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CH 7

VIEWS: 3 PAGES: 27

									             CHE-240
               Unit 3
Physical and Chemical Properties and
 Reactions of Alkenes and Alkynes

        CHAPTER SEVEN

          TERRENCE P. SHERLOCK
       BURLINGTON COUNTY COLLEGE
                  2004
     Orbital Description
•   Sigma bonds around C are sp2 hybridized.
•   Angles are approximately 120 degrees.
•   No nonbonding electrons.
•   Molecule is planar around the double bond.
•   Pi bond is formed by the sideways overlap of
    parallel p orbitals perpendicular to the plane
    of the molecule.
                                                =>

                       Chapter 7                     2
          Pi Bond
• Sideways overlap of parallel p orbitals.
• No rotation is possible without breaking
  the pi bond (63 kcal/mole).
• Cis isomer cannot become trans without
  a chemical reaction occurring.




                                             =>
                  Chapter 7              3
          Elements of
          Unsaturation
• A saturated hydrocarbon: CnH2n+2
• Each pi bond (and each ring) decreases the
  number of H’s by two.
• Each of these is an element of unsaturation.
• To calculate: find number of H’s if it were
  saturated, subtract the actual number of H’s,
  then divide by 2.
                                              =>
                      Chapter 7              4
  Propose a Structure:
               for C5H8

• First calculate the number of elements of
  unsaturation.
• Remember:
   A double bond is one element of unsaturation.
   A ring is one element of unsaturation.
   A triple bond is two elements of unsaturation. =>


                        Chapter 7                       5
        Heteroatoms
• Halogens take the place of hydrogens, so
  add their number to the number of H’s.
• Oxygen doesn’t change the C:H ratio, so
  ignore oxygen in the formula.
• Nitrogen is trivalent, so it acts like half a
  carbon.
                   H H             H
                   C C N C             =>
                   H H H H
                       Chapter 7                  6
  Structure for C6H7N?
• Since nitrogen counts as half a carbon,
  the number of H’s if saturated is
  2(6.5) + 2 = 15.
• Number of missing H’s is 15 – 7 = 8.
• Elements of unsaturation is 8 ÷ 2 = 4.

                                     =>

                   Chapter 7                7
 IUPAC Nomenclature
• Parent is longest chain containing the
  double bond.
• -ane changes to -ene. (or -diene, -triene)
• Number the chain so that the double
  bond has the lowest possible number.
• In a ring, the double bond is assumed to
  be between carbon 1 and carbon 2.
                                          =>
                   Chapter 7              8
 Name These Alkenes
 CH2    CH CH2      CH3

        1-butene
                                               CHCH2CH3
                                         H3C
  CH3   C CH CH3
        CH3                   2-sec-butyl-1,3-cyclohexadiene
 2-methyl-2-butene

              CH3



3-methylcyclopentene                  3-n-propyl-1-heptene
                                                             =>
                          Chapter 7                          9
    Name these:

     H         CH3                 Br         Br
         C C                            C C
CH3CH2         H                    H         H

   trans-2-pentene               cis-1,2-dibromoethene


                                                   =>

                     Chapter 7                       10
   E-Z Nomenclature
• Use the Cahn-Ingold-Prelog rules to
  assign priorities to groups attached to
  each carbon in the double bond.
• If high priority groups are on the same
  side, the name is Z (for zusammen).
• If high priority groups are on opposite
  sides, the name is E (for entgegen).
                                         =>
                   Chapter 7              11
    Example, E-Z

  1           1     2           Cl
H3C                H            1
              Cl                 CH CH3
        C C         C C
    H         CH2                H
              2 1               2
2
        2Z              5E

(2Z, 5E)-3,7-dichloro-2,5-octadiene
                                          =>

                    Chapter 7             12
Commercial Uses:
   Ethylene




                   =>

       Chapter 7   13
Commercial Uses:
   Propylene




                   =>
       Chapter 7   14
Other Polymers




                  =>

      Chapter 7   15
    Substituent Effects
• More substituted alkenes are more stable.
 H2C=CH2 < R-CH=CH2 < R-CH=CH-R < R-CH=CR2 < R2C=CR2
 unsub. < monosub. < disub.         < trisub. < tetra sub.
• Alkyl group stabilizes the double bond.
• Alkene less sterically hindered.




                                                              =>
                        Chapter 7                            16
 Cycloalkene Stability
• Cis isomer more stable than trans.
• Small rings have additional ring strain.
• Must have at least 8 carbons to form a
  stable trans double bond.
• For cyclodecene (and larger) trans
  double bond is almost as stable as the
  cis.
                                        =>
                   Chapter 7                 17
       Alkene Synthesis
           Overview

•   E2 dehydrohalogenation (-HX)
•   E1 dehydrohalogenation (-HX)
•   Dehalogenation of vicinal dibromides (-X2)
•   Dehydration of alcohols (-H2O)
                                           =>


                      Chapter 7             18
  Removing HX via E2
• Strong base abstracts H+ as X- leaves
  from the adjacent carbon.
• Tertiary and hindered secondary alkyl
  halides give good yields.
• Use a bulky base if the alkyl halide
  usually forms substitution products.
                                       =>

                  Chapter 7             19
 Some Bulky Bases
     CH3          CH(CH3)2
           _
H3C C O           N   CH(CH3)2
                  H                 H3C   N        CH3
     CH3
 tert-butoxide   diisopropylamine   2,6-dimethylpyridine



                  (CH3CH2)3N :
                  triethylamine
                                              =>


                       Chapter 7                    20
     Hofmann Product
• Bulky bases abstract the least hindered H+
• Least substituted alkene is major product.
      H CH3                 _    H3C            CH3 CH3CH2          H
                 CH3CH2O                  C C                C C
  CH3 C C CH2
                 CH3CH2OH             H         CH3    H3C          H
      H Br H
                                           71%                29%


    H CH3               _       H3C          CH3 CH3CH2            H
                (CH3)3CO                                     C C
 CH3 C C CH2
                CH3CH2OH
                                      C C
                                             CH3                   H
                                                                            =>
                                  H                    H3C
    H Br H
                                          28%                72%
                            Chapter 7                                  21
  E2: Cyclohexanes




Leaving groups must be trans diaxial. =>
                 Chapter 7             22
 E2: Vicinal Dibromides
• Remove Br2 from adjacent carbons.
• Bromines must be anti-coplanar (E2).
• Use NaI in acetone, or Zn in acetic acid.
      -
   I
          Br
  H            CH3
                                  H          CH3
                                       C C
                                 H3C         H
               H                                   =>
  CH3
          Br
                     Chapter 7                     23
    Removing HX via E1
•   Secondary or tertiary halides
•   Formation of carbocation intermediate
•   Weak nucleophile
•   Usually have substitution products too
                                             =>



                     Chapter 7               24
      Dehydration of
        Alcohols
• Reversible reaction
• Use concentrated sulfuric or phosphoric
  acid, remove low-boiling alkene as it
  forms.
• Protonation of OH converts it to a good
  leaving group, HOH
• Carbocation intermediate, like E1
• Protic solvent removes adjacent H+
                   Chapter 7
                                        =>25
          Dehydration
          Mechanism
                                            H
H O H          O                        H O H            _
C C        H O S   O H                  C C           HSO4
               O




      H
 H O H               H
                                     H2O:
 C C                 C C                        C C            +
                                                             H3O        =>


                         Chapter 7                                 26
              POWER POINT IMAGES FROM
           “ORGANIC CHEMISTRY, 5TH EDITION”
                      L.G. WADE
    ALL MATERIALS USED WITH PERMISSION OF AUTHOR

PRESENTATION ADAPTED FOR BURLINGTON COUNTY COLLEGE
             ORGANIC CHEMISTRY COURSE
                        BY:
 ANNALICIA POEHLER STEFANIE LAYMAN    CALY MARTIN




                       Chapter 7                     27

								
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