Monday, December 9, 2013

Sapling: Chapter 8, Problem 22

In this problem, 2-chloro-3-dimethyl-pentane is treated with H2O and -OH to show the results of both an SN1 and SN2 reactions. SN1 reactions react with weak bases, so the compound treated with H2O will cause SN1. SN2 happen in presence of strong bases so -OH will cause an SN1 reaction. First in the SN1 reaction, unlike an SN2 reaction, a carbocation can form. Because of this, when the OH goes to substitute for the Cl, the positive charge on the second carbon is moved to the third carbon with a methyl shift from 3 to 2 (moves from secondary to tertiary to form most stable carbocation). The OH is formed on the third carbon to form 2-methyl-3-methyl-3-pentanol. The SN2 reaction with -OH doesn not have a carbocation, so the OH will form on the secondary second carbon. However, SN2 reactions invert the stereochemistry on the carbon it attaches to (it has a back side attack and comes in on the side opposite the leaving group). because of this, the OH is shown attached at the bottom of the C below where the Cl left.

 I'm really sorry this isn't better drawn.


  1. The SN reactions gave me the most problems. I kept looking for a carbo cation when I didn't need one. The OH on the bottom moves to the most substituted Carbon

  2. I felt the opposite, and had more trouble with the elimination problems. I had trouble remembering where the double bond went. That is why I did a problem on this blog concentrating on both e1 and sn1 reactions.