Chapter 6: Alkenes: structure and reactivity 6.2 Degree of unsaturation Alkenes are unsaturated because they have fewer than Alkenes contain a C=C double bond (also occasionally the maximum number of hydrogens in a hydrocarbon called olefins). Alkenes are very common in natural and (C nH2n + 2 ). synthetic organic compounds. Each π bond or ring takes the place of two hydrogens.
If a formula is C 6H10 , it is 4 H's short of the maximum (the saturated compound would be C 6H14 ), so its DOU (degree of unsaturation) is 2. Ethylene and propylene are the starting materials for Two double bonds • Two rings hundreds of synthetic organic compounds and plastics. One triple bond • One ring and one double bond
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. Small alkenes are produced by the thermal breakdown of 2C-8C hydrocarbons from petroleum (" cracking ")
ch6 Page 1 ch6 Page 2 6.3 Naming alkenes 6.4 Cis-trans isomerism in alkenes Alkenes use the suffix -ene . Recall when C is double -bonded, it is sp 2-hybridized
1. Name the parent hydrocarbon (which must contain The unhybridized p orbitals on the adjacent carbons the double bond - even if there's a longer chain combine to make the π bond. somewhere else!) If there's a tie, use the chain with the double bond and the most branches.
2. 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 While rotation occurs along σ bonds, the π bond's shape t- and sec -) (above and below the σ) does not allow 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: 3. Use the double -bond carbon with the lower number . cis: the two groups point the same direction With multiple double bonds, use diene, triene, etc as trans: the two groups point opposite directions the suffix. Add the prefix cyclo - if the double bond is in a ring.
ch6 Page 3 ch6 Page 4 6.6 Stability of alkenes 6.5 The E/Z designation In general, tetrasubstituted alkenes are the most stable, For trisubstituted or tetrasubstituted alkenes, the due to an effect called hyperconjugation which involves cis/trans designation does not work. more overlap of the π electrons with other bonds. The fewer groups attached, the less stable. The E/Z desgination uses a series of sequence rules to assign priorities to groups on each of the double -bonded In disubstituted alkenes, cis are less stable than trans - carbons. this can be rationalized by imagining steric strain between the two groups on the same side of the double 1. On one of the double -bonded carbons, rank the bond in the cis stereoisomer. atoms attached to it by their atomic number. Then rank the atoms on the other double -bonded carbon. Any reaction that can interconvert the stereochemistry E (Entgegen, apart) - the of a double bond will favor the trans isomer because of high priority groups are on its stability. opposite sides, like trans Z (Zusammen, together) - the high priority groups are on ze zame zide, like cis
2. If the 2 atoms attached to the C are tied, look at the next atoms down the line. Keep going until there's a difference. 3. 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 6.8 Orientation of electrophilic addition: Markovnikov's rule Electrophilic addition to an alkene Markovnikov's rule: in addition of HX to an usually follows a two-step mechanism, unsymmetrical alkene, like the previous reaction, the as we saw in chapter 5. halogen is added to the more substituted carbon.
1. π bond (weak nucleophile), attacks This is a regiospecific reaction - addition to one atom in the H of HBr (strong electrophile) the molecule is favored over another. which makes a new C -H bond and breaks the H-Br bond. The other alkene carbon is now a carbocation. 2. The bromide ion (nucleophile) attacks the carbocation (strong electrophile) to form a new Br -C bond.
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 6.10 The Hammond postulate Carbocation : C with 3 bonds and a + charge We know these two facts experimentally: In electrophilic addition to an unsymmetrical alkene, the more highly substituted carbocation Planar structure intermediate forms faster sp 2 hybridized (120 o More substituted carbocations are more stable o o o + bond angle) than less substituted. 3 > 2 > 1 > CH 3 One vacant p orbital perpendicular to the But, we learned in chapter 5 that rates are related to activation energy (ΔG‡), and stability is related to Gibbs three hybrid orbitals free energy change ( ΔGo) between reactants and products.
The Hammond Postulate : the transition state resembles Alkyl substituents stabilize carbocation by an inductive the nearest stable species in energy and structure. effect - the polarizable alkyl groups are able to shift a. In endergonic processes, the transition state electron density toward the + charge. resembles the product b. In exergonic processes, the transition state resembles the reactant
ch6 Page 9 ch6 Page 10 Analysis of electrophilic addition 6.11 Carbocation rearrangements More stable products (or intermediates) tend to form Some reactions that involve carbocation intermediates faster! 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 rearrangements was conclusive evidence towards the existence of carbocations themselves.
Hydride shift : a hydrogen atom and its pair of electrons slides over to an adjacent carbocation, in order to form a more stable carbocation. Because the first step is the rate -limiting step in this reaction (higher ΔG‡ than the second step), it determines Methyl or alkyl shift : an alkyl group will shift with an which product will be formed. electron pair to an adjacent carbocation.
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