CHE202 Structure & Reactivity in Organic Chemistry: Reduction Reactions and Heterocyclic Chemistry Semester A 2016 Office: 1.07 Joseph Priestley Building Office hours: 9.30-10.30 am Monday 1.30-2.30 pm Thursday (by appointment only) Dr. Chris Jones [email protected]
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CHE202 Structure & Reactivity in Organic Chemistry: Reduction Reactions and Heterocyclic
§ Reduction:- Definition (recap.)- Reduction of carbon-carbon double (C=C) and triple (CΞC) bonds
- Heterogeneous hydrogenation- Homogeneous hydrogenation, including stereoselective hydrogenation- Dissolved metal reductions- Other methods of reduction
- Reduction of carbon-heteroatom double and triple bonds- Reduction of carbonyl derivatives, addressing chemoselectivity- Stereoselective reduction of carbonyl derivatives- Reduction of imines and nitriles
- Reductive cleavage reactions
- Reduction of heteroatom functional groups
- Hydrogenolysis of benzyl and allyl groups- Dissolved metal reduction- Deoxygenation reactions
e.g. azides, nitro groups, N-O bond cleavage
Selectivity is a key theme of this course
H H
H H
electron transfer 2nd ET
+ 2H
Reduction: definition (recapitulation)
§ Reduction of an organic substrate can be defined as:
- The concerted addition of hydrogen.e.g.
Catalytic hydrogenation (e.g. H2(g) & Pd)O O HH2 (g)
H
Hydride addition then protonation(e.g. LiAlH4 then acid w/up)
O O H"H " then HH
- The ionic addition of hydrogen NB: Dr. Lebrasseur’s & Dr Bray’s carbonyl lectures.e.g.
Dissolved alkali metals(e.g. Na in liquid NH3)
one electron added
- The addition of electrons.e.g.
Overall two electrons added
‘OIL RIG’: a helpful mnemonic...
§ Consider the reaction from the point of view of the electrons:
OIL Oxidation Is Loss
RIG Reduction Is Gain
A. All of themB. 3, 4 and 5C. 1, 3, 4 and 5D. 1, 3 and 5E. 1, 4 and 5
§ Which of these transformations represent reductions?
Question: reduction or not?
1
NO
NO2
N
NO2
NO
NO2
NO
NH2
2 3
NNH2
4
CO2Me CO2Me5
✔
N
O
NH2
H H
N
O
NH2
HO
CO2H CO2H
OHH
pyruvic acid lactic acid
enzymatic
reaction+ +
NADH NAD+
Reductions: why are they so important?
§ Reductions appear in almost all synthetic chemistry:
- in Nature:e.g. important metabolic pathways (e.g. reduce pyruvic acid to lactic acid).
- in pharmaceutical synthesis.e.g.
NNMe
O2N
O
H2N
n-Pr
NNMe
H2N
O
H2N
n-Pr
NNMe
n-Pr
HN
N
O
O2SN
NMe
OEtreduction
nitro group amine group Sildenafil
§ Reduction of carbon-carbon double and triple bonds
Reduction of carbon-carbon double and triple bonds
§ Catalytic hydrogenation- Concerted addition of hydrogen across a π-bond.- Use hydrogen gas: H2(g).- Transition metal (TM) catalyst promotes the reaction.- Catalyst can be heterogeneous or homogeneous.
Metal Catalyst
H2 (g)
HH
H
H
- Hydrogenation has a different mechanism of reduction compared to hydride reducing agents (e.g. NaBH4), therefore different chemoselectivity is often observed.e.g.
Reduction of carbon-carbon double and triple bonds
§ Heterogeneous hydrogenation- Catalyst insoluble in reaction medium.- TM (e.g. Pt, Pd, Rh) adsorbed onto a solid support, typically carbon or alumina (Al2O3); e.g. Pd/C.
(Z)-alkene
PhPh
PhPh
PtO2 (cat.)
H2 (1 atm.)H
H(S)(R)
PhPh
PhPtO2 (cat.)
H2 (1 atm.)H
HPh
(E)-alkene
(S)(S)
- Need a polar reaction solvent to dissolve sufficient hydrogen (e.g. methanol, ethanol, acetic acid).
§ Reduction of alkenes:- Reactions generally proceed at room temperature (r.t.) and 1 atmosphere (atm.) H2 pressure, however, reaction rates increase when elevate T and/or P.- Hydrogenation is typically selective for syn-addition.e.g.
syn-addition(i.e. H atoms added to
same face of C=C bond)
i). H2 dissociatively adsorbed onto metal surface.ii). Alkene π-bonds coordinated to catalyst surface.iii). Alkene π-bonds adsorbed onto catalyst surface.iv). A hydrogen atom is added sequentially onto both carbons.v). Reduced product can dissociate from catalyst surface.
Reduction of carbon-carbon double and triple bonds
§ Mechanism:- Complex and difficult to study (reaction occurs on metal surface and each catalyst is different).- Working model without curly arrows (explains syn-selectivity):
syn-addition
H H
metal catalyst surface
R
R R
RH H
ADSORPTION
H HR R
R RR
R R
R
COORDINATION
H2
ADSORPTION
H
R R
H
R R
ADDITIONthenDISSOCIATION
- Syn-selectivity increases with increased hydrogen pressure.- Reactivity decreases with increased alkene substitution (more steric hindrance).
Reduction of carbon-carbon double and triple bonds
§ Complete reduction of aromatic compounds:- Lose aromaticity so more forcing conditions required c.f. isolated alkenes.- Rh, Ru & Pt are most effective catalysts (i.e. Pd less active so use Pd when require chemoselectivity for alkene reduction in presence of aromatic ring).
NB: carboxylic acid untouched
(±)-monomorine
NB: syn-addition of hydrogen
NB: increased H2 pressures
Carbocyclic ring
Heterocyclic ring
- Carbocyclic and heterocyclic aromatic rings amenable.
OHOOH
O
O
OH
OHOOH
O
O
OH
Pt/C (cat.), H2 (4 atm.)
AcOH
N N
H10% PtO2 (10 mol%)H2 (5 atm.), HBr, r.t.
Me BunMe74%Bun
A. 1B. 2C. 3D. 4E. 5
§ Which of these structures 1 to 5 is the correct reduction product?
Question: reduction of carbon-carbon double and triple bonds
OPd/C (10 mol%), H2 (1 atm.)
MeOH, r.t.?
O
HH H
O
H
OH
HH
O
HH H
O
H
1 2 3
4 5
✔
PhOMe
O Ph OMe
O
Pd/C (10 mol%)
H2 (1 atm.), MeOH H H
H H
Reduction of carbon-carbon double and triple bonds
§ Reduction of alkynes to alkanes:- Standard heterogeneous hydrogenation results in complete reduction to alkanes.e.g.
2 moles of H2 added overall
- Require chemoselectivity to differentiate between reduction of alkyne and alkene.
- Require control over cis- or trans- geometry.
- Alkynes to alkenes, is it possible?
Ph CO2Me
H H
alkene intermediate
1 mole of H2 added
PhOMe
O Pd/CaCO3 (10 mol%)
H2 (1 atm.), quinoline,Pb(OAc)4, EtOAc
Ph CO2Me
H H
Reduction of carbon-carbon double and triple bonds
§ Reduction of alkynes to cis-alkenes- Lindlar’s catalyst: affords cis-alkenes.- Pd is poisoned with Pb and an amine – makes a less active catalyst that is more reactive towards alkynes than alkenes (NB: alkene reduction is still possible so reactions often require careful monitoring).e.g.
cis-(Z)-alkeneQuinoline – competitive binding to catalyst surface
Solid support – CaCO3 or BaSO4
Pb(OAc)4 – deactivates catalyst
Herbert Lindlar
N
- Mechanism: two hydrogens added to same face of alkyne, leading to syn-addition.
O
O
RTBSO
O
O
TBSO
O
RO
Pd/CaCO3 (10 mo%)
H2 (1 atm.), quinolinePb(OAc)4, hexane, r.t.
86%
e.g. in the synthesis of a complex molecule
Z
Reduction of carbon-carbon double and triple bonds
§ Homogeneous hydrogenation:- Metal-ligand complex is soluble in reaction medium.- Phosphines are common ligands (i.e. good electron donors).
e.g. Wilkinson’s catalyst, (Ph3P)3RhCl- Stereospecific syn-addition of hydrogen across alkene.- Less substituted and least sterically hindered double bonds reduced most easily.- Chemoselectivity (i.e. ketones, carboxylic acids, esters, nitriles, ethers and nitro groups all inert to these conditions.
O
OO
O
OO(Ph3P)3RhCl (cat.), H2 (1 atm.)
benzene/EtOH95%
NB: This mechanism is beyond the scope of the course – for a full explanation see ‘Ox & Red in Org Synth’ p. 54
Least hindered alkene
NB: Heterogeneous hydrogenation (e.g. H2, Pd/C) is non-selective and leads to over reduction
Sir Geoffrey Wilkinson(Nobel Prize in Chemistry 1973)
PPh2PPh2
(S)-BINAP
Reduction of carbon-carbon double and triple bonds
§ Enantioselective hydrogenation: (FYI only, beyond scope of course)- If metal-ligand complex (MLn) is chiral, then possible to control to which face of alkene H2 is delivered.- Discrimination between enantiotopic faces of alkene can lead to single enantiomer of product.
ANSWER: use chiral ligand.e.g. BINAP is a chiral & bidentate ligand for TM.- Commercially available as both (R)- and (S)- enantiomers (i.e. chose desired product enantiomer).
NB: this is a ‘protected’ version of the amino acid - alanine
- Beyond hydrogenation, homogeneous catalysis is an extremely important area of organic chemistry and you will encounter numerous examples in future lecture courses…
CO2H
NHCOPh
Rh(I), (S)-BINAP
H2 (1 atm.)
CO2H
NHCOPhH
Very high facial selectivity (98% ee)
top face
H2
H2
WHICH FACE?MLn +
CO2H
NHCOPhH
Rh(I), (R)-BINAP
H2 (1 atm.)
Bottom face
Pre-Lecture Diagnostic Test – Feedback
§ Average mark: 18 / 20.
§ Excellent – no obviously weak areas.- individuals are strongly encouraged to revise any specific areas of weakness that the test revealed.
FEEDBACK: each week please use QMplus “Lecture feedback quiz” to comment on which area of the last two lectures was least clear – I will then revise that topic at the start of the next lectures.
Lowest scoring questions were on hybridisation state and pKa values (these concepts are fundamental to understanding organic chemistry, so please revise if necessary).
Revision:N
O
O
sp3sp
sp2
Put the following species in order of increasing acidity (i.e. least acidic first)
pKa =
O
OHF
FF
O
OHMe
O
OH MeS
O
OH
OOHHCl
10.0 4.8 4.2 –2.6–0.3 –8.0
> > > > >
Reduction of carbon-carbon double and triple bonds
§ Dissolved metal reductions:- Alkali metals (e.g. Na, Li, K) dissolved in liquid ammonia (T < –33 ˚C).- ‘Free electrons’ add to low-lying π*-orbitals.
Na Na + eNH3 (l) Consider the solution as a
source of ‘free electrons’
§ Reduction of alkene to alkane:- Selectively reduces electron deficient alkenes to alkanes (i.e. no reaction with standard alkenes).
O Oi). Li, NH3 (l), t-BuOH (1 equiv), –78 ˚C
ii). NH4Cl
Reduces electron deficient alkene only
NB: Low temperature as NH3 is a gas at T > –33 ˚C.NB: Stoichiometry of alcohol is crucial (see mechanism next page).
O
O
O OH
OHO
+ e
PROTON TRANSFERH
H Ot-Bu
+ e
H3NCl
Reduction of carbon-carbon double and triple bonds
§ Mechanism of alkene reduction:i). Addition of electron to π-bond gives radical anion.ii). Radical anion protonated by 1 equiv. of t-BuOH to give neutral radical.iii). Addition of 2nd electron yields allylic anion.iv). Proton transfer affords enolate (NB: no proton source so enolate stable under these conditions).v). Addition of more acidic proton source, NH4Cl (aq), protonates enolate to give ketone product.
Single electron transfer
No t-BuOH remaining – enolate stable until add NH4Cl (aq)
Single electron transfer
–ve charge on oxygen
(radical)
(radical)
(ionic)
(ionic)(ionic)
allyl anion
H
+ e
H NH2 + e
H NH2
Reduction of carbon-carbon double and triple bonds
§ Reduction of alkynes to trans-alkenes:- Dissolved metal reduction forms trans-double bond with high levels of stereoselectivity.e.g.
c.f. Lindlar’s catalyst gives cis-alkene
Vinyl anions sufficiently basic to deprotonate NH3 (c.f. enolate
basicity on previous slide)
Na, NH3 (l), –33 ˚C
trans-(E)-alkene
- Provided it is not electron deficient, product alkene will not be reduced.
Vinyl anions are geometrically unstable and (E)-geometry is
§ Put these anions in order of decreasing stability (i.e. most stable first and least stable last).
Question: some pKa revision
O
N
1 2 3
4 5
✔
H HH H
+ e
H Ot-Bu + e
H Ot-Bu
H H
Reduction of carbon-carbon double and triple bonds
§ Partial reduction of aromatic rings:- Birch reduction: dissolved metal reduction of aromatic rings.- Affords non-conjugated diene product.- Similar mechanism to previous slides: electron transfer, radical anion protonation etc.e.g.
Pentadienyl anion has the highest electron density at the central position, hence kinetic protonation (at low T) affords the non-conjugated product regioselectively.
H H
H H
Na, NH3 (l), t-BuOH, –33 ˚C
Arthur Birch
+ e
H Ot-Bu + e
H Ot-Bu
CO2Me CO2MeCO2Me CO2Me
Reduction of carbon-carbon double and triple bonds
§ Regioselectivity of Birch reduction:- Electron withdrawing groups promote ipso, para reduction. e.g.
- Electron withdrawing groups stabilise electron density at the ipso and para positions.- Protonation favoured at C-4 as leaves a radical stabilised by conjugation with the carbonyl group.
H H
Na, NH3 (l), t-BuOH, –78 ˚C
CO2Me CO2MeH
ipso, para
ipso, paraortho, meta
+ e
H Ot-Bu + e
H Ot-Bu
OMe OMeOMe OMe
Na, NH3 (l), t-BuOH, –78 ̊ C
OMe OMeH
HH
H
Reduction of carbon-carbon double and triple bonds
§ Regioselectivity of Birch reduction:- Electron donating groups promote ortho, meta reduction.e.g.
- Electron donating groups stabilise electron density at the ortho and meta positions.- Protonation favoured at C-2 as leaves a radical stabilised by conjugation with the methoxy group.
ortho, meta
ipso, para ortho, meta
Reduction of carbon-carbon double and triple bonds
§ Birch reduction of heterocycles:- Heterocycles bearing an electron withdrawing group can also be partially reduced.e.g.
- Heterocycles are electron rich – electron withdrawing group required for reduction.- Electron withdrawing group also stabilises radical anion formation, controls regioselectivity.
Na, NH3 (l), t-BuOH, –78 ˚COi-PrO
O
Oi-PrO
O
+ e
H Ot-Bu + e
H Ot-Bu
Oi-PrO
O
Oi-PrO
O
Oi-PrO
O
A. 1B. 2C. 3D. 4E. 5
O MeOCN
1 2 3
4 5
§ Which of these substrates will be reduced the fastest under Birch conditions (i.e. dissolved metal, NH3 (l), –78 ˚C)?
Question: Birch reduction
✔
OMe3Si Al
H HH
H AlH3Li
LiMe3Si
OH2Al
HLi
Me3Si
H
Li
OH2Al
H2OH
Reduction of carbon-carbon double and triple bonds
§ Reduction of propargylic alcohols to trans-allylic alcohols:- Use LiAlH4.- Stereoselectivity due to complexation of reducing agent to oxygen.e.g.
OHMe3Si
Me3SiOH
LiAlH4, THF, 70 ˚C
LiAlH4 deprotonates alcohol then complex
forms between alkoxide and resultant
Lewis acidic AlH3
Intramolecular delivery of hydride
Intramolecular coordination of vinyl anion to Lewis acidic Al species locks geometry of alkene (i.e. Al
and alcohol chain on same side of C=C)
stereospecific protonation of C–Al bond during w/up
(E)-alkene
H2 (g) evolution
Reduction of carbon-carbon double and triple bonds
§ Summary:- C-C π-bonds typically reduced by hydrogenation or dissolved alkali metal.
- Heterogeneous catalysts of Pd, Pt, Rh, Ru etc. are effective at C-C π-bond hydrogenation (alkene, alkyne, aromatic).
- Hydrogenation of C-C π-bond involves stereoselective syn-addition of hydrogen.
- Pd is less active and can be used for chemoselective alkene reduction in presence of aromatic ring.
- Mechanism of action means hydrogenation can be chemoselective for C-C π-bond over other FGs (e.g. ketone, nitrile, ester, amide etc.).
- Homogenous hydrogenation can offer greater regioselectivity w.r.t. heterogeneous catalysis (e.g. alkene reduction more sensitive to steric requirements) and opportunies for enantioselectivity.
- cis-Alkenes formed stereoselectively from alkynes using Lindlar’s catalyst.
- trans-Alkenes formed stereoselectively from alkynes using dissolved alkali metal or LiAlH4 (if O atom).
- Regioselectivity of Birch reduction determined by substituents (EWG gives ipso, para; EDG gives ortho, meta).
Me Li, NH3 (l), –78 ˚C, t-BuOH?
Q1. What is the product of this Birch reduction? Draw a curly arrow mechanism to describe the reaction.
Q2. Provide two different sets of reagents and conditions that would carry out this transformation. Draw curly arrow mechanisms for both.
OHPh Ph
OH?
Q3. Which two types of selectivity (i.e. chemo, regio, stereo) are observed in this Lindlar reduction and how do the reagents and conditions control these selectivities?
Q4. Which of the two following hydrogenations would require the greater pressure of hydrogen, and why?
Pd/C (cat.), H2 (g), MeOH
Pd/C (cat.), H2 (g), MeOH
Ph CO2Me PhCO2MeLindlar reduction
Questions: end of Lectures 1 & 2
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H H
+ e
H Ot-Bu + e
H Ot-Bu
Na, NH3 (l), t-BuOH, –78 ̊ C
CO2Me CO2MeH
CO2Me CO2MeCO2Me CO2Me
Birch reduction – recap.
§ Initial electron addition – consider the stability of the subsequent radical anion.- These curly arrow mechanisms are discussed in all general organic chemistry textbooks.e.g.
Radical is closest to stabilising group (i.e. ester)
NB: pentadienyl anion has largest coefficient of electron
density at central position (“kinetic-controlled protonation”
occurs here)
Other questions from L1&2
§ Why only reduce electron-deficient alkenes (e.g. Na, NH3 (l))?- adding electrons to alkene, so electron deficient alkenes more susceptible.
§ Why only reduce electron-deficient pi-bonds (e.g. LiAlH4)?- adding hydride (i.e. “H–”) to multiple bond, so electron deficient pi-bonds more susceptible.
§ Birch reduction rates.- adding electrons to aromatic ring, so electron deficient pi-systems more susceptible.
§ Reduction of propargylic alcohol.- covered in Workshop A9 (see workshop feedback and textbooks).
§ Reduction of carbon-heteroatom double and triple bonds
Aims: at the end of this section you will be able to…
§ Understand the origin and reactivity of C-O and C-N multiple bonds.
§ Draw curly arrow mechanisms for standard C-O and C-N multiple bond reductions.
§ Understand the origin of (and predict) chemoselectivity of different hydride reducing agents (e.g. LiAlH4, NaBH4, LiBH4, BH3, DIBAL etc).
§ Understand the origin and predict the stereochemical outcome of ketone reduction using appropriate Felkin-Anh models of 1,2-stereoinduction.
Reduction of carbon-heteroatom double and triple bonds
§ Catalytic hydrogenation:- Possible to reduce C-heteroatom π-bonds with H2 (g) and TM catalyst…- BUT chemoselectivity is big issue (i.e. many substrates contain C-C π-bonds that are also reduced).e.g.
NB: Pt more reactive metal than Pd
H
OPtO2 (5 mol%)
H2 (g) (1 atm.)OH
§ Ionic hydrogenation:- C-heteroatom π-bonds are polarised and susceptible to attack by nucleophilic ‘hydride’.- Opportunity to control chemoselectivity.
NB: see Dr Lebrasseur’s & Dr Bray’s lectures on carbonyl chemistry
H
O
0 ˚C, 97 %OH
NaBH4, MeOH
alkene also reduced
R
O
ClACID CHLORIDE
Reduction of carbon-heteroatom double and triple bonds
§ Functional groups:- Wide range of C-X FGs, with very different levels of electrophilicity (i.e. reactivity).e.g.
§ Hydride reducing agents:- Wide variety of reducing agents, with different levels of nucleophilicity (i.e. reactivity).e.g.
§ 1st Year revision: order these functional groups in terms of decreasing electrophilicity (i.e. most reactive first and least reactive last).
Reduction of carbon-heteroatom double and triple bonds
R
NH
HIMINE
R
O
HALDEHYDE
R
O
R'KETONE
R
O
OR'ESTER
R
O
NHR'AMIDE
R
O
OHCARBOXYLIC ACID
✔
R
NR'H
HIMINIUM ION
Reduction of carbon-heteroatom double and triple bonds
§ Revision aid: summary of FG reactivity with common reducing agents.
decreasing electrophilicity
R
O
HALDEHYDE
R
O
R'KETONE
R
O
OR'ESTER
R
O
NHR'AMIDE
R
O
OHCARBOXYLIC ACID
LiAlH4
NaBH4
DIBAL(@ –78 ˚C)
BH3
NaCNBH3
LiBH4
✓
✓
✓
✓slow
✓
✓
✓
✓
✓
✓(✗)
slow
slow
✓
✓
✓
✓(✗)
slow
slow
✓
slow
✓
✓(✓ aldehyde)
slow
✗
✓
✗
✗
✓
✗
slow
✗
✗
✓(✓)
✓
✗
✓(✓ aldehyde)
AMINEProduct AMINEALCOHOL ALCOHOL ALCOHOL ALCOHOL
N
O
O
N
O
OHNaBH4, MeOH
H
Reduction of carbon-oxygen double and triple bonds
§ Reduction of aldehydes and ketones to alcohols (recap.):- Using NaBH4.e.g.
Racemic product
NB: amide not reduced
acidic ‘work-up’H BH3
N
O
O
H
Na
HOH2
H AlH3N
O
OLi
LiAlH4
N
O
OAl
H
Li
H3O
H
HH
Reduction of carbon-oxygen double and triple bonds
§ Reduction of aldehydes and ketones to alcohols (recap.):- Use LiAlH4 (NB: stronger reducing agent than NaBH4 – reduces amide too, no chemoselectivity).e.g.
Li+ functions as a Lewis acid (i.e. precoordination activates the carbonyl by lowering the LUMO energy)
Reduce amide then work-up(NB: mechanism of amide reduction covered later)
work-up
N
O
O
N
OHLiAlH4, THF, ∆
HH
H
H AlH3N
O
OLi
LiAlH4
N
O
OAl
H
Li
H3O
H
HH
Reduction of carbon-oxygen double and triple bonds
§ Reduction of aldehydes and ketones to alcohols (recap.):- Use LiAlH4 (NB: stronger reducing agent than NaBH4 – reduces amide too, no chemoselectivity).e.g.
Li+ functions as a Lewis acid (i.e. precoordination activates the carbonyl by lowering the LUMO energy)
Reduce amide then work-up(NB: mechanism of amide reduction covered later)
work-up
N
O
O
N
OHLiAlH4, THF, ∆
HH
H
N
O
OAlH2
HH
H2O
–AlH3
N
O
OAlH2
H
H
H2O
H3O
R
NR'H
HIMINIUM ION
Reduction of carbon-heteroatom double and triple bonds
§ Revision aid: use as a quick guide to determine chemoselectivity.
decreasing electrophilicity
R
O
HALDEHYDE
R
O
R'KETONE
R
O
OR'ESTER
R
O
NHR'AMIDE
R
O
OHCARBOXYLIC ACID
LiAlH4
NaBH4
DIBAL(@ –78 ˚C)
BH3
NaCNBH3
LiBH4
✓
✓
✓
✓slow
✓
✓
✓
✓
✓
✓(✗)
slow
slow
✓
✓
✓
✓(✗)
slow
slow
✓
slow
✓
✓(✓ aldehyde)
slow
✗
✓
✗
✗
✓
✗
slow
✗
✗
✓(✓)
✓
✗
✓(✓ aldehyde)
AMINEProduct AMINEALCOHOL ALCOHOL ALCOHOL ALCOHOL
N
O
O
O
MeO2C
HOH
MeO2C
H
Al(Oi-Pr)3, i-PrOH, ∆
78 %+
OMeO2CMeO2C
O
MeO2C
HOH
MeO2C
H
Al(Oi-Pr)3, i-PrOH, ∆
78 %
OO AlOi-Pr
Oi-Pr Al
i-PrOHAl(Oi-Pr)3
OH
Me Me
H
Oi-PrOi-Pr
O
+O
MeO2CMeO2C
Reduction of carbon-oxygen double and triple bonds
§ Reduction of aldehydes and ketones to alcohols (alternative):- Meerwein-Ponndorf-Verley (‘MPV’) reduction.- High chemoselectivity for aldehydes and ketones over esters, alkenes etc.- Works via hydride transfer from the isopropoxy group (i.e. different mechanism c.f. NaBH4 & LiAlH4). e.g.
6-membered transition state
Reaction completely reversible
NB: α,β-unsaturated carbonyls reduced to allylic alcohols
Al is Lewis acid w/up
hydride transfer
N.B. esters & alkene untouched
what is the ratio?59 : 41 without CeCl3
99 : 1 with CeCl3
OH
+
O O[1,4]-attack
H BH3
tautomerisation
H BH3
- Luche reduction:- Excellent regioselectivity for [1,2]-attack by using CeCl3 in addition to NaBH4.- Ce(III) activates carbonyl & promotes formation of alkoxyborohydrides (“harder” nucleophiles) from NaBH4 and MeOH (N.B. this mechanism is beyond scope of course).
Reduction of carbon-oxygen double and triple bonds
ONaBH4
MeOH
OH[1,2]-attack
- [1,2]-attack leads to allylic alcohols.- [1,4]-attack leads to fully reduced alcohol (i.e. via conjugate addition, enol tautomerism, reduction).
§ Reduction of α,β-unsaturated ketones:- Two potential sites of attack leads to two possible products.
e.g.
A. At room temperature LiAlH4 is a solidB. At room temperature LiAlH4 is a liquidC. At room temperature NH3 is a liquidD. At room temperature NH3 is a gasE. NaBH4 will not reduce an esterF. NaBH4 will reduce an ester
§ Identify which statements are correct (more than one statement may be correct).
Question:
✔
✔
✔
Reduction of carbon-oxygen double and triple bonds
§ Reduction of esters to alcohols:- N.B. esters less reactive than aldehydes & ketones so need a strong reducing agent.- NaBH4 not reactive enough, use LiAlH4 (or LiBH4 for a milder & chemoselective alternative).e.g.
aldehyde intermediatereduced rapidly
NB: DIBAL at r.t. will also reduce esters to alcoholsNB: aluminate species produced in reaction, but these are less reactive than parent LiAlH4.
O
O
OHLiAlH4, THF, 0 ˚C
O
OLi
H AlH3O
OH3Al
H
Li
H
O
LiAlH4LiAlH4
–EtOAlH3
Li+ activates carbonyl group
aldehyde reduced to alcohol
(see previous slides)
tetrahedral intermediate collapses
aluminate
Can we stop the reduction at the aldehyde? Require tetrahedral intermediate not to collapse in situ.
H3O
O
OH
H
H3O
O
OAl
O
OAl
H
i-Bu
i-Bu
i-Bu
Hi-Bu
Ali-BuH
i-Bu
Reduction of carbon-oxygen double and triple bonds
§ Reduction of esters to aldehydes:- Use 1 equivalent of Lewis acidic DIBAL (diisobutylaluminium hydride) at low temperature.- DIBAL forms Lewis acid-base complex with carbonyl group in order to become a reducing agent (i.e. reduces more electron rich C=O groups more quickly). e.g.
Intermediate collapses to aldehyde upon w/up, but excess DIBAL has been quenched by H3O+ so no further reduction can occur
DIBAL reduces esters at –70 ˚C
tetrahedral intermediate stabilised at low T
O
O
ODIBAL, Toluene, –78 ̊ C
NB: alkene not reduced
hemiacetal
w/up
DIBAL (NB: Al is Lewis acid)
Reduction of carbon-oxygen double and triple bonds
§ Reduction of amides to amines:- Amides poor electrophiles, require powerful reducing agent LiAlH4.e.g.
iminium ion reduced rapidly by LiAlH4
(N.B. iminium ion more reactive than aldehyde)
tetrahedral intermediate collapses by expelling best
LG (i.e. O better than N)
NMe2
ONMe2LiAlH4, Et2O
88 %
NMe2
OLi
NMe2
H
NMe2
OH3Al
H
Li
H AlH3
H AlH3LiAlH4
–H3AlOLi
Li+ activates carbonyl group
Drawback: if molecule contains aldehydes, esters, etc. then LiAlH4 will reduce everything (i.e. no chemoselectivity).
then w/up
N
MeO2C
O
Bn
B HH H
N
MeO2C
O
Bn
B HH
H
N
MeO2C
H
BnH
H2B
N
MeO2C
H
Bn
H
BH2
BH
H
H
–BH2O
H2O
BH3•THF
N
MeO2C
O
Bn
N
MeO2C
Bn
Reduction of carbon-oxygen double and triple bonds
§ Reduction of amides to amines (alternative):- Use BH3 (chemoselective for amide over ester, ketone, etc.).e.g.
BH3 reduces reactive
iminium ion
Borane (B2H6) is a gas; stabilised as a liquid in BH3 complex with THF
B OH
HH
Borate donates hydride to
reactive imidate
w/up
NB: ester not reduced here
Like DIBAL, BH3 forms Lewis acid-
base complex
(c.f. 1st step of hydroboration mechanism)
Reduction of carbon-oxygen double and triple bonds
§ Reduction of amides to aldehydes?- Require stabilised tetrahedral intermediate following hydride addition (c.f. partial reduction of ester).e.g.
- Tetrahedral intermediate is difficult to stabilise and collapses in situ to iminium ion which is further reduced to amine (c.f. reaction with LiAlH4 on previous slide).- No reagent for general amide reduction to aldehyde… BUT…
w/up to reveal hemiaminal
stabilise tetrahedral intermediate?
hydrolysis to aldehyde
NMe2
OOM-[H]
NMe2
OM
NMe2
OH
NMe2
OM
H
[H]
M-[H]
H3O
generic metal-hydride species
w/up
N
OAl
N
OH
N
O
H
H3O
H Ali-Bu
i-Bu
H
i-Bui-Bu
Ali-Bu
i-Bu
OMe OMe OMeH
HNOMe
–
Reduction of carbon-oxygen double and triple bonds
§ Reduction of amides to aldehydes:- Weinreb’s amides (-CON(OMe)(Me) form stable chelated intermediates at low T.e.g. with DIBAL
w/up reveals hemiaminal
stabilised chelated tetrahedral intermediate
hydrolysis to aldehyde
N
OODIBAL, tolueneOMe
0 ˚C, 74 %
Al is Lewis acid
Borate donates hydride to reactive imidate
w/up
Question: apply this understanding to a new situation
§ Weinreb amides and organolithium reagents (a quick aside):- Considering the previous slide, determine the product of the following reaction:e.g.
stabilised chelated tetrahedral intermediate
NB: also works with Grignard reagents
elimination of MeNHOMe
ketoneN
OMeLi, THFOMe
0 ˚C? Me
O
N
O
Me
Li
OMe N
OH
MeOMe
HNOMe
–
H3O
Me Li
Q: What would happen if this was a simple amide, rather than a Weinreb amide?
w/up
OH
O
O
OOHLiAlH4
Liheat
–H2 (g)
Reduction of carbon-oxygen double and triple bonds
§ Reduction of carboxylic acids to alcohols:1). Use LiAlH4 but require very forcing conditions, as form quite unreactive carboxylate salts.e.g. 1
quite unreactive
2). Use BH3: Lewis acidic, so chemoselective for most electron rich carbonyl groups (e.g. acids and amides).- Mechanism: complexation with C=O lone pair forms active reducing agent (see amide reduction).- Milder reduction conditions compared to using LiAlH4.e.g. 2
O
OHOHBH3•THF
NB: for further explanation see ‘OC’ textbook p. 619.
mixed anhydride
BocHNO
O
OEt
O
H
BocHNH
O
NaBH4H BH3
–CO2 (g)
–EtO
BocHNOH
OBocHN
O
O
Cl
O
OEtBocHN
OHOEt
O
Et3N, CH2Cl2
NaBH4
MeOH, 0 ˚C
97 % yield over 2 steps
Reduction of carbon-oxygen double and triple bonds
3). Via more reactive intermediate: convert carboxylic acid to mixed anhydride, then reduce with NaBH4.- 2 steps, but generally quick and high yielding procedures.e.g. 3
NB: most reactive at C=O that originated from carboxylic acid, as other C=O π-system has overlap from two oxygens.
Eliminate CO2 (g) and EtO-
alcoholacid
Questions from L3&4
§ Reduction of amides to amines:- can use LiAlH4 (powerful nucleophilic hydride source – poor chemoselectivity).- BH3 or DIBAL (initially Lewis acidic so target most Lewis basic C=O groups).
iminium ion reduced rapidly by LiAlH4
(N.B. iminium ion more reactive than aldehyde)
tetrahedral intermediate collapses by expelling best
LG (i.e. O better than N)
NMe2
ONMe2LiAlH4, Et2O
88 %
NMe2
OLi
NMe2
H
NMe2
OH3Al
H
Li
H AlH3
H AlH3LiAlH4
–H3AlOLi
Li+ activates carbonyl group then w/up
Aims: at the end of this section you will be able to…
§ Reduction of C-N multiple bonds.
§ Understand the origin and predict the stereochemistry of the enantioselective CBS reduction of ketones.
§ Understand the term ‘reductive cleavage’ and be familiar with suitable reagents and conditions.
§ Understand the different mechanisms and methods to reduce heteroatom functional groups.
§ Note: Recalling reagents & conditions in questions – reagents are the most important part of a reaction to remember/apply (solvent is less important to answers, unless it has been specifically mentioned in lecture notes e.g. polar & protic solvents required for heterogeneous hydrogenations)
§ Understand the origin and predict the stereochemistry of reductions of cyclohexanones (i.e. axial or equatorial attack?).
§ Understand the origin and predict the stereochemical outcome of ketone reduction using appropriate Felkin-Anh models of 1,2-stereoinduction.
Reduction of carbon-nitrogen double and triple bonds
§ Reduction of imines to amines:- Use NaBH4, LiAlH4, etc. (NB: imines more electrophilic than aldehydes).e.g.
iminium ion(highly reactive)
NB: analogous to mechanism of aldehyde reduction
§ Reductive amination:- Convert carbonyl group to amine through in situ imine formation.- Add acid to increase imine reactivity, can now use much weaker hydride reducing agent (e.g. NaCNBH3) to chemoselectively reduce iminium ion in presence of aldehyde.e.g.
O
HBnNH2
MeOH, HCl
NHBnNBn
H
HH BH2NaCN
H
N HNNaBH4, MeOH
0 ˚C, 95 %
Aldehyde starting material not reduced
by NaCNBH3
Iminium ion formation(see 1st Yr for mechanism)
H H
O
(CH2O)n (aq.)
Hint:
Question: reductive amination
§ Eschweiler-Clark reaction:- Introduced in 1st year.- Can you draw the mechanism for this reductive amination?e.g.
CO2 (g) evolution
NH
N(CH2O)n (aq.)
HCO2H
N
HH
H O
O
H H
O
NB: iminium ion formation
NLi
H3OLiAlH4
HN
Li
H AlH3
LiAlH4, then
Reduction of carbon-nitrogen double and triple bonds
§ Reduction of nitriles to amines:- Complete reduction using LiAlH4 (i.e. powerful reducing agent).e.g.
Overall 2 equivalents of hydride added
NNH2LiAlH4, THF
reduction of intermediate imine followed by w/up
(mechanisms analogous to
previous slides)
Li+ activates nitrile group
imine intermediate
NAl
H
i-Bu
i-Bu
NAli-Bu
i-Bu
H
H3OH Ali-Bu
i-Bu
i-Bu
H
i-BuAl
N
Reduction of carbon-nitrogen double and triple bonds
§ Reduction of nitriles to aldehydes:- Partial reduction using DIBAL (note: 1 equivalent of reducing agent is crucial to avoid over-reduction).e.g.
iminoalane
NB: Lewis acidic DIBAL coordinates to nitrile lone pair and ‘ate’ complex is hydride donor
w/up
NODIBAL (1 equiv.), Toluene, –78 ˚C
then H3O H
no DIBAL remaining so iminoalane
reduction cannot occur
R
NR'H
HIMINIUM ION
Reduction of carbon-heteroatom double and triple bonds
decreasing electrophilicity
R
O
HALDEHYDE
R
O
R'KETONE
R
O
OR'ESTER
R
O
NHR'AMIDE
R
O
OHCARBOXYLIC ACID
LiAlH4
NaBH4
DIBAL(–78 ˚C)
BH3
NaCNBH3
LiBH4
✓
✓
✓
✓slow
✓
✓
✓
✓
✓
✓(✗)
slow
slow
✓
✓
✓
✓(✗)
slow
slow
✓
slow
✓
✓(✓ aldehyde)
slow
✗
✓
✗
✗
✓
✗
slow
✗
✗
✓(✓)
✓
✗
✓(✓ aldehyde)
AMINEProduct AMINEALCOHOL ALCOHOL ALCOHOL ALCOHOL
§ Revision aid: summary of FG reactivity with common reducing agents.
Questions: end of Lectures 3 & 4
O OH
Al(Oi-Pr)3, i-PrOH
∆
Q1. Provide a curly arrow mechanism for this reaction. What is the name of this transformation?
CO2Me
CONMe2
LiBH4 LiAlH4A B
Q2. What are the structures of A and B? Draw mechanisms for both reactions and rationalise the observed selectivity.
Q3. Put these FGs into their order of increasing reactivity with DIBAL (i.e. least first, most reactive last).
ESTER ALDEHYDEKETONE AMIDE
Q4. What is the product of this reaction? Draw a mechanism to describe the formation.
O , HCl, MeOH1.
2. NaCNBH3
NH2?
Q5. What is the stereochemistry of the product in this reaction? Explain how you came to your choice.
PhO
EtNaBH4
MeOHPh
OH
Et
* *
§ Stereoselective reduction of carbonyl groups
Aims: at the end of this section you will be able to…
§ Understand the origin and predict the stereochemical outcome of ketone reduction using appropriate Felkin-Anh models of 1,2-stereoinduction.
§ Understand the origin and predict the stereochemistry of the enantioselective CBS reduction of ketones.
§ Understand the term ‘reductive cleavage’ and be familiar with suitable reagents and conditions.
§ Understand the different mechanisms and methods to reduce heteroatom functional groups.
§ Note: Recalling reagents & conditions in questions – reagents are the most important part of a reaction to remember/apply (solvent is less important to answers, unless it has been specifically mentioned in lecture notes e.g. polar & protic solvents required for heterogeneous hydrogenations)
§ Understand the origin and predict the stereochemistry of reductions of cyclohexanones (i.e. axial or equatorial attack?).
Diastereoselectivity with hydrides: 1,2-stereoinduction
§ Felkin-Anh model (recap.):e.g.
NB: see Dr Bray’s 1st year lectures notes
Convert to Newman projection
PhO
PhOH
LiBH(s-Bu)3THF
‘Felkin’ product
Look down highlighted C-C bond
Ensure configuration around chiral centre
is correct
- Drawing Newman projections (recap.).
PhO
MePh
H
OO
MeH
add substituents
Diastereoselectivity with hydrides: 1,2-stereoinduction
§ Felkin-Anh model (recap.):e.g.
NB: see Dr Bray’s 1st year lectures notes
Newman projection
PhO
PhOH
LiBH(s-Bu)3THF
MePh
H
O
R3BH
- Reactive confirmation (i.e. least sterically hindered).- Nu approaches along Bürgi-Dunitz trajectory (107˚ O=C…Nu).- Least hindered approach is past ‘H’ and away from large group (e.g. Ph).
‘Felkin’ product
MePh
H
O HPh
Me
O
(i.e. place largest group 90˚ to C=O)
Lowest energy conformations of chiral starting material
or
O
SMei-Pr
H O
SMei-Pr
H
new LUMO (π* + σ*)
C=O π*
C-S σ*
combine
BR3H
i-PrMeS
H
O
Diastereoselectivity with hydrides: 1,2-stereoinduction
§ Felkin polar model:- Use when α-substituent is electron withdrawing.e.g.
Newman projection
- Electronegative group 90˚ to C=O (even if it isn’t the largest group).- Reactive conformation (i.e. π* and σ* combine to lower LUMO).- Approach along Bürgi-Dunitz trajectory (107˚).- Least hindered approach is past ‘H’ and away from electronegative group.
‘anti-Felkin’ product
(i.e. most reactive in this conformation)
π* C=O overlaps with σ* C-S; creates lower energy LUMO
i-Pr
SMe
Oi-Pr
SMe
OHLiBH(s-Bu)3
THF (i.e. place eWG group 90˚ to C=O)
Me
OMe
OMe
OMe
OHZn(BH4)2THF
Diastereoselectivity with hydrides: 1,2-stereoinduction
§ Felkin chelation model:- Can reverse selectivity when (i) α-substituent contains lone pairs and (ii) use chelating metal.- Require Lewis acid metal that chelates to more than one heteroatom at once (i.e. to the carbonyl group and α-substituent).e.g.
Newman projection
- Electronegative group almost eclipses C=O to enable chelation with the metal (even if electronegative group is the largest group).- Reactive conformation (i.e. Lewis acid coordination lowers LUMO).- Approach along Bürgi-Dunitz trajectory (107˚).- Least hindered approach is past ‘H’ and away from group at 90˚.
‘Felkin’ product
Li+ (sometimes)Mg2+
Zn2+
Cu2+
Ti4+
Ce3+
Mn2+
Chelating metals
MeOMe
H
O
R3BH
Zn2+
NB: Na+ and K+ do not chelate
H
NBn2
O
H
CO2Me
Bn2NH
O
H CO2Me
HNBn2
O
HBn2N
H
O
HOMe
O
NBn2H
O
MeO
ONBn2
OHCO2Me
NBn2
OHCO2Me
+
Diastereoselectivity with hydrides: 1,2-stereoinduction
§ Examples:e.g. 1 Key step in synthesis of anticancer agent, dolastatin.
most reactive conformation
Aldol reaction
Visualisation aid: draw product in same orientation, then rotate to put longest chains in same plane
1 20diastereoselectivity
- How do we explain diastereoselectivity? Consider reactive conformation.1. No chelation so electronegative group goes 90˚ to C=O (Felkin polar)2. Approach alongside H
Diastereoselectivity with hydrides: 1,2-stereoinduction
§ Examples:e.g. 2 Chelation or not?
Felkin polar model
42 58
diastereoselectivity
- How do we explain diastereoselectivity? Consider reactive conformation.
R = OSi(i-Pr)3
R = Me 99 1
ROMe
H
O
Ph
Mg2+
Nu
1Me
ROH
O
Ph Nu
‘R’ = Me
Chelation model
‘R’ = OSi(i-Pr)3
Large group disrupts efficient chelation
Ph
O
OR
PhOR
HO Me
PhOR
HO MeMe2Mg
THF, –70 ̊ C+
Diastereoselectivity with hydrides: 1,2-stereoinduction
§ Summary: which model to use?
Yes
α-chiral carbonyl compound
XR
O
ZY
Use Felkin chelation model (i.e. conformations with the heteroatom and
C=O almost eclipsed)
Yes
No
No
Use Felkin polar model (i.e. conformations with
the electronegative group 90˚ to C=O)
Use Felkin-Anh model (i.e. conformations with the largest group 90˚ to C=O)
Is there a metal ion capable of
chelation?
Is there a heteroatom at the chiral centre?
A. 1B. 2C. 3D. 4
§ What is the stereochemistry of the product of this reaction?
Question: 1,2-stereoinduction
PhO
EtNaBH4
MeOHPh
OH
Et
* *
PhOH
Et
PhOH
Et
PhOH
Et
PhOH
Et1 2 3 4
✔
§ What is the stereochemistry of the product of this reaction?
Question: 1,2-stereoinduction
H
H
HH
HH
H
H
HH
Reduction of cyclohexanones
§ Conformations of 6-membered rings (recap.):- Cyclohexane is not flat… puckered to enable tetrahedral carbon atoms.- Two most common conformers are the ‘chair’ (lowest energy) and ‘boat’.e.g.
Chair
Half chair
- Cyclohexene adopts a ‘half chair’ in its lowest energy conformation.
Boat
NB: for further explanation see ‘OC’ textbook p. 370-374.
NB: equatorial
NB: axial
H
H
H
H
H
H H
H
Cyclohexane
Cyclohexene
H
Small hydride
H
HH
H OH
HH
HH
HH
HH
OH
OH
H'EQUATORIAL
ATTACK''AXIAL
ATTACK'
Reduction of cyclohexanones
§ Axial or equatorial attack?- When reducing cyclohexanones, the hydride can end up in an axial or equatorial position.
e.g. LiAlH49:1 ax:eq
less hindered approach
- ‘Axial attack’ favoured with a small hydride source (e.g. LiAlH4, NaBH4).
more hindered approach (i.e. 1,3-diaxial interactions)
§ Why does ‘axial attack’ occur at all?- If ‘axial attack’ is more hindered then why is it even favoured with small hydride sources?- Need to consider the transition state leading to the alkoxy intermediate from ‘axial attack’:
NB: for further explanation see ‘OC’ textbook p. 471.
(i.e. less hindered approach)
(i.e. more hindered approach)
- ‘Equatorial attack’: oxy substituent moves towards neighbouring C-H bond, leading to higher torsional strain in T.S (i.e. disfavoured).
H
O
H
OH
H
A. 1B. 2C. 3D. 4
§ What is the stereochemistry of the product of the following reduction?
Question: reduction of cyclohexanones
O
BnO
NaBH4
MeOH?
OH
BnO
2
OH
BnO
1OH
BnO
4
OH
BnO
3
OBnO
t-Bu
H
H
H
axial attack(small hydride source)
✔
OR
R'
Hydride approaches from top face
Hydride approaches from bottom face
RR'
OH
H
RR'
H
OHor
Enantioselective reduction of ketones
§ In general:- Selective synthesis of one enantiomer of secondary alcohol over the other.- Require a chiral reagent in order to introduce a chiral environment for the reduction.- Reagent determines from which face of carbonyl group the hydride approaches.- Highly desirable but difficult to achieve in practice.
top face
- Can the reagent be used in catalytic amounts (i.e. chiral and non-racemic compounds are often more expensive than racemic mixtures, so want to use as little as possible)?
bottom face
NO
H
H3B
Ph
PhB
Me
Ph Me
O
Ph Me
OHON B
Ph
PhB O
HHH
Me
Ph
Me10 mol% catalyst
BH3, THF
Enantioselective reduction of ketones
§ Corey-Bakshi-Shibata (‘CBS’) reduction:- Reduce unsymmetrical ketones to chiral secondary alcohols.- Uses a chiral reducing agent.- CBS reagent used in catalytic quantities (also need BH3 as the stoichiometric source of hydride).- Catalyst binds to both BH3 and substrate in an ordered manner, resulting in high enantioselectivity.
Boron in catalyst is Lewis acidic and activates ketone C=O group
‘CBS’ reagent with BH3 bound
NB: for further explanation see ‘Ox and Red’ textbook p. 69.
Hydride delivered from ‘top’ face
BH3 used to regenerate active catalyst
Binding of BH3 and ketone to exo face of catalyst
Boat-like arrangement controls facial selectivity
(R)-alcohol,97 % ee
e.g.
?Ph
OHON B
Ph
PhB O
HHH Ph
Me
ClCl
Ph
O 1 mol%
BH3, THFCl
NB O
H Ph
Ph
Me
1 mol%
BH3, THF
NB O
H Ph
Ph
MeTBSO
Me
O
Enantioselective reduction of ketones
§ Predict the stereochemistry of the products of these ‘CBS’ reductions:e.g. 1
NB: require good size differentiation between two ketone substituents for high levels of enantioselectivity.
96 % ee
e.g. 2
? TBSOO
N BPh
PhB O
HHH
MeMe OH
Me
94 % de
smaller substituent pseudo axial
§ Reductive cleavage reactions
Reductive cleavage reactions
§ Definition: Break single bonds between carbon and electronegative elements and replace with bonds to hydrogen.
§ Hydrogenolysis- Reductive cleavage of a carbon-heteroatom (C-X) single bond through the addition of hydrogen.- Most commonly used to cleave benzylic (PhCH2–) groups attached to oxygen or nitrogen.- Pd is usual choice of metal catalyst as it reduces the C-X bond faster than the aromatic ring π-bonds. (c.f. Pt, Rh, Ru – see Lecture 1).e.g.
NB: This cleavage mechanism is beyond the scope of the course – for an explanation see ‘Ox & Red in Org Synth’ p. 77
Ease of formation and removal means that the benzyl group is commonly used as a ‘protecting group’ in organic synthesis for more reactive FGs such as amines, amides, alcohols and carboxylic acids.
C X C HREDUCTION
X = N, O, S, halogen
NPh Pd/C (10 mol%), H2 (1 atm.)
EtOHNH
O
O
Ph OH
OPd/C (10 mol%), H2 (1 atm.)
EtOAc
Reductive cleavage reactions
- Reductive dehalogenation:- Csp3-Hal and Csp2-Hal bonds are amenable to hydrogenolysis.- Typically use Pd and a base (NB: produce HX acid as a byproduct which may retard reaction if it isn’t neutralised by base).e.g.
NB: This mechanism is beyond the scope of the course – for an explanation see ‘Ox & Red in Org Synth’ p. 76
Weaker bonds reduced more easily. Most reactive:C-I > C-Br > C-Cl > C-F
NB: This mechanism is beyond the scope of the course – for an explanation see ‘Ox & Red in Org Synth’ p. 76
N
OBn N
BnH
H
HN
OHN
H
H
Na, NH3 (l), t-BuOH, –78 ̊ C
§ Dissolved metal: - An alternative method to cleave N-benzyl and O-benzyl groups.e.g.
BrOMe
OMeOMe
Pd/C (10 mol%)
H2 (1 atm.), Et3NMeOH, r.t., 97%
HOMe
OMeOMe
O H
H H H1). LiAlH4
2). TsClOTs 3). H AlH3Li
THF
Reductive cleavage reactions
§ Deoxygenation – ketones and aldehydes to alkanes: e.g. 1 Reduction to alcohol, make into good LG, displace with hydride.
SN2 reaction(i.e. reactivity 1˚ > 2˚ >> 3˚)
3 steps overall
e.g. 2 Wolff-Kischner reduction:- Use hydrazine (NH2NH2) and KOH.- Requires elevated temperatures (up to 200 ˚C).
NB: mechanism beyond scope of this course, for more details see ‘Ox & Red in Org Synth’ p. 82
§ Which set of reagents and conditions A to E will deliver the reduction product?
Question
HO
O
OBn HO
O?OH
✔
§ Reduction of heteroatom functional groups
Reduction of nitrogen-containing functional groups
§ Reduction of azides (-N3) to amines (-NH2):1). Use H2 & Pd/C to reduce azide.e.g. use Lindlar’s catalyst if alkene present in substrate and don’t want it reduced, chemoselectivity:
OTs N3 NH2NaN3
DMF
Pd/BaSO4H2 (1 atm.)
Pb(OAc)4
alkene intact
azide amineSN2
- Strategic advice: azide is an excellent nucleophile (better than NH3 and easier to handle ), especially for SN2 reaction (i.e. charged, small size, low basicity), so good method of introducing an amine into a molecule is via an azide SN2 displacement reaction then a reduction.
N3 H2N
PPh3, THF
Reduction of nitrogen-containing functional groups
2). Staudinger reduction: - use PPh3 as another chemoselective method for azide reduction in the presence of an alkene.e.g.
NB: This mechanism is beyond the scope of the course, for further details see ‘Ox & Red in Org Synth’ p. 47
NB: Pd/C & H2 would also reduce alkene
amineazide
Reduction of nitrogen-containing functional groups
§ Reduction of nitro groups (-NO2) to amine (-NH2):- H2 & Pd/C is most common method.- Aromatic and aliphatic nitro groups are reduced.e.g.
FYI (but beyond scope of this course): many other reagent combinations will reduce nitro groups(e.g. Fe & HCl; FeCl3 & H2O; SnCl2 & HCl; Zn & HCl).
NB: For details of this mechanism via SET see ‘Ox & Red in Org Synth’ p. 46
CF3
NO2
OEt
CF3
NH2
OEt
H2 (1 atm.), Pd/C (10 mol%)
EtOH, r.t.
A. 1B. 2C. 3
§ Identify the correct product of this reaction.
Question: reduction as a tool in total synthesis
N
OMeO
ON3
N
H2N Me
O
OMe
OMeOMe
OBn
Br
10% Pd/C, H2 (1 atm.)
MeOH:EtOAc?
N
OMeO
OH2N
N
H2N Me
O
OMe
OMeOMe
OBn
Br
1
N
OMeO
OH2N
N
H2N Me
O
OMe
OMeOMe
OH2
N
OMeO
ON3
N
H2N Me
O
OMe
OMeOMe
OH
Br
3
✔
Questions: end of Lectures 5 & 6
O
OPhNaBH4, MgBr2
MeOH
Q1. What is the product of this reduction? Assign the stereochemistry of the new stereocentre and use Newman projections to explain how you arrived at this assignment. What happens if NaCl is used instead of MgBr2?
?
O
THF
OH
*
Q2. What is the stereochemistry at C*? Use conformational drawings to explain your answer.
Na(s-Bu)3BH
OF I
OHF I
Q4. What is the stereochemistry at C*. Use conformational drawings to explain your answer.
*
BH3.THFN
O
H Ph
PhB
Me
Catalyst A (1 mol%)
Q3. Provide reagents and conditions to complete the following transformation in 2 steps.Br
NO2
O
NH2
Reagents?
2 steps
A
Heterocyclic Chemistry
Aims for this session
§ Heterocycles:- Definition.- Why heterocycles are important.- Nomenclature of common heterocycles.- Recap. of key reactions of ketones, enols, imines and enamines.- Recap. of aromaticity.- Look at pyridine: structure and reactivity (how does it compare to benzene?).- Look at pyrrole: structure and reactivity (how does it compare to benzene?).- Summary & comparison of pyrrole and pyridine structure & reactivity.- Consider epoxide ring opening under acidic and basic conditions.- Synthesis of pyridine and pyrrole.
What is a heterocycle?
§ A cyclic system (or ‘ring’) that contains one or more heteroatoms.- Can be aromatic or non-aromatic.e.g.
§ Heterocycles are an extremely common motif.- Found in almost half of all known organic compounds.- Key constituents of natural products, DNA, amino acid proline (i.e. proteins), medicines, agrochemicals, materials and dyes.e.g.
Caffeine
N
NN
N
NH2
O
OH
HO
N
HN
MeOSO
NMe
Me OMe
O
N SMe
Me
NH
NH
NO2
Me
N
MeO
HO N
N
N
N
N
MeO
MeO Me
N
NMe
NicotineOmeprazole Ranitidine
Deoxyadenosine
Quinine
§ How many heterocycles are in this pharmaceutical compound? (NB: count any fused rings separately)
A. 0B. 1C. 2D. 3E. 4F. 5
NN
Me
SO
O
OEt
N
HNN
N
O
i-Pr
Me
Sildenafil(‘Viagra’)
Question: identify the heterocycles
✔
Top 20 global best-selling drugs in 2009
NB: Drugs containing heterocycles
indicated in bold
HN O
HN O
HN
O
Heterocycles: nomenclature
§ Five membered rings with one heteroatom:- NB: we will only study the highlighted compounds in more detail.e.g.
pyrrole furan pyrrolidine tetrahydrofuran (‘THF’)
thiophene(X = S)selenophene (X = Se)
etc.
indole benzofuran (X = O)benzothiophene (X = S)
etc.
isoindole (X = N)isobenzofuran (X = O)
etc.
X HN X
X
pyrrolidinone
Heterocycles: nomenclature
§ Five membered rings with more than one heteroatom:- NB: we will not study these heterocycles in detail, but you should be aware of these compounds.e.g.
imidazole pyrazole 1,2,3-triazole tetrazole
N
HN
N
HN
NN
HN
NN
HN
N NN
HN
O
HN
S
HN
O
HN
S
HN
O
N
1,2,4-triazole
oxazole thiazole isoxazole benzoxazoleisothiazole
HN O
HN
HN O
Heterocycles: nomenclature
§ Six membered rings with one heteroatom:- NB: we will only study the highlighted compounds in more detail.e.g.
§ Six membered rings with more than one heteroatom:- NB: we will not study these heterocycles in detail, but you should be aware of these compounds.e.g.
pyridazine pyrimidine 1,3,5-triazine quinazoline
piperazine morpholine 1,4-dioxane
NB: aromatic
NB: non-aromatic
NN
N
N
N N
N
N
N
NH
HN
O
HN
O
O
(C, T & U nucleobases)
(common heterocycle in pharmaceuticals)
Heterocycles: nomenclature
§ Beyond five and six membered rings:- NB: we will only study the highlighted compound in more detail.e.g.
epoxide aziridine oxetane azetidine
Penicillins
O
O
OH OH OH OH OH
HO
OH
O NHO NH NH
O
N
HNPhO
O
O
S
CO2H
MeMe
(+)-Roxaticin
β-lactam
NB: macrocycle
R
OH
R
ONi-Pr2Li
R
OH
R
OH
R
OH
HH OH2 H2O
Heterocycles: key reactions (recap.)
§ Enol formation:- NB: acidic conditions.e.g.
protonation
enolketone
deprotonation
NB: consider pKa of α-proton
§ Enolate formation:- NB: basic conditions.e.g.
enolateketone
NB: consider pKa of α-proton
NB: for a comprehensive list of pKa values, see: http://evans.harvard.edu/pdf/evans_pka_table.pdf
NB: need a strong base to effect complete
irreversible deprotonation of a ketone (i.e. use LDA)
R
OH
R
OH
R
OH
HH OH2 H2O
§ What are the pKa values of the two α-protons highlighted below?
A. They are both the sameB. 43 and 26C. 26 and –6D. 43 and –6
Question: pKa (again!)
✔
Heterocycles: key reactions (recap.)
§ Formation of imines:- NB: typically require catalytic acid.e.g.
imineketone
§ Enamine formation:- NB: occurs under acidic or basic conditions.e.g.
enamineimine
c.f. enol and enolate formation
R
O
R
NRRNH2
cat. H++ H2O
R
NRH
R
HNRtautomerisation
NB: enamines react with electrophiles on carbon and not nitrogen
§ Mannich reaction:- NB: electrophile is imine rather than aldehyde in mechanistically similar Aldol reaction.- Amine is poorer LG (c.f. alcohol in Aldol reaction), therefore intermediate does not eliminate.e.g.
amine product imine
§ Conjugate addition:- NB: orbital-controlled [1,4]-nucleophilic addition.- NB: called a “Michael addition’ when carbon-based nucleophile such as enol or enolate.e.g.
enolate intermediateenone
NB: favoured when using ‘soft’ nucleophiles (i.e. uncharged and/or diffuse orbitals)
enol
R
OH NR
PhH R
O
Ph
cat. H++
NHR
Aromaticity (recap.)
§ Hückel’s Rule:e.g. aromatic compounds
Planar structures which have a cyclic array of [4n + 2] delocalisable π-electrons
PYRROLEisoelectronic and isolobal
with cyclopentadiene anion
2π electrons(3 x p orbitals)
NHN
6π electrons(5 x p orbitals)
6π electrons(6 x p orbitals)
6π electrons(7 x p orbitals)
6π electrons 6π electrons
PYRIDINEisoelectronic and isolobal
with benzene
Heterocycles
Question: heteroaromatic?
§ Identify the heterocycle(s) that are not aromatic
A. AB. BC. CD. DE. EF. F
HN O
A
O
HN
B
N
C
D
HN
F
HN
E
✔
HC H
Pyridine
§ Structure:- Isoelectronic with benzene (i.e. 6π electrons, one in each of 6 parallel p orbitals).e.g.
one electron in each p orbital
NN
one electron in each p orbital
sp2 hybridised carbon: one electron in sp2 orbital σ-bonding to H, orthogonal to
plane of π-system
sp2 hybridised nitrogen: lone pair of electrons in sp2 orbital, orthogonal to plane of π-system
pyridine
benzene
Nitrogen atom has major effect on reactivity and properties of pyridine w.r.t benzene
N
§ 1H NMR spectroscopic chemical shifts (i.e. δH in ppm):- i.e. how deshielded is each environment?e.g.
Pyridine: is it really aromatic?
§ 13C NMR spectroscopic chemical shifts (i.e. δC in ppm):- i.e. how deshielded is each environment?e.g.
128.5149.8
123.6135.7
N
H HH
H7.2
8.5
7.1
7.5δH in aromatic region of 1H NMR spectrum
(N.B. more deshielding in pyridine than benzene due to electronegative N atom)
YES, pyridine is aromatic
δC in aromatic region of 13C NMR spectrum (N.B. more deshielding in pyridine than benzene due to electronegative N atom)
§ In which hybrid atomic orbital is the nitrogen lone pair located?
A. pB. spC. sp2
D. sp3
Question: hybridisation states in heterocycles
N
✔
§ In which hybrid atomic orbital is the nitrogen lone pair located?
A. pB. spC. sp2
D. sp3
Question: hybridisation states in heterocycles
NH
✔
§ Lone pair reactivity:- Basic (i.e. can be easily protonated).- Nucleophilic (i.e. can be alkylated or acylated).- N-Oxidation (forms N-oxides, see later).
Pyridine: general reactivity trends
§ Benzene-like reactivity:- Attack on π-system.- Electrophilic aromatic substitution (SEAr).
§ Imine-like reactivity:- Susceptible to nucleophilic attack (e.g. displacement of halo-pyridines) due to electron-deficiency.
N E H X R
O
RXe.g. E =
acylationalkylationprotonation
N
E
N
Cl Nu
N
Cl Nu
N
Nu
NB: behaves like an imine
NB: SEAr is less likely than with benzene due to less electron-density in pyridine ring
(N.B. remember deshielded signals in NMR)
NN + H
HH HN
NH
+ H
§ Basicity:- Consider pKa of the conjugate acid (i.e. pKaH)…
Pyridine: lone pair reactivity
NB: for a recap. of pKa, see ‘OC’ p 197.
- The weaker the base, the stronger its conjugate acid (i.e. low pKaH).
- The stronger the base, the weaker its conjugate acid (i.e. high pKaH).
- Pyridine lone pair in sp2 orbital, hence weak base.
PYRIDINEpKaH 5.2
PIPERIDINEpKaH 11.5
(Stronger base)
NH
e.g.
- Piperidine lone pair in sp3 orbital, hence stronger base (i.e. greater ‘p’ character so higher energy hybrid atomic orbital and therefore more reactive).
N N NMe Cl
1 2 3
§ Which pyridine is the strongest base?
A. 1B. 2C. 3
Question: pyridine basicity
pKaH 5.2 pKaH 6.8 pKaH 0.7
✔
NN N
90˚
§ Nucleophilicity:- NB: lone pair cannot be delocalised around the ring (i.e. sp2 orbital orthogonal to the p orbitals).
Pyridine: lone pair reactivity
- Typically, the more basic a pyridine the more nucleophilic it becomes, as lone pair is raised in energy.- Except when sterically crowded.
X
sp2 orbital
p orbitals
- Hence lone pair is a good nucleophile as it is available (e.g. readily alkylated or acylated).e.g. acylation
N
Cl R
O
N R
O
Cl
N2,6-di-tert-butyl pyridine
(good base but poor nucleophile as only small
electrophile (i.e. proton) can reach lone pair)
acylated
§ N,N-Dimethyl-4-amino pyridine (DMAP) is a useful pyridine-based catalyst for acylation reactions. What type of catalysis is this an example of? (hint: what is the pyridine doing and why might DMAP be better than pyridine?)
A. Acid catalysisB. Base catalysisC. Nucleophilic catalysis
Question: DMAP as a catalyst
N
Cl Ph
O
N Ph
O
ClMe2N
Me2NOHO
PhO
N
Me2N+
DMAP
✔
§ Electrophilic aromatic substitution (SEAr):- Readily occurs on benzene, but much less favourable on unsubstituted pyridine.- Electronegative nitrogen atom lowers energy of p orbitals within pyridine ring (w.r.t. benzene), hence pyridine ring less nucleophilic (i.e. lower HOMO) but more electrophilic (i.e. lower LUMO).- Pyridine: nucleophilic lone pair leads to reaction with E+ on nitrogen rather than carbon.e.g.
Pyridine: benzene-like reactivity
nitrogen makes pyridine ring electron deficient (i.e. less nucleophilic).
- Yields from SEAr reactions are typically poor:
e.g. 1. Friedel-Crafts reactions normally fail.
e.g. 2. Nitration with HNO3 + H2SO4 at 300 ˚C gives < 5% of 3-nitropyridine.
e.g. 3. Halogenation with Cl2 + AlCl3 at 130 ˚C gives only moderate yields of 3-chloropyridine.
NB: benzene reacts readily under these conditions (see 1st Year lecture notes).
NN
EE
N
NO2
N
Cl
pyridines preferentially react with electrophiles on nitrogen
N N N N N
δ+
δ+δ+
δ-
2
4
6
§ Electrophilic aromatic substitution:- Electronegative nitrogen atom withdraws electron density; C-2, C-4 and C-6 most affected.e.g.
Pyridine: benzene-like reactivity
overall
- C-3/5 substitution is more favourable (but still unlikely to occur, see previous slide).e.g.
- C-2/6 or C-4 substitution results in δ+ localised on nitrogen atom (highly disfavoured).e.g.
N N N NE
EH
EH
EH
N N N N
EEH H
EH
E
N NMe2N Me2N
NO2HNO3
H2SO4
§ Electrophilic aromatic substitution:- Does occur if electron donating group present (i.e. increases electron density within heterocycle). e.g.
Pyridine: benzene-like reactivity
nitrogen lone pair is easily oxidised
- If no activating groups present, can oxidise pyridine into pyridine N-oxide.e.g.
Pyridine N-oxides readily undergo SEAr
N N
mCPBA
O
NB: stable solid
NB: This mechanism is beyond the scope of the course
§ Electrophilic aromatic substitution:- Negative charge on oxygen delocalised into pyridine π-system (i.e. makes ring more electron rich).- Reaction with electrophiles occurs at C-2 or C-4 (but mainly at C-4 due to sterics).e.g.
NO2
N
HNO3
O
H2SO4NO
H NO2
NO
NO2
Pyridine: benzene-like reactivity
C-4 attack
- Easily cleave N-oxide with P(III) compounds (e.g. P(OMe)3) to regain pyridine.e.g.
NN
P(OMe)3
O
NO2NO2
THFPO
OMeMeOOMe
+
Formation of strong P=O double bond drives reaction
NB: This mechanism is beyond the scope of the course, for further details see ‘OC’ textbook p 730)
Rearomatisation
Reduction
§ Nucleophilic substitution (SNAr):- C-2 and C-4 halo-pyridines react easily with nucleophiles (i.e. electron poor ring readily accepts e-).- C-3 halo-pyridines less reactive (i.e. negative charge cannot be delocalised onto nitrogen).e.g.
Pyridine: imine-like reactivity
‘Activated’(i.e. N-protonated)
- Unsubstituted pyridine will react with strong nucleophiles (e.g. LiAlH4, Grignard reagents etc.).- Activated pyridine (e.g. protonated) will react with weaker nucleophiles (e.g. NaBH4, -CN).e.g.
NB: 2-chloropyridine is 3 x 108 times more reactive than chlorobenzene
N Cl NOMe
Cl N OMeMeOH
NaOMe
NNH
H CNNH
CN
N
Me Me
MeN
Me Me
MeO
N
Me Me
MeO
NO2
N
Me Me
MeO
OMe
H2O2
NaOMeMeOH
N
HN
MeOSO
N
Me OMe
Me
H2SO4
HNO3
steps
§ Synthesis of Omeprazole:- $Billion-selling drug: proton pump inhibitor to treat stomach ulcers.e.g.
Pyridine reactivity in action
oxidation
N-oxide
SEAr
nitration
SNAr
Omeprazole(proton pump inhibitor)
- Electronic structure of pyridine.
- Difference between pyridine and benzene w.r.t. structure & reactivity.
- Basicity: strength relative to non-aromatic amines (pyridine is weaker base; effect of substituents).
- Nucleophilicity: reacts through lone pair.
- SEAr: position of substitution; rate of reaction w.r.t. benzene; why unreactive?- Pyridine N-oxides: reactions, methods for synthesis & removal.
- Nucleophilic attack: easy with halo-pyridines & activated pyridines.
Pyridine: summary
C H
Pyrrole
§ Structure:- Isoelectronic with cyclopentadiene anion (i.e. 6π electrons, one in each of 4 parallel p orbitals on the carbons and a negative charge or lone pair in a parallel p orbital located on final C or N atom).e.g.
one electron in each neutral
carbon p orbital
one electron in each carbon p
orbital
sp2 hybridised carbon: negative charge in p
orbital, parallel to plane of π-system
sp2 hybridised nitrogen: lone pair of electrons in
p orbital, parallel to plane of π-system
pyrrole
cyclopentadieneanion
NH
N H
Pyrrole lone pair required for aromaticity
Pyrrole vs pyridine
§ Structure:- Lone pair on pyridine in sp2 orbital (mild base) c.f. lone pair on pyrrole in p orbital (non-basic).e.g.
one electron in each carbon p
orbital
sp2 hybridised nitrogen: lone pair of electrons in
p orbital, parallel to plane of π-system
pyrrole
Location of nitrogen lone pair has major effect on reactivity and properties of pyrrole w.r.t pyridine
NH
N H
NN
one electron in each p orbital
sp2 hybridised nitrogen: lone pair of electrons in sp2 orbital, orthogonal to plane of π-system
pyridine
mild base
non-basic
N
NH
N N H
Imidazole: more than one type of nitrogen in the ring
§ Structure:- Imidazole has 6π electrons, hence one nitrogen is pyridine-like and one nitrogen pyrrole-like.e.g.
pyrrole-like nitrogen
imidazolepyridine-like nitrogen
non-basic
mild base
N
§ 1H NMR spectroscopic chemical shifts (i.e. δH in ppm):- Pyrrole is electron rich from lone pair donation, hence greater shielding.e.g.
Pyrrole: NMR spectroscopic properties
§ 13C NMR spectroscopic chemical shifts (i.e. δC in ppm):- Pyrrole is electron rich from lone pair donation, hence greater shielding.e.g.
128.5149.8
123.6135.7
N
H HH
H7.2
8.5
7.1
7.5
NH
NH
H
H6.2
6.5≈10
107.7
118.0
NB: not as big a difference in δH around ring c.f. pyridine
electron rich c.f. benzene electron poor c.f. benzene
§ In which (hybrid) atomic orbital is the nitrogen lone pair located?
A. pB. spC. sp2
D. sp3
Question: hybridisation states in heterocycles
NH
✔
§ In which (hybrid) atomic orbital is the nitrogen lone pair located?
A. pB. spC. sp2
D. sp3
Question: hybridisation states in heterocycles
NH
✔
NH
NH
HH
H
§ Lone pair reactivity:- Non-basic (i.e. pKaH –3.8), in strong acid pyrrole protonates on carbon not nitrogen (c.f. pyridine).- Nucleophilic on carbon rather than nitrogen (i.e. lone pair delocalised into ring).- Does not undergo N-oxidation (c.f. pyridine).
Pyrrole: general reactivity trends
§ Electrophilic substitution:- Electron rich ring is highly reactive to SEAr (more so than benzene).
§ Nucleophilic substitution:- Electron rich ring unreactive towards nucleophilic attack.- Require electron withdrawing substituent to activate ring.
pKa –3.8
NB: SNAr on an activated pyrrole is beyond the scope of the course, for an example see ‘OC’ textbook p 738.
NH
NH
EH
E
NH
NH
E EX
nucleophilic on carbon
§ Basicity:- Pyrrole lone pair lowered in energy via delocalisation around ring, hence non-basic.- Pyrrolidine lone pair in sp3 orbital, hence stronger base.
Pyrrole: lone pair reactivity
NH
NH
HH
+ H
NN + H
HH H
PYRROLEpKaH –3.8
PYRROLIDINEpKaH 11.0
N H Ndeprotonation
NH
NBase
§ Acidity:- Pyrrole has N-H present (c.f. no N-H in pyridine).- Weakly acidic (i.e. N lone pair delocalised around ring so N already less electron dense & is therefore better able to stabilise an additional charge following deprotonation).e.g.
Pyrrole: lone pair reactivity
pKa 16.5
N-H bond uses nitrogen sp2 orbital
negative charge in nitrogen sp2 orbital, whilst N lone pair is still delocalised into aromatic ring
NH
NBase
- Pyrrolidine is much less acidic.e.g.
negative charge in nitrogen sp3 orbital (N.B. higher energy c.f. sp2), plus lone pair fully localised on nitrogenpKa 44.0
NB: sp2 orbital orthogonal to π-system so no
resonance stabilisation (solely provided by
electronegative nitrogen)
§ SEAr:- Pyrrole is electron rich, reacts readily with electrophiles (NB: more reactive than benzene).- Nitrogen lone pair delocalisation increases electron density at each carbon in ring.e.g.
Pyrrole: electrophilic substitution
- Is substitution at C-2 or C-3 favoured?
NB: adding another heteroatom into the ring deactivates towards SEAr as this is ‘pyridine-like’ (i.e. electron-withdrawing).
overall
NH
NH
NH
NH
NH
NH
δ+δ-
δ-δ-
δ-
N
HN
O
HN
§ SEAr at C-2 position:- Generally favoured.- Cation resulting from electrophilic attack is more stabilised (i.e. 3 resonance forms).- Linear conjugated intermediate (i.e. both double bonds conjugated with N+).e.g.
Pyrrole: electrophilic substitution
linear conjugated intermediate
NH
NH
EH
E NH
EH
NH
EH
NH
E
– H
C-2 product
3 resonance forms of intermediate
NH
NH
E NH
NH
– H
E
EH
EHE
§ SEAr at C-3 position:- Less favoured.- Cation resulting from electrophilic attack is less stabilised (i.e. 2 resonance forms).- Cross conjugated intermediate (i.e. only one double bond conjugated with N+, less stable than linear conjugated intermediate).e.g.
Pyrrole: electrophilic substitution
cross conjugated intermediate
C-3 product
2 resonance forms of intermediate
O
H NMe2
ClPO
ClCl O
H NMe2
PO
ClCl Cl
H Cl
NMe2
ClP
O
O+
NH
NH
POCl3 DMF
H
O
H Cl
NMe2+
NH
H Cl
NMe2
NH
HH
NMe2 NH
NMe2
H
NH
H
O
e.g. Formylation in the absence of a strong Lewis acid: Vilsmeier-Haack reaction.
Pyrrole: electrophilic substitution
SEAr
NB: formation of intermediate driven by strong P=O bond formed
iminium ion hydrolysis
(e.g. Na2CO3, H2O)
Anton Vilsmeier
attack at C-2(via linear conjugated intermediate)
DMF
NH
H
ONH
NH
NH
OH
NMe2 NMe2
1 2
3 4
§ What is the product of this Mannich reaction?
A. 1B. 2C. 3D. 4
Question: electrophilic substitution of pyrrole
NH
Me2NHH H
NMe2+OH
H? ✔
§ Pyrrole anion undergoes N-alkylation and N-acylation:- Anion in sp2 orbital (90˚ to plane of π-system), so cannot overlap with ring & subsequent reaction occurs at nitrogen.- 2 steps: (1). Deprotonate pyrrole (NB: pKa 16.5), (2) Add electrophile.
Pyrrole: reaction at nitrogen
NB: without deprotonation would get reaction solely at carbon
N-acylation
N HNaH
NCl
O
PhN
O
Ph
N HLDA
N N MeMe I
N-alkylation
e.g. 1.
e.g. 2.
N
N.B.
sp2
Pyrrole: summary
- Electronic structure of pyrrole.
- Difference between pyrrole, pyridine and benzene w.r.t. structure & reactivity.
- Recognise pyrrole-like nitrogens and pyridine-like nitrogens in heterocycles.
- Basicity: pyrrole is not basic; protonate at C-2 in strong acid (leads to polymerisation).
- Acidity: pyrrole is weakly acidic; comparison to non-aromatic amines; electrophilic attack at nitrogen through pyrrole anion.
- Nucleophilicity: nitrogen lone pair delocalisation makes pyrrole ring electron rich; high nucleophilicity (at carbon).
- SEAr: position of substitution (i.e. C-2); rate of reaction w.r.t. benzene; why more reactive?- Vilsmeier-Haack reaction; Mannich reaction.
- Nucleophilic attack: unreactive, require electron withdrawing substituents to activate pyrrole ring.
Pyrrole & pyridine: schematic reactivity summary
N H Base (very easy)
E+ (very easy)
PYRROLE
H
E+ (easy)
Nu-
(difficult)
Base
(difficult)
BENZENE
N HNu-
(easy)
Base
(difficult)
E+ (difficult)
PYRIDINE
N E+ (very easy)
PYRROLE ANION
NH Nu- (very
easy)
ACTIVATED PYRIDINE
§ Epoxide ring-opening:- Requires either a good nucleophile or acid catalysis to react well.- SN2 mechanism, hence inversion of stereochemistry at reaction centre (i.e. stereospecific).e.g.
Useful reactions of another heterocycle: epoxides
NB: inversion of stereochemistry at
reacting carbon centre
SN2
- But… unsymmetrical epoxides lead to issues of regioselectivity (i.e. which end of epoxide reacts?).e.g.
O
OH
Nu
1). Nu
2). H3O
O
OH
Nu
1). Nu
2). H3O
Nu
OH
or
attack at most hindered end
attack at less hindered end
*
NB: acidic work-up to protonate alkoxylate
§ Epoxide ring-opening in base:- Attack less hindered end of epoxide (i.e. minimise steric interactions between Nu- and E+).- Pure SN2 mechanism, inversion of stereochemistry at reaction centre (i.e. stereospecific).e.g.
Useful reactions of other heterocycles
NB: epoxide oxygen is a poor LG (i.e. RO–), so needs a strong Nu- for SN2 reaction
SN2 reaction proceeds via a pentacoordinate T.S. hence minimise steric interactions
attack at less hindered end
O
OH
OMe
OMe
OH
orOMeNa
MeOH
NB: inversion
*
NB: for further details on epoxide ring opening, see ‘OC’ textbook p 438
O
OH
OMe
MeOH, HClOMe
OH
or
H
§ Epoxide ring-opening in acid:- Attack more hindered end of epoxide as epoxide oxygen protonated in acid, so build up of positive charge in T.S. stabilised at most substituted end.- ‘Loose’ SN2 transition state, still inversion of stereochemistry at reaction centre (i.e. stereospecific).e.g.
Useful reactions of other heterocycles
NB: epoxide oxygen is protonated, so it is a good LG (i.e. ROH) and does not
need a strong Nu– for SN2 reaction
loose SN2 transition state
attack at more hindered end
NB: inversion
*
NB: for further details on epoxide ring opening, see ‘OC’ textbook p 438
OH
OH
MeOH
(+)(+) O
H(+)
(+)
OMeH (+)
OH
OMeH
HH H
attack site best able to stabilise partial +ve charge
§ Identify the correct product from this epoxide ring opening.
A. 1B. 2C. 3D. 4
Question: epoxide opening
O MeOH, HCl?
OH
OMe
OH
OMe
OMe
OH
OMe
OH
1 2
3 4
✔
NR R
R'HN
R R
R'O R R
O O+ R'NH2
§ In general:- Based on simple carbonyl chemistry (see previous recap. of key reactions).- Can disconnect the main bonds to reveal simple linear precursors.
Synthesis of heterocyclic rings
e.g. Pyrrole
1,4-dicarbonyl compound
pyrrole
1,5-dicarbonyl compound
pyridine
NR R H2NRR
ORR
O O
+ NH3
e.g. Pyridine
NB: for more info on 5- and 6-membered ring synthesis beyond the scope of this course, see ‘OC’ textbook p 785 & 786.
RO
RO
R'NH2
NR RAcOH
R'R' = H, alkyl, aryl
Cl
NH2 O
O
H
Cl
N
NH OAr H
NAr
OH
H
NAr
OH2
± H
AcOH
NAr
H
§ Paal-Knorr synthesis:- Uses a 1,4-dicarbonyl compound and either ammonia or a 1˚ amine.- Requires weak acid (NB: strong acid will lead to formation of the corresponding furan).
Synthesis of pyrroles
hemiaminal
1,4-dicarbonyl compound
e.g.
Ludwig Knorr
- Mechanism:e.g. NB: enamine
formation
cyclisation
NB: overall, lose 2 moles of water
pyrrole
NB: enamine formation
§ Identify the correct pyrrole product from this reaction
A. 1B. 2C. 3D. 4
Question: synthesis of pyrroles
O
O
H2N AcOH+ ?
NN
N
N
1 2
3 4
✔
R
HOEtO2C
O
CO2Et
ONH3
NH
EtO2C CO2EtR
N
EtO2C CO2EtR
pH 8.5
EtOH
[O]oxidation
§ Hantzsch synthesis:- 4 component reaction (NB: 3 different substrates).- Initially affords a 1,4-dihydropyridine, but readily oxidises in air to give the pyridine.- Requires basic conditions.
Synthesis of pyridines
Arthur Rudolf Hantzsch
NB: aldehyde provides extra carbon for pyridine ring (c.f. pyrrole synthesis)
pyridine1,4-dihydropyridine
e.g.
NB: for further details of the Hantzsch pyridine ring synthesis, see ‘OC’ textbook p 763.
R
HOEtO2C
O
EtO2C
O
O
REtO2C
O
OH
REtO2C
O
RH
EtO2C
O
OH
RH
H OEt
EtO
EtO2C
O
RCO2Et
O
EtO2C
O
R
O
EtO2C
O
R
O
CO2Et CO2Et
Synthesis of pyridines
α,β-unsaturated product
conjugate addition
NB: for further details of the Hantzsch pyridine ring synthesis, see ‘OC’ textbook p 763.
§ Mechanism:- Condensation with aldehyde to form α,β-unsaturated product (c.f. aldol reaction).
- α,β-Unsaturated product is good electrophile (i.e. 1,4-acceptor), so Michael reaction now occurs.
NB: 2nd equivalent of enolate used
1,5-dicarbonyl compound
E1cb
Synthesis of pyridines
enamine formation
NB: The mechanism for oxidation is beyond the scope of this course, but for further details see ‘OC’ textbook p 764
§ Mechanism (continued):- Enamine formation, then intramolecular cyclisation of nitrogen onto other ketone.
- Final oxidation step is facile (energetic preference to become aromatic).- Occurs in air, or with chemical reagents such as DDQ.
Dichlorodicyanoquinone(DDQ)
NH
EtO2C CO2EtR
N
EtO2C CO2EtR
oxidationO
O
Cl
Cl
CN
CN
NH
EtO2C CO2EtR
EtO2C
O
R
O
NH3
EtO2C
O
R
H2N
CO2Et
NH
EtO2C CO2EtR
CO2EtH
enamine formation iminium ion formation
1,4-dihydropyridine
Summary of the lecture
- Epoxides open via an SN2 reaction with inversion of stereochemistry at reacting centre.
- Under acidic conditions, epoxides open at the more hindered end (i.e. loose SN2 T.S.).
- Under basic conditions, epoxides open at the less hindered end (i.e. minimise steric crowding in T.S.).
- Synthesise pyrroles and pyridines using standard carbonyl chemistry (e.g. aldol reaction, conjugate addition, imine & enamine formation etc.).
- Can use Paal-Knorr pyrrole synthesis to make pyrroles from 1,4-dicarbonyl compounds.
- Can use Hantzsch pyridine synthesis to make pyridines via a 4 component coupling reaction involving 1,5-dicarbonyl intermediates.
Questions: end of heterocyclic chemistry lectures
Q1. In what hybrid atomic orbital is the lone pair of the nitrogen atoms in both pyrrole and pyridine?
Q3. Identify the correct product A to E from the following nitration, explaining the regioselectvity.
Q4. What is the product of this ring-opening? Include a reaction mechanism with curly arrows to explain the regio and stereochemistry of the product.
HNMe 1. POCl3
2. Na2CO3, H2OH
O
NMe2
?+
Q2. Identify the product from the following Vilsmeier-Haack reaction and draw a curly arrow mechanism to describe its formation. Explain the regioselectivity observed here.
N OMe
HNO3
H2SO4?
N OMeN OMe
N OMe N OMeN OMe
NO2NO2 O2N
O2N NO2
A B C D E
O
i-Pr
MeEt
HCl, MeOH?
Exam Questions
§ Representative questions
This a relatively new course and hence there are only a few past exam papers and one mock exam paper (see QMplus) that relate specifically to this course (and the exclusion of any other). However, much of the material is what I would class as core material that will crop up across a range of examination questions.
With regards to older lecture courses, particular attention should be paid to parts of the SBC703 course, ‘Synthesis of Pharmaceutically Active Molecules’.
Access to past papers can be made via the library webpage.
For general practice, many textbooks contain questions that will relate to the topics covered.
Make sure to revise workshops, quizzes and end-of-lecture questions.