Chapter Nine Homework Question

Sapling Homework Question

Jessica Mayhew

 

Part a.)   

On the left side of the picture you will find my work for the Sn1 mechanism. First, the C-Br bond broke and created a secondary carbocation. The secondary carbocation was then converted into a tertiary carbocation, which was more stable. One thing you have to watch out for when working on a Sn1 mechanism is carbocation stability because choosing the less stable carbocation could potentially yield the incorrect final product. Next, the hydroxide ion attacked the tertiary, more stable carbocation intermediate. This resulted in the formation of 2-methyl-2-pentanol.

Part b.) 

On the right side of the picture you will find my work for the Sn2 mechanism. In order to experience an Sn2 reaction you must have a secondary alkyl halide. 3-bromo-2-methylpentane is, in fact, a secondary alkyl halide because the carbon attached to the halogen group (Br) is also directly attached to two other alkyl groups. Therefore, it is completely possible for this reaction to undergo the Sn2 mechanism. Since the Sn2 mechanism is concerted, meaning everything happens in one step, the nucleophile and substrate are both accounted for in the rate determining step. At the same time that the leaving group (Br) is leaving, the hydroxide ion is reacting with the substrate. This one step mechanism results in 2-methyl-3-pentanol.

REFERENCES

http://www.saplinglearning.com/ibiscms/mod/ibis/view.php?id=1390616

http://www.masterorganicchemistry.com/2012/08/08/comparing-the-sn1-and-sn2-reactions/