reactivity and control.pptx - the burton groupburton.chem.ox.ac.uk/reactivity-and-control.pdf ·...

Reactivity and Control for Organic Synthesis 1

Reac%vity and Control for Organic Synthesis

︎ Advanced Organic Chemistry: Parts A and B, Francis A. Carey, Richard J. Sundberg Organic Chemistry, Jonathan Clayden, Nick Greeves, Stuart Warren

Molecular Orbitals and Organic Chemical Reac<ons, Ian Fleming

Heterocyclic Chemistry, John A. Joule, Keith Mills

[email protected]

O

OO

MeMgBr

O

OO

MeCuBr•SMe2

chemoselective - enone reacts in preference to lactone (ester)regioselective - 1,4- not 1,2-addition to enone(dia)stereoselective - one major diastereomer formed

Which group reacts? Where does it react? How does it react?


︎ there are many different types of selec%vity in organic synthesis:

Chemoselec%vity – func%onal group discrimina%on

Regioselec%vity – product structural isomer discrimina%on

Stereoselec%vity – product stereoisomer discrimina%on ︎ we will be primarily concerned with:

Chemoselec%vity – i.e. selec%vity between two func%onal groups

Regioselec%vity – i.e. selec%vity between different parts of the same func%onal group

O

OO

MeMgBr

O

OO

MeCuBr•SMe2

chemoselec%ve -‐ enone reacts in preference to lactone (ester) regioselec%ve -‐ 1,4-‐ not 1,2-‐addi%on to enone (dia)stereoselec%ve -‐ one major diastereomer formed

O

OMe

O NaBH4 OH

OMe

O

Me

O

Me

Me MeMgBr

Me Me

MeMe OHCl

HNO3, H2SO4

Cl

NO2

ClNO2

+

ketone reduced in preference to ester

direct addi%on in preference to conjugate addi%on ortho and para products in preference to meta product


︎ acidity and basicity ︎ nucleophilic alipha%c subs%tu%on

nucleophilic aTack on carbonyl groups

electrophilic aroma%c subs%tu%on

nucleophilic aroma%c subs%tu%on

forma%on of rings

Hard and SoU

direct and conjugate addi%on

lithium halogen exchange, and directed lithia%on

reduc%ve amina%on


Brønsted (1879-‐1947) defini%on: an acid is a proton donor a base is a proton acceptor

HA H+ + A-‐

HA + H2O H3O+ + A-‐

acid HA is the source of a proton H+

remember that the solvent (not usually drawn) acts as the base and deprotonates the acid HA

HA + solvent solvent•H+ + A-‐

K = [H3O+][A-‐] [HA][H2O] ________ equilibrium constant water is in such large excess (as solvent)

that its concentra%on effec%vely does not change

________ Ka = [H3O+][A-‐] [HA]

What is the concentra<on of pure water?

i.e. Ka =K[H2O]

Ka > 1, equilibrium lies more to the right ∴ stronger acid i.e. HA is a stronger acid than H3O+ and A-‐ is a weaker base than H2O

Ka < 1, equilibrium lies more to the leU ∴ weaker acid i.e. A-‐ is a stronger base than H2O and HA is a weaker acid than H3O+

Ka values span a huge range ca. 1012 to 10-‐50 therefore much more convenient to use a logarithmic scale

conjugate base

conjugate acid


log10xy = log10x + log10y log10x/y = log10x -‐ log10y

Logarithms – a reminder the logarithm of a number is the exponent to which another fixed value (the base, b) must be raised to

produce that number

pKa = -‐log10Ka = -‐log10 ________ [H+][A-‐] [HA]

= -‐log10[H+] -‐ log10 ____ [A-‐]

[HA]

x = by logbx = y we will use log10 log10xa = alog10x

pKa = -‐log10Ka ∴ Ka = 10-‐pKa

higher pKa, smaller Ka, equilibrium lies more to the leU ∴ weaker acid

lower pKa, larger Ka, equilibrium lies more to the right ∴ stronger acid

pH

∴ pKa = pH -‐ log10 ____ [A-‐]

[HA]

rewri%ng the above gives the Henderson-‐Hasselbalch equa%on: ∴ pH = pKa + log10

____ [A-‐]

[HA]

if [HA] = [A-‐] then pH = pKa (remember log10 1 = 0)


pH = -‐log10[H+] at neutral pH, 7 = -‐log10[H+] ∴ [H+] = 10-‐7 M

at higher pH, [H+] < 10-‐7 and solu%on is basic (i.e. less acidic)

at lower pH, [H+] > 10-‐7 and solu%on is acidic. lower pH – more acidic higher pH – less acidic

for water pH = 7 this refers to the following equilibrium

H2O + H2O H3O+ + OH-‐

[H3O+][HO-‐] = Kw -‐ ionisa%on constant of water and is a constant in aqueous solu%on – its value is easy to find water has pH = 7 ∴ -‐log10[H+] = 7 ∴ [H+] = 10-‐7 M [H+] = [HO-‐] = 10 -‐7 M ∴ [H+][HO-‐] = 10-‐7•10-‐7 = 10-‐14 = Kw ∴ pKw = -‐log10Kw = 14

________ Ka = [H3O+][HO-‐] [H2O]


HA H+ + A-‐ pKa = -‐log10Ka ∴ Ka = 10-‐pKa

strong acids have Ka > 1 and ∴ pKa < 0 weak acids have Ka < 1 and ∴ pKa > 0 moderately strong acids Ka ≈ 1 and pKa ≈ 0 (generally view acids with pKa -‐2→+2 as moderately strong acids)

Calculate the pH of a 0.1 M solu<on of aqueous sodium hydroxide

strong acids include: pKa CF3SO3H -‐14 HCl ≈-‐7 H2SO4 ≈-‐3 H3O+ -‐1.74

weak acids include: pKa ace%c acid 4.76 NH4

+ 9.2 water 15.74 HC≡CH 25 NH3 38

Calculate the pKa of H3O+

the vast majority of organic compounds are weak acids

Note: the strongest base in aqueous solu<on is HO-‐ and the strongest acid is H3O+ any acid stronger than H3O+ is deprotonated by H2O to give H3O+

any base stronger then HO-‐ deprotonates H2O to give HO-‐


Worked example: how much acetylene would be deprotonated on treatment with hydroxide in aqueous solu9on?

in a mixture of two acids or two bases: the difference in the pKa’s gives us the log of the equilibrium constant, and the ra%o of the Ka’s gives us the equilibrium constant

HC≡CH + HO-‐ HC≡C-‐ + H2O

Keq = _____________ [HC≡C-‐][H2O] [HC≡CH][HO-‐]

Ka HC≡CH = ___________ [HC≡C-‐][H3O+] [HC≡CH]

________ [H3O+][HO-‐] [H2O]

Ka H2O =

∴ Keq = ______ Ka HC≡CH Ka H2O

10-‐25/10-‐15.74 = 1015.74/1025 = 10-‐9.3 =

i.e. only 1 in 1 billion molecules of acetylene would be deprotonated at equilibrium – to deprotonate acetylene use a solvent which does not have a pKa < 25 and use a stronger base – e.g. NaNH2 in liquid NH3

HC≡CH + H2N-‐ HC≡C-‐ + NH3

pKa NH3 = 38 pKa HC≡CH = 25 ∴ Keq = 10-‐25/10-‐38 = 1013

and


OH

H2OMeOH tBuOHCF3CH2OH

12.5(23.5)

15(28)

15.74(31.2)

9.95(18.0)

17(29.4)

Me OH

O

F3C OH

O

Ph OH

O

CO2HHO2C HO2C

CO2H

OH

O

O2N2.454.76

(12.3)-0.25 4.2

(11)3.02, 4.38 1.92, 6.23

O

HO OH

3.6, 10.3

CH4Ph Ph

PhHC CHH2

48 43 23 25 15~36

Remember: lower pKa = stronger acid; higher pKa = weaker acid

Me NH3 Me NH2

MeMe N Me

Me10.6 10.75

HMe Me

11.05

some pKa values in water and DMSO


we are going to generally look at pKa values in water

the majority of organic reac%ons are not conducted in water

generally pKa values when measured in organic solvent – typically DMSO – are higher then those measured in water

this is a consequence of the organic solvent being less good then water at solva%ng the conjugate base it is generally the case that the trend in pKa values in water and DMSO is very similar

pKa H2O in H2O = 15.74 pKa H2O in DMSO = 32; pKa AcOH in H2O = 4.76 pKa AcOH in DMSO = 12.3

when predic%ng or ra%onalising pKa values (i.e. the strengths of organic acids and bases) we need to consider three things: i)  strength of the H-‐A bond;

ii)  effect of hybridisa%on;

iii)  effect of conjuga%on/delocalisa%on

Most important factor in acid strength is the stability of the conjugate base A-‐


most important is to draw the equilibrium: then look at the stability of the conjugate base Worked example: explain why phenol is more acidic than methanol

Step 1: draw equilibria for both species

OH O+ H+

+ H+CH3OH CH3O

Step 2: evaluate stability of the conjugate base

i) delocalisa%on one of the oxygen lone pairs is in a ‘p’-‐orbital which can overlap with the π-‐system of the aroma%c ring

Remember: these structures are just different ways of drawing the same species – the charge is not actually moving around the ring

O O O O

equilibrium A

equilibrium B

Two factors work to stabilise the phenoxide anion

O


most important is to draw the equilibrium: then look at the stability of the conjugate base Predict which of the following two phenols is the stronger acid. OH

O2N

OH

NO2

ii) induc%ve effect The aroma%c subs%tuent is sp2 hybridized (vs sp3 hybridized in methanol) and hence has more ‘s’ character. The higher propor%on of ‘s’ character means that the electrons see more effec%ve nuclear charge. i.e. sp2 hybridised carbons are more electron-‐withdrawing (electronega%ve) than sp3 hybridised carbons

both of the above factors stabilise the phenoxide anion with respect to methoxide ∴ equilibrium B lies further to the right than equilibrium A and hence phenol is the stronger acid


Worked example: explain the following order of acid strengths methane (pKa = 48); benzene (pKa = 43); HC≡CH (pKa = 25)

Step 1: draw equilibria for the three species Step 2: evaluate stability of the conjugate base

CH4 CH3 + H+

+ H+

HC CH HC C + H+

eq. A anion in sp3 orbital – 25% s-‐character

eq. B anion in sp2 orbital – 33% s-‐character

eq. C anion in sp orbital – 50% s-‐character

acetylide anion more stable than C6H5-‐ which is more stable then CH3

-‐

∴ equilibrium C lies further to the right than equilibrium B which lies further to the right than equilibrium A and hence acidity order is as shown.

H HH

HC C

Explain the acidity of the following compounds

CN

H3C

pKa (DMSO) 30.8

H3C

43

CH3

44 18


how about bases? Brønsted -‐ base is a proton acceptor. There are two ways to deal with bases. Let’s start with a base A-‐

HA + HO-‐ A-‐ + H2O

________ Kb = [HA][HO-‐] [A-‐]

pKb = -‐log10Kb

stronger base – lower pKb weaker base – higher pKb

it is inconvenient to have two scales and chemists just use pKa to talk about the strengths of acids and bases i.e. look at the ability of the conjugate base to act as a base.

________ ∴ Kb = [HA]Kw [A-‐][H3O+]

________ Ka = [H3O+][A-‐] [HA]

∴ Kb =Kw/Ka

∴ pKb =14 -‐ pKa

HA + H2O H3O+ + A-‐ acid

base conjugate acid

conjugate base

Kw = [H3O+][HO-‐]


How do you find out which is the stronger base – t-‐butoxide, or acetate? look at the pKa’s – here tBuOH holds onto the proton to a much greater extent than ace%c acid, or to put it another

way tBuO-‐ much more readily accepts a proton than acetate and hence tBuO-‐ is a stronger base.

tBuOH tBuO + H+

Me OH

O

Me O

O+ H+

pKa = 17

pKa = 4.76

Higher pKa = weaker acid and hence stronger conjugate base. Lower pKa = stronger acid and hence weaker conjugate base.

Example butane is a very weak acid (pKa ~ 43) but butyllithium is a very strong base

H2SO4 is a strong acid (pKa -‐3.0) but HSO4-‐ is a very weak base.

The problem of amines what do we mean by the ques%on “what is the pKa of ammonia?”

strictly speaking this refers to the following equilibrium:

NH3 NH2 + H+ pKa ~ 38 i.e. ammonia is a very weak acid and H2N-‐ is a very strong base

but we might be asking, how good a base is NH3 – which means we need the pKa of ammonium NH4+

NH4 NH3 + H+pKa ~ 9.2 i.e. NH3 is a weaker base than HO-‐

(pKa H2O = 15.74) to get around this possible ambiguity we should be specific in asking for the pKa of the conjugate acid of ammonia (some%mes given the symbol pKaH) i.e. of ammonium


compound pKa (water)

6.95

5.21

4.76

H2CO3 3.6, 10.3

4.6

-‐0.25

CF3SO3H -‐14

Important to know some pKa’s

for excellent tabulated pKa values for a large number of organic compounds see: hVp://evans.harvard.edu/pdf/evans_pka_table.pdf and hVp://www.chem.wisc.edu/areas/reich/pkatable/index.htm


CH4 48

NH3 38

25

25

20

tBuOH 17

MeOH 15

HC CH

Me OtBu

O

Et Et

O


15

13

11

10

10

9.24

9

NH4

Me

O

NH2O

MeO OMe

O

O

EtO Me

O

CH3NO2

OH

Me

O O

Me

Me

O

OH

NH

F3C

O

OH

HN NH

NH3

comparing any 2 acids: conjugate base of acid with higher pKa will deprotonate acid with lower pKa e.g. BuLi will deprotonate HC≡CH; NaOMe will deprotonate dimethyl malonate etc.


1) Explain the following pKa orders:

2) Which of the following is more basic?

(a) (c)

NH

NHMe O-Na+ Me S-Na+

(a) (b)

(d) (b)

(c)

MeN

N

N

Me

Me Me

3) Explain the pKa’s of maleic and fumaric acid:

CO2HHO2C HO2C

CO2H

3.02, 4.38 1.92, 6.23

OHOH

NO2 NO2

OH

MeMe< <

most acidiclowest pKa

least acidichigest pKa

Me

O

Me

O

Me

O

OMe

O

MeO

O

OMe

O< <



Me

O

Me

O

Me

O

Me

Me

OMe



< <NH

NH

Me

Me

Me

Me

MeMe

NH

MeMe

> >




4) How might you carryout the following transforma<ons?

O

OH

HTBDPSO

O

OMe

HTBDPSO

Me

O O

OMe Me

O O

OMeMe

O O

OMeMe

HO

OH

O

MeO

OH

O

HO

OMe

O

(a)

(b)

(c)

(d) NH2

HO

HN

HO

Me

O

HONH2

OH

OH

OH

OHAcO

NH2OAc

OAc

OAc

OAc

(e)


PhMe

OMe

O

enantiopure

LDA, Br

5) Explain the following transforma<ons?

(a)

(b)

O O

OEtOEt

O

O

OEt

EtO, EtOH

O O

OEtEtO, EtOH

O O

OEt

MeMe

6) In water, the basicity of the amines below is as follows, explain.

Me NH2 Me NH

MeMe N Me

Me

< <

7) Predict the product of the following reac<on.


“When two molecules collide, three major forces operate. (i) The occupied orbitals of one repel the occupied orbitals of the other. (ii) Any posi%ve charge on one aTracts any nega%ve charge on the other (and repels any posi%ve). (iii) The occupied orbitals (especially the HOMOs) of each interact with the unoccupied orbitals (especially the LUMOs) of the other, causing an aTrac%on between the molecules.” Molecular Orbitals and Organic Chemical Reac9ons I. Fleming

this means that reac%ons generally have both a charge and an orbital component. Reac%ons can be predominantly charge controlled, predominantly orbital controlled, or a mixture of the two.

nucleophiles and electrophiles were classified by Pearson as HARD or SOFT – R. G. Pearson, Chemical Hardness, John Wiley & Sons, 1997. Hard nucleophiles (and electrophiles) are small and highly charged and have high electronega%vity (i.e. have a large

charge:radius ra%o) – they have a low energy HOMO.

Sob nucleophiles (and electrophiles) are larger and have lower electronega%vity and are more polarizable – they have a high energy HOMO. Bases (Nucleophiles) Acids (Electrophiles)

Hard Hard

H2O, HO-‐, F-‐, RCO2-‐, Cl-‐, ROH,

RO-‐, NH3, RNH2 H+, Li+, Na+, K+, Mg2+, BF3

Intermediate Intermediate

PhNH2, N3-‐, NC-‐, Br-‐ carboca%ons

SoN SoN

I-‐, RS-‐, RSe-‐, S2-‐, RSH, RSR, R3P, alkenes, aroma%cs, R-‐

Ag+, Pd2+, I2, Br2, radicals


Generally the case that: Hard nucleophiles tend to react well with hard electrophiles i.e. the reac<on is predominantly charge controlled Sob electrophiles tend to react well with sob nucleophiles i.e. the reac<on is predominantly orbital controlled

Hard nucleophiles have low energy HOMO’s and a high charge:radius ra%o Hard electrophiles have high energy LUMO’s and a high charge:radius ra%o charge dominates their reac%vity

SoU nucleophiles have high energy HOMO’s and they are polarizable SoU electrophiles have low energy LUMO’s and they are polarizable orbital interac%ons dominates their reac%vity

hard:hard charge control

soL:soL orbital control

hard nucleophile

hard electrophile

soft nucleophile

soft electrophile

Recently the HSAB theory has been disputed see: H. Mayr, M. Breugst, A. R. Ofial, Angew. Chem. Int. Ed., 2011, 50, 6470


Generally the case that: Hard nucleophiles tend to react well with hard electrophiles i.e. the reac<on is predominantly charge controlled Sob electrophiles tend to react well with sob nucleophiles i.e. the reac<on is predominantly orbital controlled

the principle of hard/soU acid bases has been applied to a large number of chemical reac%ons

Recently the HSAB theory has been disputed see: H. Mayr, M. Breugst, A. R. Ofial, Angew. Chem. Int. Ed., 2011, 50, 6470

HO H OH

H2H2O faster than Br BrHO HO Br + Br

BrBr

Brfaster than H O

H

HH

Bases (Nucleophiles) Acids (Electrophiles)

Hard Hard

H2O, HO-‐, F-‐, RCO2-‐, Cl-‐, ROH,

RO-‐, NH3, RNH2 H+, Li+, Na+, K+, Mg2+, BF3

Intermediate Intermediate

PhNH2, N3-‐, NC-‐, Br-‐ carboca%ons

SoN SoN

I-‐, RS-‐, RSe-‐, S2-‐, RSH, RSR, R3P, alkenes, aroma%cs, R-‐

Ag+, Pd2+, I2, Br2, radicals


Ambident nucleophiles

alkyla%on of enolates

with enolates the majority of the charge is, as expected, on the oxygen atom

charged electrophiles aTack oxygen e.g. protons, carboca%ons

soU electrophiles will generally aTack carbon – largest HOMO coefficient

in general, in enolate reac%ons the oxygen atom is associated with a metal ion and solvent and hence both of these variables affect the ra%o of C:O alkyla%on

to maximise C-‐alkyla%on use a lithium base (strong O-‐Li bond) and an alkyl halide in THF (soU-‐soU interac%ons)

to maximise O-‐alkyla%on use a highly coordina%ng solvent (e.g. HMPA), a potassium base, and an alkyl sulfonate

O :BaseH

O MRX

OR

OR

+

O-‐alkyla%on C-‐alkyla%on

OPh Br+

O Ph OH

Ph

DMF 97% CF3CH2OH 7%

0% 85%


HNO

R

CO2tBu

R'

NEtO

R

CO2tBu

R'

K2CO3, Et3O BF4

CH2Cl2

HNO

R

NaH,

O

O

H

CO2tBuMeO

ClNO

RO

O

H

CO2tBu

MeO

DMF

Ambident nucleophiles X = A B

OTs 88% 11%

Cl 60% 32%

Br 39% 38%

I 13% 71%

A. L. Kurts, A. Masias, N. K. Genkina, I. P. Beletskaya, O. A. Reutov, Tetrahedron, 27, 4777

in a similar manner, deprotonated secondary amides alkylate on nitrogen with alkyl halides

neutral secondary amines alkylate on oxygen with hard alkyla%ng agents

A. Endo, S. J. Danishefsky, J. Am. Chem. Soc., 2005, 127, 8298.

Explain these two transforma<ons

O

OEt

OK Me X O

OEt

O

Me

O

OEt

O

Me

+HMPA

A B


XMe

MeMe

NuMe

MeMe

MeMeMe

Nu

Nucleophilic SubsTtuTon at a saturated carbon two limi%ng mechanis%c cases – SN2 and SN1 – mechanis%c con%nuum between these extremes SN2 – subs<tu<on, nucleophilic bimolecular

example: MeI + NaOH → MeOH + I-‐

rate = k[substrate][nucleophile] i.e. rate dependent on both substrate and nucleophile

concerted reac%on, single transi%on state no intermediate is formed

SN1 – subs<tu<on, nucleophilic unimolecular

example: MeCl

MeMe

H2O MeOH

MeMe

rate = k[substrate] i.e. rate is independent of nucleophile

NuR

R'R''

LGR

R'R''

R

R' R''LGNu

(-) (-)

‡

Nu

stepwise reac%on, via an intermediate -‐ the 1st step is rate determining (forma%on of C+), 2nd step is fast

favoured by 1° substrates and some 2° substrates requires good nucleophile and leaving group

favoured by 3° substrates and some 2° substrates requires good leaving group and solvent that stabilises

carboca%ons


X

R R'R''

RR''

R'X

R R'R''

‡

(+)

(-)Nu Nu

R R'R'' Nu

R R'R''step 1 step 2

SN1 reac%on

SN1 reac%ons proceeds via a discrete carbenium ion and forma%on of the carbenium ion (step 1) is usually the rate determining step the lowest energy conforma%on of carbenium ions is planar

trapping of the carbenium ion by a nucleophile (step 2) is generally fast

Nomenclature of carboca%ons proposed by Olah J. Am. Chem. Soc., 1972, 94, 808.

hyperconjuga%on is the overlap of filled C-‐H (or C-‐C) σ-‐bonding orbital with the empty p-‐orbital resul%ng in a lowering in energy of the system i.e. stabilisa%on

CH3

CH3

H

HH

the nucleophile can trap the carbenium ion from either side, hence enan%oenriched substrates should be expected to give racemic products under SN1 condi%ons – c.f. SN2 reac%ons go with strict inversion of configura%on hence the rate of the reac%on is not affected by the added nucleophile

the stability of carbenium ions is in the order ter%ary > secondary > primary due to hyperconjuga%on

enanTomers


R R > R R > Rtertiary secondary primary

R

filled σ C-H orbital

empty p-orbital

energy of the bonding electrons reduced system stabilised

greater number of C-‐H (or C-‐C) σ-‐bonds the greater the extent of hyperconjuga%on and the greater stabilisa%on

CH3

CH3

H

HH

dona%on of C-‐H σ-‐bond electrons in empty p orbital

carbenium ion stability therefore goes in the order:

carbenium ions have been observed by NMR and X-‐ray crystal structure analysis

a recent X-‐ray structure of the t-‐butyl ca%on (anion is CHB11Cl11) shows the planar nature of the carbenium ion. E. S. Stoyanov, I. V. Stoyanova, F. S. Tham, C. A. Read; Angew.Chem., Int.Ed. 2012, 51, 9149

conjuga%on with alkenes, arenes and lone pairs, also stabilises carbenium ions

HyperconjugaTon


X

conjuga%on with alkenes, arenes and lone pairs, also stabilises carbenium ions

X

benzyl ca%on stabilised by delocalisa%on

X

allyl ca%on stabilised by delocalisa%on energy of

isolated p-‐orbital

ψ1

ψ2

ψ3

allyl ca%on more stable than energy of p-‐orbital – conjuga%on is stabilising

RO X RO RO

α-‐heteroatom subs%tuted ca%ons stabilised by delocalisa%on

Which orbitals are overlapping in the stabilisa<on of α-‐heteroatom-‐subs<tuted ca<ons?


the more stable the carbenium ion the faster SN1 reac%on

rates of hydrolysis of alkyl chlorides in 50% aqueous ethanol (adapted from Organic Chemistry, Clayden, Greeves and Warren, 2nd Edi%on, OUP 2012)

1° chloride ∴ SN2

2° chloride, not that stable C+ not good at SN1 1° but allylic 1° but benzylic

allylic ca%on is 2° at one end

3° chloride very good at SN1

allylic ca%on is 3° at one end

Cl

butyl

Cl

iso-propyl

Cl

tert-butyl

Cl

ClCl

Cl

Cl

benzylallyl

methallyl dimethallylcinamyl

0.07 0.12

2100

1.0 4.0

91 1300007700

1° but allylic and benzylic


as a carbenium ion is formed during an SN1 reac%on a polar solvent is required for reac%on

the best solvents for promo%ng SN1 reac%ons are polar pro%c solvents such as water and alcohols (they can readily solvate the carbenium ion as well as the leaving group (by hydrogen bonding – see later))

rela%ve rate of solvolysis (i.e. reac%on with solvent as the nucleophile) of tert-‐butyl bromide is 3 x 104 %mes faster in 50% aqueous ethanol than in neat ethanol

solvent water ethanol ace%c acid DMSO DMF

dielectric constant ε 80 25 6.2 46 38

solvent acetone EtOAc THF ether hexane

dielectric constant ε 21 6 7.5 4.3 1.9

In polar solvent the carbenium ion is solvated by polar solvent. It is easier to cluster water molecules around the carbenium ion and the leaving group than ethanol molecules

Explain the rela<ve rates of solvolysis of the following alkyl bromides in 80% aqueous ethanol

BrBr Br

1 10-6 10-14

LG

H

H

H

H

OH

OH

HO

HO

δ+

δ+

δ+

δ+δ-

δ-

δ-

δ-

+HOH H

OH

H OH

HOH

δ-

δ-δ-

δ-


H

I

HH

Nucleophilic SubsTtuTon at a saturated carbon

SN2 – subs<tu<on, nucleophilic bimolecular example: MeI + Na+OH-‐ → MeOH + Na+ I-‐

rate = k[substrate][nucleophile]

at the transi%on state the central carbon atom is bonded to 5 other atoms – hence fundamentally SN2 reac%ons are difficult reac%ons

the trajectory of approach is along the path of the bond to the leaving group – evidence from Eschenmoser’s experiments

HOMO of nucleophile (nucleophile lone pair) aVacks the back side of the carbon atom as it is pulng electrons into the C-‐I σ* orbital

reac<on shown to be exclusively intermolecular Eschenmoser, Helve<ca Chim. Acta 1970, 53, 2059

HI

HHHO

HHO

HH I

H

H H

IHO(-) (-)

‡

SO

O O

CH3

S OO

H3C

NaH

SO

O O

CH3

S OO

H3C

SOH

O O

S OO

H3C

CH3X


trajectory results in inversion of configura%on between star%ng material and product

What makes a good nucleophile? – i.e. what gives a fast reac%on with an electrophile nucleophilicity is related to basicity but is significantly more complex

if the atom we are comparing is the same then nucleophilicity does parallel basicity.

basicity is a measure of electron pair dona%on to a proton (generally under equilibra%ng condi%ons)

nucleophilicity is electron pair dona%on to another atom, frequently carbon, generally under kine%c condi%ons

factors which influence nucleophilicity include: charge, electronega%vity, solvent, size, bond strength Note: the order of nucleophilici<es is also dependent on the nature of the leaving group

HO PhOMe O

OH2O ClO4> > >>

the rate of an SN2 rec%on is influenced by the nature of the substrate, the nucleophile, the leaving group and, with anionic nucleophiles, the associated counterion, and the solvent

MeLG

EtHNu

MeNu

EtH LG

Me

Et H

LGNu(-) (-)

‡


Charge charged species are more nucleophilic than their neutral counterparts this is expected as nucleophiles are electron pair donors so the more electron rich the nucleophile the beTer donor it is

Me O

O

more nucleophilic than Me OH

O NHmore nucleophilic than

NH2

Me S more nucleophilic than Me SH BuLi is obviously more nucleophilic than butane

ElectronegaTvity nucleophilicity is related to basicity, but significantly more complex as it involves dona%on of an electron pair to any

atom, whereas basicity is dona%on of an electron pair to H+ in the same row of the periodic table more basic means more nucleophilic

∴ going from leU to right across the periodic table nucleophilicity decreases the more electronega%ve atom is the

weaker nucleophile as it holds on to its lone pairs of electrons more %ghtly and is less able to donate an electron pair to form a bond.

CH3 > NH2 HO F> > NH3 H2O HF> >

most basic most nucleophilic

least basic least nucleophilic



this does not necessarily mean we will get good yields in SN2 reac%ons with these anions as they are also very basic and hence other reac%on pathways can dominate


Solvent in polar pro%c solvents (e.g. water, MeOH, AcOH) nucleophilicity increases going down the group – again the less

electronega%ve atom is the more nucleophilic

in polar pro<c solvents the most electronega<ve atom has the highest charge density and forms the strongest hydrogen bonds with the solvent therefore the nucleophile is shielded from aVacking the electrophile and the reac<on is slower

most nucleophilic least nucleophilic F

H

H

H

H

OR

OR

RO

RO

δ+

δ+

δ+

δ+δ-

δ-

δ-

δ-

F < Cl Br I< <

in polar apro%c solvents (e.g. DMSO and DMF) the order of nucleophilicity can invert when compared with polar pro%c solvents as the solvent has weaker interac%ons with the nucleophile. Frequently reac%ons are much faster in these solvents compared with in water

for the halides under some condi%ons, nucleophilicity now decreases going down the group and again parallels basicity (here the most electronega%ve atom is the best nucleophile). Here charge control appears to be domina%ng the reac%on F > Cl Br I> >



MeI + Cl MeCl + I


with uncharged nucleophiles, nucleophilicity increases going down the group – here orbital control appears to be domina%ng – the nucleophile with the highest energy HOMO reacts the fastest

higher energy HOMO most nucleophilic

lower energy HOMO least nucleophilic

A rule of thumb is that nucleophilicity increases going down a group and increases in moving from right to leU in the periodic table

Note: nucleophilicity is complicated and the above should be viewed as guidelines

PR3 > NR3

H2Se > H2S H2O>C N O F

Si P S Cl

Ge As Se Br

Sn Sb Te I

Pb Bi Po At

increasing nucleophilicity


the shape of the nucleophile also influences its nucleophilicity in moving from the star%ng materials to the transi%on state the central carbon goes from 4-‐coordinate to 5-‐

coordinate hence sterically hindered nucleophiles react more slowly

MeO > >O O

fastest slowest

conversely, small linear anions such as N3-‐, NC-‐ and RC≡C-‐ are good nucleophiles


Leaving Group

the following is the order of reac%vity of various nucleophiles with methyl iodide in methanol – all of these anions would be considered good nucleophiles (from Chem. Rev. 1969, 69, 1-‐32) – PR3 are also excellent nucleophiles PhS-‐ > I-‐ > SCN-‐ ≈ CN-‐ > N3

-‐ ≈ Br-‐ > Cl-‐ > OAc-‐ in polar pro%c solvents PhS-‐ > CN-‐ > -‐OAc > Cl-‐ ≈ Br-‐ ≈ N3

-‐ > I-‐ > SCN-‐ in dipolar apro%c solvents

during the SN2 reac%on the bond to the leaving group is broken and the LG departs with a lone pair of electrons i.e. becomes more nega%vely charged in the transi%on state ∴ two factors generally influence the leaving group ability: i)  the strength of the bond to carbon ii)  the stability of the leaving group

MeLG

EtHNu

MeNu

EtH LG

Me

Et H

LGNu(-) (-)

‡


leaving group ability relates to pKa i.e. good LG’s are weak bases

rough order of LG ability – the LG ability depends on the nucleophile and the solvent and the above order can vary; however, very weak bases are good leaving groups

iodide is a good leaving group as it forms a weak bond to carbon as well as being a stable anion

F-‐ is a very poor leaving group in SN2 reac%ons as it forms a very strong bond to carbon

HO-‐ is a very poor leaving group in SN2 reac%ons as it is a strong base (pKa H2O = 15.74) but can be made into a

good leaving group by protona%on (pKa H3O+ = -‐1.74) or conversion into a tosylate or triflate

> Br Cl>N2OSO

OF3CMe

SO

O O

>

pKa -14 -3-10 -9 -8

> I ≈

triflate TfO-‐ tosylate TsO-‐

Common leaving groups in SN2 reac%ons tend to have a pKa < 2


Nature of the substrate

SN2 reac%ons – back to the transi%on state

RLG

R'R''Nu

RNu

R'R'' LG

R

R' R''

LGNu(-) (-)

‡

the nucleophile has to aTack carbon, hence with larger the R groups the rate of reac%on decreases

in moving from substrate to transi%on state carbon moves from being 4-‐coordinate to being 5-‐coordinate hence as the R groups become larger the rate of the reac%on decreases

SN2 reac%ons ∴ only occur with primary and some secondary substrates – not with ter%ary substrates

rela%ve rates of the reac%on of the bromides below with chloride are (Chem. Rev. 1956, 56, 571):

Me BrMe Br MeBr

Me Br

Me

MeBr

MeMe

Me

Me BrMe

relative rate 1060 6.5 0.13 0.0003 negligible

methyl ethyl propyl iso-propyl neo-pentyl tert-butyl

neopentyl bromide is par%cularly unreac%ve as the nucleophile is severely hindered from aTacking the necessary carbon atom (Note: neopentyl systems are also unreac<ve in SN1 reac<ons)

MeLG

HHNu Nu

R'R'' LG

H H

LGNu(-) (-)

‡Me

Me Me MeMe

MeMeMe


Nu LG

HH

LG

HH

Nu

H H

NuLG

(-)(-)

‡


rate of SN2 reac%ons is increased with substrates which carry an adjacent sp2 (or sp) hybridized atom

at the SN2 transi%on state the central carbon is partly bonded to both the nucleophile and the leaving group

3 atoms are sharing 4 electrons i.e. there is a 3-‐centre,4-‐electron bond the central carbon has a partly filled p-‐orbital and the electrons in this orbital can be delocalised into the adjacent π-‐

system which lowers the energy of the transi%on state and the reac%on is faster

delocalisa%on into π-‐system


Nu LG

HH

LG

HH

Nu

H H

NuLG

(-)(-)

‡


the π-‐system has to be in the correct orienta%on for efficient overlap and consequent transi%on state stabiliza%on

α-‐halo carbonyl compounds are par%cularly reac%ve under SN2 condi%ons as they contain an α sp2 hybridised atom aTached to oxygen and the C=O π* is lower in energy than for an alkene

Cl Cl O

PhClClMe Me

Cl ClNMeO Cl

relative rate 1 200 79 200 3000920 100,000

rela%ve rates for reac%on alkyl halides with KI in acetone at 50 °C are given below (from Mechansim in Organic Chemistry, R. W. Alder, R. Baker, J. M. Brown, Wiley, 1971)

delocalisa%on into π-‐system


Summary of structural varia%ons and nucleophilic subs%tu%on taken from Organic Chemistry, Clayden, Greeves and Warren, 2nd Edi%on, OUP 2012.

Electrophile

methyl primary secondary ter%ary ‘neopentyl’

SN1 mechanism? ✗ ✗ ✗✓ ✓✓ ✗

SN2 mechanism? ✓✓ ✓ ✗✓ ✗ ✗

Electrophile

allyl benzyl α-‐alkoxy α-‐carbonyl α-‐carbonyl and ter%ary

SN1 mechanism? ✓ ✓ ✓ ✗ ✗

SN2 mechanism? ✓ ✓ ✓ ✓✓ possible

X X RO XR

OX R

OX

Me X R XR

R

X R

RR

X R

RR

X

✗ = bad ✓ = good, ✓✓ = excellent, ✗✓ = poor


1) Explain why the reac<on of B with iodide is several <mes faster than the reac<on of A with iodide

O

OO2N

NO2

KIacetone

O2N

NO2OH

OI O2N

O

O

O

PhNO2

KIacetone

O2NO

OHNO2

Ph

OI

BA

2) For the reac<on below predict the effect on the rate of changing the: i) substrate to i-‐propyl chloride; ii) substrate to methyl iodide; iii) nucelophile to CH3SNa; iv) solvent to DMSO

MeCl + NaOMeMeOH

CH3OCH3 + NaCl

3) Suggest reagents for the following reac<ons:

NMe

NMe Me

I HOCl

O

H

H

OH

H

H

SMeN

OH

Br Br

(a) (b)

(c) (d)


O AlCl3

Cl

O

4) Explain the outcome of the following reac<ons

Et2NCl

MeNaOH HO

NEt2

Me

ClNBn2

EtH2O Bn2N

OH

Et

BrOH

HBrBr

BrBr

OHHBr

BrBr

OH

Cl

H NaOH, H2OOH

OH

H

(a) (b)

(c)

5) Predict the outcome of the following reac<ons

O

Cl

Cl

MeOH, Et3N

PhO MeOH, H

Ph

OMeOH

Ph

OHOMe

MeO, MeOH(d)


EliminaTon reacTons mechanis%c con%nuum from E1→E2→E1cB

E2 – elimina<on bimolecular

example:

rate = k[substrate][base] i.e. rate dependent on both substrate and base

concerted reac%on, single transi%on state no intermediate is formed, an%periplanar arrangement of proton and leaving group is most favourable for elimina%on

E1 – elimina<on unimolecular

example:

rate = k[substrate] i.e. rate is independent of base (which is EtOH in the above case)

stepwise reac%on, via an intermediate -‐ the 1st step is rate determining (forma%on of C+), 2nd step is fast

requires good base and leaving group favoured by 3° substrates and some 2° substrates

requires good leaving group and solvent that stabilises

carboca%ons

MeBr

MeMe

EtOH

MeMe+ HBr

XMe

MeMe

MeMe

B

H

HH

B

Me

Me

HBr

R+ EtO R

EtOH

Br+

HX

B

‡

HX

B(-)

(-)

BH X

HX

C-H σ to C-X σ*

3° substrates give more elimina%on than 2° substrates which give more elimina%on than 1° substrates


EliminaTon reacTons

E1cB – elimina%on from the conjugate base example:

variable kine%cs depending on substrate

requires a carbanion stabilising group

requires a base and leaving group

RO O

Me+ HO RO O

Meconjugate basebaseacid

O

Me+RO

as the carbanion (an enolate in the above example) helps to expel the leaving group, conjuga%on is developed in the transi%on state leading to the product, HO-‐, and RO-‐ can ∴ be leaving groups

SubsTtuTon versus EliminaTon

SN1 reac%ons are frequently accompanied by E1 reac%ons if there is an appropriately posi%oned proton – this is unsurprising as both reac%ons proceed through the same intermediate

SN2 reac%ons can also be accompanied by E2 reac%ons

we need to look at factors affec%ng: i)  SN1/E1 product ra%os

ii) SN2/E2 product ra%os

iii) change of mechanism i.e. SN1/E1 → SN2/E2


primary substrates do SN2 or E2 – SN2 is generally favoured but bulky bases (t-‐BuOK) allow E2 to occur

ter%ary substrates do SN1 / E1 or E2

with good ionising solvents and no added anionic base then SN1 / E1 will be favoured

with added base E2 will be favoured

secondary substrates can do SN1 / E1 or SN2 / E2

with good ionising solvents and no added anionic base or nucleophile then SN1 / E1 will be favoured

with added base E2 will be favoured

with good nucleophiles, dipolar apro%c solvents SN2 will be favoured

BrEtO

OEt

91 9

BrEtO

OEt

<0.1 >97

BrEtO

OEt

20 80

to maximise SN2 – use good nucleophile e.g. RS-‐, X-‐, N3-‐ in dipolar apro%c solvent


for all substrates moving from primary to secondary to ter%ary favours elimina%on

polar pro%c solvents and base favour E2 over SN2 [base/nucleophile is solvated so easier to aTack outside of the molecule (i,e. remove a proton) than aTack carbon in an SN2 reac%on]

heat favours elimina%on over subs%tu%on.

increased branching at the β-‐posi%on favours elimina%on

BrEtO

OEt

40 60

A good overview can be found at: hTp://www.masterorganicchemistry.com/2013/01/18/wrapup-‐the-‐quick-‐n-‐dirty-‐guide-‐to-‐sn1sn2e1e2/


1) Predict the major product (if any) from the following reac<ons giving mechanis<c reasoning

2) The rate of hydrolysis of tBuCl in water is greatly accelerated by the addi<on of hydroxide. How will the product distribu<on be affected?

Br EtOH

Br EtSNa

EtSH

Br EtONa

EtOH

Br

EtONa

EtOH

(a) (b)

(c) (d)


as we have seen, allylic systems are reac%ve under SN2 (stabilisa%on of the transi%on state for subs%tu%on) and under SN1 condi%ons

the allylic system has two posi%ons which can be aTacked leading to isomeric products – i.e. there are issues of regioselec%vity

X

Nu

Nu

X

Nu

Nu

X

Nu

Nu

Nu

Nu

SN2

SN2’ SN1 SN1’

sterics and electronics play a role in determining SN/SN’ reac%ons

O

OEt

O

EtOCl

O

OEt

O

EtOEtO O Br O+

Cl

EtO EtO

Cl

PhS PhS


SN2’ reac%ons are not very common they can also be solvent dependent, I. Fleming, E. J. Thomas, Tetrahedron, 1971, 28, 4989

MeO

Cl

Cl MeO

Cl

SPh

MeO

Cl

SPh

PhS

MeO

Cl

MeO

Cl

Cl MeO

Cl

SPh

SPh

MeO

Cl

Cl

MeO

Cl

Cl

DME

PhS

DME

PhS

MeOH

PhS

MeOH

SN2’ SN2

excellent control of SN2/SN2’ can achieved with organometallic reagents – most notably with copper, palladium and iridium

OAc

10 mol% CuCNn-BuMgBr

BuBu

THF 0 °C 94 6

Et2O, 20°C 3 97J. E. Bäckvall, M. Sellén, B. Grant, J. Am. Chem. Soc., 1990, 112, 6615. For a review on copper-‐catalysed enan%oselec%ve conjugate addi%on and allylic subs%tu%on see: A. Alexakis, J. E. Bäckvall, N. Krause, O. Pàmies, M. Diéguez Chem. Rev. 2008, 108,2 796.


main names in the field: Helmchen, Alexakis, Hartwig, Carreira, You

R O OMe

Ocat. [Ir(COD)Cl]2

MeN

MeP

O

O

Nu +additive R

Nu

*

Nu = RNH2, ArO-, malonates, enamines, silylenol ethers, indoles, PhMgBr, NH3,alkenes, vinyltrifluoroborates etc.

high yieldshigh regioselectivityhigh enantiomeric excess

and related phosphoramidites

OH Br

K2CO3, DMF

O heat OH


OH OH

2) Suggest reagents and reac<on condi<ons for the following transforma<on


AromaTcity – Electrophilic AromaTc SubsTtuTon and Nucleophilic AromaTc SubsTtuTon

“I was silng wri<ng at my textbook but the work did not progress; my thoughts were elsewhere. I turned my chair to the fire and dozed. Again the atoms were gambolling before my eyes. This <me the smaller groups kept modestly in the background. My mental eye, rendered more acute by the repeated visions of the kind, could now dis<nguish larger structures of manifold confirma<on: long rows, some<mes more closely fiVed together all twining and twis<ng in snake like mo<on. But look! What was that? One of the snakes had seized hold of its own tail, and the form whirled mockingly before my eyes. As if by a flash of lightning I awoke; and this <me also I spent the rest of the night in working out the rest of the hypothesis. Let us learn to dream, gentlemen, then perhaps we shall find the truth... But let us beware of publishing our dreams ‘<ll they have been tested by waking understanding.”

August Kekulé


retains aroma%c sextet of electrons in subs%tu%on reac%ons does not behave like a “normal” polyene or alkene benzene is both kine<cally and thermodynamically very stable

typical reac%ons of alkenes

typical reac%ons of benzene

heats of hydrogena%on

ΔHohydrog = -‐120 kJmol-‐1

ΔHohydrog= 3 x -‐120 = -‐360 kJmol-‐1

(hypothe%cal, 1,3,5-‐cyclohexatriene) ΔHo

hydrog= -‐210 kJmol-‐1

benzene ≈150 kJmol-‐1 more stable than expected – (represents stability over hypothe%cal 1,3,5-‐cyclohextriene) – termed the empirical resonance energy (values vary enormously)

Me + Br2

fast

Me

Br

Braddition

MeBr not substitution

+ Br2FeBr3 catalyst Br

substitution

+ HBr not additionBr

Br

H2/Pt catalyst H2/Pt catalyst

H2/Pt catalyst


AromaTcity Hückel’s rule holds for anions, ca%ons and neutrals

(4n +2) π-‐electrons for aroma%c compounds; 4n π-‐electrons for an%-‐aroma%c

cyclopropenium ca%on -‐ (4n +2), n = 0, 2π electrons

insoluble in non-‐polar solvents; 1 signal in 1H NMR δH = 11.1 ppm -‐ aroma%c and a ca%on

compare with cyclopropyl ca%on which is subject to rearrangement to the allyl ca%on

Cl

SbCl5(Lewis acid)

SbCl6

H

H

H

ClNu Nu

Ph

Ph

Ph

Ph

Ph

Ph

Ph

Ph

Ph

Ph

Ph

Ph

6.3 D

Ph

Ph

Ph

Ph

Ph

Ph

reduced barrierto rotation

2π-‐aroma%c 6π-‐aroma%c


N

H δ = 7.46H δ = 7.01

H δ = 8.50

1.40 Å

1.39 Å

1.34 Å

2.2 D

Benzene (4n +2), n = 1, 6π electrons δH = 7.26 ppm, planar molecule; bond length = 1.39 Å

isoelectronic with pyridine

Cyclopentadienyl Anion (4n +2), n = 1, 6π electrons

H

pKa = 16 pKa = 43 pKa < -‐2

C-‐C sp3-‐sp3 1.54 Å

C-‐C sp3-‐sp2 1.50 Å

C-‐C sp3-‐sp 1.47 Å

C-‐C sp2-‐sp2 1.46 Å

C-‐C benzene 1.39 Å

C=C 1.34 Å

C≡C 1.21 Å

H H H

CF3F3C

F3CCF3CF3

Hbase

B:


cyclopentadienide anion is isoelectronic with furan pyrrole and thiophene in each case the (one of the) lone pair(s) is parallel to the p-‐orbitals and part of the π-‐system

S O NH X

thiophene furan pyrrole

0.66 D0.55 D 1.74 D

NH

pyrrolidine

X X X

Electrophilic AromaTc SubsTtuTon

step 1 is usually rate determining because aroma%city is lost step 2 is fast as aroma%city is regained

HE E

HE

E

HE

σ-‐complex Wheland intermediate arenium ion


HE E

HE

E

HE

Electrophilic AromaTc SubsTtuTon

σ-‐complex Wheland intermediate arenium ion

step 1 step 2

Hammond’s postulate: The transi%on state resembles the structure (intermediate or substrate or product) to which it is closest in energy (i.e. transi%on state resembles intermediate arenium ion, therefore what stabilises the arenium ion stabilises the transi%on state.)

E

E

+ E

activation energy

EH (+)(+)

‡E

H(+)

(+)

H

‡


Mechanis%c Evidence isola%on of intermediates

Me

Me Me

EtF, BF3- 80 °C

Me

Me Me

Me

H BF4

stable solidmp -15 °C

heatMe

Me Me

Me

Subs%tuent Effects subs%tuent Y affects both the rate and regiochemistry of the reac%on

ortho (1,2-‐disubs%tuted)

meta (1,3-‐disubs%tuted)

para (1,4-‐disubs%tuted)

Y YE

Y

E

Y

E

E

SbF5 / FSO3H

-120 °C in SO2FCl

H H

SbF6

H H

δH = 5.6

δH = 9.7

δH = 8.6

δH = 9.3

H

HH

H H H H


electron dona%ng groups ac%vate the aroma%c ring (i.e. substrate reacts faster than benzene) and are ortho and para direc%ng

electron withdrawing groups deac%vate the aroma%c ring (i.e. substrate reacts slowed than benzene) and are meta direc%ng

halogens are mildly deac%va%ng and direct ortho and para

Y YE

Y

E

Y

E

E

ACTIVATING group means that the reac%on of the subs%tuted benzene is faster than that of benzene itself Typical ac%va%ng groups include: OH, O-‐, , ,OR, NH2, NR2, alkyl, Ph O R

O

HN R

O

DEACTIVATING group means that the reac%on of the subs%tuted benzene is slower than that of benzene itself Typical deac%va%ng groups include: R3N+, CF3, NO2, SO3H, CN, O-‐, , ,

R

O

OR

O

NR2

O

OMe

Br2, AcOH

OMe

Br

OMeBr

98 2

kanisole / kbenzene = 109


orienta%on of aTack when ring carries an electron dona%ng group, X:, which carries a lone pair (e.g. OMe) ortho and para aTack

ortho aTack

para aTack

meta aTack

X: X:

Emeta

E

X:

E

X:

EH

X:

E

in the intermediates from ortho and para aTack the carboca%ons are stabliised by overlap with the lone pair of X

in the intermediate from meta aTack in the carboca%on is not stabilised by overlap with the lone pair from X

X: X:EE

ortho

X:E

X:E

XE

HX:

E

X: X:

Epara

X: X:

E E E

X

E

X:

EH

✓✓

✓✓


X:

+ E

Energy

reac%on coordinate

TS1 TS2

meta

ortho and para similar energies

products

intermdiate

more stable intermediate(s) formed faster ∴ ortho and para products predominate benzene reacts slower than these substrates as substrates are more electron rich

Therefore the intermediates (and hence the transi<on states which lead to those intermediates) are lower in energy (more stabilised) for ortho and para aVack and hence the rate of these reac<ons is faster than for meta aVack (and faster than for benzene)

reac%on coordinate diagram for aTack on X-‐subs%tuted benzene (X = EDG)


Z Z

Epara

Z Z

E E E

Z

EH

orienta%on of aTack when ring carries an electron withdrawing group, Z, (e.g. NO2) meta aTack ortho aTack

para aTack

Z ZEE

ortho

ZE

ZE

HZ

E

meta aTack

Z Z

Emeta

E

Z

E

Z

EH

Z

E

✗✗

✗✗

in the intermediates from ortho and para aTack the carboca%ons are destabilised as next to EWG Z

in the intermediate from meta aTack in the carboca%on is never adjacent to EWG


Energy

reac%on coordinate

Z

+ E

reac%on coordinate diagram for aTack on Z-‐subs%tuted benzene (Z = EWG)

intermediate

TS1

TS2

meta

ortho and para similar energies

products

less stable intermediate(s) formed slower ∴ meta products predominate benzene reacts faster than these substrates as it is more electron rich

The intermediates (and hence the transi<on states which lead to those intermediates) are higher in energy (less stable) for ortho and para aVack and hence the rate of these reac<ons is slower than for meta aVack – meta by default (all slower than for benzene as aroma<c ring is electron poor)


Halogens

mildly deac%va%ng as they are electronega%ve and withdraw electron density from the ring through the σ-‐framework (falls off with distance) halogens direct ortho and para as they have lone pairs in high energy orbitals which stabilise the intermediates for

ortho/para aTack

MeO – overall electron dona%ng on benzene ring Cl – overall electron withdrawing on benzene ring

with chlorobenzene the σ-‐electon withdrawing of the chlorine is greater than the π-‐dona%on of the chlorine 3p lone pair and chlorobenzene is deac%vated with respect to benzene

OMe: OMe

1.2 D

Cl: Cl

1.6 D

2p –lone pair

good 2p -‐2p overlap

3p –lone pair

poor 2p -‐3p overlap

both oxygen and chlorine are electronega%ve with anisole the σ-‐electon withdrawing of the oxygen is less than the π-‐

dona%on of the oxygen 2p lone pair and anisole is ac%vated with respect to benzene

N O F

P S Cl

Se Br

Te I

Po At

increasin

g electron

ega<

vity

increasin

g siz

e of p-‐orbita

ls

increasing electronega<vity

OMe

Cl


product distribu%on % kArX/kbenzene ortho meta para

PhF 12 -‐ 87 0.18

PhCl 30 0.9 69 0.064

PhBr 37 1.2 62 0.060

PhI 38 1.8 60 0.12

Xconc. HNO3, conc. H2SO4 X

NO2

Ques<on: Nitra%on of halobenzenes

why does fluorine react faster than than the other halobenzenes?

why does fluorine give the largest amount of the para isomer?

Examples

Me is an electron dona%ng group and hence an ac%va%ng group

Wheland intermediate for ortho / para aTack is stabilised by hyperconjuga%on – σCH → π

CH3

HNO3 / H2SO4

CH3NO2

CH3

NO2

CH3

NO259% <4% 37%

H

HH

+H

O2N

NO2

H

H

HH

+


what happens if there are two subs%tuents on the benzene ring?

subs%tuents can be broadly categorised into three classes

i)  STRONGLY ac%va%ng and ortho and para direc%ng (OH, OR, NH2, NR2)

ii)  mildly ac%va%ng groups such as alkyl groups (ortho and para direc%ng) and halogens (mildly deac%va%ng)

iii)  deac%va%ng meta-‐direc%ng groups subs%tuents in group i) will dominate classes ii) and iii)

subs%tuents in group ii) will dominate class iii)

AcNH: o, p Me: o, p AcNH dominates ∴ ortho

MeO: o, p F: o, p MeO dominates ∴ para

Me2N: o, p CF3: m Me2N dominates ∴ para (sterics)

Examine the electronic effects of subs%tuents then consider sterics

HN O

MeMe

NMe

MeF3C MeO

MeO

H

O

MeO: o, p CHO: m MeO dominates ∴ para (sterics)

OMe

F


opposing -‐ if similar reac%vity will get mixtures of compounds

all other things being equal a 3rd group is least likely to enter between two groups meta to one another

Cl: o, p CO2H: m Cl dominates ∴ para

MeO: o, p MeO: o, p ∴ ortho / para

HO: o, p Me: o, p HO dominates ∴ para

Me: o, p Cl: o, p ∴ mixture

Note: these are guidelines and exact ra<os of ortho / meta / para products depend on the reac<on condi<ons and the nature of the electrophile

ClCO2H

HNO3,H2SO4

ClCO2H

NO2

OMe

OMe

Br2 / AcOH

OMe

OMeBr

OHMe

OHMe

Br

Br2, AcOH

MeCl HNO3, Ac2O

MeCl

NO2

MeCl

O2N25 75


CO2H

subs%tuent effects are important for selec%vity and efficiency when designing a synthe%c route

CO2HNO2

NO2

CO2H

or

NO2

or

Me

synthe%cally we want to prepare the target material in a clean, selec%ve and efficient fashion target material

Look at the star%ng materials

MeNO2

CO2H, deac%va%ng meta direc%ng

NO2, deac%va%ng meta direc%ng

Me, ac%va%ng ortho / para direc%ng

if possible best to introduce the most deac%va%ng group(s) last in the synthe%c sequence rela%onship of NO2 groups is ortho/para with respect to CO2H ∴ best to use toluene as star%ng material

nitro group is deac%va%ng ∴ can isolate and separate isomers if required

CO2HNO2

NO2

MeNO2

NO2

HNO3, H2SO4 KMnO4

CH3

HNO3 / H2SO4

CH3NO2

CH3

NO2


OH

OMeOMe

NO O

Br

OH

OMeOMe

Br


Cl

AlCl3, heat

2 equivalents ClSO2OH SO2Cl HMe

O

ClAlCl3

O

+

O

+ +

A B C

2) AVempted Friedel-‐Crabs acyla<on of benzene with tBuCOCl gives some of the expected ketone, as a minor product, and also some tbutyl benzene, but the major product is the disubs<tuted compound C. Explain how these compounds are formed and suggest the order in which the two subs<tuents are added to form C.

(a) (b)

(c) (d)

(e)

H2SO4, 80 °C

SO3HSO3H H2SO4, 160 °C


Ipso a\ack and reversible reac%ons electrophilic aroma%c subs%tu%on is generally an irreversible process all of the above arguments with regard to

ortho, meta and para ra%os have been based on the irreversibility of the process i.e. the reac%ons are under kine%c control -‐ but there are some excep%ons

not all electrophilic aroma%c subs%tu%on reac%ons are under kine%c control

sulfona%on -‐ usual reac%on condi%ons: conc. H2SO4 with SO3

sulfonyl group is electron withdrawing so we only have mono-‐subs%tu%on

SOO

O

H SOHO

O HO3S HHO3S

at high temperatures with dilute H2SO4 – sulfona%on is reversible

aTack by an electrophile at a posi%on which already carries a non-‐hydrogen subs%tuent is termed ipso-‐subs<tu<on

OHBr SO3H

SO3H

H2SO4

OHBr

OHBr SO3H

SO3H

H

OHBr SO3H

SH

OO

OH

OHBr SO3H

H

OHBr H

Hipso aTack

we can use an SO3H group as a blocking group


+ HMe

Me

MeCl: AlCl3

Me

Me

MeCl AlCl3

MeMe Me

Me

Me

MeMeMe

H

Me

MeMeMe

kine%c control thermodynamic control

Me

tBuCl / AlCl3

Me

low temperature

tBuCl / AlCl3

Me

high temperatureMe

MeMe

Me

MeMe

MeMeMe

thermodynamic product all groups as far apart as they can be

Friedel CraUs alkyla%on can be under kine%c or thermodynamic control

Me

H

Me

HMe Me

MeMe Me

Me

Me

Me

MeMeH

Me

Me

MeMe

Me

Me

MeMe

Me

MeMe


cat. AlCl3, MeCl Me Me

Me

MeMe

+

IntroducTon of FuncTonal Groups –Synthesis Friedel CraUs Alkyla%on polyalkyla%on and rearrangement predominate

with one equivalent of alkyla%ng agent mixtures of products result as the ini%ally formed monoalkyl arene is more reac%ve than the unalkylated arene – alkyl groups are electron dona%ng

excess MeCl, cat. AlCl3

MeMeMe

Me MeMe

ClH

HH

AlCl3 MeMe

Me Me

H

Me Me

MeMe

Me MeMe Me

more reac%ve than benzene

more reac%ve than toluene

more reac%ve than toluene


IntroducTon of FuncTonal Groups –Synthesis Friedel CraUs Alkyla%on polyalkyla%on and rearrangement predominate with primary alkyl halides rearrangement occurs

cat. AlBr3

MeBr Me Me

+

Me

major minor

of monoalkylated products

1,2-‐hydride shiU

primary carboca%ons are very unstable rearrangement to the secondary carboca%on occurs

Br AlBr3H

MeMe Me

MeBr: AlBr3 Me

MeH

Me

Me


MeCl

O

AlCl3 MeO

OMe

H

OMe AlCl3

OMe

AlCl3

IntroducTon of FuncTonal Groups –Synthesis Friedel CraUs Acyla%on requires a full equivalent of the Lewis acid mono-‐subs%tu%on predominates as introduced group is electron-‐withdrawing and deac%vates aroma%c ring

carbonyl group can then be removed if required (Clemensen reduc%on, Zn/HCl; Wolf-‐Kishner reduc%on, NH2NH2 then KOH, heat; dithiane than Raney Ni) giving products of a selec%ve Friedel-‐CraUs alkyla%on

para-‐isomer generally favoured by steric hindrance

monosubs%tu%on Me

Cl

O OMe

1 equivalent AlCl3

MeOMe

O

O Me

O+

AlCl3, toluene

MeO

Me

O

93% yield


IntroducTon of FuncTonal Groups –Synthesis Friedel CraUs Acyla%on Fries rearrangement can give access to either the ortho-‐ or para-‐isomer

Friedel CraUs summary AlkylaTon AcylaTon AlCl3 cataly<c stoichiometric Rearrangement possible no, but loss of CO from R-‐C≡O+ if R+ stable, e.g. Ph3C+ subs%tu%on order poly mono

HOMe

O

O Me

O+ pyridine

OMe

O AlCl3

OMe

O

AlCl3

O

MeO

AlCl3

O

AlCl3

O

Me

inside solvent cage - tight ion pair

non-polarsolvent

solvent-separatedion pair

polar solvent

O

Me

OH

major product in polar solvents e.g. PhNO2

workup

HO

Me

O

O

AlCl3

O

MeH

O

O

Me

Cl3Al

major product in non polar solvents

workup

HO

O

Me


IntroducTon of funcTonal groups

Blanc chloromethyla%on – related to Fiedel-‐CraUs reac%ons

OMe

O

Me

Me

(CH2O)n,conc. HCl

OMe

O

Me

Me

ClCl

halogena%on – with ac%vated aroma%cs Lewis acid ac%va%on of the electrophile is not require, with benzene and with deac%vated aroma%cs Lewis acid ac%va%on of the electrophile is required

NO2

Me

Br2, FeBr3

NO2

MeBr

halogena%on can frequently be best achieved using Sandmeyer reac%ons (par%cularly good for introducing I and F as well as Cl, Br and CN)

conc. HNO3conc. H2SO4

NO2Sn, HClor H2Pd/C NH2 N

HX, NaNO2,0 °C N

X


Ar N NX

HOHO N

N Ar

IntroducTon of FuncTonal Groups –Synthesis diazonium salts -‐ reac%ons

Ar N NX

H2O, 100 °CAr OH Ar N N

BF4

heatAr F

Ar N NX

cat. CuX,KX

X = Br, Cl, CNAr X Ar N N

XAr X

KI

Ar N NX

Ar HH3PO2

Ar N NX

Me

O

OR

O

Me

O

OR

O

NNHAr

SN1 reac%on via carbenium ion

NH2

NaNO2,HX 0 °C N

N -N2

slow

v. high energy intermediate, offset by the forma%on of N2 carbenium not stabilised by π-‐system as is orthogonal to π-‐system

empty sp2 orbital

SN1 reac%on via carbernium ion – Balz Schiemann reac%on

radical reac%on

radical reac%on

radical reac%on

electrophilic aroma%c subs%tu%on

via enol


Nucleophilic AromaTc SubsTtuTon SN2: alipha%c vs aroma%c AliphaTc

nucleophile aTacks C-‐I σ* resul%ng in inversion of configura%on

AromaTc no possibility of nucleophile aTacking backside of C-‐LG σ* (transi%on geometry impossible)

Lowest Unoccupied Molecular Orbital (LUMO) is π* not σ*

aTacking electron rich arene with electron rich nucleophile

SN1: alipha%c vs aroma%c AliphaTc carbenium stabilised by hyperconjuga%on

AromaTc possible but very high energy intermediate (see Sandmeyer reaca%ons)

IEtHMe

NuMe

Et HNu I(-) (-)

‡

NuEt

HMe

+ I+

v. high energy intermediate, offset by the forma%on of N2 carbenium not stabilised by π-‐system as is orthogonal to π-‐system

empty sp2 orbital

remember SN2 reac%ons at sp2 hybridised centres (i.e. alkenes and arenes are incredibly rare)

LG

XMeMeMe Nu

MeMeMe

NuMeMeMe

X Nu Nu


Nucleophilic AromaTc SubsTtuTon SNAr – Addi%on – Elimina%on Mechanism

nucleophile aTacks LUMO, electron withdrawing groups lower energy of LUMO and stabilise the nega%ve charge in the intermediate best to have electron withdrawing group(s), ortho and / or para to the leaving group

Evidence isola%on of intermediates

LG

Nu

LG Nu LG Nu LG Nu Nu

-LG

rate determining step Meisenheimer intermediate

H. Ueda, M. Sakabe, J. Tanaka, Bull. Chem. Soc. Jpn., 1968, 41, 2866-‐2871.

O2N NO2

N

OMe

MeOO2N NMeO OMe

NOOO O

K

ClO2N NO2

NO2

MeO K O

O

O2N NMeO OMe

NOO

O

O

the nega%ve charge is delocalised ortho and para to leaving group


Nucleophilic AromaTc SubsTtuTon SNAr – Addi%on – Elimina%on Mechanism for halogens as leaving groups, rate of reac%on usually follows kF > kCl > kBr (c.f. rate of SN2 reac%ons kI > kBr > kCl > kF)

rate determining step is generally aTack of nucleophile on aroma%c ring therefore bond strength to leaving group is not so important in influencing the rate fluorine is the most electronega%ve element and enhances the electrophilicity of the carbon being aTacked

increasing the rate of aTack by the nucleophile

MeO

NO2

50 °C

NO2X OMe X = F Cl Br I

krel 2810 3.1 2.1 1

1st step usually rate determining

NF

OO

Nu

NF

OO

Nu

NNu

OO

leaving group ability does depend on the nucleophile, nevertheless leaving groups can broadly be divided into three classes: taken from Physical and Mechanis<c Organic Chemistry, R. A. Y. Jones, CUP, 1979

good: F, NO2, Me3N+, OTs, Me2S+ medium: Cl, Br, I, OR, OAr, SR, SO2R poor: NMe2, H

rate = k[substrate][nucleophile]


General Trends in Oxida<ve Addi<on order of reac%vity is generally I > OTf > Br >> Cl.

with bidentate phosphines rate of oxida%ve addi%on increases with decreasing bite angle.

low oxida%on state metals are electron rich (nucleophilic) therefore good donor ligands i.e. H-‐, R-‐, R3P.

promote oxida%ve addi%on oxida%ve addi%on to alkyl halides is slow as precomplexa%on is less favourable.

bulky ligands can be good as they lead to dissocia%on and more reac%ve metal complex.

metal is oxidised and hence substrate is reduced therefore electron deficient substrates react faster than electron

rich substrates. reac%on proceeds with reten%on of olefin geometry for sp2 electrophiles.

PdPR2R2P

θ

IPdL2+ PdL2

I‡

PdI MeO

Cl

MeO2C

Cl

reac%vity is complementary to Pd-‐catalysed cross-‐coupling reac%ons of halobenzenes where regioselec%vity is generally governed by the rate of oxida%ve addi%on into the Ar-‐X bond which depends on bond strength


Cl

Y

OMe

Y

O2N O2NMeO

Y NO2 Me3N+ SO2Ph O=CPh CF3 H

krel 114000 2130 18400 2700 800 1

Me Me

NO2

NO2

NH3, MeOH Me Me

NO2

NH2

Explain the outcome of the following reac<on

the nucleophile – typically good nucleophiles in SNAr reac%ons include: RS-‐, HO-‐, RO-‐, PO-‐, RNH2

Synthesis of fluoxe<ne ‘Prozac’ – serotonin reuptake inhibitor for treatment of depression Predict the product of the following reac<on

F

F3C Ph

NHMeHO+ NaH,

MeNMe

Me

O

Nucleophilic AromaTc SubsTtuTon the ac%va%ng group


total synthesis of vancomycin – glycopep%de an%bio%c currently the ‘last line of defence’ to treat methicillin-‐resistant staphylococcus aureus [MRSA].

OallylF

NHBoc

OMs

O

HN

HN

O

O

NH

MeHNOC

OMe

H H

HO Cl

MeOOMe

NO2

OH

OallylO

NHBoc

OMs

O

HN

HN

O

O

NH

MeHNOC

OMe

H H

HO Cl

MeOOMe

NO2

OH

OH

Na2CO3

ORO OR

N

FCl

O

O

ORO OR

Cl

NO OF

Na2CO3

D. A. Evans et al. Angew. Chem. Int. Ed. Engl. 1998, 37, 2700-‐2704

OO

NH

O HN

O

O

NH

ONHMe

NH2

OO

HN

Cl

HN

O

O

NH

HO2C

OH

OH

H H

HO Cl

HOOH

O

HOOH

OHO

MeHOH2N Me

O


OO

NH

O HN

O

O

NH

ONHMe

NH2

OO

HN

Cl

HN

O

O

NH

HO2C

OH

OH

H H

HO Cl

HOOH

O

HOOH

OHO

MeHOH2N Me

O

D. A. Evans et al. Angew. Chem. Int. Ed. Engl. 1998, 37, 2700-‐2704

allylOO

NH

OHN

O

OH

NH

ONMe

NHDdm

OO

HN

NO2

HN

O

O

NH

MeHNOC

OBn

OH

H H

HO Cl

BnO

OBn

F

Boc

CsF, DMSO

allylOO

NH

O HN

O

O

NH

ONMe

NHDdm

OO

HN

Cl

HN

O

O

NH

MeHNOC

OBn

OH

H H

HO Cl

BnO

OBn

Boc

Ddm =

MeO OMe

total synthesis of vancomycin – glycopep%de an%bio%c currently the ‘last line of defence’ to treat methicillin-‐resistant staphylococcus aureus [MRSA].


HeteroaromaTcs and Nucleophilic subsTtuTon

pyridine is electron deficient at C-‐2 and C-‐4 and it is prone to aTack by nucleophiles N N N

HOMO of pyridine is nitrogen lone pair

N

E

NE

v. slowNE

E E

pyridine undergoes electrophilic aroma%c subs%tu%on only very slowly as reac%on with electrophiles occurs on nitrogen lone pair

high energy intermediate -‐ electrophile reac%ng with posi%vely charged nucleophile

N

c. HNO3, c. H2SO4, 300 °C, 24 h

N

NO2

6% NH

1 2

3 4


HeteroaromaTcs and Nucleophilic subsTtuTon

N Cl

MeO Na

N Cl

OMeN OMeN

HO

POCl3

nucleophilic aroma%c subs%tu%on

R Cl

O MeOR OMe

Ocompare with

the leaving group needs to be posi%oned ortho or para to the pyridine nitrogen atom

below are the rela%ve rates of reac%on with MeO-‐ in MeOH at 50 °C

Cl

N

Cl

N Cl N

Cl

Cl

NO2

Cl

CF3

1 5 3,000 82,000 700,000 10-‐5

reac%on of the corresponding N-‐oxides and N-‐methyl pyridinium salts is significantly faster than for the parent chloropyridines

pyridine is electron deficient at C-‐2 and C-‐4 and is prone to aTack by nucleophiles N N N

HOMO of pyridine is nitrogen lone pair


N Cl N ClO

N ClMe

3,000 6 x 107 1.3 x 1012

N

Cl

N

Cl

ON

Cl

Me

82,000 9 x 107 4.2 x 1010

M. Liveris, J. Miller, J. Chem. Soc., 1963, 3486-‐3492

as with benzenoid aroma%cs fluoride is a beTer ac%vator (leaving group) than chloride

M. Schlosser, T. Rausis, Helv. Chimica Acta, 2005, 88, 1240-‐1249

N Cl N ClCl N Cl

F3C

N F

1 65 320 2800

rela%ve rate of reac%on with EtO-‐ in EtOH

below are the rela%ve rates of reac%on with MeO-‐ in MeOH at 50 °C


pyrimidines and related heterocycles are more reac%ve than 2-‐halopyridines towards nucleophilic aroma%c subs%tu%on

N

N

X

N

N

X NN

X

NN

X

N

N

X N XN

X

> > > > > >

increasing reac%vity toward nucleophilic aroma%c subs%tu%on

taken from “Heterocyclic Chemistry” 5th Edi%on, J. A. Joule and K. Mills, Wiley 2010.

N

N Cl

Cl

Me

O

PhLiHMDS, toluene

N

N Cl

Et

O

Ph

BuNH2

pTSA N

N BuN

Ph

Et

D. S. Chekmarev, S. V. Shorshnev, A. E. Stepanov, A. N. Kasatkin, Tetrahedron 2006, 62, 9919-‐9930

there are not always ‘back of the envelope’ explana%ons of selec%vity

R CO2Me Cl H Ph Me OMe

A:B 96:4 92:8 85:15 84:16 76:24 8:92

Y. Goto and co-‐workers, Bull. Chem. Soc. Jpn., 1989, 37, 2892

N

N

Cl

ClR N

N

OMe

ClR

MeO

N

N

Cl

OMeR+

A B


N

O

MeHNO3, H2SO4,100 °C

N

O

MeNO2

PCl3

N

MeNO2

+ POCl3

NO2

HeteroaromaTcs pyridine N-‐oxides – much more suscep%ble to electrophilic aTack at 2 and 4 posi%ons (and to nucleophilic

addi%on at 2 and 4 posi%ons)

nitra%on of pyridine N-‐oxide

promotes electrophilic subs%tu%on at 2 and 4 posi%ons

N

H2O2, CH3CO2H

N

O

2

34

N

O

N

O

N

O

PCl3

N

O

MeH NO2

N

O

MeNO2

PClCl

Cl

N

O

MeNO2

N

MeNO2

66% yield, c.f. nitra%on of pyridine in acid (6% yield)


HeteroaromaTcs pyridine N-‐oxides – N-‐deoxygena%on with rearrangement

pyridine N-‐oxides – conversion to chloro compounds

N

R

O

OP

Cl ClCl

N

R

OPO

Cl Cl

N

R

OPO

Cl Cl

H

Cl N

R

ClCl

N

Me

H

O

Me

O

O Me, 100 °C

O

N

Me

O Me

O

N

Me

H

O

Me

O

O

MeO N

Me

O

Me

O

1 23

32

1


HeteroaromaTcs pyridones

nitra%on

NH

O N OH NH

O

N

OH

N

O

HNO3, H2SO4

NH

ONO2

N

ClNO2POCl3 NaBH4

N

NO2

NO2

N

ONO2

H

H

NH

ONO2 POCl3

N

ONO2

PCl

OCl

H

Cl

N

ONO2

PCl

OCl

H

Cl

N

NO2

Cl

NaBH4

H


order of aroma%city is: thiophene > pyrrole > furan (enol ether like) sulfur is the largest atom and hence is beTer matched for bonding to sp2-‐hybridised carbon atoms in a 5-‐membered

ring leading to thiophene being the most aroma%c

HeteroaromaTcs

pyrrole, thiophene and furan all three have aroma%c proper%es

in each case the (one of the) lone pair(s) is parallel to the p-‐orbitals and part of the π-‐system

the aroma%c heterocycles are electron rich

S O NH X

thiophene furan pyrrole

0.66 D0.55 D 1.74 D

NH

pyrrolidine

1

23

αβ X X X

NNu

pyridine is electron poor


HeteroaromaTcs ReacTons electrophilic subs%tu%on – kine%c reac%on at the 2-‐posi%on is favoured over reac%on at the 3-‐posi%on

more reac%ve than benzene – e.g. pyrrole similar reac%vity to aniline

more delocalised intermediate ∴ more stable ∴ 2-‐subs%tu%on favoured

XE

XH

EXH

EXH

EXE

X

E

X

HE

X

HE

X

E

less delocalised intermediate ∴ less stable ∴ 3-‐subs%u%on disfavoured

subs%tuents already present on the aroma%c heterocycle exert less direc%ng effect than the corresponding subs%tuents in benzene

XE XH

EEDG EDG XH

EEDG X EDGE

X X

HE

XEWG EWG EWG

EE

with EWG at α-‐posi%on β’-‐subs%tu%on favoured

with EDG at α-‐posi%on α’-‐subs%tu%on favoured


HN

O

ClCl3CHN

CCl3

O

90% HNO3, -50 °CHN

CCl3

O

O2N

α'

β'

HeteroaromaTcs ReacTons nitra%on

S

O

OMeN

O

O

AcOH, 0 °C

S NO2 +S

NO260% trace

HN

O

OMeN

O

O

AcOH, -10 °C

HN NO2 +

HN

NO251% 13%

O

O

OMeN

O

O

AcOH,-25 °C

O NO2O NO2

HMe O

O

O NO2H

AcON

addi%on product acyla%on and formyla%on

β’ > α’ or β for α-‐EWG

S

O

NS

78%

HN

HN

83%

H, POCl3, 35 °CPh

H

O

O

N H, POCl3, RTMe

H

O

then hydrolysis then hydrolysisMeMe


HeteroaromaTcs indole

more enamine-‐like than pyrrole aTack of electrophiles at the β posi%on is the lowest energy pathway

aTack at β posi%on retains aroma%c sextet of benzenoid ring

NH

α

β

2

3

1 NH

O Sbenzofuran benzothiophene

NH

E

NH

EH N

H

E

NH

E

NH

NH

EH E

minor

major


HeteroaromaTcs indole

if the β-‐posi%on is blocked α-‐aTack occurs α-‐aTack can occur via β-‐aTack followed by rearrangement (● = CT2 i.e. a tri%ated methylene group)

NH

α

β

2

3

1 NH

O Sbenzofuran benzothiophene

NH

OH NH

OF3B

H NH

NH

H NH

BF3•OEt2

NH

NH

H NH

1:1 mixture direct aTack at α-‐posi%on would give solely

NH


N Cl NOH

H

Me

120 °C, 15 hours

NMeN

OH NaH,F

O

HN

MeN

O

O H

Explain the outcome of the following reac<ons

NH

NMeCO2tBu

OHH

MsCl, Et3N

NH

MeN

CO2tBu

single enantiomer racemate

N

Me

Me Me

F

O2N

NO2i)

ii) aq. NaOH

Me

Me OH

(a)

(c)

(b)

(d)

NiPr3Si

BrNO O

NiPr3Si

Br


Vicarious Nucleophilic Subs%tu%on -‐ (nucleophilic subs%tu%on of hydrogen) Reviews: M. Mąkosza, J. Winiarski, Acc. Chem. Res. 1987, 20, 282; M. Mąkosza Pure & Appl. Chem. 1997, 69, 559

mechanism

LG = Cl, Br, PhO, PhS, RO-‐ etc; EWG = SO2Ph, SO2NR2, SO2OPh, POPh2, CN, CO2Et

For VNS require a nucleophile which

carries a leaving group

LG EWG

Cl SO2Ph

KOH, DMSO

O2N

SO2Ph

O2N

H

O2NH

HPhO2S

How can we explain the following results?

N

H

O

O

Cl

SO2Ph

NO

O

HCl

SO2Ph

NO

O

SO2Ph

NO

O

SO2Ph

NO

O

SO2Ph

HO rate determining step

rate determining step is elimina%on of H-‐X (HCl) from σ-‐adduct

NO2

Cl

Cl SO2Ph

KOH, DMSO

NO2

ClSO2Ph

MeO

DMSO

NO2

MeO

69%


NO2

X SO2Ph

KOH, DMSO

NO2

SO2Ph

NO2

PhO2SR

Vicarious Nucleophilic Subs%tu%on

orienta%on of addi%on depends on: structure of the carbanion; structure of the arene; reac%on condi%ons

for aTack on nitrobenzene, as the bulk of the nucleophile increases the amount of para isomer increases

NO2

F

Cl SO2Ph

KOH, DMSO

NO2

FSO2Ph

18%

X R yield / % ortho para

F H 63 74 26

Cl H 75 53 47

Cl Et 68 100

Cl Ph 93 100

M. Makosza, J. Goliński, J. Baran, J. Org. Chem., 1984, 49 1488


AromaTc organometallics Ortho-‐directed electrophilic aroma%c subs%tu%on

ortho-‐lithia%on by lithium halogen exchange – faster then deprotona%on generally requires an organolithium base an aryl / alkenyl bromide or iodide.

OMeBr

OMeLi

+ BuBrBu Li

mechanism involves aTack of alkyl lithium at the halogen via an intermediate “ate” complex

via “ate” complex

OMeBrBu

reac%on is an equilibrium process which favours the more stable anion (remember, anion order is sp3>sp2>sp – the stability of the anion is in the order of the pKa of the corresponding hydrocarbon)

in the above example an sp3 anion (butyl lithium) gives an sp2 anion

D. E. Applequist, D. F. O’Brien, J. Am. Chem. Soc., 1963, 85, 743.

anion in an sp2 orbital anion in an sp3 orbital


OMeBr

OMeLi

NLi N

Br+X


LDA and other amide bases are good for deprotona%on but NOT for halogen lithium exchange

lithium halogen exchange with LDA would lead to the forma%on of a very weak halogen-‐nitrogen bond – reac%on is thermodynamically in the wrong direc%on

NSO2Ph

I

INSO2Ph

I

NSO2Ph

MeLDA, I2 tBuLi, MeI

Mark G. Saulnier and Gordon W. Gribble J. Org. Chem. 1982, 47, 757

Bu LiO Br O Li R X O R



in general rate of exchange I > Br >> Cl

to make aryllithiums by lithium halogen exchange – generally use n-‐butyllithium; tert-‐butyllithium may also be used

to make vinyllithiums and alkyllithiums one frequently uses tert-‐butyllithium

with primary alkyl halides it is necessary to use two equivalents of tert-‐butyllithium

O O

PMP

MeI

Me Me

OTBS

2 equiv. tert-BuLi

O O

PMP

MeLi

Me Me

OTBSMeI

MeMeLi

MeMe

HMeMe

Me

MeMeH

+

with one equivalent of tert-‐butyllithium protodeiodina%on is likely to occur O O

PMP

MeH

Me Me

OTBS

lithium halogen exchange is a very fast reac%on which can outcompete deprotona%on of OH groups and addi%on to C=O groups

O

OBr

ONRO

nBuLi, THF, -78 °C

O

OO

RHN

L. Ollero, L. Castedo, D. Dominguez, Tetrahedron, 1999, 55, 4445-‐4456



synthesis of morphine – J. E. Toh, P. L. Fuchs, J. Org. Chem. 1987, 52, 473–475 Provide a mechanism for the reac<on below.

PhO2S

OOH

OMeBr

Br

2.2 equiv. BuLi, -78 °C

SO2Ph

O OH

OMe

O

N OH

HH

Me

OH

H

morphine

I

NMe

OOMe

MeO

MeO

t-BuLi MeO

MeO

O

annula%on forming benzocyclobutanes – I. A. Aidhen, J. R. Ahuja, Tetrahedron LeV. 1992, 33, 5431-‐5432.

selec%ve halogen metal exchange is possible

N

BrBr

OMe

n-BuLi, Et2O, -100 °C

N

LiBr

OMe N

Br

OMe

OH

Cl

Ar

O

H

73%


CO2Me

I

iPrMgCl, THF, -10 °C

CO2Me

MgCl

CO2Me

PhCHO

OHPh


it is also possible to make Grignard reagents by lithium halogen exchange generally using iPrMgBr or iPrMgCl For reviews see: P. Knochel et al. Angew. Chem. Int. Ed. 2003, 42, 4302; Chem. Commun. 2006, 583; Heterocycles 2014, 88, 827.

Br

Br

Br iPrMgCl•LiCl,THF -50 °C

Br

Br

MgClBr

Br

tBu H

O

tBu

OH

PhS

CN

O O

N

Br iPrMgCl•LiCl,THF -50 °C

N

MgCl

NBr Br BrBr Br Br

CN iPrMgCl•LiCl,THF -50 °C

NBr MgCl

CN


S Br

I

S Br

MgCl

S BrEtMgCl. THF, RT

O

HMe2N

OH

AromaTc organometallics for aroma%c heterocycles metal halogen exchange shows the following selec%vity:

with 5-‐ring heterocycles the 2-‐posi%ons undergoes exchange faster than the 3 posi%on

with 6-‐ring heterocycles, the 3 posi%on undergoes exchange faster than the 2-‐posi%on

iodine metal exchange is faster than bromine metal exchange

summary 5-‐ring 2>3; 6-‐ring 3>2; I>Br

X Br

Br

R M X M

Br N

Br

Br

R M

N

M

Br

S Br

Br

EtMgCl. THF, RT S MgCl

Br

S

Br

tBuN C OO

NHtBu

76%

49%

l. Christophersen, M. Begtrup, S. Ebdrup, H. Petersen, P. Vedsø J. Org. Chem., 2003, 68, 9513


OMe

NMe2

BuLi

OMe

NMe2

LiBuLi,

Me2NNMe2

OMe

NMe2

Li

OMeH BuLi

OH

Me LiBu O

Me

Li

O

NMe2H

DMF

MeO

NMe2

H OLi MeO

NMe2

H OH, H2O

H

H

MeO

H

O

AromaTc organometallics directed ortho-‐metalla%on reviews: V. Snieckus, Chem. Rev. 1990, 90, 879-‐933; J. Clayden in “Organolithiums: Selec<vity for Synthesis, Pergamon, Oxford 2002.

use amides as electrophiles: reac<ve formyl group is not unmasked un<l the reac<on is

quenched with acid

the posi%on of metalla%on can depend on the reac%on condi%ons

various direc%ng groups can be used

COClHO NH2

MeMe

then dehydrate

NO

Me Me

BuLi

NO

Me Me

Li Br R

NO

Me Me

R


O O

NR2

SR O

OR2N O

SR2N

OO

RN O

NO

MeMe

O NR2

O

RNStBuO

CF3

X

NR2

OR

NR2( )n

O O

N

O

OtBu

increasing ability to direct ortho-‐lithia%on

temperature (°C) of ortho-‐lithia%on with RLi in THF or ether

-‐78 °C -‐78 °C -‐78 °C -‐78 °C -‐50 °C -‐20 °C 0 °C >0 °C >20 °C

condi%on dependent order

O-‐carbamates 3° amides

sulfoxides

sulfonamides sulfones

2° amides imines

oxazloines MOM ethers ethers halogens benzylic alkoxides

anilines

aminomethyl

trifluoromethyl

remote amines

N-‐carbamates

Adapted from: J. Clayden in “Organolithiums: Selec<vity for Synthesis”, Pergamon, Oxford 2002.

most powerful directors


with ortho-‐directors which are also electrophiles, the precise reac%on condi%ons including: the nature of the base, addi%ve and order of addi%on can influence the outcome of the reac%on see: P. Beak, R. A. Brown, J. Org Chem., 1982, 47, 34-‐46

O NEt2OMe

n-BuLi s-BuLi, TMEDA

O NEt2

Li E

O NEt2

E

Electrophile Yield (%)

D2O 88

MeI 77

EtI 70

B(OMe)3; H2O2 (adds an OH) 56

acetone 54

PhCHO 79

CH2=CHCH2Br 60

NEt2

O

F

s-BuLi, TMEDA NEt2

O

FLi

Me3SiCl

NEt2

O

FSiMe3

s-BuLi, TMEDA

then MeI

NEt2

O

FSiMe3

Me

R. J. Mills, N. J. Taylor, V. Snieckus, J. Org. Chem., 1989, 54, 4372


O

s-BuLi

OMeOLi Cl NEt2

O ONEt2

O t-BuLi

O OMeNEt2

O

Li

MeI

O OMeNEt2

O

Me

TBDMSO

TBDMSO TBDMSO TBDMSO

NMeMe

MeMe

Li

O OMeNEt2

O

TBDMSO

NEtO

EtO

O OMe

TBDMSO

HN

O

EtO

OEtHN

OMeO

O

O

HOO

Me

O

OMe

TBDMSO

MeO


fredericamycin

synthesis of Fredericamycin – T. R. Kelly, S. H. Bell, N. Ohashi, R. J. Armstrong-‐Chong, J. Am. Chem. Soc. 1988, 110, 6471-‐6480


HN NaNH2 N Na Me I N

MeNMe

BuLi Et I NMe

Et

HN EtMgBr N N

HN

MgBr O

H OEtO

H

H O

Hworkup

lithia%on of 5-‐membered heterocycles

lithia%on occurs preferen%ally α to the heteroatom due to induc%ve effect of heteroatom with, in some instances a DOM effect

furan and thiophene can be readily metalled α to the metal

S n-BuLi S Li O n-BuLi O Li

-10 °C, ether ether, reflux

with pyrrole itself, the N-‐deprotona%on occurs first – the more ionic the N-‐metal bond the greater the percentage aTack at nitrogen

with a more covalent N-‐M bond C-‐aTack occurs.


N

OtBuONLi

MeMe

MeMe

then Me3SnCl

N

OtBuO

SnMe3

with pyrroles bearing an N-‐EWG on nitrogen α-‐metalla%on occurs

SO2PhN

then HCl, water

SO2PhN B(OH)2

LDA, then B(OMe)3

S

Br

Bu LiS

Li

S

E

ES

Br

LiN

Li

iPriPr

H

selec%vity can be achieved using LDA or butyllithium

no lithium halogen exchange as would make weak N-‐Br bond most acidic proton removed by directed metalla%on


pyridines

pyridines are electron deficient aroma%cs and pyridines which carry a direc%ng group (halogen, CN, CO2H etc.) undergo ready metalla%on

2, and 4-‐subsistuted pyridines metallate in the 3 posi%on

3-‐subsituted pyridines generally metallate in the 4 posi%on

N CO2H

1 eq. BuLi, 3 eq.

NLi

Me MeMe Me

N CO2Hthen CO2

CO2H

N

1 eq. BuLi, 3 eq.

NLi

Me MeMe Me

Nthen CO2

CO2HCO2HCO2H

N

1 eq. BuLi, 3 eq.

NLi

Me MeMe Me

Nthen PhCHO, then H2SO4

CO2H OO

Ph

F. Mongin, F. Trécourt, G. Quéguiner, Tetrahedron LeV., 1999, 40, 5438


“Halogen dance” term introduced by BunneT for isomerisa%on reac%ons which can accompany deprotona%on of halogenated aroma%cs J. F. BunneT, Acc. Chem. Res. 1972, 5, 139.

Br

BrBr

PhNHK, NH3

Br

BrBr

40-‐60%

S Br NaNH2, NH3S Br

SBrS Br

Br

+S S

Br

S Br

S

Br

H

H+

N

I

Cl

LDA, THF - 70 °C

N

I

Cl

Li

N

Li

Cl

IH

O

OEt

N Cl

I O

H70%

F. Guillier, F. Nivoliers, A. Conchennee, A. Godard, F. Marsais, G. Queguiner Synth. Commun. 1996, 23, 4412-‐4436

l. Brandsma, R. L. P. de Jong, Synth. Commun. 1990, 20, 1697-‐1700

66-‐72%

for halogen dance to be synthe%cally useful the isomerisa%on must be thermodynamically favourable


reac%ons of alkenes -‐ depending on subs%tuents alkenes can be:

electron rich and hence nucleophilic

electron poor and hence electrophilic

E E

OR

E

NR2

E

O

alkene enol ether enamine enolate


EE

HOMO = π bond of alkene LUMO = σ* or π* on electrophile

ONu

NO

ONu

OR

ONu

NR2

ONu Nu

N

increasing electrophilicity

HOMO = lone pair on nucleophile LUMO = π* on alkene


Me Br2, H2O OH

Br

Me

BrMe

Br

Me(+)

SN2 / SN1borderline

H2O:

OH

Br

Me

Overview of reacTons of (electron rich) alkenes

Br2 Br

Br:Br Br Br

Br

SN2

Br

Br

long, weak bond

build up of par%al posi%ve charge

stereospecific an%

stereospecific an%

bromina%on

halohydrin forma%on

Me OsO4

OHO N

Me

O

OHMe

Me OOs

O

O

O OOs

OMe

O

O

MeH2O

OH

OH

+

HOOs

HO O

O

O NMe

OOOs

O

O

O

stereospecific syn

dihydroxyla%on

Os(VI)

Os(VIII)


Me O3, then PPh3MeO

OMe

OO

O OO

OMe

OO

O

O

OO

Me

MePPh3

OO

Me

O

PPh3O

OMe+O PPh3ozonolysis

Overview of reacTons of alkenes

Me m-CPBAOMe Me

OH O

O ArOMe

stereospecific syn

epoxida%on


Me HBr, water MeBr

Me

H

Me

Br

MeBr

ionic reac%on with HBr

Overview of reacTons of alkenes

Me Me MeBH3

then H2O2, NaOH

OH

H

BH H

H

B

Me

R

R

O OH

HMe

BO

OH

RR

HMe

OBR2

Me

OH

H2O2, NaOH

stereospecific migra%on with reten%on of configura%on

hydrobora%on / oxdia%on

Me H2, Pd/C

Me

Me

Me

H

Hstereospecific

syn

hydrogena%on


1) How would you carry out the following transforma<ons?

OH

OH

HSO

OEt

O

OEt+S

OEtO

O

O

H2O2, NaOH, MeOH

O

O

3) Explain the following: Treatment of the enolate A with B at -‐78 °C followed by quenching the reac<on at -‐78 °C provides C; however, if the reac<on mixture is first warmed to -‐25 °C before being quenched the ketone D is formed as the major reac<on product

2) Explain the following transforma<ons.

(a)

(b)

MeO

OLiOPh

O HOMe

OPhCO2Me

Me

O

PhO

MeCO2Me

A B C D


O Nu O NuH

HO Nu

O Nu

H

ONu ONu

direct addi%on 1,2-‐addi%on

1 2

conjugate addi%on Michael addi%on 1,4-‐addi%on

1 2 4 3

Conjugate AddiTon vs Direct AddiTon

conjugate addi%on requires the presence of an electron-‐withdrawing group which results in the lowering of the energy of all of the π-‐orbitals of the system and hence the alkene is less nucleophilic and more electrophilic

O Oi.e. alkene is electron poor

Evidence of delocalisa%on

O O

1678 cm-‐1

1628 cm-‐1 1712 cm-‐1 1653 cm-‐1

infra red – remember ν ∝ √k/μ i.e. higher stretching frequency = stronger bond

13C NMR

O143 ppm

133 ppm 133 ppm

118 ppm


examples of conjugate addi%on

regenerates cyanide anion – cataly%c HCN too weak an acid to protonate carbonyl group ∴ need a good nucleophile, cyanide anion, to aTack neutral substrate

HClO:

Cl

OH

OClH

O

Cl

H

HO

Cl

H

H

HCl protonates carbonyl oxygen making the whole system more electrophilic

KCN (cat)., HCNOMe

O

OMe

O

NC

NC

OMe

O

NC

H CN

H

enolate generated by conjugate addi%on reacts with the alcohol to regenerate alkoxide for conjugate addi%on

NaOH cat.OH

H

OOO

H

O

H

OO

OH

HOH

NaOH


O Nu ONu ONu

conjugate addi%on

the beTer the ability to stabilise the nega%ve charge the beTer the conjugate acceptor is and hence the faster the reac%on. This can be related to pKa

ONu

NO

ONu

OR

ONu

NR2

ONu Nu

N


pKa 10 20 24 25 25

O2NEWG HR

O

RO

O

R2N

ONC

faster conjugate addi%on

slower conjugate addi%on

for some nucleophiles conjugate addi%on is the major pathway, for other nucleophiles direct addi%on is the major pathway whereas for others slight varia%on in condi%ons can alter the course of the reac%on


Ph OEt

O

Ph OEt

O

NH

NH

++Ph N

O

Ph OEt

ONheat heat

irreversible reac%on but if given the choice amines do conjugate addi%on

O BuMgBr Bu OH O BuMgBr O Bu

1 % CuCl

irreversible reac%on 1,2-‐addi%on

irreversible reac%on 1,4-‐addi%on

O NaCN, HCN

5-10 °CNC OH O NaCN, HCN

80 °C

O

NC

formed faster kine%c product

lower ac%va%on energy

more stable thermodynamic product

examples of conjugate addi%on


conjugate addi%on product is generally the thermodynamically more stable product with respect to the direct addi%on product.

a rough comparison of bond energies supports the above conjecture

O NaCN, HCN

5-10 °CNC O

H

lost C=O gained O-‐H

gained C-‐C

O NaCN, HCN

80 °C

O

NCH

lost C=C

gained C-‐C gained C-‐H

lost kJmol-‐1 gained kJmol-‐1 overall gain kJmol-‐1

C=Oπ 370 C-‐C 350 90

O-‐H 460

C=Cπ 270 C-‐C 350 130

C-‐H 400

conjugate addi%on product is generally the thermodynamically more stable product with respect to the direct addi%on product because it retains the strong carbonyl double bond – this is general for most α,β-‐unsaturated systems

in the above example: the direct addi%on product is the kine%c product i.e. the fastest formed product and hence the product formed by the pathway with the lowest ac%va%on energy


ac%va%on energy

kine%c control lower ac%va%on

energy

O

NC OH

O

NC

O

NC

NC O

NaCN + HCN

why is the direct addi%on product formed fastest?

O OO delocalisa%on indicates that the carbonyl carbon (and one of the alkenyl carbons) bear par%al posi%ve charge

carbonyl carbon carries the larger posi%ve charge as it is closer to the electronega%ve oxygen atom

charged nucleophiles will aTack the carbonyl carbon faster than the β-‐carbon (Hard-‐Hard interac%on)

more electron deficient carbon

α β

charged nucleophiles will aTack the β-‐carbon but more slowly

kine%c product formed faster

thermodynamic product lower in energy more stable

energy

extent of reac%on

thermodynamic control

able to reverse at 80 °C

difficult to reverse even at 80 °C

at 80 °C cyanohydrin is reversible at 0 °C cyanohydrin is irreversible

at 80 °C kine%c product is s%ll formed first but reverts to star%ng materials and slower conjugate addi%on occurs

if direct addi%on is reversible conjugate addi%on will result


HClO:

Cl

OH

OClH

O

Cl

H

HO

Cl

H

H

what is actually happening in the addi%on of HCl to methyl vinyl ketone

thermodynamically controlled reac%on most stable product is formed charged nucleophiles usually do direct addi%on

O

Cl

HHO Cl

Ph OEt

O

Ph OEt

O

NH

NH

++Ph N

O

Ph OEt

ON

not all products arising from conjugate addi%on are the result of ini%al reversible direct addi%on

for certain nucleophiles conjugate addi%on is the kine%cally most favoured pathway

in these instances the kine%c product also happens to be the thermodynamic product

irreversible reac%on irreversible reac%on


if the nucleophile has a high energy HOMO this will be close in energy to the LUMO of the α,β-‐unsaturated system

therefore conjugate aTack will occur at the β-‐carbon – the reac%on is under orbital control

for fast reac%on we require a small HOMO / LUMO gap

O

Nu

Nu: O

Orbital Controlled Reac%ons (SoU-‐SoU interac%ons)

LUMO = π*

HOMO = Nu lone pair

Ph OEt

O

NH

+Ph OEt

ON

for amines: uncharged, direct addi%on not favoured (lone pair of intermediate energy)

conjugate addi%on is the major pathway in the above example

the reac%on is under orbital control 1. Generally 2nd row elements (e.g. P, S) favour conjugate addi%on as they have high -‐ energy 3s/3p lone pairs that are a good energy match for the LUMO of the substrate. 2. If the nucleophile is uncharged then conjugate addi%on oUen results.


Predict the outcome of the following reac<ons – ra<onalise your predic<ons

O PhSH, base O LiAlH4

O BuMgBrO BuMgBr

1% CuCl

the conjugate acceptor

the more electrophilic the carbonyl group the more likely to undergo direct addi%on – charge control dominates

Cl

O R2NH

NR2

O very reac%ve carbonyl group, charge control therefore en%rely 1,2-‐addi%on

ONO

O

OR

O

NR2

ON

H

O

Cl

O

mainly direct addi%on

nearly always conjugate addi%on

always conjugate addi%on


sterics can influence the outcome of the reac%ons

Summary: more reac%ve nucleophiles (RLi, RMgBr, LiAlH4) prefer direct addi%on

more reac%ve electrophiles prefer direct addi%on

less reac%ve nucleophiles prefer conjugate addi%on

less reac%ve electrophiles prefer conjugate addi%on

watch out for: reversible direct addi%on which leads to conjugate addi%on

i.e. kine%c versus thermodynamic control

O

OMeMgBr

O

OMeconjugate addi%on even though hard Grignard reagent


Carbonyl Groups – difference in reacTvity towards nucleophiles

Structure of the carbonyl group C-‐O σ-‐bond and C-‐O π-‐bond

R H

O

R Me

O

R OMe

O

R OH

O

R NR2

O

R Cl

O

R

O

O

O

R

Nu:Nu O

Nu:X

Nu OO

X

O

Nu

O

in general there are two types of mechanism

with aldehydes and ketones addi%on occurs – frequently followed by subsequent reac%on

with esters, acids, amides etc. addi%on and subsequent elimina%on occurs

OO O

π-‐bonding orbital

π*-‐an%bonding orbital

the π-‐bonding orbital is polarised towards oxygen the more electronega%ve atom ∴ the π* an%bonding orbital is polarised towards carbon i.e. the large lobe is on carbon

in the mechanism of aTack the HOMO of the nucleophile overlaps with the LUMO (π*) of the carbonyl group

O

NuHOMO

LUMO


typical addi%ons reac%ons of aldehydes and ketones are as follows:

hydra%on – aldehydes are prone to hydra%on – ketones far less so

R

O

H+ H2O

R H

HO OH

Keq Keq

hexafluoro-‐acetone 1.2 x 106 acetaldehyde 1.06

formaldehyde 2280 acetone 0.001

chloral 2000

OH

H

OH

Me

OH

Cl3C

OF3C

F3C

OMe

Me

O

How hydrated would you expect the following compounds to be?

O

O

O


O MeOH

H

OH MeO OHH

±H MeO OH2 OMe

MeOH

MeO OMeH

MeO OMe+H

in a similar manner, aldehydes and ketones react with alcohols under acid catalysis to give acetals

the mechanism of this reac%on is very frequently drawn incorrectly!

the intermediates in blue boxes are closely related – they are much more electrophilic versions of the original carbonyl group and so are readily aTacked by MeOH which is a very weak nucleophile

acid catalysis is required so that the intermediate in the red box can expel water – if no acid were present HO-‐ would be the leaving group – MeO is not a good enough ‘pusher’ to kick out hydroxide

electrophilic electrophilic

the whole process is in equilibrium and the most stable product is therefore formed the reac%on can be readily reversed using acidic water

aldehydes readily form acetals with simple alcohols

with ketones the equilibrium greatly favours the ketone – using diols allows efficient acetal forma%on why is this?

loss of water

O HOOH

H

O O


acetal forma%on is mechanis%cally closely related to numerous other reac%ons of carbonyl compounds

aldehydes and ketones readily react with primary amines and related nitrogen nucleophiles to give imines and related compounds – these reac%ons are generally catalysed by acid

HN OH

PhO PhNH2 O N O ±H

H

H H

PhH H

N OH2Ph

NH PhN

Ph+

acid catalysed

Explain shape of the pH rate profile for oxime forma<on between acetone and hydroxylamine

O + NH2OHNOH rate

pH 4 6 82


secondary amines also react with aldehydes and ketones to give iminium ions and subsequently enamines

O MeNHMe N

Me Me

HN

Me Me

iminium ion enamine aldehydes and ketones are more electrophilic than the corresponding imines – oxygen is more electronega%ve

than carbon

iminium ions are more electrophilic than than the corresponding aldehyde or ketone

R H

O

R H

NR'

R H

NR'R''

R R

O

R R

NR'

R R

NR'R''

least electrophilic more electrophilic

Provide a mechanism for the following reac<on (more of this later)

NH2+

O NaBH4, AcOH, NaOAc HN


Carboxylic acids and related groups

mul%ple types of delocalisa%on are possible

R

O

X R

O

XC O

X

R

X p-‐type lone pair → C-‐O π*

evidence: esters and amides are planar i.e. there is substan%al double bond character between X and the carbonyl carbon

O

NMeMe

Me

most important for esters (X= OR) and amides (X = NR2)

O and N have 2p orbitals which are a good size and energy match for C=O π* orbital (itself made up of overlap of two 2 p orbitals)

nitrogen is less electronega%ve than oxygen and hence with amides there is greater p-‐type lone pair → C-‐O π* dona%on than with esters

∴ amides are less reac%ve towards nucleophiles than esters

Me OEt

O

Me N

OMe

Me

1743 cm-‐1 1646 cm-‐1

rough order of importance NR2 > OR > Cl


117.4 ° 126.4.4 °

R

O

X R

O

XR

CX

O

O sp2 lone pair → C-‐X σ* implica%on is carbonyl group has par%al triple bond character and is also very electrophilic

νmax = 1715 cm-‐1 1766 cm-‐1 1771 cm-‐1

O OCl

ClCl

OF

FF

E. J. Corey, J. O. Link, S. Sarshar, Y. Shao, Tetrahedron LeV. 1992, 33, 7103

evidence from IR stretching frequencies and X-‐ray crystal structure analysis

Me OEt

O

Me N

OMe

Me1743 cm-‐1 1646 cm-‐1

oxygen is more electronega%ve than nitrogen and hence the C-‐O σ* is lower in energy than the C-‐N σ* and a beTer energy match for O sp2 lone pair

rough order of importance X = F > OR > Cl > NR2

CF3

O

the balance of these effects determines the reac%vity of the system


mul%ple types of delocalisa%on are possible



third type of delocalisa%on affects the conforma%on of esters

R O

OR

RC

O

O

R

O sp2 lone pair → C-‐O σ*

esters predominantly exist in the (Z)-‐conforma%on

A .A. Yakovenko, J. H. Gallegos, M. Yu. An%pin, A. Masunov, T. V. Timofeeva, Cryst.Growth Des. 2011, 11, 3964

O

OMe

Ph

O

OPhMe

Z-‐ester E-‐ester

in terms of reac%vity towards nucleophiles -‐ rough order of reac%vity is:

conversely in terms of reac%vity towards electrophiles – amides are the most reac%ve

R

O

NR2 R

O

NR2

EE


R Me

O

R OMe

O

R OH

O

R NR2

O

R Cl

O

R

O

O

O

R


chemoselec%vity in the reduc%on of carbonyl compounds.

O OOH H2, Pd/CNaBH4

for chemo and regioselec%ve reduc%on it is important to choose the correct reagent

O

O

EtOLiAlH4 NaBH4

CeCl3

OH

O

EtOOH

HO

MnO2

OHO


iminium ion aldehyde ketone ester amide acid

NaCNBH3

NaBH4

LiBH4

LiAlH4

BH3

summary of reducing agents for carbonyl groups adapted from Organic Chemistry, Clayden, Greeves and Warren, 2nd Edi%on, OUP 2012.

R H

NR

R H

O

R R

O

R OR

O

R NR2

O

R OH

O

R NHR R OH R R

OH

R OH R NR2 R OH

reduced

reduced slowly

not usually reduced

amine 1° alcohol 2° alcohol 1° alcohol 1° alcohol amine

R H

O

R H

O

R OH

aldehyde 1° alcohol aldehyde

DIBAL via acid chloride

BH3•NH3, LDA

R H

NRH


selec%vity in reduc%on

why are esters reduced with LiBH4 but only slowly with NaBH4?

R OR

O Li BH4R OR

OLi

HBH

H H

R OR

OLi

HR H

O

HBH

H H

R OH

Li+ has a higher charge/radius ra%o compared with Na+ as it is smaller

∴ Li+ is a more potent Lewis acid than Na+

Li+ serves to ac%vate the ester carbonyl for reduc%on

how can we reduce an ester to an aldehyde?

HAl

HAlAl H

exists as an H-‐bridged dimer but reacts as a monomer Al has an empty p-‐orbital, the monomer is electrophilic

DIBAL-‐H – diisobutylaluminium hydride

DIBAL-‐H only becomes a good reducing agent once it has been ac%vated by complexa%on by a Lewis base


R OR

O DIBAL-H

Al HR

R

R OR

OAl

H

R R

R

OAl

R

R

HOR

MeOHthen H

R

OHOR

H

HR H

O

DIBAL-‐H – diisobutylaluminium hydride – commercially available as solu%ons in various organic solvents

work-‐up of DIBAL-‐H reac%ons can be complicated by gela%on due to the amphoteric nature of AlIII salts

par%%oning the reac%on mixture between an organic solvent and aqueous Rochelle salt (Na+K+ tartrate) coupled with vigorous s%rring is a useful method of solubilising these gels

tetrahedral intermediate stable at low temperature (-‐78 °C)

addi%on of acid destroys excess hydride and protonates tetrahedral

intermediate

ester reduced to aldehyde with DIBAL-‐H at low temperature

at higher temperature, DIBAL-‐H reduces esters to alcohols

R OR

O DIBAL-H

Al HR

R

R OR

OAl

H

R R

R

OAlR

R

HOR R H

O DIBAL-H

R OH

tetrahedral intermediate not stable at RT

aldehyde much more reac%ve than ester

rapid reduc%on to alcohol


N

O

OO

O

DIBAL-H, toluene N

OH

O

O it can be very difficult to reduce an α,β-‐unsaturated ester to an aldehyde with DIBAL-‐H

lactols are very readily prepared by reduc%on of lactones with DIBAL-‐H

OH

H

ODIBAL-H, -78 °C O

H

H

OHOH

H

O

DIBAL is also very useful for reducing nitriles to aldehydes what is the mechanism of this reac<on?

C

H

H NDIBAL-H

then H2O, HH

HO

H

DIBAL-‐H – diisobutylaluminium hydride – commercially available as solu%ons in various organic solvents


if we add a Grignard reagent or organolithium to an aldehyde or ketone, monoaddi%on occurs but with esters double addi%on is the general outcome

R OR'

OR'' MgBror R'' Li R OR'

O R''

R R"

OR'' MgBrR R"

O

R"R R"

OH

R"

H2O, H

ketone more electrophilic than ester ter%ary alcohol

in order to obtain mono-‐addi%on use amides as the electrophile

OMeH BuLi

OH

Me LiBu O

Me

Li

O

NMe2H

DMF

MeO

NMe2

H OLi MeO

NMe2

H OH, H2O

H

H

MeO

H

O the most versa%le solu%on is to use Weinreb amides

R

O

OMe

NH

OMeMe

• HCliPrMgCl, or AlMe3

R

O

NOMe

MeWeinreb amide

note: aluminium and magnesium amides are par<cularly nucleophilic towards esters

for use of iPrMgCl in the synthesis of Weinreb amides see: J. M. Williams, R. B. Jobson, N. Yasuda, G. Marchesini, U.-‐H. Dolling,E. J. J. Grabowski, Tetrahedron LeV., 1995, 36, 5461-‐5464


Weinreb amides – a very reliable ketone synthesis. For a review see: M. Mentzel and H. M. R. Hoffmann, J. Prakt. Chem. 1997, 339, 517-‐524.

R NMe

OOMe

R' MgBr

R' Lior NMe

OMeMgO

RR'

H , H2OR N

OHOMe

R'H Me

R R'

O

stable chelated tetrahedral intermediate

tetrahedral intermediate protonated on work up and collapses to generate ketone

VITAE PHARMACEUTICALS, INC. Patent: WO2007/117560 A2, 2007.

O

BocN

O

NO

MeMe

ClMgOMe

THF, -20 °C to RT, then HCl (aq)

O

BocN

O

OMe

93%

O

NOTBDPS

OH

Me

MeO

Me

H MgBrO

OTBDPSOH

MeHTHF, 77%

D. A. Evans, J. T. Starr, Angew. Chem. Int. Ed., 2002, 41, 1787-‐1790


DIBAL-‐H reduc%on of esters to aldehydes can, at %mes, be difficult to control – using a Weinreb amide overcomes this problem

MeON

O

Me Me

OTBDMSMe

Br

OTBDMSTBDPSO

TBDMSO

Me

DIBAL-H, THF

H

O

Me

OTBDMSMe

Br

OTBDMSTBDPSO

TBDMSO

Me

D. A. Evans, J. T. Starr, Angew. Chem. Int. Ed., 2002, 41, 1787-‐1790

enolates will also add to Weinreb amides

O

NMe

OMe

OLi

OtBuO

CO2tBuO

LiOMeN

Me

H, H2O

83%

O

OtBu


R OR

O Li AlH4R OR

OLi

HAlH

H H

R OR

OLi

HR H

O

HAlH

H H

R OH

lithium aluminium hydride – LiAlH4 -‐ all four hydrides are ac%ve

powerful and frequently non-‐selec%ve reducing agent: will reduce aldehydes, ketones, esters to primary alcohols and amides to amines

LiAlH4 both in solu%on and as a solid is highly flammable – requires anhydrous solvents

work-‐up of LiAlH4 reduc%ons can be tricky due to the amphoteric nature of AlIII salts

a useful ‘anhydrous’ work up introduced by Feiser involves, for n grams of LiAlH4 adding dropwise n mL of water, n mL of 15% NaOH solu%on, and then 3n mL of water. In favourable cases a granular precipitate is produced which can be filtered. L. F. Fieser, M. Fieser, M. Reagents for Organic Synthesis 1967, 581-‐595.

another safe method for neutralising excess LiAlH4 involves quenching the reac%on with EtOAc

reduc%on of esters tetrahedral intermediate

collapses to give an aldehyde


R OR

O Li AlH4R OR

OLi

HAlH

H H

R OR

OLi

HR H

O

HAlH

H H

R OH

R NR2

O Li AlH4R NR2

OLi

HAlH

H H

R NR2

OLi

H

AlH3

R NR2

OAlH3

HR H

NR2

HAlH

H H

R NR2

reduc%on of esters

reduc%on of amides

tetrahedral intermediate collapses to give an aldehyde

tetrahedral intermediate collapses to give an iminium ion

why this difference in reac%on outcome?

RO-‐ is a beTer leaving group then R2N-‐

lone pair of amine is higher in energy than O and hence R2N is a beTer ‘pusher’ than RO


O

O

O

H

H

N Me

O

LiAlH4, THF

O

OH

H

H

N Me

H H

O

MeO

MeMe

LiAlH4, THFO

HO

MeHO

MeMe

amide reduced to amine 1,2-‐reduc%on of α,β-‐unsaturated ketone

(hard nucleophile) lactone (ester) reduc%on leads to diol

lithium borohydride – will reduce esters to primary alcohols – see above (can be prepared from cheap NaBH4 and LiCl, LiBr or LiI)

sodium borohydride – frequently used in alcoholic solvents such as MeOH or EtOH

generally does not reduce esters, epoxides, lactones, nitriles. All four hydrides are ac%ve

NaBH4 reacts with pro%c solvents to generate alkoxy borohydrides

O

OH

H

O

Cl

NaBH4, MeOH

O

OH

H

HO

Cl

87%

Explain the stereoselec<vity exhibited by the following reac<on


Luche reduc%on NaBH4 is not selec%ve for 1,2-‐ versus 1,4-‐reduc%on – addi%on of CeCl3•7H2O increases the amount of 1,2-‐reduc%on

1,2-‐reduc%on 1,4-‐reduc%on

NaBH4, MeOH 51% 49%

NaBH4, CeCl3•7H2O, MeOH 99% trace

O

MeOH

NaBH4or NaBH4, CeCl3•7H2O

OH OH

+

it appears that CeCl3 accelerates the reac%on of pro%c solvents with NaBH4 to generate alkoxy borohydrides NaBH(4-‐n)OMen which are harder reducing agents

CeCl3 acidifies the MeOH allowing it to ac%vate the carbonyl oxygen making the carbonyl carbon more posi%ve

Reagent is harder, substrate is harder, therefore 1,2-‐reduc%on -‐ A L. Gemal, J. –L. Luche, J. Am. Chem. Soc., 1981, 103, 5454-‐5459

OHOHOMe

CeIIIMeO

BOMe

MeO H

MeMe Me Me

Me

ONaBH4, CeCl3•7H2O, MeOH MeMe Me Me

Me

HO

58%Givaudan Roure (Interna%onal) SA Patent: US5929291 A1, 1999


in a related manner the use of anhydrous cerium(III) chloride in the presence of Grignard reagents and organolithium reagents allows the addi%on of organometallics to highly enolisable aldehydes and ketones T. Imamoto, N. Takiyama, K. Nakamura, T. Hatajima, Y. Kamiya, J . Am. Chem. Soc. 1989, 111 , 4392-‐4398; N. Takeda, T. Imamoto Org. Synth. 1999, 76, 228

methods to dry CeCl3•7H2O -‐ W. H Bunnelle, B. A. Narayanan, Org. Synth., Coll. Vol. VIII, 1993, 602.

product recovered star%ng material

BuLi 26% 55%

BuLi, CeCl3 92-‐97%

BuMgBr 28% 23%

BuMgBr, CeI3 96%

O

BuM, THFHO Bu


other modified borohydrides: NaBH3CN, NaBH(OAc)3 reagents of choice for reduc%ve amina%on

O

N

Ph

MeH

CH2O, NaBH3CN

pH 5

O

N

Ph

MeMe

NH

O

TBDMSO

HR

O

H

MeAcO+

NaBH(OAc)3, SnCl2

NO

TBDMSO

HR

MeAcO

in each case, reduc%on of the intermediate iminium ion is more rapid than the reduc%on of

the corresponding aldehyde

this is one method to solve the problem of polyalkyla%on when aTemp%ng to alkylate amines

R NH2MeI R NHMe MeI R NMe2

MeI R NMe3 I

at least as nucleophilic as star%ng amine

at least as nucleophilic 1° and 2° amine polyalkyla%on occurs


R

O

NR2

HBH

H

R

O

NHR2

BH H

H O

R NR2

BH

H

H

work-up

R H

NR2HBRR

R NR2BR2

R NR2

borane complexed to a Lewis base, THF•BH3, Me2S•BH3 is a good reducing agent for carboxylic acids and amides

R

O

OH

HBH

H

R

O

OH

BH H

H

R

O

O

BH

H

:LB

R

O

O

BH

HLB

O

R O

BH

LB

H

R

O

RBH

R

HR

O

H

BR R

HR OBR2work-up

R OH

P. C. Lobben, S. S. –W. Leung, S. Tummala, Org. Process Res. Dev. 2004, 8, 1072–1075.

EtO2C

HO2C

Ar

Ar' BH3•THF, THF, 0 °C

EtO2C

HO

Ar

Ar'

98%


O

MeO

O

OH

BH3•THF

O

MeO

OH

O. Hoshino,Y. Mizuno,M. Murakata, H. Yamaguchi Chem. Pharm. Bull., 1999, 47, 1380-‐1383

Me

HO2C

O

OMe Me

MeO2C

O

OMe

H H

H

H H

H

Me

O

O

OOMe

H H

HHO

Me Me

H H

HHO

O

O

CH2N2 HOOH

pTSA

Me

MeO2C

O

O

OOMe

H H

H

LiAlH4

water, pTSA

D. N. Kirk, M. S. Rajagopalan, M. J. Varley, J. Chem. Soc., Perkin 1, 1983, 2225-‐2228

ReducTon examples


O

CO2HO

MeO

OMe

O

O

MeO

OMe

HO

(COCl)2, DMF O

O

MeO

OMe

OCl

NaBH4, THF

T. P. O'Sullivan, H. Zhang, L. N. Mander, Org. Biomol. Chem., 2007, 5, 2627-‐2635

ReducTon examples

NH

OTf2O,

N F

then Et3SiHthen acid workup

O

H90%MeO2C MeO2C

NH

OTf2O,

N F

then Et3SiHthen basic workup

N

H95%MeO2C MeO2C

Tf2O, then

NH

H HCO2EtEtO2C

MeMe

N

O

Ph

O

N

O

Ph

86%

Recently ChareVe reported the chemoselec<ve reduc<on of amides – explain these results J. Am. Chem. Soc., 2008, 130, 18-‐19; J. Am. Chem. Soc., 2010, 132, 12817-‐12819;


Me

H H

H

Rings and ring strain

classifica%on of rings

classifica%on small normal medium large

number atoms in ring 3, 4 5, 6 7-‐12 >12

types of ring strain

angle strain (Baeyer strain) – distor%on of angles from the idealised values

Me Me

H H108°

111°

120°

larger to relief of strain between geminal methyl groups

angle strain – most important in small rings

O

~60° 88°

OH Na2CO3, MeOH O

HMeO

O

H


torsional strain arises from devia%on from an ideal staggered arrangement

major feature of 3 and 4-‐membered rings

H H

H

H

H HH

H

H

H

H H

H

HH

HH

H

H

HH

HH

H

in its chair form cyclohexane has no torsional strain

H

H H

HH

H

H

H

H

H

H

H

the lowest energy conforma%on of cyclopentane is an envelope which has some torsional strain

H

H H

H

H

HH

HH

H

H

HH

H

bond length strain – arises from devia%on of bond lengths from their ideal values Me Me

H H

C-‐C = 1.54 Å

C-‐H = 1.09 Å

transannular strain– arises from proximity on non-‐bonded atoms frequently important in medium rings

H H

as a result of torsional strain 6-‐membered rings are generally more stable than 5-‐membered rings


YX:

K or k

X

general features of ring closure

if ring closure is kine%cally controlled (k) then the energy of the transi%on state will be important

ΔG‡ = ΔH‡ -‐ TΔS‡

if ring closure is thermodynamically controlled (K) then energy of product will be important

ΔG = ΔH -‐ TΔS

ΔH‡ -‐ enthalpy of ac%va%on includes bond breaking/making enthalpic considera%ons and the change in strain energy in reaching the transi%on state

ΔS‡ -‐ reflects the difference in the levels of organisa%on between star%ng material(s) and transi%on state

ring strain considera%ons mean K is generally only favourable for 5-‐ and 6-‐ membered ring

kBT h

e-‐ΔG‡/RT k =


-‐6

-‐4

-‐2

0

2

4

6

2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22

Small rings: 3-‐membered rings are formed fast; 4-‐membered rings more slowly ΔS‡ favourable as liTle preorganisa%on is required – the ends are already close to one another ΔH‡ unfavourable due to developing strain

ring size

log k

es%mated value

rates of cyclisa%on of ω-‐bromo malonates: M. A. Casadei, C. Galli, L. Mandolini, J. Am. Chem. Soc., 1984, 106, 1051-‐1056

increasing ΔS‡ kBT h

e-‐ΔG‡/RT k =

CO2Et

EtO2C ( )nBr

NaH, DMSO

EtO2C CO2Et

( )n

k

small


-‐6

-‐4

-‐2

0

2

4

6

2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22

Normal rings: 5-‐membered rings generally formed fastest ΔS‡ is becoming less favourable as more preorganisa%on is required – the ends are less close to one another ΔH‡ is fairly consistent – 5-‐7 membered rings are rela%vely unstrained

ring size

log k

es%mated value increasing ΔS‡ kBT h

e-‐ΔG‡/RT k =

normal medium

Medium rings: ΔS‡ is s%ll increasing but propor%onally less as ring size increases – the ends are less close to one another ΔH‡ becomes dominant – transannular strain reflected in TS and hence rate of cyclisa%on

large

Large rings: ΔS‡ is unfavourable as the ends are unlikely to meet – similar to an intermolecular reac%on. Solu%on do reac%ons under high dilu%on ΔH‡ no ring strain so not important -‐ large rings are similar to acyclic compounds

CO2Et

EtO2C ( )nBr

NaH, DMSO

EtO2C CO2Et

( )n

k

◾ rates of cyclisa%on of ω-‐bromo malonates: M. A. Casadei, C. Galli, L. Mandolini, J. Am. Chem. Soc., 1984, 106, 1051-‐1056


correct alignment of orbitals is also key for efficient ring forma%on

O

Br

OX

OHOMO enolate

LUMO C-‐Br σ*

O Br

BrO

poor overlap so no C-‐alkyla%on good overlap O-‐alkyla%on occurs

similarly

SO

O O

CH3

S OO

H3C

NaH

SO

O O

CH3

S OO

H3C

SOH

O O

S OO

H3C

CH3X

Sir Jack Baldwin proposed a set of guidelines (Baldwin’s rules) to asses the likelihood that a given, kine%cally controlled cyclisa%on would be feasible


“Ring-‐forming reac<ons are important and common processes in organic chemistry. I now adumbrate a set of simple rules which I have found useful in predic<ng the rela<ve facility of different ring closures. I believe these will be useful to organic chemists, especially in planning syntheses.” J. E. Baldwin, Chem. Commun., 1976, 734-‐736.

Sir Jack Baldwin

Waynflete Professor of Organic Chemistry, Oxford 1978-‐2005 Baldwin’s Rules

Biosynthesis of penicillins

modes of cyclisa%on

Nu:X Y

NuX

Y

Nu:X

NuX

Y

endo – bond being broken is inside the ring being formed

exo – bond being broken is outside the ring being formed

XNu: XNu: Nu:X

exo-‐tet sp3 hybridised

exo-‐trig sp2 hydridised

exo-‐dig sp hybridised

X

Nu:

X

Nu:

X

Nu:

endo-‐tet sp3 hybridised

endo-‐trig sp2 hydridised

endo-‐dig sp hybridised


ring size 3 4 5 6

TET exo ✓ ✓ ✓ ✓

endo -‐ -‐ ✗ ✗

TRIG exo ✓ ✓ ✓ ✓

endo ✗ ✗ ✗ ✓

DIG exo ✗ ✗ ✓ ✓

endo ✓ ✓ ✓ ✓

in general all exo-‐tet and exo-‐trig cyclisa%ons are favoured

5-‐exo-‐trig is faster than 6-‐endo trig

Baldwin’s rules

SO

O O

CH3

S OO

H3C

NaH

SO

O O

CH3

S OO

H3C

SOH

O O

S OO

H3C

CH3X

How would you assign this reac<on according to Baldwin’s rules?

SO O

S OO

H3C

NaH SO O

S OO

H3C

I

Is this likely to be an efficient transforma<on?


HOO H

O

HO

H

1) The following amine undergoes cyclisa<on – predict the product

H2N OMe

O

2) Explain the contras<ng outcomes of the following reac<ons

HOO

Me

HO

HOH

Me

3) Explain the following reac<on

O

Me

NaOH, H2O

O

Me

Me


Thorpe-‐Ingold effect; M. E. Jung, G. Piizzi, Chem. Rev. 2005, 105, 1735−1766

the increased rate of cyclisa%on when pu�ng a geminal dialkyl group in the cyclising chain is known as the Thorpe Ingold effect.

krel 39 1

a good explana%on for the Thorpe Ingold effect concerns reac%ve rotamers

Br O

OO

OR R R = HR = MeO

O

MeMe

H H

H H

O O

Br

major conforma%on, ends held far apart, cyclisa%on

cannot occur

for gem-‐dimethyl-‐subs%tuted substrate all of the staggered conforma%ons are of similar energy and in two of the conformers cyclising groups are in close proximity

Me Me

H H

O O

Br

Me

MeH H

O O

Br

Me

MeHH

OO

Br

reac%ve rotamers


Explain the following rates of cyclisa<on

NH2BrNH2Br

NH2BrNH2Br

NH2Br

iPr iPr

krel 1 2.2 158 0.16 9190

O

OMo PhN

iPriPr

OOF3C

F3C

F3CCF3

oligomerscat. neat 25 °C

cat. neat 25 °C

O

Examples


O

O

O

O O O O

O O

OHHO

O O

O

O

O

HOOH TsOH

PPh3

OsO4

TsCl. pyridine

O O

OHTsO

LiClO4

HCl (aq)

From Corey’s synthesis of longifolene, J. Am. Chem. Soc., 1964, 86, 478. Explain the various aspects of selec<vity in the following reac<ons


HH

H

C

H

Appendix Hybridisa%on and bonding -‐ a brief recap

Hybridisa%on is a useful concept used by organic chemists to describe the bonding in organic molecules

A quick method to work our the hybridisa%on of an atom is to count the number of subs%tuents on that atom (including lone pairs of electrons), remembering that in the bonded environment first row elements generally have 8 electrons around them 4 subs%tuents = sp3 hybridised, 3 subs%tuents = sp2 hybridised, 2 subs%tuents = sp hybridised

sp3 hybrid orbitals are made up from one s orbital and three p orbitals giving four hybrid orbitals which point to the corners of a regular tetrahedron. This is the bonding arrangement found in methane (bond angle = 109°) where the sp3 hybrid orbitals overlap with the hydrogen 1s orbitals (not shown)

HH

H

H

sp3 hybrid orbitals

x

y

z

py

x

y

z

px

x

y

z

pz s

+ + +


H

H H

H

Similarly, the nitrogen atom in ammonia can be viewed as sp3 hybridised as can the oxygen atom in water although the H-‐X-‐H bond angle is slightly less than 109° due to lone pair–bond pair repulsion

For sp2 hybridisa%on we mix two p-‐orbitals and one s-‐orbital to give three sp2 hybrid orbitals (and leave one p-‐orbital)

The three sp2 hybrid orbitals are arranged 120° apart This is the bonding arrangement found in ethene with the sp2 hybrids overlapping with the hydrogen 1s orbitals (not shown) the remaining pz orbital(s) overlap to form the π-‐bond

sp2 hybrid orbitals

N-‐atom is sp3 hybridised

HH

N

H

lone pair in sp3 orbital

C-‐atom is sp2 hybridised

HH

N

H HO

H

HO

H

O-‐atom is sp3 hybridised

x

y

z

py

x

y

z

px

x

y

z

pz s

+ +

pz orbital H

HH

H


x

y

z

py

x

y

z

px

x

y

z

pz s

+

Similarly, this is the hybridisa%on in carbonyl compounds and imines

For sp hybridisa%on we mix one p-‐orbitals and one s-‐orbital to give two sp hybrid orbitals (and leave two p-‐orbital)

The two sp hybrid orbitals are arranged 180° apart This is the bonding arrangement found in ethyne with the sp hybrids overlapping with the hydrogen 1s orbitals (not shown) the remaining p orbitals overlapping to form the two π-‐bonds

sp hybrid orbitals

O and C-‐atoms are sp2 hybridised

lone pair in sp2 orbital

C-‐atom is sp2 hybridised

N and C-‐atoms are sp2 hybridised

p orbitals

N

H

HH

O

H

H

HH NH

in nitriles the N and C-‐atoms are sp hybridised

lone pair in sp orbital

H H


So to recap, in general, a quick method to work our the hybridisa%on of an atom is to count the number of subs%tuents on that atom (including lone pairs of electrons), remembering that in the bonded environment first row elements generally have 8 electrons around them -‐ sp3 = 4 sp2 = 3 sp = 2 What is the hybridisa%on of the red atoms in the following examples?

H3O+NH4+ CO2C C CH

H H

HNCO

O

O

Let’s look at the bonding in amides. All other things being equal, amides are planar molecules and we are happy to draw the delocalisa%on of the nitrogen lone pair as shown to indicate the par%al double bond character of the C-‐N bond

Following the discussion above the hybridisa%on of the C and O atoms is sp2 but what is the hybridisa%on of the nitrogen atom? Again, following the above discussion, and looking at the form of the amide on the leU hand side (A), the nitrogen atom has 4 subs%tuents, 2 x R, C=O and a lone pair and ∴ is sp3 hybridised However, most organic chemists would say the N atom is sp2 hybridised. Why is this?

R

O

N R

O

NHR2R

R

A B


The curly arrows above represent the overlap of the nitrogen lone pair with the C-‐O π-‐orbitals (the an%bonding π* orbital). The best overlap therefore is if the N-‐atom is sp2 hybridised resul%ng in the N-‐lone pair being in a p-‐orbital with excellent overlap with the p-‐orbitals of the C–O π-‐system If the N-‐atom were sp3 hybridised then the N-‐lone pair would be in an sp3 orbital which would result in poorer overlap with the adjacent C-‐O π-‐system – Generally beTer overlap = greater stabilisa%on In general if a π-‐system has an adjacent atom which carries a lone pair then most organic chemists would view the hybridisa%on of the adjacent atom as sp2 with the lone pair in a p-‐orbital to maximise overlap with the adjoining π-‐system. What is the hybridisa%on of each of the heteroatoms in the following molecules?

C ON

R

RR

C ON

R

RR

N-‐sp2 hybridised N-‐lone pair in p-‐orbital

N-‐sp3 hybridised N-‐lone pair in sp3-‐orbital

R

O

N R

O

NHR2R

R

A B

Perhaps we should not be too concerned about this as some molecules, for example anilines, are frequently not perfectly planar and the hybridisa%on of nitrogen is somewhere between perfectly sp2 and perfectly sp3 Addi%onally we should be aware that other effects (e.g. sterics) can override electronic effects and hence the hydridisa%on may not be as expected

O

R OR

O NMe2OMe

OMe

NO

reactivity and control.pptx - the burton groupburton.chem.ox.ac.uk/reactivity-and-control.pdf ·...

Documents