<<

CHE202 Structure & Reactivity in : Reduction Reactions and Heterocyclic Chemistry Semester A 2016

Dr. Chris Jones [email protected]

Office: 1.07 Joseph Priestley Building

Office hours: 9.30-10.30 am Monday 1.30-2.30 pm Thursday (by appointment only) Course structure and recommended texts

§ Coursework: Semester A – week 9 5% (‘Coursework 3’) Semester A – week 11 5% (‘Coursework 4’)

§ Test: Semester A – week 12 15% (‘Test 2’)

§ Recommended text books:

‘Organic Chemistry’, Clayden, ‘Oxidation & Reduction in ‘Heterocyclic Chemistry’, Greeves & Warren, OUP, 2012. Organic Chemistry’, Donohoe, Joule & Mills, Wiley, 2010. OUP, 2000.

Don’t forget clickers Overview of Reduction Chemistry lecture material

§ Reduction: - Definition (recap.) - Reduction of -carbon double (C=C) and triple (CΞC) bonds - Heterogeneous - 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 and

- Reductive cleavage reactions - Hydrogenolysis of benzyl and allyl groups - Dissolved metal reduction - Deoxygenation reactions

- Reduction of heteroatom functional groups e.g. azides, nitro groups, N-O bond cleavage

Selectivity is a key theme of this course Reduction: definition (recapitulation)

§ Reduction of an organic substrate can be defined as:

- The concerted addition of hydrogen. e.g.

O H2 (g) H O H Catalytic hydrogenation (e.g. H2(g) & Pd)

- The ionic addition of hydrogen NB: Dr. Lebrasseur’s & Dr Bray’s carbonyl lectures. e.g.

O "H " then H addition then protonation H O H (e.g. LiAlH4 then w/up)

- The addition of electrons. e.g. H H

electron transfer 2nd ET Dissolved alkali metals (e.g. Na in liquid NH ) + 2H 3 H H one electron added

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 Question: reduction or not?

§ Which of these transformations represent reductions? A. All of them ✔ 1 B. 3, 4 and 5 C. 1, 3, 4 and 5 D. 1, 3 and 5 E. 1, 4 and 5 NO2 NO2 NO2 NH2

2 3

N N N N O O O

N 4 NH2

CO Me CO Me 2 5 2 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).

H H O H O O enzymatic H OH NH2 + + NH2

CO2H reaction CO2H N N

NADH pyruvic acid lactic acid NAD+

- in pharmaceutical synthesis. e.g. O O O Me Me Me N N N H2N reduction H2N OEt HN N N N N O2N H2N n-Pr n-Pr n-Pr

nitro group group O2S Sildenafil N N Me § 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.

H H Metal Catalyst H

H2 (g) H

- Hydrogenation has a different mechanism of reduction compared to hydride reducing agents (e.g. NaBH4), therefore different chemoselectivity is often observed. e.g.

O O Pd/C (10 mol%) H H H2 (g) not reduced

(c.f. NaBH4) NB: 10 mol% = 0.1 molar equivalent (or 10:1 substrate:catalyst molar ratio) 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.

§ Reduction of : - 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. Ph Ph PtO2 (cat.) (R) Ph Ph (S) H H2 (1 atm.) H (Z)- syn-addition (i.e. H added to same face of C=C bond)

PtO2 (cat.) (S) Ph Ph Ph (S) Ph H H2 (1 atm.) H (E)-alkene

- Need a polar reaction solvent to dissolve sufficient hydrogen (e.g. , , acetic acid). 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):

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 is added sequentially onto both . v). Reduced product can dissociate from catalyst surface. syn-addition R R R R H2 H H

ADDITION ADSORPTION then DISSOCIATION R R R R R R R R H H H H R R H H R R metal catalyst surface COORDINATION ADSORPTION

- 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). - Carbocyclic and heterocyclic aromatic rings amenable.

HO O HO O OH OH

Carbocyclic ring Pt/C (cat.), H (4 atm.) NB: carboxylic O 2 O acid untouched OH AcOH OH

O O NB: increased

H2 pressures

H 10% PtO2 (10 mol%) H2 (5 atm.), HBr, r.t. Heterocyclic ring N NB: syn-addition N 74% of hydrogen Bun Me n Me Bu

(±)-monomorine Question: reduction of carbon-carbon double and triple bonds

§ Which of these structures 1 to 5 is the correct reduction product?

A. 1 O B. 2 ✔ Pd/C (10 mol%), H2 (1 atm.) ? C. 3 MeOH, r.t. D. 4 E. 5

OH O O

H H H H H H 1 2 3

O O

H H H H 4 5 Reduction of carbon-carbon double and triple bonds

§ Reduction of to : - Standard heterogeneous hydrogenation results in complete reduction to alkanes. e.g.

H H O Pd/C (10 mol%) Ph OMe Ph 2 moles of H added overall OMe H2 (1 atm.), MeOH H H O 2

1 mole of H2 added Ph CO2Me

H H

alkene intermediate

- Alkynes to alkenes, is it possible?

- Require chemoselectivity to differentiate between reduction of and alkene.

- Require control over cis- or trans- geometry. 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. Herbert Lindlar

O Ph CO Me Pd/CaCO3 (10 mol%) 2 Solid support – CaCO3 or BaSO4 Ph H2 (1 atm.), quinoline, Pb(OAc)4 – deactivates catalyst OMe H H Pb(OAc)4, EtOAc Quinoline – competitive cis-(Z)-alkene binding to catalyst surface

e.g. in the synthesis of a complex molecule

O N O Z

Pd/CaCO3 (10 mo%) O O H2 (1 atm.), quinoline TBSO O Pb(OAc)4, hexane, r.t. TBSO R R O 86%

- Mechanism: two hydrogens added to same face of alkyne, leading to syn-addition. Reduction of carbon-carbon double and triple bonds

§ Homogeneous hydrogenation: - Metal- complex is soluble in reaction medium. - Phosphines are common (i.e. good electron donors).

Sir Geoffrey Wilkinson (Nobel Prize in Chemistry 1973) 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. , carboxylic , , nitriles, and nitro groups all inert to these conditions.

Least hindered alkene

O O (Ph P) RhCl (cat.), H (1 atm.) O 3 3 2 O O /EtOH O 95%

NB: Heterogeneous hydrogenation (e.g. H2, Pd/C) is non-selective and leads to over reduction

NB: This mechanism is beyond the scope of the course – for a full explanation see ‘Ox & Red in Org Synth’ p. 54 Reduction of carbon-carbon double and triple bonds

§ Enantioselective hydrogenation: H2 (FYI only, beyond scope of course)

- If metal-ligand complex (MLn) is chiral, then possible ML + to control to which face of alkene H2 is delivered. n WHICH FACE? - Discrimination between enantiotopic faces of alkene can lead to single enantiomer of product.

H2

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).

Bottom face top face

CO2H CO2H PPhRh(I), (R)-BINAP CO2H Rh(I), (S)-BINAP H 2 H PPh NHCOPh 2H2 (1 atm.) NHCOPh H2 (1 atm.) NHCOPh

Very high facial selectivity (98% ee) NB: this is a ‘protected’ version (S)-BINAP 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… 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.

Lowest scoring questions were on hybridisation O sp2 state and pKa values (these concepts are fundamental to understanding organic chemistry, so please revise if necessary). O sp3 sp Revision: N

Put the following species in order of increasing acidity (i.e. least acidic first)

O O OH O O O F > > OH > OH > S > HCl Me OH F Me OH F

pKa = 10.0 4.8 4.2 –0.3 –2.6 –8.0

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. Reduction of carbon-carbon double and triple bonds

§ Dissolved metal reductions: - Alkali metals (e.g. Na, Li, K) dissolved in liquid (T < –33 ˚C). - ‘Free electrons’ add to low-lying π*-orbitals.

NH3 (l) Consider the solution as a Na Na + e source of ‘free electrons’

§ Reduction of alkene to : - Selectively reduces electron deficient alkenes to alkanes (i.e. no reaction with standard alkenes).

O O i). 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 is crucial (see mechanism next page). 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 (NB: no proton source so enolate stable under these conditions). v). Addition of more acidic proton source, NH4Cl (aq), protonates enolate to give product.

–ve charge on

O O OH + e H Ot-Bu (radical) (ionic) Single electron transfer Single (radical) + e electron transfer Cl O O OH H N H PROTON 3 TRANSFER

(ionic) (ionic)

allyl anion No t-BuOH remaining – enolate

stable until add NH4Cl (aq) Reduction of carbon-carbon double and triple bonds

§ Reduction of alkynes to trans-alkenes: - Dissolved metal reduction forms trans- with high levels of stereoselectivity. e.g. c.f. Lindlar’s catalyst gives cis-alkene

trans-(E)-alkene

Na, NH3 (l), –33 ˚C

+ e pKa ~ 33 (ammonia) H NH2

H H NH2 + e

Vinyl anions are geometrically Vinyl anions sufficiently basic to

unstable and (E)-geometry is deprotonate NH3 (c.f. enolate energetically favoured basicity on previous slide)

Note: pKaH ~ 45 (vinyl anion) pKaH ~ 25 (enolate) - Provided it is not electron deficient, product alkene will not be reduced. Question: some pKa revision

§ Put these anions in order of decreasing stability (i.e. most stable first and least stable last).

A. 1, 5, 2, 3, 4 B. 1, 5, 2, 4, 3 C. 5, 1, 2, 3, 4 ✔ D. 1, 2, 5, 4, 3 E. 5, 1, 4, 3, 2

O

1 2 3

N

4 5 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. Arthur Birch H H

Na, NH3 (l), t-BuOH, –33 ˚C

H H

+ e H Ot-Bu

H H H H H H H Ot-Bu + e

Pentadienyl anion has the highest electron density at the central position, hence kinetic protonation (at low T) affords the non-conjugated product regioselectively. Reduction of carbon-carbon double and triple bonds

§ of Birch reduction: - Electron withdrawing groups promote ipso, para reduction. e.g.

CO2Me CO2Me H Na, NH (l), t-BuOH, –78 ˚C 3 ipso, para

H H

+ e H Ot-Bu

CO Me CO2Me 2 CO2Me CO2Me H Ot-Bu + e

ortho, meta ipso, para

- 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. Reduction of carbon-carbon double and triple bonds

§ Regioselectivity of Birch reduction: - Electron donating groups promote ortho, meta reduction. e.g.

OMe OMe H Na, NH (l), t-BuOH, –78 ˚C 3 H ortho, meta H H

+ e H Ot-Bu

OMe OMe OMe OMe H Ot-Bu + e

ipso, para ortho, meta

- 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 . 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.

O O

i-PrO O i-PrO O Na, NH3 (l), t-BuOH, –78 ˚C

+ e H Ot-Bu

O O O H Ot-Bu + e i-PrO O i-PrO O i-PrO O

- Heterocycles are electron rich – electron withdrawing group required for reduction. - Electron withdrawing group also stabilises radical anion formation, controls regioselectivity. Question: Birch reduction

§ Which of these substrates will be reduced the fastest under Birch conditions (i.e. dissolved metal, NH3 (l), –78 ˚C)?

CN MeO O

1 2 3

4 5

A. 1 B. 2 ✔ C. 3 D. 4 E. 5 Reduction of carbon-carbon double and triple bonds

§ Reduction of propargylic to trans-allylic alcohols: - Use LiAlH4. - Stereoselectivity due to complexation of reducing agent to oxygen. e.g.

LiAlH4, THF, 70 ˚C OH Me3Si Me3Si OH (E)-alkene

LiAlH4 deprotonates alcohol then complex stereospecific forms between H AlH3Li protonation of C–Al H2O alkoxide and resultant bond during w/up H Lewis acidic AlH3 H2 (g) evolution

Li H2Al O H2Al O O Li Me Si Me3Si 3 Me3Si Al H H H Li H H

Intramolecular delivery Intramolecular coordination of vinyl anion to Lewis of hydride acidic Al species locks geometry of alkene (i.e. Al and alcohol chain on same side of C=C) Reduction of carbon-carbon double and triple bonds

§ Summary: - C-C π-bonds typically reduced by hydrogenation or dissolved .

- 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, , , 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). Questions: end of Lectures 1 & 2

Q1. What is the product of this Birch reduction? Draw a curly arrow mechanism to describe the reaction.

Me Li, NH3 (l), –78 ˚C, t-BuOH ?

Q2. Provide two different sets of reagents and conditions that would carry out this transformation. Draw curly arrow mechanisms for both.

OH ? OH Ph Ph

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?

Lindlar reduction CO2Me Ph CO Me 2 Ph

Q4. Which of the two following would require the greater pressure of hydrogen, and why?

Pd/C (cat.), H2 (g), MeOH

Pd/C (cat.), H2 (g), MeOH

This year’s Chemistry World science communication competition is inspired by the results of our public attitudes to chemistry survey and it’s open to all ages.

Closing date: 8 January 2016

So why not encourage your students to enter? They could win £500, get their article published in the magazine, take part in a Chemistry World special assignment and even present at the Royal Institution’s iconic Faraday Theatre. All this would look great on job applications. http://www.rsc.org/chemistryworld/2015/09/chemistry-world-science- communication-competition-2015-2016 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.

CO2Me CO2Me H

Na, NH3 (l), t-BuOH, –78 ˚C

H H

+ e H Ot-Bu

CO Me CO2Me 2 CO2Me CO2Me H Ot-Bu + e

Radical is closest to stabilising NB: pentadienyl anion has group (i.e. ester) 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.

O PtO2 (5 mol%) H OH H2 (g) (1 atm.)

NB: Pt more reactive metal than Pd alkene also reduced

§ Ionic hydrogenation: - C-heteroatom π-bonds are polarised and susceptible to attack by nucleophilic ‘hydride’. - Opportunity to control chemoselectivity.

O NaBH4, MeOH H OH 0 ˚C, 97 %

NB: see Dr Lebrasseur’s & Dr Bray’s lectures on carbonyl chemistry 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. O O O

R N R NHR' R R' R Cl NITRILE AMIDE KETONE ACID CHLORIDE

O NH O O

R H R H R OH R OR' ALDEHYDE ESTER

§ Hydride reducing agents: - Wide variety of reducing agents, with different levels of nucleophilicity (i.e. reactivity). e.g.

LiAlH4, NaBH4, BH3, LiBH4, DIBAL, NaCNBH3, Al(Oi-Pr)3.

Achieve chemoselectivity through careful selection of reducing agent for desired FG reduction. Reduction of carbon-heteroatom double and triple bonds

§ 1st Year revision: order these functional groups in terms of decreasing electrophilicity (i.e. most reactive first and least reactive last).

O NH O

R NHR' R H R R' AMIDE IMINE KETONE O O

R H R OH ALDEHYDE CARBOXYLIC ACID O

R OR' ESTER

A. Aldehyde, Ketone, Imine, Ester, Amide, Acid B. Imine, Ester, Amide, Ketone, Aldehyde, Acid C. Acid, Amide, Ester, Ketone, Aldehyde, Imine D. Imine, Aldehyde, Ketone, Ester, Amide, Acid ✔ E. Imine, Aldehyde, Ketone, Ester, Acid, Amide Reduction of carbon-heteroatom double and triple bonds

§ Revision aid: summary of FG reactivity with common reducing agents.

NR'H O O O O O

R H R H R R' R OR' R NHR' R OH IMINIUM ALDEHYDE KETONE ESTER AMIDE CARBOXYLIC ACID

decreasing electrophilicity

✓ ✓ ✓ ✓ ✓ slow LiAlH4

✓ ✓ ✓ slow ✗ ✗ NaBH4

✓ ✓ ✓ ✗ ✗ LiBH4 ✓

DIBAL ✓ ✓ ✓ ✓ ✓ ✓ (@ –78 ˚C) slow (✗) (✗) (✓ aldehyde) (✓ aldehyde) (✓)

BH3 ✓ slow slow slow ✓ ✓

✓ slow slow ✗ ✗ ✗ NaCNBH3

Product AMINE ALCOHOL ALCOHOL ALCOHOL AMINE ALCOHOL Reduction of carbon-oxygen double and triple bonds

§ Reduction of and ketones to alcohols (recap.): - Using NaBH4. e.g. O OH

NaBH4, MeOH N N Racemic product H O O

acidic H BH OH2 3 H ‘work-up’

Na O

N NB: amide not reduced H O 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. O OH

LiAlH4, THF, ∆ N N H H O H

LiAlH 4 work-up H3O

H Li H O Al O H H AlH 3 Li N N H O O

Li+ functions as a Lewis acid (i.e. precoordination activates the Reduce amide then work-up carbonyl by lowering the LUMO energy) (NB: mechanism of amide reduction covered later) 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. O OH

LiAlH4, THF, ∆ N N H H2O H O –AlH H 3 H AlH2 O AlH2 O H3O LiAlH4 work-up H O N 3 H N H O H2O H Li O H H O Al O H H AlH 3 Li N N H O O

Li+ functions as a Lewis acid (i.e. precoordination activates the Reduce amide then work-up carbonyl by lowering the LUMO energy) (NB: mechanism of amide reduction covered later) Reduction of carbon-heteroatom double and triple bonds

§ Revision aid: use as a quick guide to determine chemoselectivity.

NR'H O O O O O

R H R H R R' R OR' R NHR' R OH IMINIUM ION ALDEHYDE KETONE ESTER AMIDE CARBOXYLIC ACID

decreasing electrophilicity

✓ ✓ ✓ ✓ ✓ slow LiAlH4

✓ ✓ ✓ slow ✗ ✗ NaBH4

✓ ✓ ✓ ✗ ✗ LiBH4 ✓

DIBAL ✓ ✓ ✓ ✓ O ✓ ✓ (@ –78 ˚C) slow (✗) (✗) (✓ aldehyde) (✓ aldehyde) (✓) N BH ✓ slow slow slow ✓ ✓ 3 O ✓ slow slow ✗ ✗ ✗ NaCNBH3

Product AMINE ALCOHOL ALCOHOL ALCOHOL AMINE ALCOHOL 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. N.B. esters & alkene untouched

H H O OH Al(Oi-Pr)3, i-PrOH, ∆ O + MeO2C MeO2C 78 %

MeO2C MeO2C Reaction completely reversible

Al is Lewis acid Al(Oi-Pr)3 i-PrOH w/up

Oi-Pr O Oi-Pr O Al Oi-Pr Al Oi-Pr hydride transfer H H O O Me Me

6-membered transition state

NB: α,β-unsaturated carbonyls reduced to allylic alcohols Reduction of carbon-oxygen double and triple bonds

§ Reduction of α,β-unsaturated ketones: - Two potential sites of attack leads to two possible products. - [1,2]-attack leads to allylic alcohols. - [1,4]-attack leads to fully reduced alcohol (i.e. via conjugate addition, tautomerism, reduction). e.g.

O OH OH [1,2]-attack NaBH4 + 59what : 41 iswithout the ratio? CeCl3 MeOH 99 : 1 with CeCl3

H BH H BH3 3 [1,4]-attack O O

tautomerisation

- Luche reduction: - Excellent regioselectivity for [1,2]-attack by using CeCl3 in addition to NaBH4. - Ce(III) activates carbonyl & promotes formation of alkoxyborohydrides (“harder” )

from NaBH4 and MeOH (N.B. this mechanism is beyond scope of course). Question:

§ Identify which statements are correct (more than one statement may be correct).

✔ A. At room temperature LiAlH4 is a solid

B. At room temperature LiAlH4 is a liquid

C. At room temperature NH3 is a liquid

D. At room temperature NH3 is a gas ✔

E. NaBH4 will not reduce an ester ✔

F. NaBH4 will reduce an ester 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. O

LiAlH4, THF, 0 ˚C O OH

+ Li activates LiAlH LiAlH aldehyde reduced to carbonyl group 4 4 alcohol Li (see previous slides) H3Al aluminate Li O O O H AlH3 –EtOAlH3 O O H H

tetrahedral intermediate aldehyde intermediate collapses reduced rapidly

NB: DIBAL at r.t. will also reduce esters to alcohols

NB: aluminate species produced in reaction, but these are less reactive than parent LiAlH4.

Can we stop the reduction at the aldehyde? Require tetrahedral intermediate not to collapse in situ. 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. NB: alkene not reduced O DIBAL, Toluene, –78 ˚C O O

H i-Bu DIBAL Al H3O (NB: Al is Lewis acid) i-Bu

i-Bu i-Bu i-Bu OH Al Al H3O H O i-Bu O w/up O H O O H hemiacetal

DIBAL reduces tetrahedral intermediate esters at –70 ˚C stabilised at low T

+ Intermediate collapses to aldehyde upon w/up, but excess DIBAL has been quenched by H3O so no further reduction can occur Reduction of carbon-oxygen double and triple bonds

§ Reduction of to : - Amides poor , require powerful reducing agent LiAlH4. e.g. O

LiAlH4, Et2O NMe2 NMe 2 88 %

Li+ activates LiAlH4 then w/up carbonyl group H AlH3

Li Li H3Al O O H H AlH3 –H3AlOLi

NMe2 NMe2 NMe2 H

tetrahedral intermediate iminium ion reduced

collapses by expelling best rapidly by LiAlH4 LG (i.e. O better than N) (N.B. iminium ion more reactive than aldehyde)

Drawback: if molecule contains aldehydes, esters, etc. then LiAlH4 will reduce everything (i.e. no chemoselectivity). Reduction of carbon-oxygen double and triple bonds

§ Reduction of amides to amines (alternative): - Use BH3 (chemoselective for amide over ester, ketone, etc.). NB: ester not reduced here e.g. MeO C MeO2C BH3•THF 2 O N H N H B O Bn Bn H H H Like DIBAL, BH3 B Borane (B2H6) is a gas; forms Lewis acid- stabilised as a liquid in H2O w/up base complex H BH3 complex with THF H H MeO2C H MeO2C B H H O N N BH2 Bn Bn

Borate donates BH3 reduces hydride to reactive reactive imidate iminium ion H MeO2C MeO2C H B H –BH2O H O N H N Bn H B Bn 2 (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. O M-[H] O NMe2 generic metal- hydride species

M-[H] to aldehyde

M M O O OH [H] H3O

NMe2 NMe2 NMe2 H w/up

stabilise tetrahedral w/up to reveal intermediate?

- 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… 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

O OMe DIBAL, toluene O N 0 ˚C, 74 %

i-Bu OMe Al is Lewis acid H Al – HN i-Bu hydrolysis to aldehyde i-Bu i-Bu i-Bu Al Al i-Bu H O O OH H O OMe OMe 3 OMe N N w/up N H H

Borate donates hydride stabilised chelated w/up reveals to reactive imidate tetrahedral intermediate hemiaminal 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.

O O OMe MeLi, THF N ? Me ketone 0 ˚C

OMe NB: also works with Me Li – HN elimination of Grignard reagents MeNHOMe

Li O OH OMe H3O OMe N N Me w/up Me

stabilised chelated tetrahedral intermediate

Q: What would happen if this was a simple amide, rather than a Weinreb amide? 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

O O

LiAlH4 heat OH OH O Li –H2 (g)

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 forms active reducing agent (see amide reduction).

- Milder reduction conditions compared to using LiAlH4. e.g. 2 OH

BH3•THF OH O

NB: for further explanation see ‘OC’ textbook p. 619. Reduction of carbon-oxygen double and triple bonds

3). Via more : convert carboxylic acid to mixed anhydride, then reduce with NaBH4. - 2 steps, but generally quick and high yielding procedures. e.g. 3

acid alcohol

O O O O BocHN BocHN Cl OEt BocHN NaBH4 OH OH O OEt MeOH, 0 ˚C Et3N, CH2Cl2 97 % yield over 2 steps mixed anhydride H BH3 NaBH4 NB: most reactive at C=O that originated from carboxylic acid, O as other C=O π-system has O O overlap from two . –CO2 (g) BocHN BocHN H O OEt H –EtO

- Eliminate CO2 (g) and EtO 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).

O

LiAlH4, Et2O NMe2 NMe 2 88 %

Li+ activates LiAlH4 then w/up carbonyl group H AlH3

Li Li H3Al O O H H AlH3 –H3AlOLi

NMe2 NMe2 NMe2 H

tetrahedral intermediate iminium ion reduced

collapses by expelling best rapidly by LiAlH4 LG (i.e. O better than N) (N.B. iminium ion more reactive than aldehyde) Aims: at the end of this section you will be able to…

§ Reduction of C-N multiple bonds.

§ 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 reductions of cyclohexanones (i.e. axial or equatorial attack?).

§ 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) Reduction of carbon- double and triple bonds

§ Reduction of imines to amines: - Use NaBH4, LiAlH4, etc. (NB: imines more electrophilic than aldehydes). e.g.

N NaBH4, MeOH HN NB: analogous to mechanism 0 ˚C, 95 % H 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.

H O NBn NHBn BnNH2 H BH2NaCN H MeOH, HCl H

Iminium ion formation iminium ion Aldehyde starting (see 1st Yr for mechanism) (highly reactive) material not reduced

by NaCNBH3 Question: reductive amination

§ Eschweiler-Clark reaction: - Introduced in 1st year. - Can you draw the mechanism for this reductive amination? e.g.

N (CH2O)n (aq.) N H HCO2H

Hint: O O O

H H H O CO2 (g) evolution H H (CH O)n (aq.) 2 N

NB: iminium ion formation H H Reduction of carbon-nitrogen double and triple bonds

§ Reduction of nitriles to amines: - Complete reduction using LiAlH4 (i.e. powerful reducing agent). e.g.

NH2 LiAlH4, THF N

reduction of intermediate imine Li+ activates LiAlH4, then LiAlH followed by w/up nitrile group 4 H O (mechanisms 3 analogous to previous slides) H AlH N 3 Li N Li H

imine intermediate

Overall 2 equivalents of hydride added 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.

O DIBAL (1 equiv.), Toluene, –78 ˚C N H then H3O

i-Bu H Al H3O w/up i-Bu

N i-Bu N i-Bu Al Al H i-Bu H i-Bu iminoalane

H N no DIBAL remaining Al so iminoalane i-Bu i-Bu reduction cannot occur

NB: Lewis acidic DIBAL coordinates to nitrile lone pair and ‘ate’ complex is hydride donor Reduction of carbon-heteroatom double and triple bonds

§ Revision aid: summary of FG reactivity with common reducing agents.

NR'H O O O O O

R H R H R R' R OR' R NHR' R OH IMINIUM ION ALDEHYDE KETONE ESTER AMIDE CARBOXYLIC ACID

decreasing electrophilicity

✓ ✓ ✓ ✓ ✓ slow LiAlH4

✓ ✓ ✓ slow ✗ ✗ NaBH4

✓ ✓ ✓ ✗ ✗ LiBH4 ✓

DIBAL ✓ ✓ ✓ ✓ ✓ ✓ (–78 ˚C) slow (✗) (✗) (✓ aldehyde) (✓ aldehyde) (✓)

BH3 ✓ slow slow slow ✓ ✓

✓ slow slow ✗ ✗ ✗ NaCNBH3

Product AMINE ALCOHOL ALCOHOL ALCOHOL AMINE ALCOHOL Questions: end of Lectures 3 & 4

Q1. Provide a curly arrow mechanism for this reaction. What is the name of this transformation?

Al(Oi-Pr)3, i-PrOH

O ∆ OH

Q2. What are the structures of A and B? Draw mechanisms for both reactions and rationalise the observed selectivity.

CO2Me LiBH4 LiAlH4 A B CONMe2

Q3. Put these FGs into their order of increasing reactivity with DIBAL (i.e. least first, most reactive last).

ESTER KETONE ALDEHYDE AMIDE

Q4. What is the product of this reaction? Draw a mechanism to describe the formation.

O 1. NH2 , HCl, MeOH ?

2. NaCNBH3

Q5. What is the stereochemistry of the product in this reaction? Explain how you came to your choice. Et OH NaBH 4 Ph Ph * * MeOH O 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 reductions of cyclohexanones (i.e. axial or equatorial attack?).

§ 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) Diastereoselectivity with : 1,2-stereoinduction

§ Felkin-Anh model (recap.): NB: see Dr Bray’s 1st year lectures notes e.g.

O OH LiBH(s-Bu) Ph 3 Ph THF

‘Felkin’ product

- Drawing Newman projections (recap.).

O O O add substituents Me Ph Ph H H Me

Look down highlighted Convert to Newman Ensure configuration C-C bond projection around chiral centre is correct Diastereoselectivity with hydrides: 1,2-stereoinduction

§ Felkin-Anh model (recap.): e.g.

Lowest energy conformations of chiral starting material

O OH O O LiBH(s-Bu)3 Me H Ph Ph Ph or Ph THF H Me

‘Felkin’ product (i.e. place largest group 90˚ to C=O)

O - Reactive confirmation (i.e. least sterically hindered). Me … - Nu approaches along Bürgi-Dunitz trajectory (107˚ O=C Nu). Ph R B - Least hindered approach is past ‘H’ and away from large 3 H group (e.g. Ph). H

Newman projection

NB: see Dr Bray’s 1st year lectures notes Diastereoselectivity with hydrides: 1,2-stereoinduction

§ Felkin polar model: - Use when α-substituent is electron withdrawing. e.g. O i-Pr MeS BR3 O OH H H LiBH(s-Bu) i-Pr 3 i-Pr THF (i.e. place eWG group 90˚ to C=O) SMe SMe

‘anti-Felkin’ product Newman projection

π* C=O overlaps with σ* C-S; creates lower energy LUMO

- Electronegative group 90˚ to C=O (even if it C-S σ* isn’t the largest group). SMe SMe - Reactive conformation (i.e. π* and σ* combine combine to lower LUMO). i-Pr i-Pr H O H O - Approach along Bürgi-Dunitz trajectory (107˚). - Least hindered approach is past ‘H’ and away C=O π* new LUMO from electronegative group. (π* + σ*) (i.e. most reactive in this conformation) 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. Chelating metals

Li+ (sometimes) O OH Mg2+ Zn(BH ) Me 4 2 Me Zn2+ THF Cu2+ Ti4+ OMe OMe Ce3+ 2+ ‘Felkin’ product Mn

NB: Na+ and K+ do not chelate

- Electronegative group almost eclipses C=O to enable chelation with the metal (even if electronegative group is Zn2+ the largest group). O - Reactive conformation (i.e. Lewis acid coordination MeO Me lowers LUMO). R B 3 H - Approach along Bürgi-Dunitz trajectory (107˚). H - Least hindered approach is past ‘H’ and away from group at 90˚. Newman projection Diastereoselectivity with hydrides: 1,2-stereoinduction

§ Examples: e.g. 1 Key step in synthesis of anticancer agent, dolastatin.

O

NBn2 NBn2 NBn2 MeO H + CO2Me CO2Me O OH OH

1 diastereoselectivity 20

- How do we explain diastereoselectivity? Consider reactive conformation. O H CO2Me 1. No chelation so electronegative group goes 90˚ to C=O (Felkin polar) 2. Approach alongside H H NBn2

O H O NBn O 2 Bn N O 2 Bn2N H H H H H CO2Me OMe Visualisation aid: draw product in most reactive conformation same orientation, then rotate to put longest chains in same plane Diastereoselectivity with hydrides: 1,2-stereoinduction

§ Examples: e.g. 2 Chelation or not?

O HO Me HO Me Me2Mg Ph + Ph Ph THF, –70 ˚C OR OR OR diastereoselectivity R = Me 99 1

R = OSi(i-Pr)3 42 58

- How do we explain diastereoselectivity? Consider reactive conformation.

‘R’ = Me ‘R’ = OSi(i-Pr)3

Mg2+ O O Me RO RO 1 Large group disrupts Me H Nu efficient chelation H Ph Nu Ph

Chelation model Felkin polar model Diastereoselectivity with hydrides: 1,2-stereoinduction

§ Summary: which model to use?

O Is there a heteroatom at the chiral centre? X R Y Z No Yes α-chiral carbonyl compound

Use Felkin-Anh model Is there a metal (i.e. conformations with the ion capable of largest group 90˚ to C=O) chelation?

No Yes

Use Felkin polar model Use Felkin chelation (i.e. conformations with model (i.e. conformations the electronegative with the heteroatom and group 90˚ to C=O) C=O almost eclipsed) Question: 1,2-stereoinduction

§ What is the stereochemistry of the product of this reaction?

Et OH NaBH 4 Ph Ph * * MeOH O Et

OH OH OH OH A. 1 Ph Ph Ph Ph B. 2 C. 3 ✔ Et Et Et Et D. 4 1 2 3 4 Question: 1,2-stereoinduction

§ What is the stereochemistry of the product of this reaction? 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. NB: axial H H H H H H H NB: equatorial H H H

Cyclohexane Chair Boat

- Cyclohexene adopts a ‘half chair’ in its lowest energy conformation.

H H H H H H H H

Cyclohexene Half chair

NB: for further explanation see ‘OC’ textbook p. 370-374. Reduction of cyclohexanones

§ Axial or equatorial attack? - When reducing cyclohexanones, the hydride can end up in an axial or equatorial position. - Can we achieve this selectively?

more hindered approach (i.e. 1,3-diaxial interactions)

H H H H H OH O H 'AXIAL H 'EQUATORIAL H OH ATTACK' ATTACK' H Small hydride Bulky hydride H H H H H H

H less hindered approach e.g. LiAlH4 e.g. Na(s-Bu)3BH 9:1 ax:eq 4:96 ax:eq

- ‘Axial attack’ favoured with a small hydride source (e.g. LiAlH4, NaBH4).

- ‘Equatorial attack’ favoured by bulky hydride sources (e.g. ‘L-selectride’ Na(s-Bu)3BH). Reduction of cyclohexanones

§ 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’:

(i.e. more hindered approach)

O 'AXIAL ATTACK'

H H oxy substituent H rotates H H O O H H

'EQUATORIAL alkoxy intermediate ATTACK' from ‘axial attack’

(i.e. less hindered approach)

- ‘Axial attack’: oxy substituent moves away from neighbouring C-H bond. - ‘Equatorial attack’: oxy substituent moves towards neighbouring C-H bond, leading to higher torsional strain in T.S (i.e. disfavoured).

NB: for further explanation see ‘OC’ textbook p. 471. Question: reduction of cyclohexanones

§ What is the stereochemistry of the product of the following reduction? O A. 1 ✔

NaBH4 B. 2 ? BnO MeOH C. 3 D. 4

OH OH

H BnO BnO H O BnO t-Bu OH OH 1 2 H axial attack (small hydride source) BnO BnO

3 4 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.

Hydride approaches from top face

H OH R R R or R' O R' R' OH H

top face bottom face

Hydride approaches from bottom 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)? 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.

H Ph Boron in catalyst is Lewis acidic and activates ketone C=O group N Ph BH3 used to regenerate active catalyst H3B B O Me Boat-like arrangement controls facial selectivity

‘CBS’ reagent with Binding of BH3 and ketone to exo face of catalyst

BH3 bound

e.g. Ph O Me O 10 mol% catalyst B OH Ph N Me BH , THF Ph Me 3 B O Ph Me H H H Ph

Hydride delivered from ‘top’ face (R)-alcohol, 97 % ee

NB: for further explanation see ‘Ox and Red’ textbook p. 69. Enantioselective reduction of ketones

§ Predict the stereochemistry of the products of these ‘CBS’ reductions: e.g. 1 H smaller substituent Ph pseudo axial N Ph B O Ph O Me O Me 1 mol% B OH N Cl Cl Ph ? BH , THF Cl Ph 3 B O Ph H H H Ph 96 % ee

e.g. 2 H Ph N Ph B O O Ph O Me OH Me 1 mol% B TBSO Ph ?N Me TBSO Me BH , THF Me 3 B O H H H 94 % de

NB: require good size differentiation between two ketone substituents for high levels of enantioselectivity. § Reductive cleavage reactions Reductive cleavage reactions

§ Definition: Break single bonds between carbon and electronegative elements and replace with bonds to hydrogen. REDUCTION C X C H X = N, O, S, halogen

§ 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.

Ph Pd/C (10 mol%), H2 (1 atm.) N NH EtOH

O O

Pd/C (10 mol%), H2 (1 atm.) O Ph OH EtOAc

Ease of formation and removal means that the is commonly used as a ‘’ in organic synthesis for more reactive FGs such as amines, amides, alcohols and carboxylic acids.

NB: This cleavage mechanism is beyond the scope of the course – for an explanation see ‘Ox & Red in Org Synth’ p. 77 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. Br H OMe OMe Pd/C (10 mol%) Weaker bonds reduced more easily. Most reactive: H2 (1 atm.), Et3N OMe MeOH, r.t., 97% OMe C-I > C-Br > C-Cl > C-F OMe OMe

NB: This mechanism is beyond the scope of the course – for an explanation see ‘Ox & Red in Org Synth’ p. 76

§ Dissolved metal: - An alternative method to cleave N-benzyl and O-benzyl groups. e.g. O Bn O H H H Bn N Na, NH (l), t-BuOH, –78 ˚C N N 3 HN

H H

NB: This mechanism is beyond the scope of the course – for an explanation see ‘Ox & Red in Org Synth’ p. 76 Reductive cleavage reactions

§ Deoxygenation – ketones and aldehydes to alkanes: e.g. 1 Reduction to alcohol, make into good LG, displace with hydride.

H H H 1). LiAlH4 OTs 3). H AlH3Li O H 2). TsCl THF

need to convert

3 steps overall OH to good LG SN2 reaction (i.e. reactivity 1˚ > 2˚ >> 3˚)

e.g. 2 Wolff-Kischner reduction: - Use (NH2NH2) and KOH. - Requires elevated temperatures (up to 200 ˚C).

O H H

NH2NH2, KOH Nicolai Kischner glycol, ∆

is a solvent with high b.p.

NB: mechanism beyond scope of this course, for more details see ‘Ox & Red in Org Synth’ p. 82 Reductive cleavage reactions: summary

§ Which conditions break which bond?

Benzyl group cleavage

H X Ph X Use Pd/C, H2; or dissolved metal X = O or N

Reductive dehalogenation

X H C C Use Pd/C, H2

X = Halogen (i.e. I, Br, Cl, F)

Deoxygenation

O H H Wolf-Kischner or 3 step sequence R R (i.e. reduction, activation, reduction) Question

§ Which set of reagents and conditions A to E will deliver the reduction product?

O ? O HO OH OBn HO

A. H2 (1 atm.), Pd/C, MeOH, r.t.

B. H2 (1 atm.), Pd/BaSO4, Pb(OAc)4, quinoline, r.t.

C. Na, NH3, r.t.

✔ D. Na, NH3, –78 ˚C

E. H2 (1 atm.), Pd/C, MeOH, 50 ˚C § 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:

alkene intact

Pd/BaSO4 NaN H (1 atm.) OTs 3 N3 2 NH2

DMF Pb(OAc)4

SN2 azide amine

- Strategic advice: azide is an excellent (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. 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.

PPh3, THF

N3 H2N

azide amine

NB: Pd/C & H2 would also reduce alkene

NB: This mechanism is beyond the scope of the course, for further details see ‘Ox & Red in Org Synth’ p. 47 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.

CF3 CF3

H2 (1 atm.), Pd/C (10 mol%) EtOH, r.t. NO2 NH2 OEt OEt

NB: For details of this mechanism via SET see ‘Ox & Red in Org Synth’ p. 46

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). Question: reduction as a tool in

§ Identify the correct product of this reaction.

O MeO O N N3 N OMe 10% Pd/C, H2 (1 atm.) O ? H N Me 2 MeOH:EtOAc OBn

Br OMe OMe A. 1 B. 2 ✔ C. 3

O O O MeO MeO MeO O O O N N N H2N N OMe H2N N OMe N3 N OMe O O O H2N Me H2N Me H2N Me 1 OBn 2 OH 3 OH

Br OMe OMe Br OMe OMe OMe OMe Questions: end of Lectures 5 & 6

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? OPh NaBH4, MgBr2 ? MeOH O Q2. What is the stereochemistry at C*? Use conformational drawings to explain your answer. O OH

Na(s-Bu) BH 3 * THF

Q3. Provide reagents and conditions to complete the following transformation in 2 steps. Br Reagents? O 2 steps NH NO2 2 Q4. What is the stereochemistry at C*. Use conformational drawings to explain your answer. H Ph Catalyst A (1 mol%) * F I F I N Ph . O BH3 THF OH B O A Me Heterocyclic Chemistry Aims for this session

§ Heterocycles: - Definition. - Why heterocycles are important. - Nomenclature of common heterocycles. - Recap. of key reactions of ketones, , imines and enamines. - Recap. of aromaticity. - Look at : structure and reactivity (how does it compare to benzene?). - Look at : structure and reactivity (how does it compare to benzene?). - Summary & comparison of pyrrole and pyridine structure & reactivity. - Consider 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.

CARBOCYCLIC HETEROCYCLIC

H H N N N H

CYCLOPENTANE PYRROLIDINE 2,3-DIHYDROIMIDAZOLE (non-aromatic) (non-aromatic) (non-aromatic)

N N

N

BENZENE PYRIDINE (aromatic) (aromatic) (aromatic)

one heteroatom multiple heteroatoms Why are heterocycles important?

§ 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. NH2 N Me N N O N N HO N N HO N MeO Me N O Quinine O Me N Caffeine OH Deoxyadenosine

H N O Me OMe S N MeO N Me Me N N NO2 Nicotine Omeprazole Ranitidine Me Me N Me N S N H H O Question: identify the heterocycles

§ How many heterocycles are in this pharmaceutical compound? (NB: count any fused rings separately)

O Me Me N N O HN N N S N O i-Pr OEt

Sildenafil (‘Viagra’) A. 0 B. 1 C. 2 D. 3 ✔ E. 4 F. 5 Top 20 global best-selling drugs in 2009

NB: Drugs containing heterocycles indicated in bold Heterocycles: nomenclature

§ Five membered rings with one heteroatom: - NB: we will only study the highlighted compounds in more detail. e.g.

H H H N O N O N O

pyrrole furan pyrrolidine tetrahydrofuran pyrrolidinone (‘THF’)

H X N X X

thiophene(X = S) indole benzofuran (X = O) isoindole (X = N) selenophene (X = Se) benzothiophene (X = S) isobenzofuran (X = O) etc. etc. etc. 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.

H H H H H N N N N N N N N N N N N N N

imidazole pyrazole 1,2,3-triazole 1,2,4-triazole tetrazole

H H H H N N N N N O S O S O

oxazole thiazole isoxazole isothiazole benzoxazole Heterocycles: nomenclature

§ Six membered rings with one heteroatom: - NB: we will only study the highlighted compounds in more detail. e.g.

8 N N O 7 1 2 N NB: aromatic 6 3

5 4

pyridine quinoline isoquinoline pyrylium cation

H H H N N N O O 6 1 2 NB: non-aromatic 5 3

4

piperidine 1,2,3,4-tetrahydropyridine pyridone tetrahydropyran Heterocycles: nomenclature

§ 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.

N N N N N NB: aromatic N N N N

pyridazine pyrimidine 1,3,5-triazine quinazoline (C, T & U nucleobases)

H H N N O NB: non-aromatic N O H O

piperazine morpholine 1,4-dioxane (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.

O O NH NH O NH epoxide aziridine oxetane azetidine β-

OH OH OH OH OH H PhO N S Me O O HO N Me O CO H O OH 2

(+)-Roxaticin Penicillins

NB: macrocycle Heterocycles: key reactions (recap.)

§ Enol formation: - NB: acidic conditions.

e.g. NB: consider pKa of α-proton

H H O H OH2 O H2O O H protonation H deprotonation R R R

ketone enol

§ Enolate formation: - NB: basic conditions.

e.g. NB: consider pKa of α-proton

Li NB: need a strong base O Ni-Pr2 O to effect complete irreversible deprotonation H R R of a ketone (i.e. use LDA) ketone enolate

NB: for a comprehensive list of pKa values, see: http://evans.harvard.edu/pdf/evans_pka_table.pdf Question: pKa (again!)

§ What are the pKa values of the two α-protons highlighted below?

H H O H OH2 O H2O O H H R R R

A. They are both the same B. 43 and 26 C. 26 and –6 ✔ D. 43 and –6 Heterocycles: key reactions (recap.)

§ Formation of imines: - NB: typically require catalytic acid. e.g.

O NR RNH2 + H2O cat. H+ R R

ketone imine

§ Enamine formation: - NB: occurs under acidic or basic conditions. e.g. c.f. enol and enolate formation

R NB: C=N bond weaker NR tautomerisation HN than C=O, therefore easier H to form enamine from R R imine compared to forming enol from ketone imine enamine

NB: enamines react with electrophiles on carbon and not nitrogen Heterocycles: key reactions (recap.)

§ Aldol reaction: - NB: formation of thermodynamically favoured α,β-unsaturated product drives reaction. e.g.

H O O cat. H+ O + R H Ph R Ph

enol aldehyde α,β-unsaturated product Heterocycles: key reactions (recap.)

§ : - NB: 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.

H O NR cat. H+ O NHR + R H Ph R Ph

enol imine amine product

§ Conjugate addition: - NB: orbital-controlled [1,4]-. - NB: called a “Michael addition’ when carbon-based nucleophile such as enol or enolate. e.g.

O Nu O O

R R Nu R Nu

enone enolate intermediate

NB: favoured when using ‘soft’ nucleophiles (i.e. uncharged and/or diffuse orbitals) Aromaticity (recap.)

§ Hückel’s Rule: e.g. aromatic compounds

2π electrons 6π electrons 6π electrons 6π electrons (3 x p orbitals) (5 x p orbitals) (6 x p orbitals) (7 x p orbitals)

Heterocycles

H N PYRROLE N PYRIDINE isoelectronic and isolobal isoelectronic and isolobal with cyclopentadiene anion with benzene

6π electrons 6π electrons

Planar structures which have a cyclic array of [4n + 2] delocalisable π-electrons Question: heteroaromatic?

§ Identify the heterocycle(s) that are not aromatic

H H N O N N O A B C

H H N N

D E F

A. A B. B C. C D. D E. E F. F ✔ Pyridine

§ Structure: - Isoelectronic with benzene (i.e. 6π electrons, one in each of 6 parallel p orbitals). e.g.

sp2 hybridised carbon: C H one electron in sp2 orbital σ- bonding to H, orthogonal to H plane of π-system benzene one electron in each p orbital

sp2 hybridised nitrogen: N lone pair of electrons in N sp2 orbital, orthogonal to plane of π-system

pyridine one electron in each p orbital

Nitrogen atom has major effect on reactivity and properties of pyridine w.r.t benzene Pyridine: is it really aromatic?

1 § H NMR spectroscopic chemical shifts (i.e. δH in ppm): - i.e. how deshielded is each environment? e.g.

H 7.1 H 7.5 H H 8.5 δ in aromatic region of 1H NMR spectrum 7.2 H N (N.B. more deshielding in pyridine than benzene due to electronegative N atom)

13 § C NMR spectroscopic chemical shifts (i.e. δC in ppm): - i.e. how deshielded is each environment? e.g.

123.6 13 135.7 149.8 δC in aromatic region of C NMR spectrum 128.5 (N.B. more deshielding in pyridine than N benzene due to electronegative N atom)

YES, pyridine is aromatic Question: hybridisation states in heterocycles

§ In which hybrid atomic orbital is the nitrogen lone pair located?

N

A. p B. sp C. sp2 ✔ D. sp3 Question: hybridisation states in heterocycles

§ In which hybrid atomic orbital is the nitrogen lone pair located?

H N

A. p B. sp C. sp2 D. sp3 ✔ Pyridine: general reactivity trends

§ Lone pair reactivity: - Basic (i.e. can be easily protonated). - Nucleophilic (i.e. can be alkylated or acylated). - N-Oxidation (forms N-, see later).

O X N E e.g. E = H R X R protonation acylation

§ Benzene-like reactivity: - Attack on π-system. E NB: SEAr is less likely than with benzene due to - Electrophilic aromatic N less electron-density in pyridine ring substitution (SEAr). (N.B. remember deshielded signals in NMR)

§ Imine-like reactivity: - Susceptible to nucleophilic attack (e.g. displacement of halo-) due to electron-deficiency.

N N N NB: behaves like an imine

Cl Nu Cl Nu Nu Pyridine: lone pair reactivity

H § Basicity: N - Consider pKa of the conjugate acid (i.e. pKaH)… e.g.

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. - Piperidine lone pair in sp3 orbital, hence stronger base (i.e. greater ‘p’ character so higher energy hybrid atomic orbital and therefore more reactive).

H H H H N N N N + H + H

PYRIDINE PIPERIDINE

pKaH 5.2 pKaH 11.5 (Stronger base) Question: pyridine basicity

§ Which pyridine is the strongest base?

N Me N Cl N

1 2 3

pKaH 5.2 pKaH 6.8 pKaH 0.7

A. 1 B. 2 ✔ C. 3 Pyridine: lone pair reactivity

§ Nucleophilicity: - NB: lone pair cannot be delocalised around the ring (i.e. sp2 orbital orthogonal to the p orbitals).

p orbitals

90˚ N N N X

sp2 orbital

- Hence lone pair is a good nucleophile as it is available (e.g. readily alkylated or acylated). e.g. acylation O Cl Cl R N R N acylated O

- Typically, the more basic a pyridine the 2,6-di-tert-butyl pyridine more nucleophilic it becomes, as lone pair N (good base but poor is raised in energy. nucleophile as only small electrophile (i.e. proton) can - Except when sterically crowded. reach lone pair) Question: DMAP as a catalyst

§ 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?) O Ph O O HO Me N Cl 2 + Me2N Cl Ph N Ph Me2N N O N DMAP

A. B. Base catalysis C. Nucleophilic catalysis ✔ Pyridine: benzene-like reactivity

§ 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.

E nitrogen makes pyridine ring electron N E N deficient (i.e. less nucleophilic).

pyridines preferentially react with electrophiles on nitrogen

- Yields from SEAr reactions are typically poor:

e.g. 1. Friedel-Crafts reactions normally fail.

N

e.g. 2. Nitration with HNO3 + H2SO4 at 300 ˚C gives < 5% of 3-nitropyridine.

NO2

e.g. 3. Halogenation with Cl2 + AlCl3 at 130 ˚C gives only moderate yields of 3-chloropyridine. N

NB: benzene reacts readily under these conditions (see 1st Year lecture notes). Cl Pyridine: benzene-like reactivity

§ Electrophilic aromatic substitution: - Electronegative nitrogen atom withdraws electron density; C-2, C-4 and C-6 most affected. e.g.

4 δ+

2 6 N N δ+ δ+ N N N δ-

overall

- C-2/6 or C-4 substitution results in δ+ localised on nitrogen atom (highly disfavoured). e.g.

E E E E N N N N H H H

- C-3/5 substitution is more favourable (but still unlikely to occur, see previous slide). e.g. E E E E H H H

N N N N Pyridine: benzene-like reactivity

§ Electrophilic aromatic substitution: - Does occur if electron donating group present (i.e. increases electron density within heterocycle). e.g.

NO2 HNO3

Me N N H2SO4 2 Me2N N

- If no activating groups present, can oxidise pyridine into pyridine N-. e.g.

mCPBA

N N O

nitrogen lone pair is NB: stable solid easily oxidised NB: This mechanism is beyond the scope of the course

Pyridine N-oxides readily undergo SEAr Pyridine: benzene-like reactivity

§ 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 NO2 H NO2

HNO3 H SO N 2 4 N N O O C-4 attack O Rearomatisation

- Easily cleave N-oxide with P(III) compounds (e.g. P(OMe)3) to regain pyridine. e.g.

NO2 NO2 O P(OMe)3 + P THF MeO OMe N OMe N O Reduction

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) Pyridine: imine-like reactivity

§ 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.

Na OMe OMe MeOH N N OMe N Cl Cl

NB: 2-chloropyridine is 3 x 108 times more reactive than chlorobenzene

- 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.

H CN N N CN N H H

‘Activated’ (i.e. N-protonated) Pyridine reactivity in action

§ Synthesis of Omeprazole: - $Billion-selling drug: proton pump inhibitor to treat stomach ulcers. e.g.

S Ar NO2 Me Me E Me Me Me Me H2O2 HNO3 N Me H SO N Me 2 4 N Me O oxidation nitration O N-oxide

NaOMe SNAr MeOH

H OMe N O Me Me S N steps MeO N Me N Me Me OMe O Omeprazole (proton pump inhibitor) Pyridine: summary

- 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. 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.

sp2 hybridised carbon: C H negative charge in p orbital, parallel to plane of π-system cyclopentadiene one electron in anion each neutral carbon p orbital

H sp2 hybridised nitrogen: N N H lone pair of electrons in p orbital, parallel to plane of π-system

pyrrole one electron in each carbon p orbital

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.

sp2 hybridised nitrogen: mild base N lone pair of electrons in N sp2 orbital, orthogonal to plane of π-system

pyridine one electron in each p orbital

H sp2 hybridised nitrogen: N N H lone pair of electrons in non-basic p orbital, parallel to plane of π-system

pyrrole one electron in each carbon p orbital

Location of nitrogen lone pair has major effect on reactivity and properties of pyrrole w.r.t pyridine 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.

H N N N H pyrrole-like nitrogen N non-basic

imidazole pyridine-like nitrogen mild base Pyrrole: NMR spectroscopic properties

1 § H NMR spectroscopic chemical shifts (i.e. δH in ppm): - Pyrrole is electron rich from lone pair donation, hence greater shielding. e.g.

7.1 H ≈10 H 6.5 7.5 H N H H H 8.5 7.2 N 6.2 H

NB: not as big a difference in δH around ring c.f. pyridine

13 § C NMR spectroscopic chemical shifts (i.e. δC in ppm): - Pyrrole is electron rich from lone pair donation, hence greater shielding. e.g.

H 123.6 118.0 N 135.7 149.8 128.5 N 107.7

electron rich c.f. benzene electron poor c.f. benzene Question: hybridisation states in heterocycles

§ In which (hybrid) atomic orbital is the nitrogen lone pair located?

H N

A. p ✔ B. sp C. sp2 D. sp3 Question: hybridisation states in heterocycles

§ In which (hybrid) atomic orbital is the nitrogen lone pair located?

H N

A. p B. sp C. sp2 D. sp3 ✔ Pyrrole: general reactivity trends

§ 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).

H H H H N E N E N H N X H pKa –3.8 H

§ Electrophilic substitution: H H N E N - Electron rich ring is highly reactive to SEAr E (more so than benzene). H

nucleophilic on carbon

§ Nucleophilic substitution: - Electron rich ring unreactive towards nucleophilic attack. - Require electron withdrawing substituent to activate ring.

NB: SNAr on an activated pyrrole is beyond the scope of the course, for an example see ‘OC’ textbook p 738. Pyrrole: lone pair reactivity

§ Basicity: - Pyrrole lone pair lowered in energy via delocalisation around ring, hence non-basic. - Pyrrolidine lone pair in sp3 orbital, hence stronger base.

H H N N H + H H

PYRROLE

pKaH –3.8

H H H N N + H

PYRROLIDINE

pKaH 11.0 Pyrrole: lone pair reactivity

§ 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. H N Base N

pKa 16.5

NB: sp2 orbital orthogonal deprotonation to π-system so no N H N stabilisation (solely provided by electronegative nitrogen)

N-H bond uses negative charge in nitrogen sp2 orbital, whilst N nitrogen sp2 orbital lone pair is still delocalised into aromatic ring

- Pyrrolidine is much less acidic. e.g. H N Base N negative charge in nitrogen sp3 orbital (N.B. higher energy c.f. sp2), plus lone pair fully localised on nitrogen pKa 44.0 Pyrrole: electrophilic substitution

§ 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.

H H H H H H δ+ N N N N N δ- N δ- δ- δ-

overall

- Is substitution at C-2 or C-3 favoured?

H H NB: adding another heteroatom into the ring deactivates N N towards SEAr as this is ‘pyridine-like’ (i.e. electron-withdrawing). N O Pyrrole: electrophilic substitution

§ 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.

linear conjugated intermediate

H H H H N E N N N E E E H H H

– H 3 resonance forms of intermediate

H N E

C-2 product Pyrrole: electrophilic substitution

§ 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. cross conjugated intermediate

H H H N E N N 2 resonance forms of intermediate

E H H E E

– H

H N

E C-3 product Pyrrole: electrophilic substitution

e.g. Formylation in the absence of a strong Lewis acid: Vilsmeier-Haack reaction.

H N H NMe2 N O POCl3 + DMF H Cl H Anton Vilsmeier

O O Cl Cl P P O Cl Cl Cl O Cl NMe2 O + P H NMe2 H NMe2 H Cl Cl O DMF NB: formation of intermediate driven by strong P=O bond formed

NMe2 H H H H H Cl NMe 2 NMe O N N N 2 N H SEAr H H iminium ion H hydrolysis attack at C-2 (e.g. Na CO , H O) (via linear conjugated intermediate) 2 3 2 Question: electrophilic substitution of pyrrole

§ What is the product of this Mannich reaction? H N A. 1 H B. 2 NMe2 O + Me2NH ? C. 3 ✔ H H H D. 4

H H N N O

H H O 1 2

H H N N

NMe 2 NMe2

3 4 Pyrrole: reaction at nitrogen

§ Pyrrole anion undergoes N-alkylation and N-acylation: N.B. - Anion in sp2 orbital (90˚ to plane of π-system), so cannot overlap with ring & subsequent reaction occurs at nitrogen. N sp2 - 2 steps: (1). Deprotonate pyrrole (NB: pKa 16.5), (2) Add electrophile. e.g. 1.

O

O NaH Cl Ph N H N N Ph

N-acylation

NB: without deprotonation would get reaction solely at carbon

e.g. 2.

LDA Me I N H N N Me

N-alkylation Pyrrole: summary

- Electronic structure of pyrrole.

- Difference between pyrrole, pyridine and benzene w.r.t. structure & reactivity.

- Recognise pyrrole-like 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

E+ (easy) E+ (difficult)

Base Base (difficult) (difficult) H N H Nu- Nu- (difficult) (easy) BENZENE PYRIDINE

E+ (very easy)

N

N H Base (very easy) N E+ (very easy) Nu- (very H easy)

PYRROLE PYRROLE ANION ACTIVATED PYRIDINE Useful reactions of another heterocycle:

§ 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.

NB: inversion of OH 1). Nu stereochemistry at * reacting carbon centre O 2). H3O Nu NB: acidic work-up to protonate alkoxylate SN2

- But… unsymmetrical epoxides lead to issues of regioselectivity (i.e. which end of epoxide reacts?). e.g.

OH Nu 1). Nu or O 2). H3O Nu OH

attack at most attack at less hindered end hindered end Useful reactions of other heterocycles

§ 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.

NB: inversion

OH OMe Na OMe or * O MeOH OMe OH

NB: epoxide oxygen is a poor LG (i.e. RO–), attack at less - so needs a strong Nu for SN2 reaction hindered end

SN2 reaction proceeds via a pentacoordinate T.S. hence minimise steric interactions

NB: for further details on epoxide ring opening, see ‘OC’ textbook p 438 Useful reactions of other heterocycles

§ 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. NB: inversion

OH OMe MeOH, HCl * or O H OMe OH

NB: epoxide oxygen is protonated, so it attack at more is a good LG (i.e. ROH) and does not hindered end – need a strong Nu for SN2 reaction

MeOH H (+) H OMe OMe

(+) (+) O (+) O (+) O O H H H H H H H

attack site best able to loose SN2 transition state stabilise partial +ve charge

NB: for further details on epoxide ring opening, see ‘OC’ textbook p 438 Question: epoxide opening

§ Identify the correct product from this epoxide ring opening.

A. 1 MeOH, HCl O ? B. 2 C. 3 ✔ D. 4

OH OMe

OMe OH 1 2

OH OMe

OMe OH 3 4 Synthesis of heterocyclic rings

§ In general: - Based on simple carbonyl chemistry (see previous recap. of key reactions). - Can disconnect the main bonds to reveal simple linear precursors. e.g. Pyrrole

R R R R + N HN R R R'NH2 O O O R' R'

pyrrole 1,4- compound

e.g. Pyridine

+ NH3 R R R N R R R O H2N O O

pyridine 1,5-dicarbonyl compound

NB: for more info on 5- and 6-membered ring synthesis beyond the scope of this course, see ‘OC’ textbook p 785 & 786. Synthesis of

§ 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). e.g.

R R AcOH Ludwig Knorr 1,4-dicarbonyl O O R R compound N R'NH 2 R' = H, , aryl R' pyrrole - Mechanism: e.g. NB: enamine H formation NH2 O OH AcOH cyclisation NH O N H Cl H O Ar Ar

± H H

OH2 N NB: enamine N N formation Cl Ar Ar hemiaminal NB: overall, lose 2 moles of water Question: synthesis of pyrroles

§ Identify the correct pyrrole product from this reaction

A. 1 O H N 2 AcOH B. 2 + ? C. 3 ✔ O D. 4

N

1 2

N

3 4

N N Synthesis of pyridines

§ 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. e.g. Arthur Rudolf Hantzsch

R R R O H pH 8.5 EtO2C CO2Et [O] EtO C CO Et EtO2C CO2Et 2 2 EtOH oxidation N O O H N NH3 1,4-dihydropyridine pyridine

NB: aldehyde provides extra carbon for pyridine ring (c.f. pyrrole synthesis)

NB: for further details of the Hantzsch pyridine ring synthesis, see ‘OC’ textbook p 763. Synthesis of pyridines

§ Mechanism: - Condensation with aldehyde to form α,β-unsaturated product (c.f. aldol reaction). R EtO O OH OH R O H H H E1 EtO2C EtO2C EtO2C EtO2C cb EtO2C R H OEt R R

O O O O O

α,β-unsaturated product

- α,β-Unsaturated product is good electrophile (i.e. 1,4-acceptor), so now occurs.

NB: 2nd equivalent of enolate used R R R EtO2C CO2Et EtO2C CO2Et EtO2C CO2Et

O O O O O O

conjugate addition 1,5-dicarbonyl compound

NB: for further details of the Hantzsch pyridine ring synthesis, see ‘OC’ textbook p 763. Synthesis of pyridines

§ Mechanism (continued): - Enamine formation, then intramolecular cyclisation of nitrogen onto other ketone. 1,4-dihydropyridine R R R R H EtO2C CO2Et EtO2C CO2Et EtO2C CO2Et EtO2C CO2Et NH3 O O O N N H2N H H enamine formation iminium ion formation enamine formation

- Final oxidation step is facile (energetic preference to become aromatic). - Occurs in air, or with chemical reagents such as DDQ.

R R O EtO2C CO2Et oxidation EtO2C CO2Et Cl CN

N H N Cl CN O

Dichlorodicyanoquinone NB: The mechanism for oxidation is beyond the scope of this course, (DDQ) but for further details see ‘OC’ textbook p 764 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- 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?

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.

Me H O N 1. POCl3 + ? H NMe2 2. Na2CO3, H2O

Q3. Identify the correct product A to E from the following nitration, explaining the regioselectvity.

HNO3 ? N OMe H2SO4

NO2 NO2 O2N

N OMe N OMe N OMe O2N N OMe N OMe NO2 A B C D E

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.

O Me HCl, MeOH Et ? i-Pr 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.