Chapter 6: Alkenes: structure and reactivity Alkenescontain a C=C double bond (also occasionally called olefins). Alkenes are very common in natural and synthetic organic compounds. 6.2 Degree of unsaturation Alkenes are unsaturatedbecause they have fewer than the maximum number of hydrogens in a hydrocarbon (C n H 2n+ 2 ). Each πbondor ringtakes the place of twohydrogens. Ethyleneand propyleneare the starting materials for hundreds of synthetic organic compounds and plastics. If a formula is C 6 H 10, it is 4 H's short of the maximum (the saturated compound would be C 6 H 14 ), so its DOU (degree of unsaturation) is 2. Two double bonds Two rings One triple bond One ring and one double bond Small alkenes are produced by the thermal breakdown of 2C-8C hydrocarbons from petroleum ("cracking") Halogens are the same as H when calculating DOU Oxygens have no effect If you're given a formula, try drawing it in a straight chain with all single bonds. The number of empty spots is twice the DOU. ch6 Page 1 ch6 Page 2
6.3 Naming alkenes Alkenes use the suffix -ene. 6.4 Cis-trans isomerism in alkenes Recall when C is double-bonded, it is sp 2 -hybridized 1. Name the parent hydrocarbon (which mustcontain the double bond -even if there's a longer chain somewhere else!) If there's a tie, use the chain with the double bond and the most branches. The unhybridized p orbitals on the adjacent carbons combine to make the πbond. 2. 3. Number from the end nearest the double bond. If it's a tie, number from the end nearest the first branch point, then second, etc. If that's a tie, differentiate the branch points alphabetically (ignoring the prefixes t-and sec-) Use the double-bond carbon with the lower number. With multiple double bonds, use diene, triene, etc as the suffix. Add the prefix cyclo-if the double bond is in a ring. While rotation occurs along σbonds, the πbond's shape (above and below the σ) does notallow for rotation. In a 1,2-disubstituted alkene(has one non-hydrogen group attached to each double-bonded carbon), the stereochemical descriptors cis and trans can be used: cis: the two groups point the same direction trans: the two groups point opposite directions ch6 Page 3 ch6 Page 4
6.5 The E/Z designation For trisubstituted or tetrasubstitutedalkenes, the cis/trans designation does not work. The E/Z desginationuses a series of sequence rules to assign priorities to groups on each of the double-bonded carbons. 1. On one of the double-bonded carbons, rank the atoms attached to it by their atomic number. Then rank the atoms on the other double-bonded carbon. E(Entgegen, apart) -the high priority groups are on opposite sides, like trans Z(Zusammen, together) - the high priority groups are on ze zame zide, like cis 6.6 Stability of alkenes In general, tetrasubstitutedalkenes are the most stable, due to an effect called hyperconjugation which involves more overlap of the πelectrons with other bonds. The fewer groups attached, the less stable. In disubstituted alkenes, cis are less stable than transthis can be rationalized by imagining steric strain between the two groups on the same side of the double bond in the cis stereoisomer. Any reaction that can interconvert the stereochemistry of a double bond will favor the transisomer because of its stability. 2. 3. If the 2 atoms attached to the C are tied, look at the next atomsdown the line. Keep going until there's a difference. Double bonds count twice. A C=O bond is like two C-O bonds. ch6 Page 5 ch6 Page 6
6.7 Electrophilic addition reactions of alkenes Electrophilic addition to an alkene usually follows a two-step mechanism, as we saw in chapter 5. 6.8 Orientation of electrophilic addition: Markovnikov's rule Markovnikov's rule: in addition of HX to an unsymmetrical alkene, like the previous reaction, the halogen is added to the more substituted carbon. 1. 2. π bond(weak nucleophile), attacks the Hof HBr (strong electrophile) which makes a new C-H bond and breaks the H-Br bond. The other alkene carbon is now a carbocation. The bromide ion(nucleophile) attacksthe carbocation(strong electrophile) to form a new Br-C bond. This is a regiospecificreaction -addition to one atom in the molecule is favored over another. The product is lower in energy than the reactant, so it is spontaneous overall. The reaction has two steps, each with a transition state and activation energy: This regioselectivity originates from production of the more substituted carbocation intermediate. ch6 Page 7 ch6 Page 8
6.9 Carbocation structure and stability Carbocation: C with 3 bonds and a + charge Planar structure sp 2 hybridized (120 o bond angle) One vacant p orbital perpendicular to the three hybrid orbitals Alkyl substituents stabilize carbocation by an inductive effect-the polarizable alkyl groups are able to shift electron density toward the + charge. 6.10 The Hammond postulate We know these two facts experimentally: In electrophilic addition to an unsymmetrical alkene, the more highly substituted carbocation intermediate forms faster More substituted carbocations are more stable than less substituted. 3 o > 2 o > 1 o > CH + 3 But, we learned in chapter 5 that rates are related to activation energy (ΔG ), and stability is related to Gibbs free energy change (ΔG o ) between reactants and products. The Hammond Postulate: the transition state resembles the nearest stable species in energy and structure. a. In endergonic processes, the transition state resembles the product b. In exergonic processes, the transition state resembles the reactant ch6 Page 9 ch6 Page 10
Analysis of electrophilic addition More stable products (or intermediates) tend to form faster! 6.11 Carbocation rearrangements Some reactions that involve carbocation intermediates give an unexpected mix of products that can't be accounted for just by using Markovnikov's rule (sometimes the nucleophile ends up on a carbon that didn't have the double bond!) The discovery of carbocation rearrangementswas conclusive evidence towards the existence of carbocations themselves. Because the first step is the rate-limiting step in this reaction (higher ΔG than the second step), it determines which product will be formed. Hydride shift: a hydrogen atom and its pair of electrons slides over to an adjacent carbocation, in order to form a more stable carbocation. Methyl or alkyl shift: an alkyl group will shift with an electron pair to an adjacent carbocation. ch6 Page 11 ch6 Page 12
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