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Chapter 4

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University of Florida
CHM 2210

Chapter 4 Friday, July 12, 2013 3:21 PM 1. Alkanes a. Stability of alkanes is determined by the heat liberated when they undergo combustion b. As the molecular mass increases, both the boiling point and melting point increases c. As the branching increases, the boiling point decreases d. Lowestdensity out of all groups e. They undergo a minimal number of reactions, and the few they do undergo involve free radical chem f. Larger moleculeshave larger surface areas, resulting in increased interolecular van der waals attractions i. These increased attractions must be overcomefor vaporization and boiling to occur ii. Thus a largermoleculewith greater surface area and greater van der waals attractions boils at a highter temperature g. In general, a branched alkane boils AT LOWER TEMPERATURESthan the n-alkane with the same number of carbons i. This is because branched alkanes are more compact, with less surface area for london force interactin h. Like boiling points, melting points increase with increasingmolecular weight (so it is a solid for a longer because it must reach a greater tempearture to melt) i. Branched alkanes, however, have a highter melting point then the n-alkane with the same number of carbon atoms i. Branching gives it a more compact 3d strcuture,which packs more easily into a solid structure and increases th emelting point (it wants to stay solid for longer because its better that way) 2. Free radical chemistry a. Carbocations are sp hybridized and carbanions are sp hybridized i. ii. The difference in geometryresults from the difference in the number of nonbonding electrons iii. A carbocationhas 0 nonbonding electrons while a carbanion has two nonbonding electrons b. A carbon radical is in-between the two cases, and has for all intents and purposes trigonal planer geometry i. c. Like carbocations,radicals follow the same trend in which tertiary is more stable than secondary… i. d. Radicals can be formed from homolytic cleavage which requires a lot of energy either in the form of heat or light i. 1. A free radical halogenation starts with the initiation of light/heat a. 1. Once a free radical is present, it will attack an alkane and abstract a hydrogen from the alkane to leave behind an alkyl free radical, called propagation a. The first propagation step is hydrogen abstraction, and the second is halogen abstraction b. 2. When excess radical is present, the reaction will continue in a chain reaction and continue to add to the alkane a. 3. Termination can happen in a few ways a. 2. Free radical halogenation energetics 1. ΔH and ΔG values determines how much product is formed while E actictates speed a. ΔG bascailly tells uf if the rxn can occur b. In order for a rxn to favor products over starting materials, the rxn must exhibit a -ΔG c. If ΔG is positive,than starting materials will be favored 2. ΔG = ΔH - TΔS a. For halogenations of an alkane, the entropy component(TΔS) is negligible because two moleculesof starting material are convertedinto two moleculesof products i. b. We can therefor say i. c. The enthalpy term is determined by a number of factors, but the most important factor is bond strength i. We can estimate ΔH by cimparing the energy of the bonds borken and the energy of the bonds formed ii. iii. Rememberbonds broken require energy (+ΔH) and bonds formed release energy (- ΔH) d. The first propagation step (hydrogen abstraction) is the rate determining step 3. Free radical halogenation selectivity 1. Sometimes,there are two possible products formed a. 2. Because tertiary radicals are more stable than other free radicals, halogenation occurs preferentially at tertiary, than secondary, then primary 3. This is evident for chlorination of propane a. b. This happens because a 2' radical has a lower activation energy than a 1' radical i. 4. For bromination, the 2 radical is selected for much more a. b. This is because for bromination,the amount of radical character is much larger, so the transition state will be more sensitive to the nature of the substrate i. 5. As a general rule, slower reactions are more selectivethan faster reactions!!! a. As the reaction speeds up (say by increasingtemperature), the result is formation of products based more on random probability rather than selection 6. The reactivity of halogens from most reactive to least reactive is F, Cl, Br, I 7. Selectivityfrom most selective to least selective is I, Br, Cl, F a. Selectivityis how selectivea halogen radical is when choosing a position on an alkane 4. Free radical halogenation Stereochemistry 1. If halogenation creates a new chirality center, the products formed create a racemic mixture a. 2. If halogenation occurs at an existing chirality center, we also expect a racemic mixture of products a. 5. Elimination reactions 1. One way to form alkenes is by elimination 2. In elimination reactions, a proton from the β position is removedtogether with the leaving group, forming a double bond 3. This type of elimination is called β elimination or 1,2-eliminationand can be accomplished with any good leaving groups 4. Some types of β eliminations are named on the basis of the leaving group a. When the leaving group is a halide, the rxn is called a dehydrohalogenation b. When the leaving group is water, it is called dehydration 5. The elimination rxn can either occur in a stepwise fashion (like SN1) or simultaneously(SN2) 6. The E2 mechanism (CARRIEDOUT UNDER BASIC CONDITIONSAT HIGH TEMPS!!!) 1. 2. Exhibits second order kinetics a. 3. Unlike SN2 reactions, E2 reactions use tertiary substrates quite rapidly a. This is because the key difference from a substitution reactions and E2 reaction is a substitution reactions occurs when the reagent functions as a nucleophile and attacks an electrophilic position, while an elimination reactions occurs when the reagent functions as a base and abstracts a proton b. With a tertiary substrate, steric hindrance prevents the reagent from functioning as a nucleophile at an appreciable rate, but the reagent can still function as a base without encountering much steric hindrance c. d. Tertiary substrates react readily in E2 reactions; in fact, they react even more rapidly than primary substrates i. This is because the transitions state has two R groups as alkyl groups for a tertiary. ii. For a primary substrate, both of these R groups are hydrogen atoms iii. In the transition state, a C=C is forming iv. For a tertiary substrate, the transition state exhibit a partial double bound that is more highly substituted, and there for the transition state will be lower in energy v. vi. e. This doesn’t mean that primary susbtrates don’t undergo E2 reactions;they do. Just slower than tertiary with the same conditions 7. Regioselectivityof E2 reaction 1. In many cases, elimination reactions can produce morethan one possible product a. b. The β positions are not identical, so the double bond can form in two different regions c. This is an example of Regiochemistry 2. The more substituted alkene is generally observed to be the major product a. 3. The more substituted alkene is called the Zaitsev product 4. The less substituted alkene is often called the Hofmann product 5. The regiochemicaloutcomeof an E2 reactions can often be controlled by carefully choosing the base!!! a. Sterically hindered bases are used to produce Hofmann products b. Non sterically hindered products are used to produce Zaitsev products c. 8. Stereoselectivityof E2 Reactions 1. 2. This compound has two identical β positions so regiochemistiryis not an issue. 3. Deprotonationof either β positions produces the same results 4. However,two possible stereoisomericalkenes can be obtained a. 5. The tran isomer usually dominates 9. Stereospecificity 9. Stereospecificity 1. If there is only one β hydrogen to be asbstracted, a mixture of steroisomersis not obtained. There is only ONE steroisomericproduct a. 2. The reason there is only one product obtained is because of the π bond that is being formed a. A π bond consists of overlapping p orbitals b. In order for these orbitals to overlap, the proton at the β position, the leaving group, and the two carbon atoms that ultimatery bear the π bond must be coplaner i. c. There are two coplanar arrangements, syn and anti i. ii. d. We can see the anti-coplanar is staggered and syn is eclipsed. Therefore, the syn requires a transition state of higher energy e. In fact, in mostcases, elimination is observedto occur exclusivelyvia the anti-coplanar conformation,which leads to one specific stereoisomericproduct f. To rotate, draw the Newman projection i. 3. The E2 reactions is said to be stereospecific,because the stereoisomerismof the product is dependent on the stereoisomerismof the substrate a. The sterospecificity of an E2 reaction is only relevant (forms only one product) where the β position has only one proton i. b. If there are two β hydrogens, then either of these two protons can be arranged so that the it is antiperiplaner to the leaving group, resulting in both steroisomericprodcuts i. 10. StereoselectiveE2 reaction 1. The substrate itself is not necessarily stereoisomeric;nevertheless, this substrate can produce two stereoisomericproducts, and it is found that one stereoisomericproduct is formed in a higher yield 2. 11. StereospecificE2 reactions 1. The substrate is stereoisomeric,and the stereochemicaloutcomeis dependent on which stereoisomericsubstrate is used 2. StereospecificReaction: A reaction in which the stereochemistryof the reactant completelydetermines the stereochemistryof the product without any other option. StereoselectiveReaction:A reaction in which there is a choice of pathway, but the product stereoisomer is formed due to its reaction pathway being more favorable than the others available. From 12. E1 Reactions (acidic conditions, also high temps) 1. Exhibit first order kinetics a. 2. Tertiary substrates are the most reactive due to the same reason they are the most reactive in sn1 reactions (stabilize cation) 3. E1 reactions are generally accompanied by a competing Sn1 reaction, and a mixture of products is generally obtained 4. If the substrate is an alcohol, a strong acid will be required to protonate the OH group 5. Because carbocations are commonintermediates,these reactions can undergo rearrangementsby having hydrides or alkyl group shifts 13. Elimination reactions vs. Substitution reactions 1. To determine which mechanism dominates,first look at the attacking species a. When the reagent can only function as a nucleophile, only substitution can occur i. The substrate will determine which mechanism operates (SN2 vs. SN1) i. The substrate will determine which mechanism operates (SN2 vs. SN1) 1) ii. SN2 reactions can further be enhanced by using a
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