E1 elimination PDF

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E1 elimination reaction...

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E1 Elimination (unimolecular) - Rate = k [R3LG] The rate only depends on the substance with the leaving group. The mechanism is very closely related to SN1. 15/01/2019 C-O = 358 kJ mol-1 CGW: p388 C-C = 614 kJ mol-1 43

(1) Reagents (3) Products reaction co-ordinate 44 22 E1 Not surprisingly E1 and SN1 are in competition. H2O/EtOH + 15/01/2019 Stereochemistry – adjacent bulky groups repel. H3O+ + 64% SN1 product 36% E1 product 45 reflux E-isomer major product Z-isomer minor product Z-isomer

CGW: p391-2 46 E-isomer Steric clash is minimised 23 Regiochemistry – influence of adjacent groups:

CGW: p393-4 E1 E1 major -Ha formed -Ha + + Minor -Hb Not formed -Hb Why? Conjugation is only possible with adjacent pi bonds 47 Why? Internal alkene has more opportunity for hyperconjugation 15/01/2019

Test yourself – explain the following reactions, include mechanisms H+ heat

Revision note – you need to be able to draw mechanisms of reaction involving unseen reagent combinations, predict reactivity and provide detailed overview of the mechanism 48 24 E2 Elimination (bimolecular) - Rate = k [RLG][Base] The rate depends on both: the concentration of the substance with the leaving group and the concentration of the base. Both are present in the transition state ‡ 15/01/2019

r.d.s Strong base Good leaving group Concerted B-H bond forms C-H s bond breaks C-C p bond forms C-LG s bond breaks energy An E2 Reaction Profile (2) = ‡ 49

(1) Reagents Activation Energy DH (3) Products reaction co-ordinate 50 25 CGW: p395-8

abbreviation: Ph = phenyl i.e. 51 CGW: p395-8 52 Conformational requirements of E2 These different diastereoisomers eliminate to give different alkene stereoisomers AND This is because only antiperiplanar HBr is eliminated s s* Why? p E2 requires alignment of: 1. (i) Filled s orbital of C-H bond being broken by base 2. (ii) Empty s* orbital of C-LG bond This leads to formation of new p bond. Orbitals are best aligned when they are antiperiplanar (180°) 15/01/2019 26

Revision note – you need to be able to draw mechanisms of reaction involving unseen reagent combinations, predict reactivity and provide detailed overview of the mechanism 53 80% Why only 20% terminal alkene? 20% 15/01/2019 Test yourself – explain the following observations Compare the reactions above Me Me Na+ -OEt THF Br

E-isomer major product Z-isomer minor product Less hyperconjugation

CGW: p395 54 27 Can we select the terminal proton? = tBuOK Normally: H Br H Internal alkene would normally be the major product because it has more hyperconjugation (s to p*) HH H H Thermodynamic product Is the most stable one Kinetic product formed because the terminal proton is readily accessible How can we increase the amount of terminal alkene? HH EtOtBuOK Why? 15/01/2019 HH H H CGW: p398 Steric hindrance slows down removal of internal (blue) proton

Terminal (red) proton is readily available Not formed because of steric clash abbreviation: tBu = tert-butyl 55 H H Br Sometimes E2 is impossible Because antiperiplanar arrangement cannot be achieved

This molecule is locked in this conformation in which there is no proton antiperiplanar to the leaving group 56 28 E1 ‡1 ‡2 2 Revision - Eliminations (e.g. alkene formation) E2 ‡2 Rate = k [R3Br] time 1 Rate = k [EtO-][RBr] time Br slowMe Mefast Meb H3 Me Et BH EtO CH ‡ H H CH3 H HH Br 123 •Like SN1 – favoured for tertiary leaving group. •A Base attacks beta proton causing elimination. •There is always competition between SN1 and E1:

H2O Cl OH 64% 36% SN1 E1 HH Br •LikeS 2–bothbaseandLGinone‡ H CH3 HH Br 15/01/2019

Energy 1 Energy 3H3

Test yourself – explain the following observations What would be the main product from this reaction? N •Simultaneous breaking of C-H and C-LG, and formation of alkene. •Donation of electrons from filled C-H s (bonding) into s* (antibonding orbital of C-LG bond). •Must be antiperiplanar (Newman) EtO H3C Newman H H HH Br

Revision note – you need to be able to draw mechanisms of reaction involving unseen reagent combinations, predict reactivity and provide detailed overview of the mechanism 58 29 15/01/2019 Test yourself – explain the following observations

Revision note – you need to be able to draw mechanisms of reaction involving unseen reagent combinations, predict reactivity and provide detailed overview of the mechanism 59

MeO HO ALDEHYDE H O KETONE (b-keto alcohol) O H Compare the reactions above Carbonyl Chemistry MeO HO

HO MeO Carbonyl containing compounds are fun! HN O AMIDE 60 O

30 CGW: p126-7 61 Chemistry of the Carbonyl Group p Carbonyls are flat (sp2) Bond angles are 120° p* - LUMO Nucleophile approaches at an angle of 107◦ approach dictated by LUMO and repulsion from the attached carbonyl substituents O 3.5 C 2.5 Carbonyl reactivity is dictated by the presence of electronegative oxygen atom. Carbonyl compounds have a dipole. BQrgi-Dunitz trajectory 15/01/2019

Example 1: Reduction of a ketone into an 2° alcohol Carbonyls (sp2) react with nucleophiles to form tetrahedral products (sp3) Reduction In this example the reagent sodium borohydride Its compatible with ethanol, so protonation of the alkoxide occurs at the same time This reaction is a REDUCTION abbreviation: R =any substituent 2° = secondary; 3°62= tertiary 31 Carbonyls (sp2) react with nucleophiles to form tetrahedral products (sp3)

Example 2: Addition of nucleophilic alkyl to a ketone gives a 3° alcohol Grignards are not stable to protons so we add them in a ‘work-up’ after the alkoxide is formed Grignard addition 15/01/2019

Cyanohydrins Mechanism: This reaction is reversible until acid is added to the alkoxide [H+] d- d+ Me Br Me MgBr Grignard reagents can be formed from the reaction of magnesium and alkyl bromides/iodides abbreviation: R =any substituent 2° = secondary; 3°63= tertiary Reaction: NaCN, H2O then HCl cyanohydrin

64 32 pKa ~40 (poor leaving group) pKa ~10 (good leaving group) pKa ~35 (poor leaving group)

Notice that we just drew collapse of a tetrahedral intermediate to regenerate a carbonyls This is a common feature of carbonyl chemistry that we will revisit several times. We can use the relative pKa of each substituent to predict which is the best leaving group Cyanohydrins Some animals secrete a cyanohydrin as a cyanide releasing defence toxin Lotaustralin Contained in defensive droplets, cyanide only released upon ingestion by a mammal in the acidic environs of the stomach Insects 2018, 9(2), 51 65 15/01/2019

66 33 HOMO p* - LUMO The fastest reactions occur when the HOMO and LUMO are closest in energy (slide 23) Reactivity of Carbonyl Groups If X = electron withdrawing group (e.g. chloride) lower the LUMO energy bringing it are closer in energy to the nucleophile HOMO If X = electron donating group (e.g. nitrogen in a amide) then the LUMO energy is raised and the energy gap to the nucleophile HOMO is raised >>>>>>> We can estimate relative reactivity of carbonyl containing function groups on this basis 15/01/2019 CGW: p529,534 67 Most reactive Least reactive Look at the relative electronegativity values d+ d- 2.04 2.20

BH d+ d- 1.61 2.20 Al H Lithium Aluminium Hydride A strong reductant that catches fire in protic solvents >>> Selective Reactions on Carbonyl Compounds H HBH H Na H Li H Al H H Most reactive Sodium Borohydride A mild reductant that can be used in protic solvents (EtOH) Least reactive

>>>> NaBH4 only reduces these carbonyls CGW: p529,534 68 34 Look at the relative electronegativity values d+ d- 2.04 2.20 BH d+ d- 1.61 2.20 Al H Lithium Aluminium Hydride A strong reductant that catches fire in protic solvents Selective Reactions on Carbonyl Compounds 15/01/2019 H Li H Al H H H HBH H Na Most reactive Sodium Borohydride A mild reductant that can be used in protic solvents (EtOH) Least reactive >>>>>>> CGW: p529,534 69 Test yourself – predict the products of the following reactions NaBH4 LiAlH4 LiAlH4 LiAlH4 LiAlH4 reduces all carbonyls

70 35 Carbonyls revision summary so far

Nucleophiles add to carbonyls to form tetrahedral products (or intermediates) Electrons add to the LUMO p* and follow an approximately tetrahedral trajectory Carbonyl reactivity can be predicted by the electronic influence of substituents on the carbonyl Test yourself – predict the products of the following reactions NaBH4 LiAlH4 LiAlH4 LiAlH4 ? 71 15/01/2019 72...