wade_ch01_answered

wade_ch01_answered - |Chapter 1 Introduction and Review...

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Unformatted text preview: |Chapter 1: Introduction and Review: Organic Chemistry: The stud},r of compounds ( = molecules) of carbon. Synthesis: The preparation of larger, more complex molecules from :Pi—e All“ i- re 5 u; + precursors. I Synthesis for us will mean making and breaking bonds. 7 Mechanisms: An orderly and reasonable description of which bond break 7 Lift l: £1 oriental-rant: and form, and why. 7 7 Structure: How the atoms in a molecule are connected and how this ? rw -l- ,- :5 affects the properties of the molecule. fiohler. Germany, 1828 D i heat H a} i '53. - NH4 ricera —"' c H—ifiu + ‘N: 11::le HEN" “MHz I ‘ ammonium cyanate urea H t it” . i ___A I- : ‘9 II inorganic organic semi lim- [92 ‘min car :5 U ' fth f' t d d rea is oneo e no "man me e" or anic com con 5 'lLt-L and Star "Uri ‘Elmn g p Structural Theor}r (Lewis, 1915): l) Atoms seek to fill their valence ( = outermost) electron shells and thus attain the confi uration of a noble gas. g fliflM‘L flcaoa-Qen'l' 2} Atoms do this by either transferring or sharing electrons {a shared electron pair = a covalent bond). {he'- A_'E TLJ-Ir'llci "-—" re Ct shared Fair” ‘15 ‘2' Electronic Configuration of Atoms: Electrons reside in orbitals. An orbital, then, is a region of space where we might find up to two electrons. l—l TIM. Mclwfi t5. 6—} Cl betrch Relative ener ies and sha es: eflac-lrmm‘": fir-E (7-) CLO—nagé Closer to the nucleus = lowereriergy Tit-44.114 liLE "lb la"?— ‘35,) = more stable Clause :39, HJ’SLM'E-Mr'gq: rug-lab w: (2.3, fit—D": areas.) 4— + — 2px, Epy, 2;}: 2p orbitals ls orbital nucleus I:I 2 eff,- QQ flail 25 2f”; “HE is (£1 Bond Formation: The atoms in a molecule are held together by bonds. l Atoms will transfer ( = ionic) or share ( = covalent} electrons so that they attain noble gas confi urations. : _, u u '—— _.g_._ signal] bola-l- rule 1) lonic bonds ——._— (1 lc * 'leC't a) Fonned between atoms of very different e *.‘.‘-tronegatii.ritj,rr EN). EU is as Meser a”? bah-i swam” CSlQlale) can Co‘le f5; C“) Llwaarccfiia b) The result is complete electron transfer from the low EN atom to the high EN atom. In the Periodic Table1 the EN trends are. .. WHY? Sherl- answer: "TL,a {Lhasa—r L-[oq 4:er +3 an BL'L'E‘L ‘Hm. Hear: (ion ma-l— ‘l‘c: again L4". loci: G'I‘E... Uttflenca a‘ ' Let’s consider lithium {a Group I element]. .. I V ammic "1335 E1 1, _ electron configuration is ls22s1 or [l—le]2sl 3 LI atomic number fl EN =1.o (vexq lea; EU) Lithium is between the noble gases He (2 electrons} and Ne (10 electrons). Lithium loses one electron and it looks like He (versus gaining seven! electrons to look l1ke He). .1 U a be £_ '19 - For fluorine, 9 F [He]2s22p’ EN = 4.1 {highest in the RT!) Fluorine gains one electron to look like Ne (versus losing seven electrons to look like He} A from 1:. Thus: When lithium and fluorine combine... Jamil a mi. 16. rim '5 ‘5" a haircare Li on ‘13:: —n- LiGfl + :F:@ . are formed 1 E: altar-«324 Because the electron transfer is 10-13%, charged sgecies { = ions) result. Attractive forces between oppositely charged ions are veg high. As a result, ionic solids have veg}r high melting points and are soluble only veg: golar solvents such as water. In [i‘m'irncé Gauge :35} i italic-ma ens +La. €- eb re: Shae-=5 4 M b} The two atoms will share electrons. And is thus 5 hated o lectrons. We rig a line; when an atolectrons we d - 2) Covalent bonds a) Formed between atoms of similar EN’s. a “lone pair" of electrons or u / The fluorine molecule, F; :F F t t E.../:{/. “fl L-acovalenthontl SEA 151—) 1-3 | tX~ 14 w b‘l'qr'io 'e'i‘fitc'iu {Tit-utint f6 4;. Given a molecular formula, how do we generate a m? +5 @3611 “11.1 mm ,5 1c l)L-earntocounte1ectrons “3 % Ll “Enid 2) Give each atom a noble gas configuration ‘5 {39'4- "I' girqtiuw} a] For C, N, O, F use the “octet rule”. gagged. {-I- egrtl log, b) For H, only two electrons are needed. Drawing Lewis Structures: ideas and Concegts: How do we count electrons? We need to know the number of valence electrons in each atom. The number of valence electrons is the same as the groug number in the Periodic Table. When checki; for the noble as configuration, we pretend the atom “owns” a_ll of t e lone pai bondin air electrons. We get these typical results... # electrons GFDUD # Valence “ewe” “3 # Bonds Result Atom Number electrons oat octet formed 1 igal - c- 4 4 + 4 as 4 | ' {"tetravalent") _£|'3“ . N. 5 5 + 3 = 9 s -- "' ("trivalent") “T— - é: ~ 5 e + 2 = s? 2 .. ("divalent") —9— Halogens (F, Cl, Br, ll T ? -i- 1 :3 1 .. {"X" in general) (“monovalent”) = H 1 1 i —H Atoms can also fill their valence I I \ / shell bv making multiple _C_C_ CZC _CEC_ bonds... |T| / T \ 1 Male. [In em: in a single bond double triple Ce; ea CHE-tr born has. Ll louflcl'fil Polar Covalent Bonds: Ionic bond complete electron transfer Covalent bond = equal sharing ofelectrons if lrmflrncs Craer When two atoms of different EN’s form a bond, the electron distribution will be unegual. E a." =- " EN = 2.2 Exampleflvdrochloric acid1 HCl H (21: EN; = 32 ll-l4 (6:3; The higher EN of chlorine means Cl “wants” electrons, more than H‘ ‘U" H C—N The H—Cl bond is polarized — the bonding electron pair is closer to the Cl than to the H, and partial charges {333 t 59] result. 3 = “Far 'lt‘cLQ“ The HCl molecule is better -l—I-— , described as: 5'3 H a: fie Cm ‘l’m‘l lrnw " Ext“ 'lltf laws The arrow shows the direction of the dipole moment, p. MC l-‘t ) The dipole moment is measured in Debeves {D}. It is a vector quantity (a vector has maggitude and direction}. A molecule with it ~ 1 D and a bond length of» 1 ft (1 A = ltl'”J m) has 5&3 h +t}.l e.” 311515 0-1' Decdlflt at oee-ler lnes. magma'lu 5-4”— ciavtd diceceltua . Formal Charge: Hen-E) cliple Mem‘lvt}. = Céfifje’f £43116 "I . . . at EU a lead {my We have made a Simplifying assumption that atoms share equally.r electrons in bonds and own completelz electrons in lone pairs. Formal charge is a bookkeeping device that lets us keep track of our electron count. Formal charge = “starts” — “owns” “Start” means how many electrons a given atom starts with. This is the same its fill-Ume E E‘“ or %?mlg,{3 5H: m Eficredfll @1th 1-5 “Oums” means the number of electrons an atom can call entirely its own. We say that atoms own it" of their lone pair electrons and one-half the number of electrons inbonds. \r’fibl-fLuE i5 :1 Eu‘l‘ *9 Pi‘t‘lura . it: bat—let'- Exampleflarbonate ion. C032— I ‘1 D We will draw the electrons and assign the charges in class II 55) J‘G/ Fe.er la: fi‘lfir‘l‘a— hearse. = G- -' E: + ifiajj \\ 5 9—7 EeHer-t roll-'1'le ge'mo.‘m57 :“i Ionic Structures:Coordinatc Covalent Bonds: 6 fol-ta 12' at”; ELLE" 7E3” rte. I,an Some compounds have both covalent bonds and are also ions. These can be tricky. Example: methylammonium chloride, CH3NH3C1 FEM” “d C I'm rfléiQ. H H 1 he “bond” (better: attractive force) between | | E) u (3 the N3 and the Cle’ arises because oppositely H—‘f— I _H [ =2” 1 charged ions attract each other. H H THERE 18 fl SHARED PAIR OF ELECTRONS BETWEEN N AND Cll Guideline: For our purposes, the low EN atoms — the group 1 and 2 elements — will almost always exist as ions when found as a component of a compound. We should NOT I connect these ions to other atoms with a bond {a line]. i | This is especially true for Li, Na. and K — these will almost always be found as Lia; Nag, andK‘B- “flux-Lats? woe-l“ algae-1.94111 SJGWUafl-‘I ' H H L‘ a I II@ II '1’” ’k H—c—p: [Nae] H—tls—p—Na H H t Which one is correct? {1) - Na Clare: E4 “alanine E.— Resonance: Electrons are not {always} as confined to atoms as a Lewis structure suggests. If possible, electrons will delocalize. or spread out. a"; re E‘H—Ifi Gil-e4" "-ll-Gti min-l wh? ' F "lb ePi—ecsd saw” (2) Example: Carbonate ion, (3033— Draw the electrons and assi gn the charges in class Given: Experimentally it is known that the bond lengths in the carbonate ion are all equal. Whv is this oxygen 'b" I a: E) 1 E; : a uncharged? {ll} (it OR (I: C9 :15” “one: ‘23in “he‘- ",0"” HELE- Whv are thEse Why not draw instead: - flaw: [usdfijss‘f v All three structures are equally good, vet none are “correct”. oxygens charged? 3* The “true” structure is a combination (or hybrid) of all three possible resonance structures. *a The individual resonance structures are n_ot interconverting. *r We sapr the electrons can delocalize. How do we show this? are l ‘3: r: J 1" iii-SH. -2is are Individual resonace structures the Pestilence h rid What do the arrows labeled "a" and "b" mean? a} make: convert lone pair to bond EGHUH+ lsaa 'I' it; sans-m5 paw {3) b} break: convert bond to lone pair canflE-r-l‘ legacies-a3 Emil" ‘i‘e: lien-m flir- I 1—? t—I—fl—‘u—I ! l The list description on the carbonate ion is the resonance hybrid. In the hybrid we expect... U) +3 All the bond lengths are equal (in between that of a (3—4:) and {3:0}. (9 v3 The net minus two charge is spread out over all three oxygens (hence —2t3 on each oxygen). (9 +3 Matches experimental data. -— 4.1% Land 5 CC":- rE i a ‘E-e fi'i sa. ME. Le 1a gih The great advantage to Lewis {i.e. the individual resonance) structures is that they' show all the electrons, show them as easy to count pairs. and show where they are. +5 BEAR IN MIND THAT SOMETIMES ELECTRONS ARE NOT AS LDCALIZED AS A GIVEN STRUCTURE SUGGESTS! Far fiance... ‘i‘de ice [Jr ‘Er $Tfifi4eén . . 5115:}? i'hcr-i- ct Aer-thine ['an (a. I n "gt/“- Eher'i'fid— +1.15“ :3 St'mzzfla. (Dong. .‘C'- I If!“ a @_thfi_@ u, I. This Ff.“ invariance $4me‘iura lung? (amen-S- '5 Eiudfl-LI aha-s) ?l:-E_é£'i$ eon-a fisher-i ++LUD [tang intends. - we [chow (9&2. qflfl iinra heads are. i'Le. Seam. lea 3-H]. LOH it: maCLuejeewvw) @ 4:: great: a Lag-9nd and make ahead. G‘l‘ fiamnfitmfi? I l .' E: ' a. 21'. ' ' I 2M '. E II I: #4.. Cf.- I E) (3:5) '1br’. N 'i $0: ‘ E C. we a are f Rules for Drawing Lewis Structures Given a molecular formula, how do we generate a Lewis structure? a) Learn to count electrons b) Give each atom in the molecule a noble gas configuration... For B (sometimes), C, N, O, F we want 8 electrons in the valence shell (this is the “octet rule"); for H we need only 2 electrons. c) Penna] charge Ideas and Concepts: I d) Resonance The Procedure: 1} Add up all the valence electrons from each atom in the molecule. The number of valence I electrons is simply the group number of that atom in the Periodic Table. Where applicable, add one electron for every negative charge 1|Where applicable, subtract one electron for every positive charge Example: CH3F (Lrn " iere is the idea that we have n bonded to 3 H's and one F) it valence e— w one C X 4 = 4 i 3 H x l = 3 | one F x T" = E 14 e-, or T pairs (Remember, H only gets two electrons} 2] Find the central atont (C in this example] and attach the remaining atoms to H it with one hand each... I _ , H—c—H Used four of our seven pairs of electrons, so three pairs are left. | . . . . F 3} Give the nonwhydrogen, non-central ato1n(s} an sing the remaining H electron pairs... I H—ClI—H F I... II Fl This example is done. lI'CJther examples will require rules 4—1 4} Place any remaining electrons on the central atom, even if this exceeds the octetl [This will never happen with C, N, O, and F). 5] Calculate the formal charges. The sum of the formal charges must equal the net charge on the W~ 6} lt'the centnil atom doesn’t have an octet, try using lone pairs on the non—central atom to Form multiple bonds to the central atom. T} Calculatefieheek the formal charges again The sum of the formal charges must equal the net charge on the molecule or ion. ii} Check that you have the best resonance structure { maximize the number of bonds}. Enm tints, new 1-9 G) fi‘ccum“. 9“ “+7” 32:”: w oc4e¥3?'/ Exampjes: Dichloromathana, CHgClg @L’méfi *5 (9 {I E) arc? cenhafl Ch'luln “1 zero H _C .4; i all ,C, l' ('ij SJ Pmpanmfifi‘m 91””? +5 f2+5=20¢r 45:524- _____.. RSI-e: Usually, atoms belong to the C on lafi : H H H i. I l 7 . H_C~"‘Cl-C”_.--H “3P”?- #3/ J H 1); (Lil?— @c4c—15?/ 3‘ Ethoxideiun,CH3CHgOe{'fED CA“ ~‘- {45! GM— ‘5 74“ “m” @1 magi aim-x639 / H H {53 O 7 l tit-act (E) '1 a .. _ l Fla. 5 Hhc—cuc: __.@_>.H-C-—c;—9. i t I 7 F” P: 1 Moa1le¥al/' H H —{ Ethanol, CH3CH20H H H H ‘C—vT—ii—H / L in H H F ldh d ,CH20 11 a“ Q _ 011113. 3 3" 2/ J a) e l' "b ' If 2' ' ii' HI‘Q‘CLl' I c‘P-E” E - H ' ® Hr: H @639 “3%” H 65 H FA. H/ r? . new} 3P”, «gang-H, ("Pa .__|I Frem this epeeifle example, let us induce a rule: Usually, 1nq§¢ifi1+LaH is an flee! Curr C4 “‘fi‘%‘$"”m+ @563 E —§r—: a H 5:2: H had zero 5 r‘: a CHf—LhcaaeHB M Delete? 4":‘5 Maui '67 E A if“ ham-id! “if-5: lgPrb‘g/ _.m._ CHE—I 6? "—C'HE 344' “‘3 firm-El .3 r; CHE—g—EBEQHE .0. '1 " g I 0' (Name not given): NH4C1 (j g 5+L1 +1:f€:€*/?Prs *H' adgfigfi7/' H ' ¥ H a: “we-fl [m] 1 L. H (in (but. Cheapoan Tr? NIL-g- on your awn _ h i k EClecé re" cit—~qu 71 C3) mrmwfzi £75193Chacé £5:- fl¢1Z€7f5/V:b/a1flbu5 Fif— I—l '._ ""‘-——~ Rules for Resonance 1/ 3:3: -1r2 l) Resonance structures (R.S.‘s) exist ll: h J: only on paper. The real molecule R” “:6: o R! “a (5 m is a hybrltl [average] ofall RFS‘ #1 R3 #2 E + 5 PIJSSIIJIE RS. 5. [resonance The doubier-headersI arrow indicates Ft.S.‘s. [It] NOT confuse it with equilibrium arrows. .S.‘s we only move I t t r trample: H—CEC—CHEGHg and c cc rous- never a otns. H2t;=CH—EH=CH2 are NOT resonance structures . A] . . s must be proper (“good”) Lewis structures — for example, no 5 bonds to carbon. All R.S.*s must have the same number of unpaired electrons... bleed {virtually} complete overlap of all p-orbitals —- this means all the p-orbitals must be parallel. The energy of the actual (“1 WE W2 . ' - fl fl E -_ molecule Is lower than any Hflcx’ CH2 H HEC “mew HECfiCHZ single RS. would suggest. is o _ _ The} "extra" is 133.]le Taken amnel [hESE appear to be l3 tapprox'm?lalllr 1 . . o . assa eino moreso resonance stabilisation. primary (1 learbocations.._. than a 2D carbonation especially large when _ equivalent R.S.’s are All!” rad'calsl involved. Equwalent R.S_s Beanne (Ch. .16} Equivalent R.S_'s The resonance . i ll stabilizatiouis HECfiCHg "" 3" HZCfiCHQ H Non—equivalent a . :3. + + iiiajafi‘dkfi “ficagf‘egug Highly/firsts = Hacbfigrffi‘ecfinz contributions to (3H3 H CH3 __ ,3 H3 . lgfiflrdstliirle 1” ailrlic 3” allrlic hybrid the individual - the more stable - more 5+ on the more subs. carb r SlFUcture structure, the larger ‘ts contribution to the hybrid. It) 0 1 DE) E Eye—fie client‘s metreé -—.—.—.. A} The less fermal charge, the more stable the individual resenanee structure. Evaluating Resonance Structures and Estimating Stabilities I Usually this means maximizing the number ef eevalent bends... REA/Te id H |-[ { 13:0: 1% IE ,I | . C a” “fete R’eijpzm Better RS. Werse RS. fermal eherge = -1 L/ Cemaéjflbofl J 7i 3 formal eharges 3 ceuo€647igoflcfj an”; 3 carriers .. .but semetimes the number of eevaleni bends is unehanged. .. CI .. jeeUaficzm‘i‘ Better RS. ne fen-n a] charge B} Fill eetets whenever pessiblel El in E‘Qfil'i E} ii e ii H3CXflCH2 H Hac/ ekeCHz A lone pair is n t te an emetyr p-erbital en t sp2 hybridized earben - it wii eatiin be denaled! All Erwin eetet 1—13 Exceptionsi'Unusual Structures: a) Odd electron species (“radicals”) Examples ' '. H " : M=o , I H—o ' - H—c-—H H Electrons drawn etth radical hvdrox radical In class -’i'-?-e|ectrons? 4H" el 1 7 active! Why might this he ni so radical electrons Radicals re very he: oaiei b) Incomplete octets Examgle: BF3, boron tfifluoricle (electrons in class): . we as a ’2? electrons? (vlla / OH H in. " B a all wig/[2?” I II In structure 1, boron does not have an octet {= not good} but has no formal charges (= Jflflfl . In structure ll, boron does have an octet ( m4) but has formal charges [ :9 J, has a negative charge on the less EN oron, and has positive charge on the mureENfluorintt=J1eLaaed)- 6:) fan F [a o “learn wreck” “ lineal kan- claq'l al . . had ‘35“ N4:- qu-‘EL‘LM-AlLO-F. 05:: J 5 can: fa tit-36hr c4107 The consequences of not having an octet are simple: Boron will readily make a bend {a coordinate covalent bond) to a species willing to donate electrons... Which is “correct”? BF: *1? H '« :E’Q“F= Li. I , =r= II I. g. E.) II Hi * + _n/B.H --_ .' —El—_lf_'- electrons and .P H. " I IL. arrows in class no octet B has octet 1—14 c) More than 3 electrons (“exceeding the octet”) *:* THIS 1WILL NEVER HAPPEN WITH 13, C, N, U, AND F! Generally, we need a 3rd period element or higher. Phosphorus and sulfur are common examples; actually P and S can obey the octet or exceed it (see below). Examples: PC13, PC15, SF5 {electrons in class) .. . “In: . :91 We :Eeew x .. :91: Ice 91* Phosphorus Phosphorus _ trichlorioe pentachloride hexflflflflnde ??electrons‘? S+2I=2éfi3y. 5+ 35":170/2‘0/9L ‘5 1‘ '7'2= 957’ 21/ obeys octet? 7 '5‘": n o ~Duflr on P Where do the “extra” electrons go? Why do we need a third period or higher Element? Bril- WI‘DA Wis Lone cl—orlarl afie- Example: 804 2", sulfate ion {electrons in class) (how many electrons?) G) . =o 'o t I... l .. oi u 9:0 39-39%: Efio s—ora u I Ir .. i. :q: to, G) obeys octet“? Li ea, he“ Qual— mn Q what is good? OC‘lf‘l‘ {of} e whimd FL whatis bod? late Fc‘. Exmfléa £33,184 con ‘5 544 on g} +l1q+l5 .\ Shapes of Molecules: Valence Shell Electron Pair Repulsion Theory: Key.r concept: Electron pairs, whether in bonds or as lone pairs, being negatively charged, repel each other. Thus, when pairs of electrons are distributed about a central atom they seek to be osfrtr oportfmm each other as possible. a) The closer two pairs of electrons are, the stronger the repulsion. (I. p, A Two “electron groups” (electron In“ I ‘3 ,1 HQ 3 glroup = a lone pair, bprlid, sinfgl: X_.B x c ectron or atom] are c oser 1 t e KB A, Ev {‘3 1" angle between them is 90” rather than ct=QDD ot=12ElD o=1800 | AIEclm-e— 120” or 180“. i b) Lone pairs require more room than do bonded pairs. In other words, the order of repulsion is... | lone pair—lone pairl‘e lone pair—bond pair 3* bond pairmbond pair !! 'Iu——__.l' has? fed”; rifyau/srbo Rules: i 1} Draw the Lewis structure 2} Find the “central” atom (if more than one; do each sequentially) 3) Count the number of electron groups (usually lone pair and atoms) 4) This determines the base structure or “electron group geometry” | 5) Describe the molecular structure based only on the atoms {the “molecular geometry“) 6) Keep in mind points a and b above for finishing touches I 7) VSEPR lets us predict hybridization (later) lwlfi Electron ldeal # Electron group # lone VSEPR Molecular bond Specific grougs geometry gairs notation geometry angles examnle Structure 2 AK; a AXE linear I 30“ co2 ::0:r;:u_ _ : Fr 3 AX; [1 AK; trigonal 1 20" BF; | planar B 4 A1414 0 AX4 tetrahedral I [39.5‘1 4 AK; 1 M3E trigonaI ] 09. 5” pyramid 4 AK; 2 AXgEg angular 1 09.5” {Fl-bent? 3‘] 5 M5 1) fig trigonal 9t)”, 1213“1 bipyramid 5 AK; 1 quE “seesaw” 90“, 120” (“sawhorse”) a AXfi fl AXF, octahedral 90“ -r “s uare bi ramid” Hale cat-3 tele- csot CH.1 fiENth-nHLcad q W } TLa. H-—-~‘-‘t'l'oM——H angfia gal el-fshllu minifrfiggeéz In“; Fat-r5 m4 more. room“ There are a few other cases; see your text. Let‘s work one below: For t "l {,— . _ F5. : '5 5:3" Examgle: Bromine pentafluorldej Eing F FHBIer Ll?- tfi ; 2’ Ft!" Lewis structure: Fr“ ""31: # electr oups ' mm (5V 6’ 7/le i I l electron geometry: . .. u A- Xgl F . 52¢: a LEM 1t Molecularge met : i use enué 21-95: 51:64 r-e' [lynch-u J J REMEMBER: It is the atoms that describe the geometry! Ways to Draw Structures: I Where to place the lone pair? fl go‘jf-{LM‘S .. l Examgie: Fropyl alcohol, C3H30 I | Lewis "dot" Lewis "dash" Condensed P L‘ {Tlectrons in class} CH CH CH DH Tu -- {215, H H I? I? I? s 2 2 I'M ! " " . I‘ll-- _ _ _ _II— Lamsuuaiug H {g g g: '2 H H (l: (I: (l; a H (GFCHsiCHalEDHl H H H H H H _E_- _|-_ 'Fth-ai- 6"" Er A very fast wa a draw structures - I use it all the ti . [email protected] assumed to be C H‘s on C's are not 5 r n ...- enough H's to each C to make it -%I All other atoms are shown i H H \ / :I H x... Cxcx {RH Missing H's in class d Ail-rs. /O r ‘3. i C [Dot-i- ill/J; H/N'f énuflf‘r/fi? gamma {afreséaafqzlrb l fiLaLl'C ._.-——-—*""' [5 75,449? r'rsIL 5374-44 A? aflofirdbme/ l 1.13 jg ‘ ! Acid-Base Chemistry: $ A ejcu‘n l~ 5 Arrhenius: Acid = H30 ‘ Base = l-[C'e m m _ - Bronstedeowry: Ame] = Hf:B donor 6 {Aloft Base = H acceptor a Lewis: Acid = Electron pair aecegtor came. I a" Base = Electron pair donor ‘05”: H no H aflbfi’ “at” no General Form: + Base :_—— I'Ijug + conjugate acid base ! m _—=_— H—B + A3 Stronger + Stronger ..—._- Weaker + Weaker acid base acid base Example: (ta—(8| + H09 T...” ?? H—D—® 4 (Le (5} Stronger + Stronger we“: mm LE... acid base as afi- Loam CMfiqq'J Conj. H—CI + H25: fl H3O$ + Ole U At equilibrium, onlyr IIIOUD ofthe Clean be found as HCl; 999E1Dflfl are (319. Lute: I? oo. ream; aw. egg) pg, 858—- ‘Hw. s rHL+UTflE Prf‘i‘i‘f} whart Lot'“ “QPP‘MW I ExarnglezHC] in H20 i I | 8* -- 3” 5k "3.— ?3 n E.) | H-—-:C..I‘E + H-«no‘ fl +4-6)- +13: - QUH KH 5r 1 H ‘ .. H The “curved airows” show the electron movement needed to make and break bonds. These arrows ALWAYS start at the eleotron-rioh site (the arrow‘s tail) whereas the head of the arrow shows where the electrons will go. A V 06min" Siar‘il _ ‘- 3 o ' if Peta:— Aeid Strengths, Details: 19" rich 8 = e: 3* 93/ i" ! Aeidtbase reactions are reactions — we can in asurefealeulate equilibrium constants ' (Kc‘s or Kai’s]. Examgle: Acetic acid (“AeDH”) in water 0 O /U\ + H20 1...— HSQE’ + A OH 0 El In a reasonably dilute solution of acetic acid in water ' only 1% of the AeDH is ionized. We could write the - [Halt] [5:209] K = —— F'ure H20 is ~ 55.5 M. ‘ eq [H20] [ACOH] The AeOH eoneentratio | Pun out the constant Therefore [H20] will not 0 .. a - eh [H20] to create a new Therefore treat [H20] as a constant equilibrium constant, K3... [H30$] [Acoe] K3 = KEN] = [HA] Fri-Edie“ S+rbfi3 rag-{J lg; Hr} Ar; l.'i+ieHA~ ‘35" e (Generalform: K3: m]) For ACDH, to, = LTfi x 1o'5. IE: 6439‘“ a“. (13$ ice-‘I‘LH weal: car—{A 4. .. Haas] Leia HA; IxiiLs H it 59% 1% Hf! _ —---..___I i As a rough estimate: 1% ionized x (0301?;0301?’ 1f]Hes _5 0.1 Metal-ting AcOH=0.001M. Ka "- ml— : E3 10 = .l fiance We. Emil-exi- ' Giant} O M Clog: n.:si),__fH"] : Cal amen = co. the} In general, a large K, {a l) = a strong acid :Efle’] In many ways, the pKa scale is easier... | PKa = 409 Ka I Table opr,’s page 25 in text Remember the pH scale? pH 14 (high) = basic i pH 1 (low) = acidic ! The [leEl scale is similar: low pK,L = stronger acids high pKa = stronger bases Examples: AcOH + H20 :Acoe + mod K,=l.75xlfl'5ande,,=+4.T6 ago + Hgo sacs + Hart K,=i.99xio"fianapi<,=+is.r .——|———|—|—|———|—|—» -7 4.? c r 14 15.? I—_L_____l___l__l | ‘ HCI H30$ Hoe Strong acids Weak acids Weak bases Strong bases ‘ Strong acids will thus completely donate a proton to H20 and ‘5}; 1 pica, flag a ihon 1 strong bases will thus completer remoye aproton from H10. an, 1 Fl“ :3. M | The pKa is also the pH at which a compound is half—ionized. . . When the pH of a solution is greater than the pK,, for a given compound, that | compound will be mostly (or completely) deprotonated. When pH is less than pKa, the compound is mostly {or completely) protonated. I _._—_._II 6E5 ALfi-b THETHJE‘J STF’ hfiflflfifié] I Structure and Acidity: I Structural theory allows us to correlate relative acid strengths. . . v . . _ F? M“ I) t::t§.‘ifi2t§if$§“§im . have. A W L‘ 10% A , _n¢g,gfih [Gill “5" ii) How stable are an},F charges that are involved? (_ chi asacifii ) (B {Ideally, we want to stabilize anyr charges in the products) 7% If: H oncepts we Elect'ronegativit)»r will use: Size of the atom Resonance Inductive effects — electron withdrawingfdonating groups {later} Hybridization {later} i Example: Hydrogen Halides, H—X (7) l E on L—er “3 Hx pK. . . . i . - — Question. Why is H] a 6 runggg ac1d than 111*? HF 3.2 _ L L _ I This is a relative comparison! HCI -7 Answer: The reaction HX ‘3» H":B + X6 HBr '9 I I I HI H1 0 requires lat-m long; the H—X bond. The 64m“? U 5‘1 L43: this bond is, the stronger the acid. as." I It turns out the shortest (and thus strongest) t bonds are formed when the two atoms are both small and similar in size. {Finish picture in class) .I ' 'pFfiAfi-L‘i' ‘- [tq 2" [H aim-g]: 6°inbe “Ht-i . 15.4": 1—22 [cm :1- i Jr] ' ca ' 4—- 1 a 5 a}? a L‘ 13—— Example: Going across the Periodic Table H3c-—i H HQRQH HQQH H “a: i l Finish inches Hbcza +H® alliem“ H—Egfi‘n-tfl Hes-f: pKa 48 33 ‘15.? 3.2 The changing atoms (C, N, O, F), being second row elements, are all approximately the same 5415 . This means the bond lengths will be very E: :35 g,‘ [Q r- . (3} i This time, variations in the bond strength arise due to polarization of the I-l-A bond. ‘ Rep-Egg : EN in: Feta-3.2%. 15—! 1Q Ecrg-ss Reactants side: i Large asn = more polarized bond (HF) 8+ é, - In H—F, there is a large 5* on H; i C‘Hg-H ere is very little 5+ on H. H _'i" 1:"- H—F is already...r well on its wa}F to 13 coming HE and Fe lE'rodncts side: (2— H Bound fliuflcbl Eiuafi . '5 ima Ft‘hta Of all the anions, which one is “happiest” with negative charge? (9} I Tin—a. Mar-E EN F- is Resonance: Resonance allows us to spread out and thus stabilize charges. When we can stabilize charges in the product, we will have a stronger acid. ___..—.-- — Example: Ethanol vs. acetic acid (Finish in class) i CH3“CH2'6‘H — -- i‘ Ha * . "\J E" " : (5.“ IO: *0”:- l .- A H —— “53+ a e- a H3C 9:) ans \ésfl cat,” “’59—. ‘ No resonance stabilization possible in ethoitide, so harder to create charged oxygen. so {am imam-n ca. Hot, Ham, H :, H25.th 1H Alt—“lg } fir—[OH J HEX—'3‘,L . let-acacia 6E1 H 03 1—23 | 1) 2} 3) 4) 6} 7} 3} '3'} The structure drawn as an ionic cempennd is cerrect. Na will not make a cevalent bend. There are a number of reasens. but letis sag,r resenance lets us Spread out the charge en the molecule. This is equivalent in raising the entropy ef the system — increasing its diserricr — and this is geed. Arrew “a” means make a (new) bend frem O te C. Arrow “b” means te break ene ef the existing bends ef the C — O deuble bend. Radicals de not have octets — their valence shell is net filled. The precincts are H10 the weaker acid) and Cla (the weaker base). Strenger. i??? shew weaker te stranger acid. The blanks are then strenger. breaking, weaker, sizefsimilar Fe is. because it is the mest EN 1-24 ...
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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_ch01_answered - |Chapter 1 Introduction and Review...

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