Friday, December 6, 2013

Sapling Ch8 #24

In an Sn1 mechanism, the leaving group leaves before the nucleophile attaches. This is the first step to the mechansim. The second step is when the nucleophile reacts rapidly with the electrophile, the carbocation. The rate of an Sn1 reaction only depends on the concentration of the alkyl halide. The first product shown reacted with a methanol, a poor nucleophile. The first step was the dissociation of bromide. The carbocation is then formed. A tertiary carbocation is more stable and is formed more easily than a secondary or primary carbocation. Therefore, a 1,2 methyl shift will occur. After the carbocation is settled, the nucleophile will attack from either the front or back side. But in this problem, it does not show whether it attacks from front or back.

A good nucleophile and a strong base proceeds to an Sn2 reaction. For the second product, it involves methoxide ion. This is a good nucleophile. There is no rearrangement needed for this mechanism, the leaving group, bromide dissociates, and the nucleophile attacks giving the second product.



First product

Second product




1 comment:

  1. Thanks for this great explanation with pictures. This really helped me while studying for the finals.

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