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E2 elimination of (E) and (Z) Acids?

  1. Nov 25, 2012 #1
    1. The problem statement, all variables and given/known data
    Both (E)- and (Z)-2-chlorobutenedioic acids dehyrochlorinate to give acetylene dicarboxylic acid. This reaction proceeds through the addition of a base and an E2 elimination. The (Z) acid reacts about 50 times faster than the (E) acid. Explain why.

    2. The attempt at a solution
    My reasoning on why the (Z) acid reacts faster is through the geometry of the substituents. I am to believe, through the (Z) configuration of the acid, that the acid contains an anti periplanar conformation, where the Chloride and Hydrogen substituents are not on the same side. In contrast, the (E) acid would have a syn conformation, where the Chloride and hydrogen substituents are on the same sides.

    In my textbook, it states that an E2 reaction is more favored if the molecule of interest is in an anti periplanar geometry. This is because the anti periplanar utilizes a staggered conformation, totaling less energy (more stable). The syn periplanar utilizes an eclipsed conformation, totaling higher energy (less stable). However, the example used was for a typical two carbon alkane, an ethane molecule with different substituents. Through E2, the alkane becomes an alkene.

    The molecule described in the problem is an alkene, and through E2 becomes an alkyne. As thus, how does the syn and anti periplanar explanation work for alkenes? Regardless of geometry, the alkene will assume an eclipsed conformation, where all the substituents lie on the same plane.

    This is where I am stuck. If the alkenes do not possess varying conformations and only utilize an eclipsed form, then what is there to account for the kinetic difference proposed by the question?

    EDIT: I have a new hypothesis: Through the (E) acid where the Chloride and Hydrogen substituents are on the same side, could the presence of the base be hindered by Chloride? Meaning, the base wants to attack Hydrogen, but is hindered by Chloride because it is on the same side as Hydrogen (they're all near each other). Now in contrast, the (Z) acid would not be hindered as much because the Chloride and Hydrogen substituents are on different sides. So when the base attacks Hydrogen, Chloride won't hinder the base as much. Can someone confirm if this has any relevance to the initial question?
     
    Last edited: Nov 25, 2012
  2. jcsd
  3. Nov 27, 2012 #2

    chemisttree

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    It doesn't have anything to do with steric hinderance of the base with chlorine. In the E isomer, the bulky carboxylic groups are on the same side. Add a base and you get a carboxylate. Add another base and you get a dianion. The third equivalent pulls off the hydrogen.

    Two groups, cis, and negatively charged as well. Ugh! How repulsive!
     
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