Copy of Organic Chemistry Jonh Mc Murry119

Copy of Organic Chemistry Jonh Mc Murry119 - 150 CHAPTER 5...

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Unformatted text preview: 150 CHAPTER 5 An Overview of Organic Reactions The electrophilic sink must be able to accept an electron pair, usually because it has either a positively charged atom or a positively polarized atom in a func- tional group. For exam ple: Nu: NH: Nu; NU: Electrons usually flow \ t \ , ., ,:,_ ,-x..; ‘x, \ A... D, to one of these 7""{'i :35." T Halogen ":—~l-I “Hit: 0 i / / i _? _.\ m— / i J: electrophiles. ,2O_ :4 Rule 2 The nucleophiie can be either negatively charged or neutral. If the nucleo- phile is negatively charged, the atom that gives away an electron pair becomes neutral. For example: Negatively charged Neutral \i\ \ \‘n’fflimr \, .. .. , CH3—O: + ii—ligii: % 04370: + :E;.,: .. ,. , .. If the nucleophile is neutral, the atom that gives away an electron pair acquires a positive charge. For example: Neutral Positively charged H H _ H t" \ r / x. i ‘3 \ l CTC + ll :: ——> +C—C—I—l + / \ / | H H H H Rule 3 The electrophile can be either positively charged or neutral. If the electro phile is positively charged, the atom bearing that charge becomes neutral after accepting an electron pair. For example: Positively charged Neutral H H ‘\ H - , / v , \ | C=C -I 1' <= ——> +C—C—H + it / \ / j - H H a H H ‘--i If the electrophile is neutral, the atom that ultimately accepts the electron pair acquires a negative charge. For this to happen, however, the negative charge must be stabilized by being on an electronegative atom such as oxygen, nitro gen, or a halogen. For example: Neutral Negatively charged H l-l , i H --! \ i / _" \ l CmC + H , li'I 7—— +C—C—H + in / \ ‘- / i H H H H Thomson Rule 4 Click Organic Interactive to practice writing organic mechanisms using curved arrows. Strategy Solution 5.6 Using Curved Arrows in Polar Reaction Mechanisms 151 The result of Rules 2 and 3 together is that charge is conserved during the reaction. A negative charge in one of the reactants gives a negative charge in one of the products, and a positive charge in one of the reactants gives a positive charge in one of the products. The octet rule must be followed. That is, no second-row atom can be left with ten electrons (or four for hydrogen). If an electron pair moves to an atom that already has an octet (or two for hydrogen), another electron pair must simul- taneously move from that atom to maintain the octet. When two electrons move from the C=C bond of ethylene to the hydrogen atom of H30+, for instance, two electrons must leave that hydrogen. This means that the H—0 bond must break and the electrons must stay with the oxygen, giving neutral water. This hydrogen already has two electrons. When another electron pair moves to the hydrogen from the doubie bond, the electron pair in the H—0 bond must leave. H /H H d \ I! / \ i CiC + —» +C7C~H + or / \ / 1 = H H H H '7 Worked Example 5.2 gives another example of drawing curved arrows. Using Curved Arrows in Reaction Mechanisms Add curved arrows to the following polar reaction to show the flow of electrons: 0 ll ll H\ /Br C I H + C —> C CH + Br— Ific/ \c/ /\ Ific/ \c/ 3 i H H /\ H H H First, look at the reaction and identify the bonding changes that have occurred. in this case, a CeBr bond has broken and a C—C bond has formed. lite formation of the C—C boncl involves donation of an electron pair from the nucleophiiic carbon atom of the reactant on the left to the electrophilic carbon atom Ol’ CH3BI‘, so we draw a curved arrow originating from the lone pair on the negatively charged C atom and pointing to the C atom of CI-I3Br. At the same time the CiC bond forms, the C’Br bond must break so that the octet rule is not violated. We therefore draw a second curved arrow from the C—Br bond to Br. The bromine is now a stable [31" ion. 0 R": ("N 0 (ll ‘ ' H ‘L \,l3l' g i + a; —- —’ , iii” 2- + / \f/ ‘ I H H3C / \ C/ . H \ H C Br— 3 /\ 152 CHAPTERS An OverviewofOrganic Reactions Problem 5.8 Add curved arrows to the following polar reactions to indicate the flow of electrons ineach: (a) :61: .. .. i — :gi—gt: + HfitTlfiH —> H~qILH + Cl H H lb) H l _ | .. .. _ capo: + H—IC—Br: 7- CH3—Q—CH3 + :B‘ri H (C) :6? :0: i H ..u c 77*- c + :c1: H303 \OCH3 H3C/ \OCH3 Problem 5.9 Predict the products of the following polar reaction, a step in the citric acid cycle for food metabolism, by interpreting the flaw of electrons indicated by the curved arrmvs: _, :ot-t2 , _ .\ — ‘ C\ n 7. 02C\CHZV“/ C02 5.7 Describing a Reaction; Equilibria, Rates. and Energy Changes Every chemical reaction can go in either forward or reverse direction. Reactants can go forward to products, and products can revert to reactants. As you may remember from your general chemistry course, the position of the resulting chemical equilibrium is expressed by an equation in which Keq, the equilibrium constant, is equal to the product concentrations multiplied together. divided by the reactant concentrations multiplied together, with each concentration raised to the power of its coefficient in the balanced equation For the general- ized reaction aA it [)8 ( CC + {D we have _ ICI"[Dl" K, — “I IAIN LBW 547 Describing a Reaction: Eqrrrlibria. Rates, and Energy Changes 153 The value of the equilibrium constant tells which side of the reaction arrow is energetically favored. ll: Km] is much larger than 1, then the product concen- tration term [CIF lDl" is much larger than the reactant concentration term Ml” |B|b, and the reaction proceeds as written from left to right. it Keq is near 1, appreciable amounts of both reactant and product are present at equilibrium. And it KQq is much smaller than ’1, the reaction does not take place as written brrt instead goes in the reverse direction, from right to left. 0 1 l.- k.) can, u. Is.\\..LJ..LLll LIKHL librium constant at room temperature is approximately 7. C ino Pnnilihrinm D\’f\l'13CCl/\l1 and um r‘nn r‘lrfi'i'r—irrhintl pvnpriinanfal .u cc tau-uncut; c,\ r .u “cl .vc ccui L .c .uuc LL . HQC CH2 + z—rrsr- 9% CH3CH28: K _ [Cl-ISCHzBI'l _ 71 X 107 6“ [HBrllH2C=CH2l ' Because Kuq is relatively large, the reaction proceeds as written and greater than 99.999 99% of the ethylene is converted into bromoethane. For practical purposes, an equilibrium constant greater than about 103 means that the amount of reactant left over will be barely detectable (less than 0.1%). What determines the magnitude of the equilibrium constant? For a reaction to have a favorable equilibrium constant and proceed as written, the energy or the products must be lower than the energy of the reactants. in other words, energy must be released. The situation is analogous to that of a rock poised pre- cariously in a high-energy position near the top of a hill. When it rolls down» hill, the rock releases energy until it reaches a more stable low-energy position at the bottom. The energy change that occurs during a chemical reaction is called the Gibbs free-energy change (AG). For a favorable reaction, AG has a negative value, meaning that energy is lost by the chemical system and released to the surroundings. Such reactions are said to be exergonic. tor an unfavorable reac- tion, AG has a positive value, meaning that energy is absorbed by the chemical system from. the Surroundings. Such reactions are said to be endergonic. You might also recall from general chemistry that the standard free-energy change fora reaction is denoted AG”, where the superscript ° means that the reaction is carried out under standard conditions, with pure substances in their most stable form at 1 atm pressure and a specified terrrperature, usually 298 K. KW, L“ l: energy out: .\G‘ negative l‘ 1 it '1 munch: in. _\.i'_r" [frtjiaii'ivw 154 CHAPTERS An Overview of Organic Reactions Because the equilibrium constant, Keg, and the standard free-energy change, AG°, both measure whether a reaction is favored, they are mathemati» cally related by the equation AC-e _— —R'I"ln K9Cl or Keq : ,3“ACEIIR-{i where R : 8.314J/(K - moi) : 1.987 cal/(K - inol) T : Kelvin temperature 6 : 2.718 In Keq : natural logarithm of Keq The freeeenergy change AG is made up of two terms, an enl’iialpi-r term, AH, and a tempcrature-dependent entropy term, TAS. Of the two terms, the enthalpy term is often larger and more dominant. s0 : AH" — TAS° For the reaction of ethylene with H Br at room temperature (298 K), the approximate values are AG° = -44.8 kJ/mol AH" = 784.1 kJ/mol AS“ = —O.132 kJ/‘(K v moll Keq = 7.1 x 107 H2c:cn2 + liar : CH3CI-Izl3r The enthalpy change, AH, also called the heat of reaction, is a measure of the change in total bonding energy during a reaction. it AH is negative, as in the reaction of l-llsr with ethylene, the bonds in the products are stronger (more sta- ble) than the bonds in the reactants, heat is released, and the reaction is said to be exothermic. if AH is positive, the bonds in the products are weaker (less sta- ble) than the bonds in the reactants, heat is absorbed, and the reaction is said to be endothermic. For example, it a reaction breaks reactant bonds with a total strength of 380 kj/mol and forms product bonds with a total strength of 400 kl/mol, then AH for the reaction is 720 kl/mol and the reaction is exothermic. The entropy change, AS, is a measure of the change in the amount 0|: molecular randomness, or freedom of motion, that accompanies a reaction. For example, in an elimination reaction of the type A % B+C there is more freedom of movement and molecular randomness in the products than in the reactant because one molecule has split into two. Thus, there is a net increase in entropy during the reaction and AS has a positive value. On the other hand, for an addition reaction ot‘ the type A + B —> C the opposite is true. Because such reactions restrict the freedom of movement of two molecules by joining them together, the product has less randomness than the reactants and AS has a negative value. The reaction of ethylene and Problem 5.10 Problem 5.11 5.8 Thomson Click Organic Interactive to use bond dissociation energies to predict organic reactions and radical stability. 5.8 Describing 3 Reaction: Bond Dissociation Energies 15 HBr to yield brornoethane, which has AS" : —0.]32 kJ/(K » mol), is an example Table 5.2 describes the thermodynamic terms more fully. Table 5.2 Explanation oiTiiermorlynaniic Quantities: AG“ 7: AH“ 7 BS" Tenn Name Explanation : i :i‘3 The energy difference hetwuen reactants and products. When AG" is negative, the reaction is e iiluiiwn rnnro-nnr fir‘r' nan nm JllJlltIHl L\|1111CIJAL,OJH.IK.ULL uu. xergonic, has a favorable equi- m are :. ll _|KJ l\ 'L inrunl’y. viii positive. the reaction is endergonic, has an unfavm‘ablc equilib- rium rainstorm. and cannot occur spontaneously. Mi" Enthalpy The heat of reaction, or difference in strength between the change honrls broken in a reaction and the bonds formed. When All“ is negative, the reaction relcases heat and is exothermic, When AH" is positive. the reaction absorbs heat and is endothermic. 1.)" 'é'ni rnpy The cl'iange in molecular randomness during it reaction. When AS: is. negative. randomness decreasm: when A5“ ix pasilive, ranrlonmrzx‘s increases. Knowing the value ofl<Cq for a reaction is useful, but it’s important to realize the limitations. An equilibrium constant tells only the position of the equilib- rium, or how much product is theoretically possible. It doesn’t tell the rate of reaction, or how fast the equilibrium is established. Some reactions are extremely slow even though they have favorable equilibrium constants. Gasoline is stable at room temperature, for instance, because the rate of its reaction with oxygen is slow at 298 K. At higher temperatures, hOwever, such as contact with a lighted match, gasoline reacts rapidly with oxygen and undergoes complete conversion to the equilibrium products water and carbon dioxide, Rates (how first a reaction occurs) and equilibria (how much a reaction occurs) are entirely different. 15 the reaction fast or slow? {arc a l:t§iillllli'li3il- — r in what direction does the reaction proceed? Which reaction is more energetically favored, one with AG° = —44 kl/mol or one with AG: 2 +44 kJ/mol? Which reaction is likely to be more exergonic, one with Keg : 1000 or one with Keq = 0.001? Describing 3 Reaction: Bond Dissociation Energies We’ve just seen that heat is released (negative AH) when a bond is formed and absorbed (positive AH) when a bond is broken. The measure of thc hear change that occurs on breaking a bond is called the bond strength, or bond dissociation energy (D), defined as the amount of energy required to break a given bond to produce two radical fragments when the molecule is in the gas phase at 23 °C. Bond dissociation A A- + energy .156 CHAPTER 5 An Overview of Organic Reactions Each specific bond has its own characteristic strength, and extensive tables of data are available. For example, a C—H bond in methane has a bond dissoci- ation energy D 2 438.4 kj/mol (104.8 kcal/mol], meaning that 438.4 kJ/mol must be added to break a (3—H bond of methane to give the two radical frag— ments CH} and ll. Conversely, 438.4 kj/mol of energy is released when a methyl radical and a hydrogen atom combine to form methane. Table 5.3 lists some other bond strengths. Talile 5.3 Some Bond Dissociation Energies, D u n a Bond 7 lkJ/mall Bond (ltd/mull Boml (kJ/nmll H 436 (CH353CW | 209 CZHE—fil-;‘_. 355 H77 570 H2Ci*-—CH - 444 lCHngCH *Cl'l» 351 H '“l 432 H2C===CH 7r?! 368 (CH3l3C [Tl—la. 33L.» Hw ‘ 2:66 HZCZCHCHZW 361 H2C==CH 70': a. 4% H77 298 H2C==CHCH2— :12: 289 H2C:CHCH27 $in 310 Ctr 243 l/ I Hzc==::.... 6'11 Br—U 103 ' | 404 _,I“.l-SI. l- 1.51 I 437 \_ CH 7H 438 f . 3 / ,. m CHE—rt 351 l 405. /CH2 (2H; \ l/ l w m CHg—Ei. 293 i . 33* \\ CHgi' 234 /CH2, H CHrn: 330 l/ i «m8 o \ H 3% CH3---“~‘l- . 335. CH3C— is 0 Hr—r 420 ,r; r 2 -> /\ / 2 HO —1-: 408 CzHr-z": 338 n 293 HO fl 7] ) \ e , CgHrer 2,85 \/ CH307H 437 Cgflsk i ‘/r "fir ‘ / CHgsil'l 371 C2H57 380 l 337 \_ C2H50'7i . 436 (CH3)2CH—' 401 “ o lCHngCHii‘I 339 A II .322 l// ' CH3C—L'.‘.l l3 (CHElZCH—i— 274 . ‘l 469 \ /. [CHglgCnl-l 390 \’ CH3CH20—r1i a1 339 (CH3J3C7C. 330 He :Cilr €52 NH2— 449 (CH3i3C"i:ll 3533 CHg—clt. 376 H ecu 5m 5.3 Describing a Reaction: Energy Diagrams and Transition States 157 Think for a moment about the connection between bond strengths and chemical reactivity. in an exothermic reaction, more heat is released than is absorbed. But since making product bonds releases heat and breaking reactant bonds absorbs heat, the bonds in the products must be stronger than the bonds in the reactants. in other words, exothermic reactions are favored by stable products with strong bonds and by reactants with weak, easily broken bonds. Sometimes, particularly in biochemistry, reactive substances that undergo highly exothermic reactions, such as ATP (adenosine triphosphate), are referred {A ~ [N r; manpg... HF Mariam.” micro} iicii Cllclsy ATP is special or different from other compounds; they mean only that ATP has relatively weak bonds that require a smaller amount of heat to break, thus lead- ing to a larger release of heat on reaction. When a typical organic phosphate such as glycerol 3-phosphate reacts with water, for instance, only 9 kJ/mol of heat is released (AH° = —9 kJ/mol), but when ATP reacts with water, 30 til/moi of heat is released (AH’ - 30 l<_l/mol). The difference between the two reactions is due to the fact that the bond broken in ATP is substantially weaker than the bond broken in glycerol 3-phosphate h n1-{l‘A-irrlr‘ r‘n-n—‘y‘nlwnr'lr CWVI‘IN Ink/xiv AnnH— mafia“ A-l-tn- LG U1 lllsll' LUUlPULUlU). JLlLlJ MIUCID UUll L lllCdll Llldi AH“ : —3 lrJ/m-‘Ji Stronger 1/! Ci ‘OH f‘.) (IDH ,/ H o WW9 0 CH2 CH CH2 OH 2 LII 35-OH + HO*CH2*CH*CH2#OH ‘ | LT [11' Glycerol 3-phosphate Glycerol AH” = —30 kJ/mol Cl] 'O~5“*OH + Ht Weaker l ,/ . + r:- l ' t , _ \ W(:l—i'.>_/O#P*07Fl’ -- .. a re 5- EtaU O O 1 i -_ | it It O 0 ‘O—P—O—P— 1 l - O’ 0' " Adenosine triphosphate (ATPI ' as Adenosine diphosphate (ADP) fleseribing a Reaction: Energy Diagrams anti Transition States For a reaction to take place, reactant molecules must collide and reorganization of atoms and bonds must occur. Let’s again look at the addition reaction of HBr and ethylene, which takes place in two steps. 158 CHAPTER 5 An Overview of Organic Reactions Figure 5.4 An energy diagram for the first step in the reaction of ethylene with HBr. The energy difierence between reactants and transition state. AG‘, defines the reaction rate. The energy differ- ence between reactants and carbonation product, AG“, defines the position of the equilibrium. 14/, r - 7 Figure 55 A hypotheti- cal transition-state structure for the first step of the reaction of ethylene with HBr. The C=C 7 bond and HrBr bond are just beginning to break, and the C—H bond is just beginning to form. Sign in at www.thomsonedu.com to see a simulation based on this figure and to take a short quiz. H H ‘1“ Hi , H ‘r | +/ 7. i l . /C:C\ Ho H— —c\~" H—l—c—ur H H r'i e H H Carbocation As the reaction proceeds, ethylene and HBr must approach each other, the ethylene n" bond and the H—Br bond must break, a new CeH bond must form in the first step, and a new C—Br bond mtrst form in the second step. To depict graphically the energy changes that occur during a reaction, chemists use reaction energy diagrams, such as that shown in Figure 5.4. The vertical axis of the diagram represents the total energy of all reactants, and the horizontal axis, called the renrtiun coordinate, represents the progress of the reaction from beginning to end. Let’s see how the addition of HBr to ethylene can be described in an energy diagram. Transition state _/ ,,,,,,,,,,,, ,Lc \iarbocation product Activation —— .'/—- CH3CH2+ "l' Bi" >. energy i D1 3 I 3 AG AG: :: I w i I. Fieactants H26 : CH2 + HBr Reaction progress At the beginning of the reaction, ethylene and HBr have the total amount of energy indicated by the reactant level on the left side of the diagram in Figs urc 5.4. As the two reactants collide and reaction commences, their electron clouds repel each other, causing the energy level to rise. if the collision has occurred with enough force and proper orientation, the reactants continue to approach each other despite the rising repulsion until the new CeH bond starts to form. At some point, a structure of maximum energy is reached, a structure called the transition state. The transition state represents the highest-energy structure involved in this step of the reaction. it is unstable and can’t be isolated, but we can never- theless imagine it to be an activated complex of'the two reactants in which both the C=C 7 bond and H—Br bond are partially broken and the new C—H bond is partially formed (Figure 5.5). the energy difference between reactants and transition state is called the activation energy, AGi, and determines how rapidly the reaction occurs at a given temperature. (The double—dagger superscript, 3*, always refers to the tran- sition state.) A large activation energy results in a slow reaction because teircoi- lisions occur with enough energy for the reactants to reach the transition state. Figure 5.6 Some hypothetical energy diagrams: la] a fast exergonic reaction {small A61, negative AG°]; [bl a slow exergonic reaction (large AG? negative AG°l; (c) a fast endergonic reaction {small A64”, small positive AG”); ldl a slow endergonic reaction (large A61, positive AG“). Sign in at www.thomsonedu.com to sea a simulation based on this figure and to take a short quiz. 5.9 Describing a Reaction: Energy Diagrams and Transition States 159 A small activation energy results in a rapid reaction because almost all collisions occur with enough energy for the reactants to reach the transition state. As an analogy, you might think of reactants that need enough energy to climb the activation barrier to the transitiori state as similar to hikers who need enough energy to climb to the top of a mountain pass. It the pass is a high one, the hikers need a lot of energy and srirmOunt the barrier with difficulty. If the pass is low, however, the hikers need less energy and reach the top easily. As a rough generalization, many organic reactions have activation energies in the range 40 to 150 kl/niol (1065 Real/mot). The reaction of ethylene with I-lBr, for example, has an activation energy of approximately 140 kj/mol (34 kcal/mol). Reactions with activation energies less than 80 kj/mol take place at or below room temperature, whereas reactions with higher activation ener- gies normally require a higher temperature to give the reactants enough energy to climb the activation barrier. Once the transition state is reached, the reaction can either continue on to give the carbocation product or revert back to reactant. When reversion to reac- tant occurs, the transition—state structure comes apart and an amount of free energy corresponding to eAGi is released. When the reaction continues on to give the carbocation, the new C—H bond forms fully and an amount of energy corresponding to the difference between transition state and carbocation prod- uct is released. The net change in energy for the step, AG”, is represented in the diagram as the difference in level between reactant and product. Since the carbocation is higher in energy than the starting alkene, the step is endergonic, has a positive value of AG", and absorbs energy. Not all energy diagrams are like that shown for the reaction of ethylene and HBr. Each reaction has its own energy profile. Some reactions are fast (small AG-ii) and some are slow (large A013); some have a negative AG", and some have a pos- itive AG“. Figure 5.6 illustrates some different possibilities. (a) (bl l Energy Reaction progress —-— (d) Reaction progress —» Reaction progress ——-— 'IED CHAPTERS Problem 5.12 5.10 Figure 5.7 An energy diagram for the overall reaction of ethylene with HBr. Two separate steps are involved, each with its own transition state, The energy minimum beIWeen the two steps represents the carbocation reac- tion intermediate. An Overview of Organic Reactions Which reaction is taster, one with soil iv +45 kJ/moi or one with AGi : +70 til/moi? Describing 3 Reaction; intermediates How can we describe the carbocation formed in the first step of the reaction of ethylene with HR)? The carbocation is clearly different from the reactants, yet it isn’t a transition state and it isn’t a iinal product. . r—. F ‘i f H i‘ H ‘ H i H \ ‘ / i i + / ’i,_.r’ i i ciec —> H7C7C'“ _—, H—C—Cii / \ | \ i | I" H H H H H Reaction intermediate We call the carbocation, which exists only transiently during the course of the multisth reaction, a reaction intermediate. As soon as the interme- diate is formed in the first step by reaction of ethylene with Hf. it reacts fur- ther with 81" in a second step to give the final product, bromoethane. This second step has its own activation energy (AGi), its own transition state, and its own energy change (AG”). We can picture the second transition state as an activated complex between the electrophilic carbocation intermediate and the nucleophilic bromide anion, in which Br— donates a pair of electrons to the positively charged carbon atom as the new CeBr bond starts to form. A complete energy diagram for the overall reaction of ethylene with HBr is sh0wn in Figure 5.7. in essence, we draw a diagram for each of the individual steps and then join them so that the carbocation product of step 1 is the reactant for step 2. As indicated in Figure 5.7, the reaction intermediate lies at an energy First transition state Carbocation intermediate ‘ Second transition state H2C=CH2 + HBr AG; Reaction progress ——r—s- ...
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