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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/ ..._ [Em—@9159 :- 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-...
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