bromo-3-methylpentane. This leads to the formation of a pentacoordinated intermediate with a positive charge on the carbon atom attached to the bromine.
- This formation of a carbocation intermediate occurs due to the departure of the bromine as a leaving group.
2. Step two: Nucleophilic attack by methanol
- In the next step, the lone pair of electrons on the oxygen atom in methanol attacks the carbocation intermediate.
- Simultaneously, the oxygen-hydrogen bond in methanol is broken, and the hydrogen ion (H+) is taken away by another methanol molecule.
- This attack by methanol on the carbocation leads to the substitution of the bromine group with a methoxy (-OCH3) group.
The overall reaction equation for the mechanism described above can be written as follows:
3-bromo-3-methylpentane + methanol → 3-methyl-3-pentanol + hydrogen bromide
Note: The reaction typically occurs under acidic conditions, which provide the necessary protonation step to form a more reactive carbocation intermediate. This is due to the fact that methanol is not a particularly strong nucleophile on its own when reacting with alkyl halides.
Remember to always consider the conditions (temperature, pressure, catalysts, etc.) and any specific requirements specified in the reaction conditions to fully understand the reaction mechanism.