Arun
Last Activity: 6 Years ago
Alcohol (R-CH2-OH) is a good nucleophile, because the oxygen atom possesses two lone-pair electrons. However, in the desired reaction for producing alkyl bromide (R-CH2-Br) from the alcohol (R-CH2-OH), it is not the alcohol that acts as the nucleophile, but rather the bromide ion (Br-). The alcoholic carbon -*CH2-OH) is the site of nucleophilic attack by Br-. In this substitution reaction, -OH group is the “leaving” group. However, there is little tendency for -OH group to leave the carbon when it can stay quite happily bonded to it. So, an inducement has to be provided to make the -OH group a “better” leaving group.
To do this, you add H2SO4, which provides H+ ions that will first protonate the alcohol.
NaBr
This is why SN2 bromination of an alcohol requires acid catalysis.
The above transition state quickly loses H2O (a stable neutral leaving group) and the negative charge of Br- neutralizes the positive charge on the carbon, and a stable neutral molecule, R-CH2-Br is formed as the product, as well as NaHSO4.
NaHSO4.Na+ + HSO4-
[Br-…CH2(R)….+OH2}Br- + R-CH2-OH2]+
The protonated alcohol has a positive charge on the oxygen that makes the entire -OH2+ group a very good leaving group. This allows the -*CH2-OH2+ carbon to become a very desirable site for nucleophilic attack by Br-.
R-CH2-OH2]+R-CH2-OH + H+
H+ + HSO4]-H2SO4
Na+ + Br-
