Chemistry 307 Chapter 7 Structure and Synthesis of Alkenes

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1 Organic hemistry Alkenes. D. oth hemistry 307 hapter 7 Structure and Synthesis of Alkenes Alkenes are hydrocarbons with (=) double bonds. Many alkenes occur in interesting natural products and others are of great industrial importance: ethylene is the largest volume industrial organic compound. 1. IUPA nomenclature ule 1: Find the longest chain containing the alkene function ule 2: Number the ene function low ule 3: Substituents are named as "prefixes" in alphabetical order 4-ethyl-1-methylcyclohexene ule 4: Geometric isomers with like substituents are differentiated by the terms cis- trans- (Latin) cis-hex-3-ene trans-hex-3-ene cyclopentene trans-cyclodecene ule 5: Geometric isomers with different substituents are designated by the terms Z- (zusammen) E- (entgegen) (German) h h h l Z- E- l l l h The pairs of geminal substituents are assigned priorities according to the Sequence ule. ule 6: the O function has priority over the ene function 3-propyl-1-octene 2 O 2-propyl-6-hepten-1-ol 1

2 Organic hemistry Alkenes. D. oth ule 7: small alkene side chains are alkenyl- trans-1-butenylcyclopropane Please, review priorities (atomic number of directly bonded atoms, work to first point of difference, count double bonds twice, triple bonds three times, use mass number for isotopes). ere is your chance to practice J 2 N O O 2. Structure and Bonding in Alkenes 3 D 3 Alkenes contain two sp2 hybridized carbon atoms connected by one σ bond and one π bond. We can construct the bonds of ethylene by combining two sp2 hybridized carbon atoms. π bond σ bond Alkenes are planar with bond angles of ~120 ; the π bond prevents free rotation about the = bond. This is the reason for cis-trans isomerism (geometric isomerism). The = bond is shorter and stronger than the bond Å Å The π bond is weaker (more reactive) than the σ bond. Thermal isomerization is possible; the energy required (Ea = 65 kcal mol 1 ; 272 kj mol 1 ) identifies the strength of the π bond. As the overall strength of the 2

3 Organic hemistry Alkenes. D. oth = bond is 173 kcal mol 1 (725 kj mol 1 ), we can assign 108 kcal mol 1 (453 kj mol 1 ) to the strength of the single bond. 3. Physical properties of alkenes i) boiling points are similar to those of alkanes; ii) melting points reflect the different packing of cis- and transalkenes: cis-isomers have lower melting points; iii) bonds between alkyl groups and sp2 hybridized carbons are polarized in the direction of the sp2 carbon; symmetrical transalkenes have no net dipole moment whereas cis-alkenes have (small) net dipole moments; the example shows the 1,2- dichloroethylene isomers, in which the l bonds are polarized toward the l. iv) l µ = 2.4 D bp = 60 l l µ = 0 D bp = 48 the acidity of an alkene bond (pka = 44) is slightly higher than that of an alkane bond (pka = 50). 4) Degree of Unsaturation A comparison of the molecular formula of a compound with that of a fully saturated compound provides the important information how many equivalents of hydrogen are missing. omparing the general molecular formula of alkanes n 2n + 2 cycloalkanes n 2n alkenes n 2n l 3

4 Organic hemistry Alkenes. D. oth we note that cycloalkanes and alkenes contain 2 fewer than alkanes. We can visualize the formation of alkenes by removing 2 from adjacent carbons and connecting them with a pi bond (shown in blue). Likewise, cycloalkanes can be thought of as formed by removing 2 from nonadjacent carbons and connecting them by a sigma bond (shown in red). 3 3 We state that cycloalkanes and alkenes have one degree of unsaturation. In general, compounds can have a) more than one ring b) more than one double bond or c) a combination of rings and double bonds We define the degree of unsaturation as the total number of rings plus double bonds A saturated hydrocarbon with n atoms has 2n +2 atoms ( = sat ). The degree of unsaturation is D(U) = ( sat ) ( actual ) 2 For example, isopentane 5 12 D(U) = 0 pentene 5 10 D(U) = 1 cyclobutane 4 8 D(U) = 1 If heteroatoms are present: you may disregard the (divalent) O and S atoms because they are inserted between and ; each halogen atom replaces a hydrogen atom - therefore, their number is subtracted; because nitrogen is trivalent, you can visualize that nitrogen is inserted between and as N therefore, each nitrogen atom increases the value of sat by 1. Overall, sat = 2n + 2 n hal + n N For example, 2-chloropentane 106l6 D(U) = 0 3-bromopentene 5 9 D(U) = iodocyclobutane 4 7 I D(U) = 1 ere is your chance to practice with D(U)s > 1 J trichloroacetone 33l3O D(U) = hexachlorodicyclopentadienene 106l6 D(U) = 1,4-dicyanobenzene 5 9 D(U) = 4

5 Organic hemistry Alkenes. D. oth 5. NM of alkenes Alkenes have deshielded 1 and 13 chemical shifts in the range ppm and ppm, respectively. The coupling of alkene protons depends significantly on their orientation. oupling between 1 nuclei on adjacent carbons is called vicinal, that between 1 nuclei on the same carbon is called geminal (typically small in alkenes) Jtrans > Jcis > Jgem. The spectrum of styrene (left) shows the three 1 resonances of the vinyl group well separated at ~5.2, 5.65, and 6.6 ppm, respectively, well deshielded, though not as much as the phenyl group (>7 ppm). The resonance at 6.6 ppm (center) has splittings of 17 and 11 z for Jtrans and Jcis, respectively; it belongs to the internal 1; Jgem is much smaller, 1.4 z; it is not noticeable in the spectrum (left). The resonance at 5.65 ppm (right) has splittings of 17 and 1.4 z; it belongs to the proton trans to the internal 1. The spectrum of divinylketone (below) shows the three 1 resonances of the vinyl groups 5.6, 6.2, and 6.6 ppm, respectively. The resonance at 6.6 ppm belongs to the internal 1, Jtrans = 16.8, Jcis = 10.0 z; Jgem again is 5

6 Organic hemistry Alkenes. D. oth much smaller, 2.1 z; barely noticeable in the spectrum. The resonance at 6.2 ppm, splittings of 16.8 and and 2.1 z belongs to the trans proton. [coupling constants in alkenes is a favorite exam topic J ] 6. atalytic ydrogenation Molecular hydrogen can be added to the alkene pi-bond with the aid of a catalyst. We measure the heat required for (or released) in this reaction (heat of hydrogenation) Pd/ PtO 2 Ni mechanism (preview) The heats of hydrogenation are important because they allow us to evaluate the relative stabilities of isomeric alkenes. We compare the heats of hydrogenation of three methylbutene isomers, which all give rise to 2- methylbutane. The isomer with the lowest heat of hydrogenation is the most stable. 6

7 Organic hemistry Alkenes. D. oth More highly substituted alkenes are more stable because alkyl substitution stabilizes double bonds. This is due to an electronic effect, hyperconjugation (or σ,π-electron donation). You will remember that alkyl substituents also stabilize radicals and carbocations by the same mechanism. The disubstituted alkenes show evidence for steric destabilization, i.e., trans smaller Δ (more stable) Overall ranking of alkene stabilities: cis larger Δ (less stable) mono-substituted tetra- > tri- > trans-di- > cis-di- > internal terminal decreasing stability 7. Formation of alkenes by E2 reactions The E2 reaction is the most general way to generate alkenes in the laboratory. You are familiar with the basic features of the E2 reaction: the reaction proceeds best when the and LG bonds are antiperiplanar to each other). 7

8 Organic hemistry Alkenes. D. oth B: B B a. egioselectivity If a proton can be removed from more than one carbon, isomeric alkenes are generated (identical composition, different constitution). The direction of the E2 reaction is called its regiochemistry (See hapter 6). In some cases, formation of the thermodynamically more stable alkene is favored by a more stabilized transition state, e.g., O Na O 70 % 30 % The removal of the hydrogen from the more highly substituted carbon atom is said to follow the Saytzev rule (where have you heard that before?). With a more hindered base this alkene yields a different result, e.g., 3 3 ( 3 ) 3 O Na ( 3 ) 3 O 27 % 73 % The more hindered base cannot reach the 2 hydrogen as easily as the 1 one. Therefore, the more stable alkene is formed via a more hindered, i.e., destabilized transition state. This regiochemistry is said to follow the ofmann rule. One additional example, 2-bromo-2-methylbutane. 8

9 Organic hemistry Alkenes. D. oth b. Stereoselectivity In some cases removal of two different protons from the same carbon can generate geometric isomers (in addition to the constitutional isomer). We have been dealing with this in detail when we discussed the E2 reaction O Na O % 18 % 2 31 % 2 5 c. Stereospecificity Optically pure haloalkanes with a single on the adjacent carbon produce alkenes stereospecifically. There is only one conformation that allows E2 elimination. * * 3 2 O Na O,-, S,S-,,S-, S,-,-, S,Sform E-,S-, S,form Z- Likewise, elimination of two vicinal bromine atoms by iodide ion (in acetone) yields one alkenes stereospecifically. This is not an important synthetic reaction, but the mechanism is analogous ot that of the E2 elimination of. I I Na + I 2 5 acetone Dehydration of alcohols You will soon learn about the formation of alkenes by dehydration of alcohols, proceeding by the E1 as well as the E2 mechanism. Dehydration via E1 may proceed with rearrangement as you may remember. The carbocation rearrangements we discussed were initiated typically by the dissociation of a good LG from a 2 carbon. If you replace LG by O and protonate, the 2 O group of the resulting oxonium ion is an excellent LG. 9

Alkanes. Chapter 1.1

Alkanes. Chapter 1.1 Alkanes Chapter 1.1 Organic Chemistry The study of carbon-containing compounds and their properties What s so special about carbon? Carbon has 4 bonding electrons. Thus, it can form 4 strong covalent bonds