Copy of Organic Chemistry Jonh Mc Murry11

Copy of Organic Chemistry Jonh Mc Murry11 - 190 CHAPTER 5...

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Unformatted text preview: 190 CHAPTER 5 Alkenes: Structure and Reactivity Electrophilic addition of HX to alkenes is successful not only with tlBr but with HCl and 1-11 as well. Note that [-11 is usually generated in the reaction mix~ tore by treating potassium iodide with phosphoric acid. Ce t ., Ether /C7CH2 + HLJ ———— —- CH3 ('2 CH3 CH3 CH3 2-:‘v’lethvipropene 2-Chloro-2~me‘hviprnpane [94%3 i Kl CH30H2CH2CH=CH2 W CH3CH26H2CHCH3 a 4 1-Pentene (HI) 2-Iodopentane "this is a good time to mention that organic reaction equations are some— times written in different ways to emphasize different points. In describing a laboratory process, for example, the reaction of Z-nietliylpropene with HCl just shown might be written in the format A + B —> C to emphasize that both reactants are equally important for the purposes ofthc discussion. The solvent and notes about other reaction conditions, such as temperature, are written either above or below the reaction arrow. Solvent ,/ H c CH3 3 \ 5.1M I C:CH2 + HCl CH3*C*C1 / 25 “‘C | H3C CH3 2-Methylpropene 2-Chloro-2-methvl- propane Alternatively, we might write the same reaction in a format to empha- size that 2-methylpropene is the reactant whose chemistry is of greater interest. The second reactant, MCI, is placed above the reaction arrow together with notes about solvent and reaction conditions. Roactant L‘\ H3c CH3 \ Hill | C:CH2 _—> CH3—C—Cl / lxi‘iot, 25 °C H30 CH3 \. Z-Methylpropene Hoar-ant 2-Chlorou2-methyl- propane 5.8 Test your knowledge of Key Ideas by using resources in ThomsonNOW or by answering end-of-chapter problems marked with Thomson Click Organic Interactive to use a web-based palette to predict products from the addition of HX to alkenes. Markovnikov's rule 6.8 Orientation of Electrophilic Additions: Markovnikov's Rule 191 In describing a biological process, the reaction is usually written to show only the structu re of the primary reactant and product, while abbreviatingjr the structures of various biological “reagents” and by-products by using a curved arrow that intersects the straight reaction arrow. As discussed in Section 5.11, the reaction of glucose with ATP to give glucose 6-phosphate plus ADP would be written as OH Ortiz-32' / / CH2 CH2 H o 0 " ("in " HO 0 HO Hexoklnase } HO \I OH OH '. OH OH Glucose Glucose 6-phosphate Orientation of Electropliilio Additions: Markovniltov's Rule Look carefully at the reactions shown in the previous section. In each case, an unsymmetrically substituted alkene has given a single addition product, rather than the mixture that might have been expected. As another example. 1-pentene might react with l-ICl to give both i-chloropentane and 2chloropentane, but it doesn’t. instead, the reaction gives only 2-chloropentane as the sole prod— uct. We say that such reactions are regiospecific (ree-jee-oh-specific) when only one of two possible orientations of addition occurs. Ci l i CH3CH2CH2CH=CH2 + no % CchHZCl-IZCHCH3 CchHZCHZCHZCHZCIl 2-Chloropentane [sole productl 1-Pentene 1-Chloropentane J (NOT formed) After looking at the results of many such reactions, the Russian chemist Vladimir Markovnikov proposed in 1869 what has become known as Markovnikov‘s rule. In the addition of HX to an alkene, the H attaches to the carbon with fewer alkyl substituents and the X attaches to the carbon with more alkyl substituents. No alkyl groups on this carbon 2 alkyl groups .. / fl \ onlhlsca’b°”\‘/C:CH2 + HCl Ether CHg—C—CH3 I I._ ' _- CH3 Z-Methylpropene 2-Chloro-2-methylpropane 192 VladimirVassilyevich Markovnikov “8354904) was born in NijniANovgorod, Russia, and received his PhD, working with A, M. Butlerov at the univer- sity in Kazan. He was a professor in Kazan (1870], Odessa ii87ii, and MOSCOW/(13734398. In addition to his work on the orien- tation of addition reactions. he was the first to synthesize a four- membered ring. Markovnikov’s rule CHAPTER 5 Alkenes: Structure and Reactivity 2 alkyl groups on this carbon W 1-Methylcvclohexene Ether + HBi 1 alkyl group on this carbon 1-Bromo-1-methylcvclohexane When both double-bond carbon atoms have the same degree of substitu- tion, a mixture of addition products results. 'ialkyl group ‘I alkyl group on this carbon on this carbon \/ r:i -;CH:CHCI*i\-., Bi Ether l 4’ CH3CH2CH2€HCH3 r g. t'rTJ + HBI + CchHZéHCi—izeH3 2-Pentene 2-Bromopentane 3-Bromopentane Since carbocations are involved as intermediates in these reactions, Markovnikov’s rule can be restated. in the addition of HX to an alkene, the more highly substituted carbocation is (restated) formed as the intermediate rather than the less highly substituted one. For L‘xairiple, addition of l-l+ to 2-methyipropene yields the intermediate tertiary carbocation rather than the alternative primary carbocation, and addition to 1-methylcyclohexene yields a tertiary cation rather than a secondary one. Why should this be? H Ci w l [F l ’—"’ CH3*C'—‘CH2 4’ CH3—CiCH3 | | CH3 ‘ CH3 CH3 tert-Butyl carbocation 2-Chloro-2-methylpropane + (tertiary; 3a] \ ix‘ H H l-Methyipropene + W I CH3—C—CH2 -"— -:r CH3A(IZMCH2C! | CH3 CH3 ‘l-Chloro-Z-methylpropane (NOT formed! Isobutyl carbocation (primary; 1°] 5.8 Orientation of Electrophilic Additions: Markovnikov’s Rule 193 . Bl 4. CH3 CH3 J‘fi r H \\H H l 'l (A tertiary carbocationl 1-Bromo-1-methv|cyc|ohexane r H CH3 1-Methylcyclo- / CH3? Br— hexane ’ a + I\H L H J Br (A secondary carbocationl ‘l-Bromo-Z-methylcy-clohexane {NOT formed) I Predicting the Product of an Eiectrophilic Addition Reaction What product would you expect from reaction of HQ with 1~etl‘rylcyclopentene? CHCH Strategy When solving a problem that asks you to predict a reaction product] begin by look- ing at the functional groupis) in the reactants and deciding what kind of reaction is likely to occur. in the present instance, the reactant is an alkene that will probably undergo an electrOphilic addition reaction with HCI. Next, recall what you know about electrophilic addition reactions, and use your knowledge to predict the procl— uct. You know that electrophilic addition reactions follow Markovnikov’s rule, so H4r will add to the double-bond carbon that has one alkyl group (C2 on the ring) and the Cl will add to the double—bond carbon that has two alkyl groups [Ci on the ring). Solution The expected product is 1-chloro-1-ethylcyclopentane. 2 alkyl groups on this carbon CHZCH3 i-ECI m L A _ 1-Chlo o-1-eth lc clo enta 9 Thomson Click Organic \ r V V p ” Interactive to practice predicting products of addition reactions according to Markovnikov's rule 1 alkyl group on this carbon 194 CHAPTER 6 Alkanes: Structure and Reactivity wonKEi) EXAMPLE 6.3 Strategy Solution Problem 6.14 Problem 6.15 Synthesizing a Specific Compound What alkene would you start with to prepare the following alkyl halide? There may be more than one possibility. cl CH3CH2cCHZCHZCH3 CH3 7 a When solving a problem that asks 110w to prepare a given product, alum-'5 work back- ward. Look at the product, identify the functional group(s) it contains] and ask your- self, “How can i prepare that functional group?” [n the present instance, the product is a tertiary alkyl chloride, which can be prepared by reaction of an all<ene with l-ICl. The carbon atom bearing the -CI atom in the product must be one of the double- bond carbons in the reactant. Draw and evaluate all possibilities. There are three possibilities, any one of which could give the desired product. ICH3 cH3 cH2 . it CH3CH:CCH2CH2CH3 or Cl'l3Cl'l2CZCl'lCH2CH3 Ol' CH3CH2CCH2CH2CH3 HCI (ii CHacHZCCH2CH2CH3 | CH3 Predict the products of the following reactions: (3) lbl $l‘l3 HB CH3C=CHCHZCH3 r 7- .7 (cl CH3 (dl CH2 l H20 ? HBr ? CH3CHCHZCH=CH2 H2504 (Addition of H20 occurs.) What alkenes would you start with to prepare the following alkyi halides? lal O/Br lbl CH2CH3 / 1 [cl Br (d) CI l CH3CH2CHCH2CHZCH3 6.9 Vacant p orbital Flex ‘ sp2 n “l- 4 R” Figure 6.9 The structure ofa carbocation. The trivalent carbon is sp2~hybridized and has a vacant p orbital perpendicular to the plane oi the carbon and three attached groups. Thomson Click Organic Interactive to rank the stability of carbocation intermediates. Figure 6.10 A plot of dissocia— tion enthalpy versus substitution pattern for the gas-phase dissocie ation of alkyl chlorides to yield carbocations. More highly substi- tuted alkyl halides dissociate more easily than less highly sub- stituted ones. 6.9 Carbocation Structure and Stability 195 Carbocation Structure and Stability To understand why Markovnikov’s rule works, we need to learn more about the structure and stability of carbocations and about the general nature of reactions and transition states. The tirst point to explore involves structure. A great deal of evidence has shown that carbocations are planar. The tri- valent carbon is splhybridized, and the three substituents are oriented to the corners of an equilateral triangle, as indicated in Figure 6.9. Because there are only six valence electrons on carbon and all six are used in the three 0' bonds, the p orbital extending above and below the plane is unoccupied. The second point to explore involves carbocation stability. 2-Methyl- propene might react with W“ to form a carbocation having three alkyl sub- stituents (a tertiary ion, 3°), or it might react to form a carbocation having one alkyl substituent [a primary ion, 1°). Since the tertiary alkyl chloride, Z—chloro- 2-methylpropane, is the only product observed, formation of the tertiary cation is evidently favored over formation of the primary cation. Thermodynamic measurements show that, indeed, the stability of carbocations increases with increasing substitution so that the stability order is tertiary > secondary > pri- mary > methyl. ‘l i? R R / / / / ‘l*C* ii—C+ R*C+ H—C* / ./ / Primary (1°) Secondary (2°] Tertiary (3") Less stable n More stable One way of determining carbocation stabilities is to measure the amount of energy required to form the carbocation by dissociation of the corresponding alkyl halide, Rex a R" + :X‘. As shown in Figure 6.10, tertiary alkyl halides dissociate to give carbocations more easily than secondary or primary ones. As a result, trisubstitnted carbocations are more stable than disubstituted ones, which are more stable than monosubstituted ones. The data in Figure 6.10 are taken from measurements made in the gas phase, but a similar stability order is found for carbocalions in solution. The dissociation enthalpies are much lower in solution because polar solvents can stabilize the ions, but the order of carbo- cation stability remains the same. >~ 5% ’3 .1: l' ,L, if: - 191 g ' — I (CH3l3_CCl A S g . r ! - .143 a = r' a e s l -l >- E 2 a — 96 .G v i if 8 . | 2 . s V 48 Q | I i O 2° 3° 196 Methyl: No alkyl groups donating electrons Figure 6.12 Stabilization of the ethyl carbocation, CH3CHZT, through hyperconjugation. Interaction of neighboring C H [T bonds with the vacant p orbital stabilizes the cation and lowers its energy. The molecular orbital shows that only the two C H bonds more nearly parallel to the cation p orbital are Oriented properly for hypere conjugation. The C—H bond perpendicular to the cation ,0 orbital cannot take part. donating electrons CHAPTER 6 Alkenes: Structure and Reactivity Why are more highly substituted carbocations more stable than less highly substituted ones? There are at least two reasons. Part of the answer has to do with inductive effects, and part has to do with hyperconjugation. inductive effects, dis- cussed in Section 2.1 in connection with polar covalent bonds, result from the shifting of electrons in a a bond in response to the electronegativity of nearby atoms. in the present instance, electrons from a relatively larger and more polar- izable alkyl group can sh itt toward a neighboring positive charge more easily than the electron from a hydrogen. Thus, the more alkyl groups there are attached to the positively charged carbon, the more electron density shifts toward the charge and the more inductive stabilization of the cation occurs (Figure 6.l1). . : CH3 c143 / / / H3C—C+ l-l3C-—C‘r i—IBC—C+ \, \ \ ' i H CH3 Primary: Secondary: Tertiary: Three alkyl groups donating electrons Two alkyl groups donating electrons One alkyl group Figure 6.11 A comparison of inductive stabilization for methyl, primary, secondary, and tertiary carbocations. The more alkyl groups there are bonded to the positively charged carbon, the more electron density shifts toward the charge, making the charged carbon less electron-poor (blue in electrostatic potential maps). Hyperconiugation, discussed in Section 6.6111 connection with the stabilities of substi tuted alkenes, is the stabilizing interaction between a vacant p orbital and properly oriented C—l—l 0' bonds on neighboring carbons. The more alkyl groups there are on the carbocation, the more possibilities there are for hyperconj ugation and the more stable the carbocation. Figure 6.12 shows the molecular orbital involved in hyperconjugation for the ethyl ca rbocation, CHgCHf, and indicates the difference between the C—H bond perpendicular to the cation p orbital and the two cell bonds more nearly parallel to the cation p orbital. Only the roughly parallel C— H bonds are oriented properly to take part in hyperconjugation. Problem 6.16 Problem 6.17 6.10 George Simms Hammond (1921—2005) was born on Hard- smahhlo Road in Auburn Maine, the son of a dairyiarmer. He received his PhD. at Harvard University in 1947 and served as professor ol chemistry at Iowa State University, California Insti- tute of Technology (1958—1972), and the University of California at Santa Cruz (19724978). He was known for his exploratory work on organic photochemistryithe use of light to oring about organic reactions. 6.?0 TheHammond Postulate 197 Show the structures of the carbocation intermediates you would expect in the fol- lowing reactions: fbl ¢CHCH3 HI ? > ? Draw a skeletal structure of the following carbocation. Identify it as primary, SQC‘ ondary, or tertiary and identify the hydrogen atoms that have the proper orienta- tion for hyperconiugation in the conformation shown. (all CH3 CH3 I ‘ lrlBr CH3CH2C :CHCHCH3 - - - The Hammond Postulate Let‘s summarize our knowledge of electrophilic addition reactions up to this point. We know that: D Electrophilic addition to an unsymmetrically substituted alkene gives the more highly substituted carbocation intermediate. A more highly substi- tuted carbocation forms faster than a less highly substituted one and, once formed, rapidly goes on to give the final product. A more highly substituted carbocation is more stable than a less highly substituted one. That is, the stability order of carbocations is tertiary > sec— ondary > primary > methyl. What we have not yet seen is how these two points are related. Why does the stability of the carbocation intermediate affect the rate at which it’s to rmed and thereby determine the structure of the final product? After all, carbocation stability is determined by the free-energy change AG°, but reaction rate is deter» mined by the activation energy A03. The two quantities aren’t directly related. Although there is no simple quantitative relationship between the stability of 3 ca rbocation intermediate and the rate of its formation, there is an intuitive relationship. it’s generally true when comparing two similar reactions that the more stable intermediate forms taster than the less stable one. 'l‘he situation is shown graphically in Figure 6.13, where the reaction energy profile in part (21) represents the typical situation rather than the profile in part (b). That is, the curves for two similar reactions don’t cross one another. An explanation of the relationship between reaction rate and intermediate stability was first advanced in 1955. Known as the Hammond postulate, the argument goes like this: transition states represent energy maxima. They are high-energy activated complexes that occur transiently during the course of a reaction and immediately go on to a more stable species. Although we can‘t 198 CHAPTER 6 Alkenes: Structure and Reactivity Figure 5.14 Energy diagrams for endergonic and exergonic steps. (a) in an endergonic step, the energy levels oftransition state and productare closer. (b) In an exergonic step, the energy levels of transition state and reactantare closer. Hammond postulate Reaction progress —~ L Slower Less Slower $33 reaction stable reaction stable _ immmediate '\ Intermediate i Faster More More reaction stable Faster stable _ intermediate reaction intermediate Reaction progress —- Figure 6.13 Energy diagrams for two similar competing reactions. In la), the faster reac- tion yields the more stable intermediate. In (bl, the slower reaction yields the more stable intermediate. The curves shown in (al represent the typical situation. actually observe transition states because they have no finite lifetime, the Ham, mond postulate says that we can get an idea of a particular transition state’s structure by looking at the structure of the nearest stable species. imagine the two cases shown in Figure 6.14, for example. The reaction profile in part (a) shows the energy curve for an endergonic reaction step, and the profile in part (b) shows the curve for an exergonic step. . . Transition state Trarlstllon state a Product Reactant Reactant Product Reaction progress Reaction progress —+ In an endergonic reaction (Figure 6.14a), the energy level of the transition state is closer to that of the product than to that of the reactant. Since the transi- tion state is closer energetically to the product, we make the natural assumption that it's also closer structurally, In other words, the transition state farm: crzcleigoriic reaction step structurally resembles the product of that step. Conversely, the transition state for an excrgonic reaction (Figure 6.14b) is closer energetically, and thus structurally. to the reactant than to the product. We therefore say that the transi- tion state for an exergonic reaction step structurally resembles the reactant for that step. The structure of a transition state resembles the structure of the nearest stable species. Transition states for cndergonic steps structurally resemble products, and transition states for exergonic steps structurally resemble reactants. How does the Hammond postulate apply to electrophilic addition reactions? The formation of a carbocation by protonation of an alkene is an cntlergonic step. Thus, the transition state for alkene protonation structurallyresembles the Figure 6.15 Energy diagrams for carbocation formation. The mnr: afal-‘ln fnrfiar\l ngrhnnatinn lI|UIK/ GLLHJIU tot HUI y uuiuuuur-uii is formed faster (green curve) because its increased stability lowers the energy of the transie tion state leading to it. 6.10 The Hammond Postulate 199 carbocation intermediate, and any factor that stabilizes the carbocation will stabi- lize the nearby transition state. Since increasing alkyl substitution stabilizes carbo» cations, it also stabilizes the transition states leading to those ions, thus resulting in a faster reaction. More stable carbocations form taster because their greater sta~ bility is reflected in the lower-energy transition state leading to them (Figure 6.15). Slewer reaction H3€\ Less stable _ _+ carbocation H Illc CH2 H C >- . 3 E’ U) I: I.“ \ ' F t ‘ i'i3C f, $631)” More stable \ 4, H C carbocation /C*CH3i 3 \ iH3c l C:CH2 .. / H3C Reaction progress —- We can imagine the transition state for alkene protonation to be a structure in which one of the alltene carbon atoms has almost completely rehybridized from sp2 to W3 and in which the remaining alkene Carbon bears much of the positive charge (Figure 6. i6). This transition state is stabilized by hyperconjuga- tion and inductive effects in the same way as the product carbocation. The more alkyi groups that are present, the greater the extent of stabilization and the faster the transition state forms. Alkene Problem 6.18 Productlike transition state Carbocation Figure 6.16 The hypothetical structure of a transition state for alkene protonation. The transition state closer in both energy and structure to the carbocation than to the alkene. Thus. an increase in carbocation stability (lower AC?) also causes an increase in transition- state stability (lower AGt), thereby increasing the rate of its formation. What about the second step in the electrophilic addition of HCl to an aikene—the reaction of chloride ion with the carbocation intermediate? is this step exergonic or endergon ic? Does the transition state for this second step resemble the reactant (carbo- cation) or product (alkyl chloride)? Make a rough drawing oi: what the tra nsition-state structure might look like. 200 CHAPTER 6 Alkenes: Structure and Reactivity 5.11 Thomson Click Organic Interactive to use a web-based palette to predict products from simple carbonation rearrangements Frank C. Whitinore (18874347) was born in North Anteboro, Massachusetts. and received his Ph.D. at Harvard working with E. L. Jackson. He was professor oi chemistry at Minnesota, North— western, and the Pennsylvania State Universny. Nicknamed "Rocky," he wrote an influential advanced textbook in organic chemistry. H3C\l 3-Methyl-1- butene 4/\ Evidence for the Meohanism of Electrophilic Additions: Carbonation Rearrangements How do we know that the carbocation mechanism for electrophilic addition reactions of alkenes is correct? The answer is that we don’t know it’s correct; at least we don’t know with complete certainty. Although an incorrect retro tion mechanism can be disproved by demonstrating that it doesn’t account for observed data, a correct reaction mechanism can never be entirely proved. The best wc can do is to show that a proposed mechanism is consistent with all known facts. If enough facts are accounted for, the mechanism is probably correct. What evidence is there to support the carbocation mechanism proposed tor the electrophilic addition reaction of all-\‘enes? One of the best pieces otevidence was discovered during the 19305 by F. C. Whitmore of the Pennsylvania State University, who found that structural rearrangements often occur during the reactiOn of HX with an aikene. For example, reaction of HCl with 3—rnethyl- 1—butene yields a substantial amount of 2-chloro-2-methylbutane in addition to the “expected” product, 2-chioro-3-methylbutane. l| l-ll I’ll If [it lit H3C .‘ H3C H H3C H :C\ ¢C\ + Ha :C\ /C< + :C\ /C: H3C C H H3C C H3C H , / \ / \ H H Cl i* H 3-MethyI-1— butene 2-Chloro-3-methylbutane (approx. 50%] 2-Chloro-2-methylbutane (approx. 50%) it the reaction takes place in a single step, it would be difficult to account for rearrangement, but if the reaction takes place in several steps, rearrangement is more easily explained. Whitmore suggested that it is a carbocation intermediate that undergoes rearrangement. The secondary carbocation intermediate formed by protonation of 3-methyl-‘l-butene rearranges to a more stable tertiary carbo- catiOn by a hydride shift—the shift of a hydrogen atom and its electron pair (a hydride ion, :H‘) between neighboring carbons. CH3 H CH3 H CH3 H l ‘ It i H3C\I i/H Hyiliitfe l l ,H /C\ + iri—Ci /C\+/C\ /+C\ /C‘/\ i C/‘\\ H H "-.,,_/C H m" H3C /C\ H A "1—! ,L l—. H l A 2“ carbonation A 3° carbocation 0- for CH3 H CH3 H H3C\l Cl/H ch\l é/H c c l-l'/ \c/ \H cr’ \c/ \H / \ / \ H CI H H 2-Chloro-3-methylbutane 2-Chloro-2—methylbutane ...
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