Give the full mechanism for the reaction of 1-bromo-3-methylpentane with methanol.
I know you cant draw on here. but can you tell write it out with words?
Thanks!
Sure! The reaction you're referring to is the nucleophilic substitution of 1-bromo-3-methylpentane with methanol. Here's the step-by-step mechanism:
Step 1: Nucleophilic Attack
Methanol (CH3OH) acts as the nucleophile and attacks the carbon atom bonded to the bromine in 1-bromo-3-methylpentane. The lone pair on the oxygen of methanol attacks the slightly positive carbon, forming a new bond between the carbon and oxygen, while simultaneously breaking the carbon-bromine bond.
Step 2: Formation of the Alkoxy Intermediate
After the carbon-oxygen bond is formed, the resulting intermediate is an alkoxy group (-OR), with the alkyl group (R) corresponding to the original 1-bromo-3-methylpentane molecule. In this case, it would be a 3-methylpentyl alkoxy group (-OCH3).
Step 3: Leaving Group Departure
The bromine atom, which now carries a negative charge, leaves as a bromide ion (Br-) due to the breaking of the carbon-bromine bond.
Step 4: Proton Transfer
In this step, a proton from the methanol molecule transfers to the oxygen in the alkoxy group, resulting in the formation of a neutral alcohol. This step restores the charge neutrality of the molecule.
Overall, the reaction of 1-bromo-3-methylpentane with methanol results in the substitution of the bromine atom with the methoxy (-OCH3) group, forming 3-methylpentyl methanol.