wade_ch04_answered

wade_ch04_answered - Chapter 4: Chemical Reactions —...

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Unformatted text preview: Chapter 4: Chemical Reactions — Mechanisms, Thermodynamics, and Kinetics Mechanisms: ( l ) ‘Which honds break and form, and in what order. can ‘5' “J [:1 Thermodflamies: (2] The energy changes associated with a chemical reactio I.‘- I . ' q- or: Mu: WELL-x 3 ‘51:“ [t Fit-H.“ Kinetics: {3] The r_ate of a reaction — how high is the barrier separating reactants and products. cal“- “Luau: $¢fi+fl Background: Read Sections 4.4 — 4.6 Bond cleavages a G) E) — - r A B —h-r A + e B “Hgtefglytltl” cleavage mum] _. m {charged} (Ch, 6, the “SN-1” reaction) h A. + .B ClBflVflgE (This chapter} neutral —t-r radicals (neutral) Thus, hfiflflli i-‘fi cleavage produces free radicals — molecules (or atoms) wr h an unpaired electron. Free radicals are very,r reactive because they i5 mi Luau: EH: ‘lE +3 (5} Note the use of arrows to show electron movement. n = move two electrons (an electron pair} a = move one electron (a “fishhook” arrow) The arrows fl = where the electrons start (in the reactant} The arrow’s head = where the electrons will go (end up in the product) 51- HR (3” 3— ilk/CF) I t.- llil For our purposes... Heterolytic cleavages occur onlyr in solution: H / +9 The charges produced are stabilized by solvent {ion-dipole and dipole-dipole forces} Homolytie cleavages occur only in the gas phase: *9 The radicals produced are not charged. Also, the absence of solvent means entropy calculations (LES) are easy. Formation ofradicals Break loan 1: Eon-rm :11 = +AH Radical Formation requires that we break a bond — we must put energy (heat, or light, hv) {rile the system. The weaker the bond, the EH 5435*" it is to break. Weak bonds we will see... a dialkvl peroxide ,0 a, :23" 2g. '5' + '5 I halogen halogen radicais X = Cl, Br, I (halogen atoms} (but not F!) Now would be a good time to review some basic thermodynamics... Basic Concepts in Thermodynamics What do we need to know to understand the concepts of equilibria and reaction outcome? ldeas and Concepts: a) Equilibria, free energy, and stability E j ; M Cane , b) Enthalpy and entropy c] Bonds and disorder For a reaction A L'- B. .. where A = reactants and E = products = C? mdacls] K [B] _ equilibrium concentration ofB ,q E — m [Qaaalaa 4w] The free energy change, M3”, is given by... 5.13“ = -2.3 RT log K where - gas cons a t= 1.93 calili mol; T = e1 erarure in ' vin “1 we re la u vii-Eerie re tile :30” is the amount of energ3lliherated mo“ :1 I], i.e. negativcfor akeu up {so 1" fl, i.e. post we when A and B, each in their standard states [1 atrn for gases, Li] M for solutions}, are mixed and we wait for the reaction to come to equilibrium. Experimentally, the easiest way to determine dG‘J is to have some way to measure the amounts offit and E at equilibrium. An alternative expression is: dG” = dH" - TdS“ where... an H dH” = the change in enthalpy [strength of bonds) d8" = the change in entropy {disorder} A favorable reaction is one that... has [13] 13- [A] [more products than reactants, and thus the sign of 5.0” is negative). When we consider that at?) = dH“ - TdS“, what will make dill" negative? A negative dH“ helps: this happens if weak bonds in the reactant are broken and strong bonds in the products are formed CIR A positive as” helps: this happens if the system has gone from more ordered to less ordered [the entropy, or disorder, has increased} or. R's-'1 p re In the ideal case, both are true. 6 +4 will lac. 'Ifi'l Lab-£43 Bond Strengths and 5H” Think of a bond as a spring... Bonds, like springs, have equilibrium lengths To stretch {and ultimately break) or compress a bond requires an input of energy Strong bonds are more stable; weak bonds are more unstable {or less stable) and tend to reform in the products as new, more stable bonds. Entropyr and disorder At its simplest level, we assume that the more molecules there are, the more disorder there is {there are twice as many places to find two molecules as there are to find one molecule}. For A —> B + C as“ is positive (one molecule becomes two molecules} l'"—_—_—‘—'__Iu For A + B m} C n5" is negative [two molecules become one molecule) For A + B —) C + D S” is more {two molecules hecome two molecules] A-tl Clo—H reins are A't-E—p (ii—D truss. Sn mmarv: A large negative AH" can overcome a small negative as", thus making so“ negative. A large positive £15" can overcome a small positive Ml". thus making so“ negative. 3G“ is guaranteed to he negative when an“ is negative AND .53“ is positive. TO PREDICT THE SIGN 0F £13", WE LOOK AT BONDS AND DISURDER Example: H! + H- —> H; This is bond formation, so we expect oil-ln to be negative. Since two atoms become one molecule, we expect a negative as”. If ltl4 kealfmol are liberated when the H3 bond is formed, then it requires 134 kcalr‘mol to break the H-I—l bond. equilibrium bond length 1'34 kcalhnol = 0.7'4 R hood length —1I- Homo]y_tic Bond Dissociation Energies (“BBB’s”) Important: These are gas Ehase measurements «w NO SOLVENT! Because there is no solvent, we can prettyr much ignore entropy has“), and thus gave iii-Go ~ til-I”. We need out},r compare BDEs to predict the reaction outcome! ' “HE—a- Olduews: Bond ne'er-n iscxothermic[fll-i°is hi3. } (r) ¥ send—gifts endothermic tau“ is a. ] (3} Us scamtw —"'ff _PL H—H —) H- + Hi fiH" = +104 kcalfmol, so the EDE is +104 kcalfmo] Question: What then is the nu“ for H- + H- a H—I—I ‘2' T Answer: "1'07 &C¢fi/Mg/ MQJL '3 “AH (9} Let’s examine Table 4-2 (p. 136]... A—B —s~ A! + E! (in kcalfmol} GAS PHASE! CH3 — H F — F +38 CH3 — F 109 C1 — C] +53 CH3 w C] 84 Br— Br 46 CH; — Br T0 1— I 36 CH3. - I 56 Find the trend: Except for F2, as atomic size increases1 “I? t 313‘“ W (10) a ? . EMT WHLE . Exglain the trend: Larger atoms have iofigg - fluxes users ker —- heart-501} 1'5 thin More ex arn les rm CH3HH —I- H 4. r- at El"""‘3f"i?sflr:ns[woman H + -H +1o4 {methylradical} -33- H I: E} CchHQCHg—H —n— “lbfiHrf-H+ 'H +93 ( l radical} H I. CH3{|3HCH3 —-I- CPIbflf—CHJ + IH +95 { 2D radical] H H" i c: {CH3]3C——-H —-- cab—c-J'J-g -H +91 ( 5 radical} CL! mask: {54* 3 inane! Cienriutl ‘Huz. [Mort C1 TLc. Mioctfl 40w. easier" The C-FH. ion-HA Synthesis of Alkyl Halides: Free Radical Substitution hese‘L . Alkane + X2 m- ? R—x + H—}( (an alkane can he symbolized by R—H) X = F, Cl, Br, but not I! “Thy? See BDE table. / f- (12} 1 rl. : H II H g =_ ‘ ' "L? fixtures 3c :4 J o :- nu r3 h Specific example: CH4 + x2 %‘- CH3X + CHEKQ + (3ng + coo + HX ht.- chlorotnethane dichlorontethane trichloromethane tetrachlorotnethane (methyl chloride} (methylene (chlorofotrn) {carbon I chloride} tetrachloride) I multiple products 1) Why multiple products? 2) How can we avoid multiple products and get only monosuhstituted RX’s? Why multiple products? We may not understand {yet} how the products are formed, but it seems likely that the reaction is seguential— Ft'l'fii' CH3 “*‘E’EHEX 3| (Lg—it" "'_"" CH2 X2 3 air. [13) This assumes, of course, that the H’s in each compound are equally reactiye. Suppose we look at the reaction at early time points... Early in the reaction... CH4 + X2 hf?” CH3X + HX + unreactedCH4 + unreacted X2 loos torso ‘2in to to $99 ‘5’?!) As time progresses, what will happen? assume: IDOC‘: MDiEqu-E't’: ai‘ Sior‘i... OH; + X2 LEE-F {3ng + HX + unreaeted CH4 + unreaeted X2 . gazdfi CH f I‘m/r heat heat (1’. £- Po;n+... | i the l i- ii]r or ' sfi. = v M *w m m. Lafi- a i nae- e. we 3% We cm {3 CH2 5‘; 1:: 5140mm to and; icofl . . .and this will continue until CE is produced. {14) | | I su y we want the meneiubstituted a} 1 hali e enlg — how can iI be done? IRA-4r! MPG-3““ X3 " ) 15f- H'Idv e: x: " .e'mr' (“E-:3?“ #11? Perhaps the easiest way is to use a [and cone“ . of X2 and (EH3 BIZ ng—fiweHa T H56. «— e—eHEC-l I isebutane ‘ [z'mfimylpmpanfl [flag-14g} chloridemssa.) + f'éufflchlofideasm 1- M : W992? care by Lt I th II aw! me. too “ten-:3 nice: I (2-methylprepane} Efith ' _'_,_,_.—o—'- relationship? + pelyehle rinated products {29%} + Cami.~i..i.~...¢g H. y”? is {aemnerfi . l __l_c_ ~ _r _ CUHMc-LI‘UIILj‘ («EC—C- Q‘ i: C— C c" C" Isebutane has 11m types c“ L | efhydregensn. Ifthe H’s were equally,r reactive... ‘ ‘i aim-u. i“ H'a “ms—=3 eases-i (m . C a “a . - .2. o: I : '15 rack more I é.) my 1 a H ‘1 “3 {a c. M i we meU& B 1' n J isebutane $2.13? In: Em Effiac‘i'eé‘ 1 l—h——l l l Lin nt‘cescn All nine 1" hydrogens are equivalent! Replacing any % of them with chlorine will give l the same iso‘uutvl chloride product. Similarly, there is c-nljlr one 3° hydrogen. Replacing it gives t—butyl chloride. Later on we will see why in chlorination reactions isobutvl chloride is the “major” product and why bromine {Bra would ive r—bu l bromide as the major product. Free radical reactions: Mechanisms: Remember that these are gas phase reactions! | Observation: A mixture of CH4 and C12 don‘t do diddlev until they are either heated to ~10!) “C or irradiated with light (hv). In fact, one photon of light allows many individual reactions to occur. This is the hallmark of a chain reaction - a reaction that, once started, is self—sustaining. . _ __ lc 1M¢rsi Whomhms The followingrnechanisnihas been proposed. ififllue [he mare final“ bib. an dim mo [con «:51 lube-*bIT-t‘; I: (+ CHECig + CHCI3 + CCI4} Overall reaction: CHg—H + Cl—Cl E35- CHg—Cl + H—Cl T r till 1‘ r brutal-c Ereqk Mo lea-e. Make Suppose we tell you the first step is to produce chlorine radical... Step 1: Chain initiation: made? none; at a- L -a + .a- ' .Il to. 'oo' .g' Broke? (Ll-“'CI he: mt‘lfi'i‘é". aHD? N + W ice-Lilwaklg) Does this help? Can we predict the next step? {mar—H + {3|—C| r;in CHa—CI + H—C! Whatdoes «Cl want? hv 1‘”?! Hm} Clo has. r -cr / __,. c. Hint H CHE .— H + - Lt " H3 1' +1 DEED? LHE-EH i- .cth—r Lr-LSFC! 1* 1‘1" +10 mad i | l H nz'i— “‘3 l, +tav ~av ! LC.“- ‘S I J . Stgp 2: First chain propagation step: {Cl- “abstracts” (removes) an H atom) (19] H H—LI: i+r~1 H methyl radical H aH°=t H +.._.L | methane HOW do we calculate an“? at bonds broke versus made. Where do we get this ittfongaggrtfl ED E in: Lie [213) i Which bond broke? CH3—H 4:: «cu3 + 0H an": *- it"‘I Which bond formed? H- + -Cl ail-{$31 are = "’ lo 3 (21] _. {Egg Net +1 kcalr’lnol Step 3: Second chain propagation stgg: \ H x i' m P l “r H—c cruel —#+ H — r'l; ——c1 1* * (fl-l} (22) H T H ion-Em 1:: -l- 5—? In «a -. -— LII fiHn = + fl __ 23L! fl tar-cal: nae-Fl e36 éfléi We}. The chlorine radical produced in Step 3 can non.r go back and participate in another round ot‘Stgp 2! Thus: Just a fit; atoms of IQ! (produced in the initiating step 1] can lead to many rounds (thousands) of Steps 2 and Steps 3. .. A Step 1 —I-* Step 2 Step 3 -—|"'* ? I_I ## l_l Initiation ‘ Propagation Termination":1 ‘finrm/mgl‘e WAcfi‘V-flfi lay.- t—q Araf‘bgfi' How long can this continue? Will the reaction stop, and it" so, why? — ..__, —'j St 4: Chain termination: When we remove the energy source (heat or light}, the reaction will soon stop. Why? I | Answer: Two radicals will combine! dHIn {kcalz’mol} M... ——-+ C;be “Cl. —34 Ham/30mg __.. (Li-{B —£H3 ass | oft/s: —~ sec: Radicals combining is a rare reaction (why?}. but once we remove {23} the energy source it is irreversible. Thus, the radical concentration [ decreases to zero and the reaction grinds to a halt. cl I lacs , ‘8 c: He not; ‘5 i The radii-Ccdafi Cm i1. [5 that” —[-L._-:r:: rodic‘cctfls-s good: Eocln a+m canal: rgcoME-énfi. are obi-la smqflfl. Overall energ changes: Thermodynamics: What is the dB“ for the reaction... CHE+ C1; —> CH3CI + H—Cl H°=‘? ! {A} Using thermodynamics: One of the central ideas in thermodynamics is that the m the m lecules follow to get from reactants to products is irrelevant {i.e.. oH” is a 6 i-e 2 function) — we need only look at the energies ofthe reactants and products. ‘ Let’s examine this reaction and decide which bonds broke and which ones formed. {24} heat J CH3—-H + Cl—Cl —:-. CHa—Cl + H—cl . fir +ioLl l l v l l P w Ear-dealt Eran-L: lineal-ca HUI. 42 a 57‘!“ {~le +313 “‘39 "2’03 25—4‘4 In other words, we don’t need to know the mechanism to calculate a m_4;lfl—__—i (B) Using the mechanism: If we know the mechanism, we can calculate 53H” for each ste. BUT: Because the chain initiatingfchain termination steps happed so infreguently relative to the chain propagation steps, we can essentially ignore the CLI’CT steps! For chlorination of methane, we get... Step 1: +1 kcall’mol Step 2: —26 14:63“le overall: —25 kcalv’mol This is the same result as Case A! By the way: Assuming —25 kcalr'mol is ~ M3", what is KW“? (25) ac“ = —2.3RT1og a” Covert no to K,q {Recall that at 293 K, 2.3)R/T = 1.35 kcalr’mol} i _ (— a a" 1.36 as '0 .2,/..5é n/ ..._ [[email protected] :- H a [LHHJECIz] [Lydia] LEG” versus 3H“: thice above that we used fiH" as if it were LEG“. How valid is this? First, an“ is much easier to measure — simva run the reaction and see ho much heat was evolved t Etto'l'lnel mic -- A H) or absorbed(€mdc:v‘l' we. )hv the system. {26] was Recall that it is no“ that determines the reaction outcome. However, free-radical halogenaticn is a gas phase reaction. So there is no s ent,.and_n omplications due to solvation energies. The reaction is also of the f m A + B phich has approximater zero entropy change. Take -can or this gas phase reaction... $8” ~ [I and thus r380 w fiH” CD Stride AGr— 5H aw ac ~ AH Lolmm M ~O ig'gr— cit-s. Lida-IP— A-arE—aCl-U Kinetics: Reaction rates: Thermodynamics only...r tells us the ratio of reactants and products at equilibrium. It tells us nothing about how long it might take to reach that equilibrium. We define the w of a reaction as (a) how fast reactants disappear or {b} how fast products appear. Reaction rates depend on many factors (concentration, temperature, etc), not all of which are intuitive. This means reaction rate laws must be determined experimental lg — they cannot be predicted from the reaction stoiehion‘tetrj.t [from the balanced equation). We can write a general expression for any rate as follows: I! A .1 "r4. e (“cow/est?! General form: Forareaetion A + B me C + D, Rate = lre[A]" [B]? Where the dependence of the rate on g and 3:; must be determined experimentally Examples: first: Woed /gJUEVt d CHgBr + H09 —) CH3ClH + Bra Rate = is. [CHfirfil—[De] So: If we double [CH3Br], the rate doubles If we double [H05], the rate doubles lfwe double BOTH, the rate TL”: . Cgbwueg :l. (Cl—13)}.CBT + H06 —} + Bl‘BL Rate = l2. (in other words, Rate = a. [(Cl-nhcsrr [1109]”) So: Ifwe double [{CHflgCBr], the rate doubles Changing [HOG] has no effect! What is the rate constant, la? la = Ara—E" "m" where... A constant (the “frequency factor", or the number of effective collisions) The energy of activation, or the height of the barrier in a reaction—energyr diagram! Ea J Guidelines: - Different rate laws imply different mechanisms. I Rate laws give us clues as to potential mechanisms. Cl-IgBr + HIDEr —a~ CI’IgOH + Bra *1 4, nail Rate = kg [CH3Er] [H09] A 1 order reaction (Ct-UH A Z orderreactionimpliesi’suggests... ‘i-usc moleculee MLL'E-‘i' Emil-(ha. i'ft “like. Elena; 5.1.1 (it; FaiE—riiQi-e—rhdfiffig E'i-elah‘mda“ ev¥ rm- [CH3)3CBr + no6 —> [CH3}3COH + Eire 1_ + Rate=|<e,[{CH3)3CBr] A ifL order reaction (GUM) d IIIII .... I! E 14 J s L 0% Zane.)th is Have deaf Fe —|—— CHSC-me iii-Jig" l‘flppfifls fiCCHJACBr £1172} - Increasing temperatures increases the number of effective collisions - More effective collisions = faster rates. For Efl in the range of w 15—25 kcalfniol ... Every 10 “C increase in temperature doubles the reaction rate. Every 1.36 kcali’rnol decrease in Edr increases the reaction rate 10X. So: If two reactions have Effs that differ by 1.36 koalr’n‘iol, the lower energy En will go 10X m. 3/!) MW [firfifi-E - AF Airfr‘fif"; £1 Egg-54- feeJ Ever zlroce — sancoé'éyu' téfifrffif‘,‘ I 'Hijfcr Technically we should be talking about the E energy of activation, fiGI, and not Ea ( = 3H1}. Once again, we are ignoring entropy — which we can get away with in the gas phase — and this allows the simplifying assumption that E, = nfl1 w no? .I The transition state: We are now ready to tie together 5.. ‘s and reaction-energy diagrams. The energy of activation, EU, is the difference in energy between reactant and the transition state — a high-energyr molecule in between (“transitioning”) reactant and product. Alternatively, the iiT is the height of the barrier in a reaction-energy diagram. Guidelines: (1} 5.. ‘s are always positive. -:-. {2) Bond making-{g1 ags behind bond breaking. Far Example: 5 mane r4 “c «.9: m at ‘ k Cl ts r-a (1' filL‘i” at g lb Nari; fila- #— g. H i H' “fl tl I 15‘. l + (L Ii2?) I u I a I I 1 1- — —-- 'I' —' + H H H» Siari fl—h Tmnstlrcw ~———-=- aha Why must bond breaking occur before bond making can start? What would happen if we let bond H . 9 lo I In makmgoccurfirst. cl gm. :94. H—C—H —I- :9. —H fi'fiflfil {a 0V0! + Ill 3g "" Man? .3” H. Once- H '— - a“ C; H H / some 1-: dairies)! (érggémj a?!” rt fkajfiezf- our: 57%,? 14:75an ) The end result is that breaking must occur first. Since bond breaking means a 12:11:35 . rilH, the Ed must be positive. (29) [n a reaction-energy diagram... ‘ CH4+ 'Cl 'CHa'l' reaction progress —1-- {reactants —I- [TS] —1I- products) Multistgp reactions: Free radical reactions require severai discrete steps. What is the M1” as displayed in a reaction-energy diagram? For the chain propagating steps. .. til-[D ISICP step CHa + ICI a ICH3 + H—Cl +1 2“‘1 CP step «3H3 + C1—C1 —.~. CH3—C] + «:1 fl [uni L314 Fume} Net —25 kcah’mol 7:" r2315 1 means "transition state" l5t C‘P step 2“ CP step h3k£ck fitnLaxefgu? The overall reaction rate is determined by the highest éoint the reaction-energy diagram! Vu’hile each step has its own transition state (TS) and rate we call the step with the highest T5 the rate-determining step; that TS is referred to as “the” TS. Summafl: la The highest barrier controls Cit-beer Wm 1‘11? . {32) The higher the hanier, the 5 L500- W the reaction. {33) The higher the temperature, the greater percentage of molecules that Imam-a «enough 'I‘D CF55?! . {34) ‘f'La Len-PC 64" Halogenatien of higher alkanes: When we leaked the halegenatien CH4 + X; -—> CH3X + H—X efraethane, we saw that there was only fl pessible menehale preduet (CH3X). Hew about ethane? CH3—CH3 + X: —> WT? + H—K Haw marl}.F pessible menehale ethanes‘?I (35) one: C'HBQHzX (KC-Hams ‘5' “ml How many pessible dihale ethanes? {36) Tm . Tlfi-z ill and CHEQHXz ‘3‘“ It: awe-=5. Lariat:er Hew abeut propane? How many pessible menehale prepanes? %" '?'?? Cl-ls—Cch Hl‘“ C-l ll-CA K: 70 a 4 Q3. 3 ru'u. It'il’i'I 2 Eat-Hi” 20H?! CHE—jr—GHB ZFCItJQmJ 7min; .~ 6:2 /er 3:: M 1'5: 25" Inc! : 2am) Clearly the 1" and 2°(ar1d as we will see, 3”] hydra-gens have different reactivities. Why is this? Log 5521+ Mar a Q _C_l F. m‘mne (663%) ‘l'lrxchn 2 it Cried». 2547:. Earn—Lee . ~. . [33) Let‘s examine again same examples from Table 4.2. .. m 1CH3_H _.. -CH3 + -H +1134 (methylradieal) b CHaCH20H2“® —"* CHacHzéHg + 'H { l radical} GHaCHCHa —I~ CH36H0H3 + 'H { 2- radical} é) a [CHalsc—“H —"" {CHslac' + ll"H +91 {3 radical} u- 2‘: H [5 3 Elite/MC}! Eda-53H 745 FQMGUE/ ('1 2‘9 C—H 35 3 lfz'eef Edie-r110 Jane'er Til/lend {flair-{J no mled- I All of these reactions are unhill — all the radicals are unstable {why‘i‘L and are highly reactive. flit: some are less uphill than others! Relatively sneaking, the more substituted radicals are less unhill and thus less difficult to form. we The more substituted the radical, the easier it is to form. The trend is as follows: Radical stability: Me I r r r r r a “I. R—tfo Z3? R—{llu 2:- H—(l3. :- H—L'l;. ET \ a . a a H Ht, H B 25 2'” J Me '5. R—H —1-- RH oH Cl ' T5,.“ + Cl '. Let's look at and compare two seemineg similar reactions... "I naminimiik-.. -. ‘ 13 l-c oro ro ane E—chloro ro ane ; (-5 l H ’s H p P p P ' a (a 1° alkyl chloride} {a 2“ alkyl chloride] l' 2 2 H ‘5 4cm song In Br: B I: CHacHgfiHa T M r \I/ + HBr - ‘tr' . .--Lu_l-£'E-i Elfinaci' Elr j 5..- z 5—- l-hromopropane 2—bromopropane Lei-.4 «as: If We see a clear preference for reaction at the more substituted carbon I {the 2° versus the 1°). Why? _ / (a) the secondary radical is easier to form (less uphill} [a 1n all: - omicle) (a 2° alkyl bromide] ,5! .. : 2 — 3’ “ n | I x m We? m 1 "a. is (b) #3131" is esEecially sensitive to this diffeten — reaction. Given a choice, molecules will react so as to follow the lowest ossiblc cner hiiv ! pat\i¢3{L‘t? CflmPkfialeclj 5,44 uni-an 14.16". melee lat—cl we. Jr ~4be Lauri . ! Bins H—Er_ GW —‘88 Lee—fl! I l | What We need is a way to estimate the heights of the transition states — the barriers of the | r f a I q '63:?! Guidelines for reclietin E or no: or TS hei ht : n 5’ 75’ These guidelines are especially useful here, for a gas fie reaction (no solvent}. .— [l) Wheno_nljgbond making m] fimfléhfl) an ooeurs, £133 (but there will always be at least a small barrier}. E i ' [2 = reocion ‘i‘ ________________________ _, . | P: T‘s: imnea'i-(en (sic-4E A, + .E a Int—B | (2) When m bond breaking ooeurs, EE w fliH“ (but Ea will always be slightly larger than fill-[0). reaetton+ 1-5 F: W —-i-r Ao+oB (3) When bond making and breaking occur in the same steg, E“ is {l and Ear is nu“. 20 reaction —:- h [ reaoliou 4-- CH4 + .31 __,+ . 6H3 + HQ: CH4 + I Elr —i-'- l CH3 -i- HBT Broke: C Had... H 1" l £3"! Broke: C H-a _ H + I‘le | Made; 1—1..“ — 103 Marie: H"EE _ _' 99 . M {w I. I J Era-oh" HE H Let s start tying this all together... H L:- ; __ 5H Chlorination of propane: +99 _lqg 3H°{kca1mml} CP'l CHSCHECHz—H +th —- CHECHECHZ + H—cl 1“ . H" CFZ cchchH2 + GITCI —-' CHSCHECHEFCI + -c: -23 ~i =39 c “‘8! H57— his?) L c n GP1 CH3'EIIQ43 + 'Cl —I- CH3EHCH3 + H—CI (g); 2 H _._... CPR . CH3CHCH3 + Cl—Cl —I- CHalCHCHa + “3| ~22 i l-I— 3D +539 0' Bromination of ro ane: +19 B. H J??? -. 0P1 H + OBI" —h- Al + H—BF 0P2 A. + sr—cr—r— W? + 'Br -22 “2’6 453 “Vi. Hi" . _ :33 W CP1 A + iBr —"' A + H_EF 20 ,__... CPE + EIr—Br—F BL. £3 + IElr '22 A we The rate-determining step (“rds”) wil] be the most uphill {or least downhill) reaction. That is the step we need to focus on. So: The important steps here are when the halogen radical abstracts the l” or 2“ hydrogen from propane! - lore-G {c.‘nfis flirfifi$ (1—H bond is 5 Ian.) Once the alk l radical is formed the site of halo enation has been determined. The actual incorporation of halogen at this site proceeds rapidly,r and exotllerrnical1:»,r downhill. 5+ . In other words, the i“ C-Lfl-lfi fii'eE in free radical (39} halogenation is the rate-determining step. r——'—' —_l Now: How high is the barrier (the transition state}? {I it :I ‘ ‘ L t l 4.; Hammond’s Postulate “HP” : ‘ {LEE 4"“ WW5 {DD LE- Endotherrnic reactions: TS is close to products in energy and structure. Exotherinic reactions: TS is close to reactants in energy and structure. ...and the more endothermic (exothermic) a reaction is, the closer the TS is to products (reactants). Again: Our question is why chlorination onropane give an ~4o:eo ratio of prirnaryzsecondaty, whereas bromination of propane gives an 4:9? ratio of primary: sec ondary?I Encla‘l‘L—A-rwa" ch bani lard? In T5) . C salmon cw let-"7 3“ T515 serfs Bromination: graham i“ 1“: _/ t The 1“ reaction has a 3‘ ""“ __' CH3IE;l_|2.["l'}:rl_|2 larger +fll-I than the . 2” reaction. HP says 1a to " CHagHCHa the 1“ reaction will CHE—[FH—tFHz a“ + thus have the larger E 1* £4 a lar or T8 height, and thus a slower reaction. (“6H +7 +to asar lEll‘ - v ~ — u n w v - v a — — _ . _ _ . _ . . . . -- [‘EECUOH progrc SS up Chlorination: Exothcrrnic reactions M am ‘ —--—- 1 have much smaller .51; m T5) 5 TS heights. HP says the difference hctween TS's (fltTS) is much closer to zero! and thus much less selectivity for [Batman FWEFESS + the Zn reaction over 2:. the 1“ reaction. , EXeHelmri 61* £457! an?” a"?! " ’L’i + H—~Cl i -"J | .1 i::H3cH2cH2 l CHacHCHa I reaction progress —:- Cl 1. Hire] HP 5H5 ri—fiar éQ-j-Lumlflq r“an Tip-axe LL35” the: ‘1 #:545743— ' r "9'": Tab (LHBCJ-iZm-L + H~Br Examination nfpmpanfit _2_ l / Ill +10 a —..—- _. reaction r0 regs. —h-- P g E... f l CI—lBEI-ILHE--¥HEf-' - l I f l l I J I \ J 1 ! I V Alternate statements of Hammond’s Postulate: 11v’ersion #2: The more stable an intermediate or product is, the lower the energy of the TS leading to that product. Tlaiiit f News that the less stable the product or intermediate, the the energy (40} of the TS. ll U. [ail-e. Ti: Version #3: In an endothermic reaction, the TS looks more like products than reactants. reaction progress + 11 uearht " T5 ‘ i P In an exothermic reaction, the TS looks more like reactants than products. 73 but-{Fifi lentil? CI F fircactionprogress++ r‘ [Elma reachai Latin .5 What is the effect of lowering the TS energy? For 5., in the range of ~ 15—25 kcalfmol (pretty typical], every 1.36 kcalfmol decrease in E, increases the reaction rate 10X. Even a modest lowering of the TS energy can have a profound effect. HP allows us to predict Egg Endothertnic reactions, with their typically higher energy TS’s, are extremel sensitive to even small decreases in TS ener l Exotherrnic reactions, therefore are much €53 sensitive to {4 1] TS energy decreases. Version #4: When cne reacticn gives twc (cr rncre] prcducts, hcw much cf the difference in product stabilities can be “seen” by the T S‘? For endothermic reactions: Mar-Q Fer exothermic reacticns: l:- ss rs Fl:- [42} And finally, Versicn #5: When one reaction gives twc (cr rncre) prcductsi’intcrmediates (c.g., a 3fl versus a 1° radical}... The mere endcthennic the twc paths are, the closer aTS will be to ddH. ..—.-.—.——. The mere exctherrnic the twc paths are, the clcser d'I'S will be to zerc! _,_,..—.——| Example: Ischutane (again) as 9H3 H3C‘_{I:_CH3 H3C_IC_CHQCI CI + H C}, t-hutyl chlcride isnhutyl chlcridc rm3 4’ [31%] (63%) Hall-(IS-CHa gss’c H 9H3 9H3. imhutane chmtll-CHa HgC"f|3"CH2Elr [2-1neth3'1prcpane} [EhTVaC Er + H r~but3rl hrcmide isflhutyl bromide (399%) (aim I__._.._._"______.._l I___,.._._I 3° 1" t-hutyl halides ischutyl halides The key ccncept here is that the rate determining step is the formaticn of an alkyl radical and Hx'" Predicted aatgkcalrmsu aTs A + H—Br CH3 /l\ + H—Etr I I eH2 O I cH2 l Hac—(E—CHE, H A + H—cl A + H—cl In other words: *3 Bromination is much more selective for the more substituted position than is chlorination! Example: methylcyclopcntane (I), Man};r monochloro products are obtained minafiw hi." on; he D b 2 b i 2 he I 25 “c b CH3 2. 1 H (E2 ,_ Cine ma'or product produced 127T saline ‘ue in!” B°H N“ Br- ! Question: What is the major monobronio product? How many monocloro products? Draw them. [43} Pain Ci an £43351 C. CHILI: ¥mr CLuEirco es CLQCl: 5113'- flifi/{fofls Reactive intermediates: Free radicals, carbanions, and carbocations: 'l i l R R—(I3El amp - R—p: >: R R R R Carbocation Radical Carbunion Carbene Hybridization: sp3 "5112“ sp‘l’ :2 Pro erties electron electron electron electron P rich deficient BC) m‘leifi strong strong base Lewis acid General trends: The carbocation, radical, and carbene do not have octets and seek “electron rich” partners with which they can make bonds. The carbanion has an octet. However, carbon is not very electronegative, so carbanions are exceptionally strong bases. Stability: For carbocations and radicals, the stability trend is: t r t r Ft—tIZII 3" R—(lili I} H—(IEI 3" H—(II' Radicals R H H H l“ i T i' R—(IJC‘B 3" R—{IEG'} 33* H—[IZEB 3" H—tljfl') Carbocations R H H H 3° 2” 1"“ Me Why? First, all radicals and carbocations are electron deficient. Clearly, the more alkyl groups attached to the electron deficient carbon, the more stable the radical or carbocation. We say the alkyl groups are “electron donating”. So how do alkyl groups donate electron density to electron deficient systems? To answer this, we need to leek at radicalicarbocation structures in detail... "‘A note that the sp2 lobe and the [3-H bond are perpendicular {no overlap possible] / H’ar..:® Haci'm@ _ _ I :HTL—LS'OWA 1-,_ C—H —CH3 1'. Hr@ Hag, "" 1 Ed [C QUE-5" Methyl radical 3n radical H ‘ I It Sp: spg gall-E. ME: Ltd "LC 6"“ REL!" m Heal Epitapfi in EICEth‘i‘r‘Lfi In the 3” radical there is the possibility Mp” et ' nt C—li bond and the electron deficient sp2 carbon. This over @gationi although somewhat poor, is better than in the methyl radical (wheret ere is none}. The carbocation is similar, except there is no lone electron in the sp2 lobe. From this. we conclude: The more alkyl groups, the better the hyperconjugation. The more hyperconjugation, the more the positive charge is spread out (which is helpful). Where else have we seen this idea that spreading out charge helps to stabilize charge? RESONANCE! Shown below is an example using a carbocation. a Gift FE +1r2 +1i2 FEiLL‘i-i 1‘15 '3' 2c‘"”'fi“‘ca2 H ng fl o o flCHg H23 rm“ CH2 P '-' CH— LI. 4105 H/ 1 %en alone. these appear to be ...yet the hybrid is approximately as stable 513 primary {1”} carbocations.... [if not more so} than 3 2° carbocation Resonance involves {most often) overlap of sp2 systems with sp2 systems. Hyperconjugation involves (most often) overlap of sp2 systems with gp: systems. Resonance provides much better overlap, but hyperconjugation. if that‘s all that can be done, is an important mechanism for stabilizing electron deficient systems. Bond Dissociation Energies: Table 4—2 (p. 136)... A—B —> H- + B- (+M-i, in kcalr’mol} GAS PHASE! H—XandX—XEflnds H ~ H 104 Warm D — D 106 (GHQECH — H 95 Cl — Cl 53 (CPR)ch - Cl 80 Br — Br 46 (CH3}2CH — Br 63 [_1 36 (CthCI-I — I 53 (CHQECH — OH 91 H — F 136 I; _ 13% Honda to Tertiafi Carbons H—[ Ti {CH3}3C —H 91 H _ OH 119 {CH3}3C — F 106 HO — OH 51 {CHshc — C1 7'9 {CH3}3C - Br 65 [CH3)3C — I 50 Methyl Bonds (CHEhC _ OH 91 CH3 — H 194 CH3 — F 199 CH3 — Cl 34 CH3 — 131‘ TU CH3. — i 56 CH31 — DH 91 Bonds 10 Primary Carbons CH3CH1 — H 98 CI-13CH1 — F III]?r CH3CH1 — Cl 8] CH3CH2 — Br 68 CH3CI‘I: — I 53 CH3CH2 — OH 91 CH3CH2CH2 — H 93 CH3CH3CH2 — F 107 CH3CH3CH2 — C1 81 CH3CH2CH3 — EI' CH3CH2CH2 —l 53 CH3'CHgCI-Ig — OH 91 1|Which bonds break and form, and in what order. The energy changes associated with a chemical reaction. The ra_te of a reaction — how high is the barrier separating reactants and products. l-lomolytie Homolytic I do not have an octet! into;r easier making»f negative hreakingi positive —I 04 kealimol bond strength weakens -lfl4 keah’mol Longer and thus weaker bonds! add heat, or light, hv to the arrow first, CH3X is fonned From CH4, then CH3}: is the “reactant” that forms ‘3ng:; CH2X1 then forms CHX3; finally (Ill-IX; tiirrn Chis. add CH2X2 and {:ng to the boxes. low concihigh cone. draw structures; they're constitutional {structural} isomers circle the nine primary and one tertiary hydrogens — methane had only one type two Ci radicals. “radicalsino oetea’reaetive”. BDE an“ = +53 kcaifmol HCl plus methyl radical. ell-IIn = +| kcalimol Table 4.2 in text. nH‘1=-1Ifl4idH”=—m3 C12 bond = +53. {CT—Cl bond 13le = +84, so change sign. dH“ = +58 + £4 = ~26 kcah‘moi. The radical concentration is very low, so the odds that two radicals encounter each other is really low. Broke: CHg—H {+lfl4}, Cl—Cl [+5 3). Made: Cl—lg—Cl {—84}, H—CI (—103). Overall = —25 Use an“ = "searingieq; get ~1 x to”. exothermic, 43H”; endothermic, +nl-i“; TS structure: {3—H bond starts to break, H-rCl bond starts to form. Products = ICH3 and H—C] 1 + 'i -* cI—H -i|:—H H H will have more than two electrons -- we can‘t have that! Once the (2—H bond starts to break — once the H has ,]o_st some electron density — then IC! can donate electron density and start to make the H—Cl bond. positive. Draw an endothermic reaction (dH° = H] with a late TS (an‘ = +4). Reaction is CH4 + M131 —3 “CH; + H—Cl 31} 32) 33} 34} 35) as) 31'} as} 39) 4e) 41) 1 3o 5- C|'+ CH4 —l|- CI"'H' 'CHg] —‘.I- C|--H+ ICHg reaction progress —:I- "—p [Ts] —:-- products} First step is endothermic reaction [.dH‘1 = +1} with a late TS (dH* = +4]; 0C1 —9 ICHg + H—Cl. Second step is exothermic; [dd-ll1 = —26} with an early TS {rill-i$ = +1). Reaction is ICH3 + C1—C1 —> CHJC]. + It’ll. Overall is —25 kealx’rnol. [reactants Reaction is CH4 + a. a. i a. a. 1 C|"'H'"CH3] [cincwcna] 1 31"]— / "transition state" 20— HCI + oCHa + (ll—Cl reaetion progress —i- -fl——-——F-fl—-——F 15' CP step 2nd CP step the overall reaction rate. slower. have enough er1erg:-,-r to cross the barrier. 0116. two. The 1.1 and 1,2 isomers two. l-chloro (4t)%] and 2-ehloro (613%). Even though there are 6 1° Hls and only 2 2“ Ha. The corresponding radicals have different stabilities and thus ease of formation. first Cl1 stop. higher. less. 42] mordlcss. 43] bl'fll'flfl = l-bmmn-I-mufl1ylcyclopentane. Chlom {six total] = 1,1, —CH2CL 1,2 (cis and trans), 1,3 [sis and trans). ...
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This note was uploaded on 10/04/2009 for the course CHE 2230 taught by Professor Graham during the Spring '09 term at St. John's.

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wade_ch04_answered - Chapter 4: Chemical Reactions —...

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