Chem 220 - Organic Chemistry
Problem Set 5, Solution
Set
Chapter 6, Alkyl Halides: Substitution and
Elimination
Due: Monday, October 12, 2009
 
Study #2 and #3 in the Alkyl Halide module and #1 in the Ether module
in ORGO.
 
   
      | 1. The inversion of configuration in an
         SN2 reaction is often called a Walden inversion,
         named after its discoverer, Paul Walden. In the cycle shown
         above, the overall conversion of one enantiomer of malic
         acid to the other one must require an inversion of
         configuration. Similarly, the same is true of the chloro
         acids. More generally, each interconversion of enantiomers
         must require an odd number of inversions. The
         PCl5 reaction requires a single inversion which
         means that the Ag2O reaction involves an even
         number of inversions of configuration, namely two in this
         instance. (-)-Malic acid is of the
         (S)-configuration. a) Show how malic acid, like any alcohol,
         might react with PCl5 and then undergo inversion
         to form a chloride. Remember that phosphoric acid is a
         strong acid and its conjugate base and analogs thereof are
         also good leaving groups. b) Silver oxide is an anhydrous form of
         AgOH. The carboxylic acid group closest to the hydroxyl
         group plays a role in the process. The reaction medium is
         mildly alkaline. Using these data, show how there is net
         retention of configuration. c) Draw these four enantiomers as Fischer
         projections with the CO2H closest to the OH or Cl
         in the topmost position. (-)-Malic acid is of the
         (S)-configuration.
 | 
 
   
      | a) An electron pair on
         the hydroxyl group of (S)-malic acid does the equivalent of
         an SN2 displacement of chloride on the phosphorus
         atom of PCl5. The proton at the positive site may
         be removed by chloride at this point and then chloride from
         dissociated HCl can effect an SN2 displacement.
         Alternatively, chloride can effect direct SN2
         displacement with inversion of configuration (1st inversion)
         to form (R)-chlorosuccinic acid, POCl3 and HCl.
         [Note: It is also likely that the
         carboxyl groups are converted to acyl chlorides during the
         reaction. Aqueous workup would rapidly reform carboxyl
         groups].  |   | 
 
   
      | b) (R)-Chlorosuccinic
         acid under alkaline conditions is converted to its
         dicarboxylate salt. Ag+1 may or may not complex
         with the chlorine atom at this point to enhance chloride as
         a leaving group. AgOH is not critical. The reaction works
         using NaHCO3 as a base. The proximate carboxylate
         acts as a nucleophile with SN2 inversion to form
         the reactive, transitory
         (R)-a-lactone.
         The strain of the a-lactone allows hydroxide to effect a
         second SN2 displacement to form (R)-malic acid
         upon acidification. This step has an even number of
         inversions -- net retention. The overall process of (R)- to
         (S)-malic acid has an odd number of inversions -- net
         inversion.  |   | 
 
   
      | c)   
 | 
 
 
   
      | 2. 
 | a) This is a primary
         bromide that is chiral by virtue of the H, D, Br and n-butyl
         group attached to carbon 1. RSH: pKa = 10;
         H2O: pKa =
         15.7. Go here for
         pKa's. RSH is a weaker acid
         than water, so the equilibrium favors RS- +
         H2O. Mercaptide ion effects an SN2
         inversion to form the (S)-sulfide.
 b) Chloride is exchanged for bromide by an SN2
         reaction. One can't see the inversion at the primary site
         (proved in 2a) but it is seen at the secondary site.
         cis-Dichloride to trans-dibromide.
 
 c) An SN2 transition state apNote that there is
         approximates a trigonal bipyramid. The apices are the
         nucleophile and the leaving group; the trigonal part is
         CH2, CH2 and H at ~120o.
         The C-C-C bond angle for cyclohexane is 111o;
         cyclobutane, 88o. It is easier for cyclohexane to
         undergo the displacement as long as elimination to
         cyclohexene is not a factor. Note that there is a limited
         amount of nucleophile.
 
 d) 1-Bromo-2-methylpentane is more hindered toward
         SN2 displacement than 1-bromo-3-methylpentane.
         Competition for limited nucleophile favors
         1-bromo-3-methylpentane.
 | 
 
   
      | 3. Show how you would convert
         (S)-2-octanol into (S)-2-octanethiol.
         [Hint: how do you make a secondary hydroxyl group a good
         leaving group for an SN2 reaction?]
         For net retention
         you need an even number of inversions of configuration.
         First, make a leaving group with retention: the tosylate.
         Now do two inversions. |   | 
4. (2S,5S)-5-Bromo-2-hexanol (A) is expected to
form optically inactive B (C6H12O) upon
exposure to aqueous NaOH. An optically active stereoisomer of
A, namely, C also forms optically inactive B
under the same conditions. Two other optically active stereoisomers
of A, namely D and E as a racemic mixture, form
optically inactive F, a diastereoisomer of B. What are
the structures of A-F? The structures D and E
are not distinguishable. Explain and illustrate with mechanisms. Why
are B and F both optically inactive?
   
      | A and
         B differ in formula by HBr. The base
         reversibly forms the alcoxide of A
         which does an intramolecular SN2 displacement
         (Williamson ether synthesis) on the C-Br bond, inverting the
         C5 configuration. B has a
         plane of symmetry. It ia achiral and necessarily optically
         inactive. 2S mirrors 5R. C is the enantiomer of
         A. It also gives B
         by the same mechanism. The diastereomers of
         A and B are
         D and E. Both are
         optically active but they are present as a racemate. They
         afford the racemate F, a diastereomer
         of B. D or
         E as a single enantiomer would give the
         respective enantiomers of F.
         D and E are not
         distinguishable in this problem. F is
         optically inactive because it is a racemate. |   | 
 
   
      | 
            
               | 5. Provide the unknown product
                  of each reaction. In all cases, provide mechanisms
                  and a rationale.
 a) Similar to #4. This time there is a double
                  displacement. Both centers are inverted. A is
                  tetrahydrothiophene.
 
 b) The chlorohydrin is a trans-decalin. Both rings
                  are in chair conformations. The methyl and hydrogen
                  are necessarily axial. The chlorine is 1,3 to the
                  methyl. It is also axial. The same is true of the
                  rlationship between the hydrogen and the hydroxyl
                  group. This means that the chlorine and hydroxyl
                  group are both well disposed for epoxide formation
                  with the epoxide on the same side of the rings as
                  the hydrogen.
     c) The same as b) except that
                  the epoxide is now on the same side of the rings as
                  the methyl group.   d) The same as 2a. The 4S
                  site is not affected. | 
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