MOP. Mechanism? Objectives. Energy hill for dehydrohalogenation. Last lecture. LQ#1) A) Draw the MOP B) Name MOP

Transcription

1 0% 0% 0% 0% bromo-,-dimethylpentane strong base salt dissolved in corresponding conjugate acid Given these conditions, select the mechanism that leads to the MP. Sodium isopropoxide Mechanism? Na isopropyl alcohol, Build -bromo-,4-dimethylhexane 1. S N 1 2. S N 2. E1 4. E2 LQ#1) A) raw the MP B) Name MP 0% 0% 1 r. Kay Sandberg 0% 0% Last lecture This lecture bjectives Alkene preparations 1) dehydration of alcohols E1 mechanism 2) dehydrohalogenation of alkyl halides E2 mechanism NMR involving pi-systems IR 2 r. Kay Sandberg Reactions: Preparation of alkenes Section 5.15 E2 mechanism: elimination bimolecular eat () E2 summary Great charge trade!! + + SB Base SB + - Increased S + Strong bases do not Weak bases (l -, -, I - ) handle do handle negative SB negative charge well charge well (relatively (relatively speaking) speaking) i E Anti conformation Lo E + R SB + 4 R Strong R 2 R 1 SB Energy hill for dehydrohalogenation Base E2 mechanism Transition State R 2 R 1 R 4 R & need to be in the anti conformation. R 2 R 4 SB + + R 1 When & are anti** R **groups that are gauche (R 2 & R 4 ) (R 1 & R ) end up cis on the alkene **groups that are anti (R 1 & R 4 ) (R 2 & R ) end up trans on the alkene 4 Which Na alkene? 2 2, MP Na 2 2, Section 5.14 Regioselective 1. A Follows Zaitsev s Rule 2. B The more substituted. B has lower E & 4. lower E TS A B 5 Rotate + + & anti** **groups that are gauche end up cis on the alkene **groups that are anti 6 end up trans on the alkene 1

2 E2 E Grab, slam, boot 2 In absence of strong base (only solvent) E1 vs E2 Section 5.15 E1 vs E2 Et + + Now a strong acid Weak base 7 - Elimination Reactions: E1 mechanism Section 5.17 Elimination reactions that exhibit 1st order kinetics + + Alkyl halide reactivity: o >2 o >1 o Rearrangements E1 kinetics 2 slow R = k[r] RI > R > Rl >> RF Weak base 2 onditions favoring E1: o R, weak base 8 - Elimination Reactions: E1 comparison E1 mechanism RS: arbocation formation Section 5.17 Enough review Spin-spin splitting rules for 1 NMR 1) hemically equivalent protons do not show spin-spin splitting Aromatic proton resonance peaks Section 1.7 ehydration ehydrohalogenation 9 Integration 2: ,ppm r. Kay Sandberg No splitting is observed between protons with identical chemical shifts. ownfield Aryl proton () Alkyl proton Aryl vs Alkyl Upfield (bonded to the aromatic ring) Section 1.7 Vinyl vs Alkyl vinyl proton () (bonded to sp 2 carbon of double bond) ownfield Alkyl proton Section 1.7 Upfield ,ppm r. Kay Sandberg ,ppm r. Kay Sandberg 2

3 o aryl proton ~ ppm shielded Protons in pi-systems o o shielded shielded vinyl proton ~ ppm A) methoxy B) ortho to ) ortho to carbonyl ) methylene LQ #2 Give the multiplicity of the following proton signals. Section alkyl proton E) methyl ~ ppm 1 14 r. Kay Sandberg methoxy methoxy 1. singlet 2. doublet. triplet 4. quartet a) methoxy 2 1. singlet 2. doublet. triplet 4. quartet 2 b) ortho to methoxy 0% 0% 0% 0% % 0% 0% 0% NMR spectrum 2 Section 1.7 What happens when protons on adjacent carbons arenot equivalent? a c Non-equivalent vicinal l protons l c J bc oupling w/ b l l J bc = 6. z b l l d J cd J cd oupling w/ J d cd =.8 z 8 o ,ppm 17 r. Kay Sandberg doublet of doublets (NT a quartet) Section

4 Nonequivalent vicinal proton coupling a J cd c oublet l l l of doublets vs quartet J bc J cd b doublet of doublets c l l d Peak intensity is the same Spacing between lines not the same Equivalent vicinal proton coupling d quartet d Section 1.11 c b Peak intensity in 1:::1 ratio a Spacing between lines the same 19 A) ow many nonequivalent vicinal protons are responsible for # responsible doublet of doublets for signal B) ow many equivalent vicinal protons are responsible for quartet A B 0% 0% 0% 0% 0% 0% 0% 0% 0% What happens when protons on adjacent carbon(s) are not equivalent? What happens when protons on the same sp 2 carbon are different? Section 1. b p downfield g J bg Section 1. = J gb = 7 z J bp J = pb = 1.5 z upfield J gp = J pg = 14 z chemical shift J bg = J gb = 7 z g p b J bp = J pb = 1.5 z b p g J gp = J pg = 14 z 14 z 7 z 14 z 1.5 z 7 z 1.5 z doublet of doublets doublet of doublets doublet of Kay doublets Sandberg b l g Section 1.11 J bg = J gb = z l p p J gp = J pg = 4 z LQ # Give the multiplicity of the following signals. b downfield z chemical shift g upfield J bg J gp p z 4 z 4 z 4 z = J gb = z = J pg = 4 z b l p g p l A) p 4 z B) g ) b Section

5 Time different? ow many different carbon environments are there in 1-chloropentane? 1 NMR spectrum -1 spectrum A separate, distinct peak is observed for each carbon. Section % 0% 0% 0% 0% l g o g 140 o r 120 b b r p 0 p p ,ppm r. Kay Sandberg 1 NMR spectrum Section 1.14 Physical basis of IR spectroscopy l g o, r, b are unresolved multiplets p g r Unresolved multiplets ow many different protons are there in 1-chloropentane? o o b b r p p p g r & b o IR excited vibration ,pp r. Kay Sandberg m 284 r. Kay Sandberg Physical basis of IR spectroscopy omparison Frequency (n) Frequency & amplitude of ground state vibration r. Kay Sandberg 5

6 Time omparison Frequency (n) is the same Symmetric stretch vibrational mode Stretching modes Absorbed E increases amplitude cm -1 1 Asymmetric stretch vibrational mode 2 r. Kay Sandberg Asymmetric in-plane bend vibrational mode (rock) symmetric out-of-plane bend vibrational mode (wag) In-plane bending modes ut-of-plane bending modes ~1400 cm cm -1 symmetric in-plane bend vibrational mode (scissor) r. Kay Sandberg 4 asymmetric out-of-plane bend vibrational mode (twist) r. Kay Sandberg ~000 cm -1 wavenumber (cm -1 ) IR spectrum - wavenumber igher E? Which is the more energetic vibration? IR spectrum of butan-2-one ~ cm bend 2. stretch ~1400 cm cm -1 c = n E = hn c = n c = n E = h c E ~ 5 r. Kay Sandberg 0% 0%

7 IR spectra Mechanism? euterium LQ #4 raw MP 2 S 4 MP 7 r. Kay Sandberg lick mechanism same chemical reactivity since both are hydrogens 1. S N 1 (only difference is in rate) 2. S N 2. E1 = 1 (protium) = 2 (deuterium) Unlike the protium isotope of hydrogen (), deuterium () is NT understood and MUST be shown explicitly. 4. E2 0% 0% 0% 8 0% euterium B euterium B o euterium B hydride shift 2 o euterium B At this point you should analyze the carbocation. What is its classification? Is there any atom/group on the beta- that can shift over to leave a more stable carbocation? 41 At this point you should analyze the carbocation. What is its classification? Is there any atom/group on the beta- that can shift over to leave a more stable carbocation? 42 7

8 methyl shift euterium 2 o B deuderide shift euterium 2 o B o At this point you should analyze the carbocation. What is its classification? Is there any atom/group on the beta- that can shift over to leave a more stable carbocation? 4 At this point you should analyze the carbocation. What is its classification? Is there any atom/group on the beta- that can shift over to leave a more stable carbocation? 44 euterium 2 o o B A 2. Which B is MP? MP?. The more substituted the db, 4. the more stable, the faster to form A B Zaitsev s Rule 46 B 8