CSUMB
ESSP 311 Organic Chemistry I
Ronald W. Rinehart, Ph.D.

Chapter 5 Alkenes: Structure and Preparation
Elimination Reactions

Chapter 5.  Structure and Preparation of Alkenes. Elimination Reactions. 

I.   Structure and Bonding. 
            A.   Double Bond:   C=C 
            B.   Planar, 120^o Bond Angles. 
            C.   Bonding 
                        1.   sp² hybrid orbitals. 
                        2.   One p orbital is perpendicular to the plane of sp² hybrids. 
                        3.   Sigma (s) bonds:  end-to-end overlap utilizing sp² hybrids. 
                        4.   Pi (p) bonds:  sideways overlap of parallel, in-phase p orbitals. 
                        5.   C-C vs C=C bond lengths/strengths: 

                        6.   In propene the C2-C3 bond length is shorter (1.50 Å) than the C2-C3 bond length in propane (1.53 Å).
                                  Why? 

                                                            C=C−C            vs       C−C−C           

                                                                sp²- sp³                 sp³-sp³ 

                                    Answer:  The greater s character sp² hybrid orbital bonds electrons more strongly than
                                    the sp³ hybrid, contracting the sigma bond and shortening the internuclear distance. 

                        7.   The pi bond: 

                         8.   The pi bond leads to stereoisomerism:  cis/trans (Z/E).   

 II.   Nomenclature of Alkenes. 

            A.   IUPAC System.  The nomenclature of alkenes is very similar to the nomenclature of alkanes. 
                        A few important differences are listed: 

                        1.   Replace -ane ending of corresponding alkane with -ene.   
                        2.   The parent chain must be the longest continuous carbon chain containing the C=C group. 
                                    a)   The parent chain is therefore not always  the longest carbon chain in the compound.
 

                        3.   Always number the parent such that the C=C  carbons are given the lowest possible numbers,
                                    and only one number is given in the name. 

                                                                        CH₃CH₂CH=CH₂ 

                                                1-butene, not 1,2-butene or 3-butene 

                                    a)   Name this alkene:  [highlight area below structure to check]

                         4.   In cycloalkenes the C=C linkage is automatically given the numbers 1 and 2, so this is not
                                    mentioned in the name   (cyclohexene). 
                        5.   If  two or more C=C groups are in a compound, the name of the compound ends with
                                    -adiene, -atriene, etc.  

                                    a)   Name the following: 

            B.   Common Names 

                        1.   Some alkenes are known by common names: 

                          2.   Just as there are alkyl groups, alkenyl groups are based upon the corresponding alkene: 

            C.   Geometric Isomerism:  cis/trans Isomers and the E-Z System                       

                        1.   Since rotation about the double bond cannot occur, geometric isomers of alkenes can exist. 
                                    a)  These are referred to as cis and trans  isomers. 

                                      b)   Is the following alkene a  cis or trans isomer? 
                                                What problem develops with this alkene in determining if it is cis or trans? 

                                                i)   To classify compounds of this type the E-Z System was developed. 

                        2.   The E-Z System. 

                                    a)   Rank the atoms directly attached to the double-bonded carbons by atomic number. 
                                    b)   Compare the rankings of the two substituents on one carbon; repeat this for the other carbon. 
                                    c)   If both higher ranking groups are cis to each other then configuration is Z (Zusammen, together) 
                                                E (Entgegen, opposite) is assigned if these groups are opposite each other. 
                                    d)   If the atoms directly attached to a C are the same, work outward from each point until the first                                                 exploitable difference is found [see examples below]

 

                        3.   Name the following alkenes: 

                         note correction to the first name above and problem 4 below

                        4.   Draw the structure for (E)-2-bromo-1-chloropropene:                                                 

III.   Physical Properties. 

            A.   Alkenes Are Similar to Alkanes.    

                        1.   Small dipole moments. 
                        2.   Melting and boiling points are similar:  C2 to C4 alkenes are gases at room temperature. 
                        3.   Alkenes are not soluble in water.  

            B.   Saturated vs Unsaturated Fats/Oils:  Melting Points. 
                        1.  Structure of fats and oils:  Triglycerides. 
                                    a)   Triglycerides are esters of glycerol and three fatty acids: 

                                     b)  The R groups consist of an odd number of carbons and can be either saturated or
                                                            unsaturated. 

                        2.  Fats are triglycerides which are highly saturated and are solids at room temperature.
                            Oils are triglycerides which are highly unsaturated and are liquids at room temperature. 

                        3.  Melting point as a function of unsaturation.       
                                    a)  Saturated fatty acids have higher melting points than unsaturated fatty acids of similar MW. 

                                    b)   Saturated fatty acids have higher melting points due to greater van der Waals interactions and
                                                 better packing ability of their carbon chains. 
                                    c)   The Z (cis) isomers of unsaturated fatty acids have lower melting points than the corresponding
                                                 E (trans) isomers due to poorer packing ability [see structures above].
                                     d)   Greater unsaturation in fatty acids results in lower melting points: 

                                       e)    Vegetable oils are partially hydrogenated in order to convert liquid oils into solid margarine. 
                                                i)   In industry this process is referred to as "hardening". 

IV.   Chemical Properties. 

            A.   Relative Stabilities. 

                        1.   Alkene Stability [the order given for the disubstituted is NOT the one in your text] *see 3 below

                          2.   Combustion Data 

                                    a)    C₄H₈  +  6 O₂  à  4CO₂  +  4H₂O 

                                      b)   Why this order is observed is not well understood.  One theory is that alkyl groups stabilize
                                            double bonds in the same way as with carbocations:  they release electrons into the pi system                                  of the alkene via hyperconjugation (overlap of C-H sigma bond of a neighboring substituent.)   

                                    c)   Another theory:  sp²-sp³ bonds are stronger than sp³-sp³ bonds.  The isomer having more
                                                sp²-sp³ bonds (compared to sp³-sp³ bonds) is more stable. 

                                    sp³-   sp² - sp²- sp³                                       sp³-   sp³ - sp²- sp²  note correction
                                    CH₃-CH=CH-CH₃                                            CH₃-CH₂-CH=CH₂
                                                2-butene                                                          1-butene
                                              more stable                                                       less stable

                    3.  The heat of hydrogenation can also be used to compare stability of double bonds, even for
                             compounds which are not isomers. Hydrogenation will be discussed more in Chapter 6.

             B.   Strain in Cycloalkenes. 

                        1.   Cyclopropene:  Angle strain is present due to 120^o sp²  hybrids vs 60^o angles of a triangle  
                                    a)   This is greater strain than is present in cyclopropane where 109.5^o sp³ hybrids exist.   
                        2.   Cyclobutene:     Less angle strain than in cyclopropene. 
                        3.   Angle strain is negligible in cyclopentenes,  cyclohexenes, etc. 
                        4.   trans-cyclooctene is the smallest ring size possible for the trans isomer. 
                        5.   In rings with less than 12 carbons, the cis isomer is more stable than the trans. 
                                    a)   In rings with more than 12 carbons, the trans isomer is more stable than cis 
                                    b)   (E)-Cyclododecene and (Z)-Cyclododecene have about the same stability.
                         6.   Bredt's rule:  NO C=C at bridgehead C in bicyclic compounds

V.         Preparation of Alkenes. 

            A.        Dehydration of Alcohols 

                        1.        

                        2.   Examples:                                   

                         3.   Examine this reaction:                                                             

                                     a)   Which product is preferred?   

                         4.   Complete the following reactions showing the possible products and indicate which are
                                    major and minor products. 

                         5.   Reactions in which more than one constitutional isomer can be formed from a reactant
                                    but one predominates are REGIOSELECTIVE. 
                        6.   If one isomer is formed exclusively (or almost exclusively):  REGIOSPECIFIC. 
                        7.   Are the following products constitutional isomers?

                                     Answer. NO:  they are stereoisomers.  The above reaction is STEREOSELECTIVE. 

                        8.   Mechanism for dehydration of alcohols. 

                                    a)   Observations:         
                                                i)   Acid Catalyzed. 
                                                ii)   Order of reactivity of alcohols:  3^o  >  2^o  >>  1^o
                                                iii)   Rearrangements can occur. 

                                     b)   Conclusions-- 
                                                i)   Carbocation intermediate 
                                                ii)   E1:  Elimination Unimolecular 

                                    c)   Mechanism:

                                     d)   Write the mechanism for this reaction: 

                        9.   E2 

                                    a)   In some cases for 1^o alcohols:  
                                                E2, Elimination Bimolecular  

                                    b)   But there is evidence that most 1^o alcohols react via carbocations:  Rearrangements ⇒ (E1). 

                                                 i)   Mechanism:  (1,2 hydride shift) 

                         10.   Complete these reactions and show the mechanism.  Also classify the mechanism as E1 or E2. 

                                                i)   A 1,2 methyl shift occurs above in part b.
                                                ii)   When do hydride and alkyl shifts  occur?  If a 1,2-shift of hydrogen or alkyl group can
                                                            form a more stable carbocation, then such a rearrangement usually takes place. 

            B.   Dehydrohalogenation of Alkyl Halides                                     

                        1.       

                        2.   Follows Zaitsev's Rule:  The most substituted alkene forms.  

                                    a)   This makes the reaction regioselective.   
                                    b)   The reaction is also stereoselective: the trans (E) product is favored over the cis (Z) product.

                         3.   Base/Solvent Systems:                                               

                                    a)   EtO¯Na⁺/EtOH   sodium ethoxide in ethanol   
                                    b)   MeO¯Na⁺/MeOH   sodium methoxide in methanol   
                                    c)   tBuO¯K⁺/ tBuOH (or DMSO) potassium tert-butoxide in tert-butanol or dimethyl sulfoxide
                                                (good for 1^o RX) 

                        4.   Complete the following: 

                         5.   Mechanism for the Dehydrohalogenation of  Alkyl Halides: 

                                    a)   Observations for dehydrohalogenation  of alkyl halides: 
                                                i)   Strong base required.
                                                ii)    2^nd order kinetics.  1^st order in RX, 1^st order in base.
                                                iii)   The reaction rate is highly dependent upon the identity of the halogen 
                                                            (to the weakness of  the C-X bond.)    

                                                                        RI  >  RBr  >  RCl  >  RF 

                                                                        Fast                              Slow 

                                                             I¯ good  leaving group        F¯ poor leaving group    

                                                iv)   The reaction rate follows the order:  3^o RX >  2^o RX > 1^o RX
                                                v)   No rearrangements (no carbocations.) 

                                    b)   Conclusion: E2-  Bimolecular Elimination 
                                    c)   Mechanism:   

                                                i)   A concerted mechanism:  Goes through Transition State but no intermediate. 
                                                                        The following all occur simultaneously: 

                                                                      ∙  C-H bond breaks 
                                                                      ∙  C=C pi bond forms 
                                                                      ∙  C-X bond breaks 

                                                No carbocation:  no rearrangement.                                                           

                                  d)   The E2 mechanism is favored by the use of a STRONG BASE.   

                        6.   Contrast this with the dehydration of a primary alcohol (E1):

                        7.   Anti Elimination in E2 Reactions. Stereoelectronic Effects.  
                                    a)   Effects due to one spatial arrangement of electrons/orbitals more stable than others. 

                                     b)   For the activated complex to be stabilized by partial pi bond formation, p orbitals developing
                                                 (due to C's sp³ hybrids changing to sp²) must be parallel:  pi-bond formation is achieved
                                                 when the H-C-C-X unit is coplanar in the transition state.   

                                                i)   Two conformations permit this: syn periplanar (boat: t-butyl group axial/unfavorable) and                                                             anti periplanar.

                        8.   Stereoselectivity in E2 Reactions. 

                                    a)   The dehydrohalogenation of alkyl halides favor the formation of the trans isomer. 

                                     b)   This stereoselectivity can be understood by looking at the Newman projections of the reactant
                                                and transition states.
                                    c)   Notice that when the reactant is in the conformation where the methyl groups are gauche to
                                                each other, this leads to unfavorable steric interactions in the transition state.

                                                i)   This leads to formation of the cis product.

                                    d)   Notice that in the formation of the trans product the transition state is more stable due to lower
                                                 van der Waals strain.

                                                 i)   The trans product is formed faster than the cis product. 

                        9.   E1 and E2 for Alkyl Halides.

                                    a)   E2 Mechanism- Bimolecular Elimination 
                                                i)   Second Order Kinetics:  Rate = k[RX][base] 
                                                ii)   Favored with STRONG BASE. 
                                                iii)   Favored in the order:  3^o RX >  2^o RX > 1^o RX                              

                                     b)   E1 Mechanism- Unimolecular Elimination 
                                                i)   First order kinetics:  Rate = k[RX] 
                                                ii)   Possible for 3^o and (some) 2^o alkyl halides WHEN: base is weak or in low concentrations. 
                                                iii)    Since E1 reactions go through the carbocation intermediate, rearrangements may occur:
                                                      E2 is often a better synthetic method when rearrangement is otherwise possible. 
                                                            (Use strong base.) 

                                                iv)   Favored in the order:  3^o RX >  2^o RX > 1^o RX
                                                v)   Favored in the order:  RI  >  RBr  >  RCl  >  RF   
                                                            since breaking the R-X bond is the rate determining step. 
                                                            The C-I bond is weaker than the C-F bond.      
                                                vi)   Example:

                                     c)   For the above reaction: 

                                                i)   Name the two products.  2-methyl-1-butene  and  2-methyl-2-butene 
                                                ii)   Try writing the mechanism for the formation of the above two products. 
                                                iii)   Why is one product favored over the other? 

                                                            Answer:  Zaitsev's Rule is followed, which results in the formation of the more stable                                                                         (more substituted) alkene. 

                                      d)   E1 reactions are normally carried out in the absence of added base:  the solvent acts as the
                                                 base and a polar protic solvent is chosen (such as an alcohol) in order to help form the
                                                carbocation.  If even small amounts of strong base present, the E2 reaction tends to occur
                                                 faster than E1.

Many thanks to Rod Oka of MPC for generously sharing his "Lecture Companion" outline,
reproduced here in edited form by permission, with web references and other goodies added by me.
Structures drawn using  MDL IsisDraw™ , CS ChemOffice ChemDraw™, and ACDLabs ChemSketch™.