Alkylhalides
Substitution reaction Sn2
In the previous blog post there was a little bit of theory explained about the substitution reaction with alkylhalides. In this post I dig deeper into the mysteries of the Sn2 reaction mechanism.
Let's start with a very basic substitution reaction of Bromomethane forming methanol.
The reaction rate law for the Sn2 reaction (and thus for the example above) is then:
As you can see, this law contains the alkyl halide AND the nucleophile, making it a 2nd order reaction (hense the 2 in Sn2). k is the 'rate constant', which is different for any reaction but remains constant.
Branched off alkyl halides will have a slower reaction rate than none-branched off alkyl halides.
Methyl halide such as CH3-Br has a relative reaction rate of 12000, a primary alkyl such as CH3CH2-Br already has a relative reaction rate of 40, and a tertiary alkyl halide such as (CH3)(CH3)CH3-C-Br is even too slow to measure the reaction rate.
Methyl halide such as CH3-Br has a relative reaction rate of 12000, a primary alkyl such as CH3CH2-Br already has a relative reaction rate of 40, and a tertiary alkyl halide such as (CH3)(CH3)CH3-C-Br is even too slow to measure the reaction rate.
Why and how does the fenomenon occur?
The answer can be found at the place of attack by the nucleophile.
The nucleophile will attack at the backside of the alkyl group that is bound to the halogen. A tertiary alkyl halide will therefore undergo no reaction. The nucleophile will not attack at the front side because of the formed bonding between Br and C, this will require to much energy and is not a logical possibility. This reaction scheme will clear it up a little
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