CHAPTER 11 - Chapter 11 Alkenes Chapter C Nam s Ending e...

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Unformatted text preview: Chapter 11: Alkenes Chapter C Nam s: Ending e Nam Ending Rule s: 1. Find longe chain wit both C 2 s in it. st 1. both sp 1 2 3 8 7 6 5 4 C ane Doublebond e ne For e ple Ethe , prope , bute , e xam : ne ne ne tc. An oct e An oct ne 2. # C rm 2. hain with C Ccloseto te inus 1 2 3 8 7 6 5 4 A 3-oct e (only thefirst of the 3- ne two C 2 s iisnam d by a #) e t wo sp s Lingo: Doublebond position position R R C2 H Te inal rm Te R R I nte rnal 3. Nam and # substitue in alphabe e nts, tical orde r 4-Ethyl-3-m thyl-3-oct e e ne 1 4. C ne 4. ycloalke s 5. S re te oisom rs: e 2 3 1 CC 3-Me cyclohe e thyl x ne 2 By de By finition R trans C H cis R R R C is/trans use for 1,2-disubstitute e ne d d the s. 6. For tri- and te trasubstitute alke s: E, Z nam UseR, S d ne E, ing. 6. priority rule at e sp2-carbon se s ach sp parate , to find highe priority groups at e e ly r ach nd. se 1 2 3 8 7 6 5 4 Oppositeside E Opposite s: S eside Z am : E-4-e thyl-3m thyl-3-oct e e ne 2 7. OH ( S H) > e ne OH 1 3 2-Prope n-1-ol 8. S ubstitue Alke nts: nyl C2 H C H C2 H C2 H C H Ethe (vinyl) nyl 2-Prope (allyl) nyl 9. Exocyclic alke s: Alkylide cycloalkane ne Alkylide ne s 9. Me thylide cyclohe ne xane Me (Me thyle cyclohe ) xane (Me ne Structure of Double Bond Structure “Ele ctron rich” The S a Bond igm ThePi Bond The Ethe ne Orbital Ene s rgie π Bond is re lative we ly ak re How Weak is the Π Bond? a Bond Strengths (kcal/mol) Bond Unusually strong be causeCuse sp hybrids s sp Unusually Acidity: Alkenyl hydrogens are Alke re lative “acidic” ly re C f. R C H C H C 3C 3 H HH H + C 3Li H C C pK ~ 44 H re , : ~ 50. The fore in principle H RC H C Li + C4 H Proble s: Re m gio-, ste ose ctivity. Be r: re le tte H H C2 C H + Li C2 C H Li Br H R H R CC + Mg CC H Br H MgBr Use ful: Re with act carbonyls carbonyls Why arealke hydroge acidic? nyl ns Why C C : : sp r. 33% characte Has 33%s characte I n contrast: sp3 has 25%s characte r characte H Ne e ct: re t ffe lative e ly -withdrawing H NMR 1 C C H δ ~ 4.5-6 ppm: deshielded! Why? 1. sp2 2. e 1. sp 2. -Flow of π cloud S ngthe H tre ns Coupling Constants Coupling De nd on ste oche istry. pe re m ste H C R C H J H trans t rans H “Vicinal” coupling: H = 11-18 Hz; Jcoupling: Hz 11-18 inal” cis= 6-14 “Ge m cis H J geminal ~ 0-3 Hz ge For cis/trans isom rs: Jtrans always Jcis. e For > Doublebond “transm longH its” longrange(ove 3-4 C coupling (1-3 r ) Hz). Hz) C NMR C 2 de lde (re shie d asons arecom x) ple sp δ 13110 – 150 ppm“le half” of total spe =3 ft ctral window 1 Alke s ne Alkane s H3C H3C 18.9 122.8 C3 H C3 H H H3C CC 34.0 C3 H C H C3 H HC H3C 19.2 132.7 22.2 H 123.7 CC H3C 12.3 20.5 C 2C 3 HH 14.0 C 3C 2C 2C 2C 3 HHHHH 13.5 34.1 (CH ) S i (CH ) S i Vibrations in Molecules: Infrared (IR) Spectroscopy Infrared C pounds re m a m chanical fram : t he “rattle Rattling is om se ble e ey ”. quantize d. quantize Excite state d E ΔE = hν ~ 1-10 kcal m -1 ol 1-10 ~ iin λ or 1/ λ = υ “wavenum rs” n be “wave Range 600-4000 cm : Range 600-4000 Ground state A ~ ~e ine B stre tching υ : De rm d by Hooke Law ’s stre D te m +m A B mm AB υ = k √f f = f orceconstant m= m ass force (re cts bond stre fle ngth) (re ~ υ goes up with larger f, smaller m goe Not only stre tching: also be nding and couple m s d ode Not C ple patte 600-1500 cm : Thefinge om x rns rprint re gion f inge re Infrared Modes Infrared The Infrared Spectrometer The Finge rprint re gion Finge ~ -1 Most use Most ful: 1. Alkyl υC 2900cm = H ~C 2 H ~ υ sp - = 3080 cm-1, υC- = 1640 cm-1, 2. Alke s ne Alke C R~ H υ = 970 cm-1 970 CC R trans H 3. R O H O 4. H H H C 3350 cm-1 3350 cm1 (broad) (broad) 1740 cm -1 Trans-2-he ne xe Mass Spectrometry Mass I onization De ction fle Them spe er s ight The ass ctrom te distinguishe ions by we we -1 m z = Mole / cular we pe charge ight r (chargeusually one ) (charge High Resolution Mass Spectrometry High-re solution m ass spe ctrom try re als e ve m cular form ole ulas Fragmentation Mole cular ions with 70 e (~ 1600 V kcal m -1) unde fragm ntation rgo e kcal ol unde f ragm The aretwo ways of re The f ragm nting a radical cation to a e radical (uncharge he d, nce unde cte ) and a cation. te unde d . + H+ CH 3 CH 4 + . CH 3 + + H . Mass S ctrumof C 4 H Mass pe Dueto 13Cnatural abundance Due 13 natural Large pe (base st Large ak pe de d as ak): fine 100% Not always . them cular ion! t he ole Mass spe re al thepre nceof isotope ctra ve se s: 13Cnatural abundanceis 1.1% the forere ; re lativehe of M+1 pe = n x 1.1% whe n ight ak , re = num r of carbons. be num Othe isotope 18O: 0.204% 35C : 37C = 3:1; 79Br : 81Br = 1:1 r s: 18O: ; 35 l 37 l Othe 79Br 81 Mass spe opropane Mass ctrumof 1-brom m z = 43; / dueto propyl Fragm ntation is m like at a highly substitute e ore ly d ce r: Follows carbocation stabilitie : t e nte s te rtiary > carbocation se condary > prim ary se Exam s: C5H12 isomers ple 12 isom Mass spe ctrumof pe ntane All C-Cbonds are All rupture with roughly d e qual probability. Note : Note Fragm nts haveodd e Fragm we ight. we Mass spe ctrumof 2-m thylbutane e Thepe at m z = 43 and aks / 57 re frompre rre sult fe d f ragm ntation e around C2 to givese condary carbocations. Mass spe e Mass ctrumof 2,2-dim thylpropane Only a ve we m cular ion pe is se n, be ry ak ole ak e causethefragm ntation to e givea t e rtiary cation is favore d. Fragm ntation also he to ide e lps ntify functional groups Alcohols: Alcohols: M+ ofte not obse d n rve Alcohol Fragmentation by Dehydration and Cleavage: C haracte ristic of wate r; fragm nt ion is e n e ve Mass spe Mass ctrumof 1-butanol Thepare ion, at m = 74, give riseto a sm pe be nt /z s all ak causeof re loss of ady wate to givetheion at m = 56. r /z Alkenes fragment to give resonance-stabilized cations Mass spe ctrumof 1-bute ne Mass spe xe Mass ctrumof 2-he ne De e of Unsaturation gre De s Mole cular form t e us how m rings and/or π bonds are ula lls any Mole pre nt in a m cule Re re is a saturate acyclic se ole . fe nce d hydrocarbon: C H2n+2. hydrocarbon: n S plee ple im xam s: C H12, not C H14. not 6 6 C H10, not C H14. not 6 6 Wene d to de rm thede te ine viation of them cular form from ole ula We e C H2n+2 (in incre e of 2H). Eve ring or doublebond take m nts ry s n away 2H, triplebond 4H fromC H2n+2. away n Effe of Pre nceof He roatom on C H2n+2 ct se te s Effe n De nds on vale of e m nt: pe ncy le e S O no e ct on count (still CnH2n+2 + S or Oy) , ffe r xo +2 CH COH Haloge n: Haloge -1; CH CX H CNR Nitroge n: +1; CH S ps: te 1. C 1. alculateHsat = 2n C+ 2 – nX + nN n = “num r of” be “num 2. C ount Hactual in give m cular form . n ole ula actual 3. De eof Unsaturation: (Hsat – Hactual)/2 gre De Exam s: ple C H16 10 1. Hsat = (2x10) + 2 = 22 2. De eof unsaturation: (22-16)/2 = 3 gre or C H5N 5 e tc. 1. Hsat= 10 + 2 + 1 = 13 2. (13 - 5)/2 = 4 de e of unsaturation: gre s Or N Pyridine CN Or? Proble m Proble C HN: How m de e of unsaturation? any gre s 3 Hsat = 2n C+ 2 – nX + nN De eof Unsaturation: (Hsat – Hactual)/2 gre A. B. C . 2 3 4 Relative Stability of Alkenes Relative Me asurehe of hydroge at nation ΔHH2 of isom rs, e bute e .g., ne Me f H2 o ΔHH2 (kcal m -1) ol cat. cat. cat. + + + S tability: H2 H2 H2 -30.3 -28.6 -27.6 I nte rnal > te inal rm , trans > cis Why? 1. Hyperconjugation: Why? Hype H : : 2. S ric hindrance(strain) te C C C C is le stablet han is le ss trans be t rans causeof ste hindrance ric Ge ral orde of stability: ne orde r C2C2 HH < RC C 2 HH < tri RC C H HR cis < RCH C HR trans < < te trasubstitute d Synthesis of Alkenes Synthesis E re visite . Best: E2 on RX. Regioselectivity? d Re C3 H C3C2CC3 HH H X S aytze v-Rule Non-bulky base : Morestable product. More base H3C CC C3 H H2C H3 C H C C2 + H C 3 H3C H m stable ore le stable ss Hofmann Rule Hofmann Bulky base Le stable te inal product is m : ss , rm ajor I s e ination ste ose ctive lim re le ? I .e will it m pre re ., ake fe ntially cis or trans product? Ye s, cis t rans but not com te ple ly. but Br +Na OC 3 H C 3OH H 51% 51% + 18% + 31% Trans pre inate (not m dom s uch) S re cificity? Ye . te ospe s Ye E or Z fromre ctivediaste om ric spe re e or haloalkane s: haloalkane * C C* H X S re cificity: te ospe Good! Each Good! diaste om r re diaste e give only one s give only ste oisom r of re e alke product ne alke Alkenes from ROH by Dehydration: Alkenes Often Messy Often Messy O conc., s quire he s at: Rprim OH + H2S 4 conc., goe by E2, re 2, re C 3C 2OH + H HH + C2C2O HH H H + H HS O : : + :: :: R , R OH : E + re arrange e m nts :: :: 4 C 2 C 2 + H2O HH + H2S 4 O Acid-Me diate De d hydration of Alcohols Acid-Me CC Acid, Δ Acid, CC+ HOH H OH Re lativeRe activity of Alcohols (ROH) in Re De hydration Re actions De R = prim < se ary condary < t e rtiary prim C 3C 2OH HH HO H C 3C H C H3 C conc. H2S 4, 170°C O 170°C HOH 50%H2S 4, 100°C O 100°C HOH C2 H C2 H HH C 3C H H C H3 HC 80% + C 2 C H2C 3 Trace H HC H (C 3)3C H OH DiluteH2S 4, 50°C O 50°C HOH C3 H H2C C C3 H 100% De hydration with Re arrange e m nt C3 H OH H2S 4, Δ O HOH HOH H3C CC H3C H 54% C C 3C C 2 C H3 H H H C3 H + C 3C H HC H H C 2C 3 HH C H3 + other minor isomers HC 8% 33%H2S 4, 1 h, 100°C O H2O OH OH α-Te rpine ol Acid-catalyze d de hydrations give m ixture s 15% Te rpinole ne + 9% Lim ne one Te ne Thesce rpe s: nt of soap of + + + 28.5% α-Te rpine ne 18.5% I sote rpinole ne 15% γ-Te rpine ne ...
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