PS2_solutions - In | Chapter 1 STRUCI'URE AND BONDING lhl...

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Unformatted text preview: In | Chapter 1 STRUCI'URE AND BONDING lhl ORGANIC MOLECULES Now we turn to methyl nitrite. Following the same procedure. we begin with only single bonds. and then arbitrarity add in the remaining electrons as lone pairs. taking care only to avoid violating the octet rule. The structure below left is the result.I and it contains a seriously electron—deficient hi. just as we found starting out with nitromeliiane. And. we “tie" it the same way. moving an electron pair “in" from the negatively charged U at the end: H .. + .._ | .. —p—Hib:h—+H—c—o—H=g: III- II-l I H H——.- I—fi—I This is pretty good: Ail nonhydrogen atoms have octets. and none are charged. Can we find any other reasonable resonance forms? There is a common panem described in the text for systems containing an atom with at least one lone pair attached to one of two atoms connected by a multiple bond. You more the lone pair “in" and move a. 11' bond “out": on + X—Y=Z 1—1 K=Y—Z Applying this pattern to methyl nitrite. we get f1 {1 ii 'i Hal—gun=o:+—+H—c~o=nwo: H H The result is the second-best resonance form—-ol-:ay lor octets. but charges are separated. which makes it less of a contributor than the Lewis structure on the left. The hybrid will more closely resemble the left—hand structure. with two noneqru'uaient NE} bonds. The contribution of the right-hand structure. while small. will tend to make the N0 bond at the end the most polar one in the molecule. with the CI at the negative end. 13. {a} Chlorine atom is igl' {seven valence electrons. neutral} Chloride ion it'- HEIE {eight electrons. aegutiycly charged} {b} Borane is planar {be around B). while phosphine is pyramidal the around P. like the N in it ammonia}: B vs. P-i_ H x a /\ H H H H to] CE. is tetrahedral. while SF... with fine electron pairs around 5. is pyramidal. Notice that all that we need to derive this answer is VSEPR. It is not necessary to first try to figure out the hybridization. {d} The procedure we follow is the same: Construct the Lewis structures. and then use VSEPR to predict geometries. Do not consent yourself with hybridisation at first. Solutions to Problem I H Nitrogen dioxide contains Ii" valence electrons to from each If.) and 5 from the Ni. and nitrite ion contains til i the additional electron gives it the — J charge}. bl is in the middle. so we have D—N—D to start We in tr bonds}. For both species we can add 12 of the remaining electrons to oxygens as lon+e pairs. The last electron {for N03} and the last two t for HOE—i can go on N. _... .._ .... + u. giving “3— N—p- for N01 and -D—N—U- for N03". Each of these Lewis structures is seriousl},r short of an octet on nitrogett and. can be improved by resonance deloeaii'sation of an electron pair front oxygen toward nitrogen: as” .._ .t. + ForNCtl. :o— -—r:t p: fl an I I+ I+ n. t. H I -t ... .. ForNDf. :big—g: “—ti9= —o: ._. ib— is}: o: So nitrogen ends Up with '3 valence electrons in ND; and ii in N‘Ul'. How about geometry? Let‘s begin with N03" because all of iLs eleetmns are paired. and VSEPR [valence shell electron pair repulsion) can be applied directly. The middle N is surrounded by two o bonding pairs and a lone pair {1r electrons are not considered for USEFR}. and three pairs leads to a bent geometryI [which may he rationalised by HF: hybridization. if you like}. In fact. the D—N—O bond angle itt nitrite is llfi“. It is a bit smaller than the nominal lid" angle for a trigonal planar structure because the lone pair. being attached to only one atom. exerts more repulsion than do the bonded pairs. closing the bond angle down a little bit. Now let Lts examine niLrogen dioxide. The N now bears a single nonbonding electrott rather than a lone pair. Doe electron exerts less repulsion titan do two. so we can predict that the D—I‘i—G bond angle should be more open in nitrogen dioxide than it is in nitrite. You do not have enough information to predict how mueh more open the angle will he. In actual fact it is 134”. The fact that it is larger than Ill]C means that the two bonding pairs exert more repulsion than does the single nonbonded electron. You'll no doubt be thrilled to lind out that nitrogen dioxide is a significant component of urban. atmosphen'e smog. Smoggy air derives ntuch of its distinctive character front this toxic, smelly, brownish gas. lei Now we compare two new dioxides. SD; and CID}. with the one we‘ve already done1 N03. Lewis structures and resonance font‘ts first? ofl'Hs— ij': «—~ :dflsi/iri: E —*s=o::—-:o=s;o: II or In- an o:_--—t_:II o: .__.;érl¢1%;._.+ -- II- II In- The structures on the extreme right both have expanded 1valence shells {more than octets}. which is UK for third—row atoms. Based on VSEFR both 30: and CEO: will be bent structures. because ol" the lone pair on 5 and the lone pair + extra single unshared electron on CI. The actual angle in SD: is IEiJ‘”. and that in CID: is l in“. the difference being due to the extra repulsion of the third nonhonding electron on (it. Clflg. despite the fact that it is smelly, toxic. and tends to blow up. is actually a major industrial chemical used to bleach wood pulp in the manufacture of paper. Prudently. it is prepared just before it is used. eliminating the need to store the stuff. I! i Chapter I STRUCI'URE AND BOND-1N6 IN ORGAl‘VliC MOLECULES 29. {a} The molecular orbitals are obtained as follows: {I f—"'—"'—'—‘—F| ," {Antibonding} ‘1, f 'I f it I S ‘1- It I .fi‘ "t l 'I E f {E sorting} Therefore. the resulting electronic configurations are H1. (trig. with two bonding electrons vs. Hf. (of. with one bonding electron. 5o H2 possesses the stronger bond. lb} some as Exercise l—IE. {cl and id] We prepare an orbital diagram similarly. How do we begin?I First. out of all the various types of orbitals presented iii the chapter. let‘s see which ones we do and do not have to consider. In multiply—bonded molecules with many atoms. such as etbenc and etltyne [Figure l-21. textbook page 315}. we needed to invoke hybrid orbitals. because we had to use them to explain geometry. In diatomic molecules such as O: and N1. however. there is no "geometry" to explain. so orbital hybridization has no purpose. and we can just use simple atomic orbitals. That's good—it makes life simpler. We note also that the Is and 2.9 orbitals itt Cl and N are completely filled. In cases such as these. it is customary to ignore the s orbitals, because their overlap will produce no net bonding {just like between two atoms of Hal—another welcome simplification. We’re down to considering for bonding just the three 2;) orbitals on each atom. because they are the only ones that are portly filled. Referring again to Figure l—El. we can visualize end-to-end overlap to bonding} between the p orbitals {one on each stout}. which happen to point toward each other. and side-by-side overlap to bonding} between the remaining p orbitals {two on each atom}. Our molecular orbital diagram will therefore include three sets of orbital interactions. one with o and rr* {bonding and antibonding} orbitals. and two with TI and 1r* (bonding and antibonding} orbitals. Because tr overlap is generally better than n overlap. the diagram is shown here with a larger energy gap between the tr and or* orbitals than between the Tr and n‘ orbitals—recall from Figure I-lE that the difference in energy between atomic and molecular orbitals is related to the strength of the bonding—the eltange in energy going from the atoms to the molecule. {Mom sophisticated forms of theoretical analysis reveal that the actual ordering of orbital energies is not quite the same as that shown here. but you don‘t need to worry about that} For is}. us. rsiiratzraftasi'rssi'. 4 net bonding electrons vs. of. rsiiiaiitsfltrri'. s not bonding electrons. So El:+ has the stronger bottd. For {d}. N2. [oiztnlzlnfl a netr bonding electrons vs. his”. lojfltnftni'. 5 net bonding electrons. So . N3 is better. . 30. {a}. {b}, and {cl Each carbon is bonded to four other atoms and. therefore. will possess approaintate tetrahedral geometry. Each carbon in these molecules is rp" hybridized. . {:1} Each carbon is attached to three other atoms [two hydrogens and the other carbon}. The bonds to hydrogen are of o- type. One of the carbondcarbon bonds is a o and the other a 11'. The result is Solution: to Problem I 13 thfll eaeh carbon has approximately U'igonal geometry [like boron in RNA} and is If}: hybridized. In other words, each carbon uses: HF: orbitals in three or bonnie. and Lin: leftover F orbital in a ‘rr bond. to] Each carbon is attached to two other atoms; {one hydrogen and the other carbon}. The C—H bonds are rr, as is one of the (3—D bonds. The other two C—C bonds [of the "triple“ bond] are 1T bonds. fieryilnetr}.r is linear (like beryllium in Eel-IE} and each carbon is JI'F hybridized, Each ufiflfi two ,Tlfil orbitals for or bonds and two F orbitals for “IT bonds. ‘I’ m flh—C—H T T ”:1 .1112 {g} Hybridization mural allow both carbons to bo doubly bonded [resonance form on the right]. Both. therefore. are spa. H H H \l/ I I I II 31. [a] H—g—CEW {h} afio—o—o—e—fi—H a H H N /-*\ ||_| H H III :tl]: I'll III flirt-nil telH—{IZ—(IZ—-(IZ—u(|3-— Id) tI—tlj—o—IEI}: H H H H H :Br: I ”I: I ”I“ II to} H—flfi—C—(lf—C—U—LI'I—H H H H I I I I If} II—fi—o—o—o—o—o—{j—H H H H H Line formulas do not :15 a rule Show Lruo bond angles. l HI It“ H H :0: H H H H]: _ H I I I l I I II J: C 'H 32. {a} H— .—C—C—C—H {b} H—C—C—C—N: | | | | l | \C—CfiH H H H H H H ”fl ]\ H H H H . H a,” I III I I \H / {e} H—c—e—e—e—e—H {d} H—CEC—C—H | | | | | \C—H H H E_E-_I1H H / 14 I Chuplar I STRUCTURE AND EDNDlN‘G IN ORGANPC MOLECULES H H H H H H {e} it—c—ij—c—CEH: m H—c—(lj—Zfi—tli—c—H A t t t It t'. 33. {a} HINCHECHENH: {h} CH3CH20CH3CN {c} CHHr, l? 34. [a] (cchHH {b} CHsiliNHCHfiIHs m CHHCHoHCHZCHaaH {d} CFaCfizflH {e} CH3CHECLCH3}: {f} CH3=CHCCHj ('5') Alternative correct answer}: extol for Several structures. in Problems 33 and 34. 35. From Problem 3!: EIJ DH ta} —C“ on on to xJ\/ NH2 0 Cl Br {a} Br\/l\fl {e} Mo” to Ho%flwflon 1' From Problem 34: o OH H {a} {NR [‘11] Affirm {c} stk H F xx fl tel W0“ to} /Y to 91/ F F H H ‘t . . t: 'o‘ H c: x” x i/ a x .x es. la} H c t; on H—C,_ 1 ~. .r -. 1t‘trl H t: H H Ct “to N_. .J'H H H H H Ht PM H fie“; ‘e" H / a. / I I I“ I I" fl'. Ii H HH H H:_E._—l-lI-IH 3?. {a} Cfil-Iu. Begin with the isomer that contains all the carbon atoms connected in a straight chain. Then shorten the chain by one carbon at a lime. connecting the rerrtot'ed atom as a substituent to interior positions ot" me remaining chain until every possibility has been drawn. There are three iaolttera: [l J CH3_CH1FCH2'—CH2—CH3 or CH}CH1CH1CH3CH3 or ETH3ECH2JI3CH3. These are all commonlyr used forms of condensed formulas for the same molecule. Bond-line: M Salulinns 10 Problems . I? 42. u ;. u :- :1 lb- n 3- After the calinn. positive: chamutur nn :nrhnn is related Ln number 111' [pulnfimcll bands to elemmncgnlim atoms. 43. [all C—D, D—H [hi Any C—C 11:} C=C [{1} C'——C Tl li‘ le} Any C wlm -'l single buncls {fl C—C=C (11' C—C'=C J T c E! H .. Elli-l1 f! H H :9: H 44.13.: H' C_' C: CI“: C_: H HIETEJN: }.;{C-L_L : L : EH1 HHl—I H HHl-IIEH E. NJ~ n 1W1." .. D C nil—Hi: 5‘ {h} and {c} fix {WK N N {$5 with . 0510‘“. Q DH {cl} “MD/m w",/’\\)K n . .r’fi‘xk T114: £5 " Cfllth‘l being “Haul-cud hjr' lhc cyanide :un 11er Lady ha; an outer. En mm: electron pair of mu IWET ll'l ll'IL' dnuhle [Hind is filmed 1n move up [:1 1h: tmygcn :me m uvnirj vinlun‘ng 1hr: mute: rule. 45. {1:} Divide the pcrcenlagc Vuluu lay the atomic weight In get the relaLive number 01" alums: carbun. 84“? = T" hydmgen. Ififl = IE 46. it} Aluminum: 3 {gmup number} —~ II] (unshared uluulmns} -- H2 {8} [shared uluulmnsl = —l 4?. in} Three. alums allauhtd in the carbon: Lr'rgnmtl planar genmfltry = 5;}: hyhridizmion 43. {cl 4?. {3} Same total number of cleutmns; all Moms In same Inuulluns [same geometric fmlncwurkl Note Title ...
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