write a step-by-step
mechanism for the reaction of 1-butanol with sodium bromide and sulfuric acid.
Sure! Here's a step-by-step mechanism for the reaction of 1-butanol with sodium bromide and sulfuric acid:
Step 1: Formation of protonated alcohol
1. Add 1-butanol (CH3(CH2)3OH), sodium bromide (NaBr), and sulfuric acid (H2SO4) to a reaction flask.
2. The sulfuric acid (H2SO4) protonates the 1-butanol molecule by donating a hydrogen ion (H+), forming a protonated alcohol intermediate. This is indicated by a positive charge on the oxygen atom of the 1-butanol.
Step 2: Formation of an alkyl bromide
3. The protonated alcohol intermediate reacts with sodium bromide (NaBr) present in the reaction mixture.
4. One of the lone pairs of electrons on the oxygen atom of the protonated alcohol attacks the partially positive bromine atom of the sodium bromide.
5. Simultaneously, the sodium ion (Na+) is released.
6. This nucleophilic substitution reaction leads to the formation of an alkyl bromide, where the bromine atom replaces the -OH group in 1-butanol.
Overall equation:
1-butanol + sodium bromide + sulfuric acid → 1-bromobutane + sodium bisulfate + water
Important Note: The above mechanism explains the reaction at a molecular level, focusing on the key steps involved. However, it is important to remember that reaction conditions and other factors may influence the reaction pathway and the actual mechanism. It is always recommended to consult reliable chemical references or literature for specific reaction conditions and details.