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 [email protected] 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[email protected] 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: [email protected]_>.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 ! &...
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