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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
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?
Reactivity and Control for Organic Synthesis 2
︎ 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
Reactivity and Control for Organic Synthesis 3
︎ 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
Reactivity and Control for Organic Synthesis 4
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
Reactivity and Control for Organic Synthesis 5
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)
Reactivity and Control for Organic Synthesis 6
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]
Reactivity and Control for Organic Synthesis 7
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-‐
Reactivity and Control for Organic Synthesis 8
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
Reactivity and Control for Organic Synthesis 9
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
Reactivity and Control for Organic Synthesis 10
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-‐
Reactivity and Control for Organic Synthesis 11
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
Reactivity and Control for Organic Synthesis 12
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
Reactivity and Control for Organic Synthesis 13
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
Reactivity and Control for Organic Synthesis 14
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-‐]
Reactivity and Control for Organic Synthesis 15
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
Reactivity and Control for Organic Synthesis 16
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
compound pKa (water)
CH4 48
NH3 38
25
25
20
tBuOH 17
MeOH 15
HC CH
Me OtBu
O
Et Et
O
compound pKa (water)
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.
Reactivity and Control for Organic Synthesis 17
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< <
most acidiclowest pKa
least acidichigest pKa
Me
O
Me
O
Me
O
Me
Me
OMe
most acidiclowest pKa
least acidichigest pKa
< <NH
NH
Me
Me
Me
Me
MeMe
NH
MeMe
> >
most acidiclowest pKa
least acidichigest pKa
Reactivity and Control for Organic Synthesis 18
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)
Reactivity and Control for Organic Synthesis 19
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.
Reactivity and Control for Organic Synthesis 20
“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
Reactivity and Control for Organic Synthesis 21
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
Reactivity and Control for Organic Synthesis 22
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
Reactivity and Control for Organic Synthesis 23
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%
Reactivity and Control for Organic Synthesis 24
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
Reactivity and Control for Organic Synthesis 25
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
Reactivity and Control for Organic Synthesis 26
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
Reactivity and Control for Organic Synthesis 27
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
Reactivity and Control for Organic Synthesis 28
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?
Reactivity and Control for Organic Synthesis 29
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
Reactivity and Control for Organic Synthesis 30
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-
Reactivity and Control for Organic Synthesis 31
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
Reactivity and Control for Organic Synthesis 32
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(-) (-)
‡
Reactivity and Control for Organic Synthesis 33
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
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
Reactivity and Control for Organic Synthesis 34
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> >
most basic most nucleophilic
least basic least nucleophilic
MeI + Cl MeCl + I
Reactivity and Control for Organic Synthesis 35
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
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
Reactivity and Control for Organic Synthesis 36
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(-) (-)
‡
Reactivity and Control for Organic Synthesis 37
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
Reactivity and Control for Organic Synthesis 38
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
Reactivity and Control for Organic Synthesis 39
Nu LG
HH
LG
HH
Nu
H H
NuLG
(-)(-)
‡
Nature of the substrate
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
Reactivity and Control for Organic Synthesis 40
Nu LG
HH
LG
HH
Nu
H H
NuLG
(-)(-)
‡
Nature of the substrate
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
Reactivity and Control for Organic Synthesis 41
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
Reactivity and Control for Organic Synthesis 42
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)
Reactivity and Control for Organic Synthesis 43
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)
Reactivity and Control for Organic Synthesis 44
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
Reactivity and Control for Organic Synthesis 45
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
Reactivity and Control for Organic Synthesis 46
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
Reactivity and Control for Organic Synthesis 47
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/
Reactivity and Control for Organic Synthesis 48
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)
Reactivity and Control for Organic Synthesis 49
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
Reactivity and Control for Organic Synthesis 50
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.
Reactivity and Control for Organic Synthesis 51
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
1) Explain the outcome of the following reac<ons
OH OH
2) Suggest reagents and reac<on condi<ons for the following transforma<on
Reactivity and Control for Organic Synthesis 52
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
Reactivity and Control for Organic Synthesis 53
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
Reactivity and Control for Organic Synthesis 54
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
Reactivity and Control for Organic Synthesis 55
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:
Reactivity and Control for Organic Synthesis 56
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
Reactivity and Control for Organic Synthesis 57
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
‡
Reactivity and Control for Organic Synthesis 58
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
Reactivity and Control for Organic Synthesis 59
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
Reactivity and Control for Organic Synthesis 60
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
✓✓
✓✓
Reactivity and Control for Organic Synthesis 61
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)
Reactivity and Control for Organic Synthesis 62
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
Reactivity and Control for Organic Synthesis 63
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)
Reactivity and Control for Organic Synthesis 64
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
Reactivity and Control for Organic Synthesis 65
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
+
Reactivity and Control for Organic Synthesis 66
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
Reactivity and Control for Organic Synthesis 67
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
Reactivity and Control for Organic Synthesis 68
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
Reactivity and Control for Organic Synthesis 69
OH
OMeOMe
NO O
Br
OH
OMeOMe
Br
1) Explain the outcome of the following reac<ons
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
Reactivity and Control for Organic Synthesis 70
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
Reactivity and Control for Organic Synthesis 71
+ 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
Reactivity and Control for Organic Synthesis 72
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
Reactivity and Control for Organic Synthesis 73
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
Reactivity and Control for Organic Synthesis 74
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
Reactivity and Control for Organic Synthesis 75
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
Reactivity and Control for Organic Synthesis 76
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
Reactivity and Control for Organic Synthesis 77
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
Reactivity and Control for Organic Synthesis 78
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
Reactivity and Control for Organic Synthesis 79
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
Reactivity and Control for Organic Synthesis 80
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]
Reactivity and Control for Organic Synthesis 81
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
Reactivity and Control for Organic Synthesis 82
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
Reactivity and Control for Organic Synthesis 83
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
Reactivity and Control for Organic Synthesis 84
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].
Reactivity and Control for Organic Synthesis 85
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
Reactivity and Control for Organic Synthesis 86
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
Reactivity and Control for Organic Synthesis 87
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
Reactivity and Control for Organic Synthesis 88
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
Reactivity and Control for Organic Synthesis 89
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)
Reactivity and Control for Organic Synthesis 90
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
Reactivity and Control for Organic Synthesis 91
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
Reactivity and Control for Organic Synthesis 92
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
Reactivity and Control for Organic Synthesis 93
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
Reactivity and Control for Organic Synthesis 94
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
Reactivity and Control for Organic Synthesis 95
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
Reactivity and Control for Organic Synthesis 96
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
Reactivity and Control for Organic Synthesis 97
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
Reactivity and Control for Organic Synthesis 98
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%
Reactivity and Control for Organic Synthesis 99
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
Reactivity and Control for Organic Synthesis 100
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
Reactivity and Control for Organic Synthesis 101
OMeBr
OMeLi
NLi N
Br+X
AromaTc organometallics Ortho-‐directed electrophilic aroma%c subs%tu%on
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
Reactivity and Control for Organic Synthesis 102
AromaTc organometallics Ortho-‐directed electrophilic aroma%c subs%tu%on
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
Reactivity and Control for Organic Synthesis 103
AromaTc organometallics Ortho-‐directed electrophilic aroma%c subs%tu%on
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%
Reactivity and Control for Organic Synthesis 104
CO2Me
I
iPrMgCl, THF, -10 °C
CO2Me
MgCl
CO2Me
PhCHO
OHPh
AromaTc organometallics Ortho-‐directed electrophilic aroma%c subs%tu%on
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
Reactivity and Control for Organic Synthesis 105
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
Reactivity and Control for Organic Synthesis 106
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
Reactivity and Control for Organic Synthesis 107
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
Reactivity and Control for Organic Synthesis 108
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
Reactivity and Control for Organic Synthesis 109
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
AromaTc organometallics Ortho-‐directed electrophilic aroma%c subs%tu%on
fredericamycin
synthesis of Fredericamycin – T. R. Kelly, S. H. Bell, N. Ohashi, R. J. Armstrong-‐Chong, J. Am. Chem. Soc. 1988, 110, 6471-‐6480
Reactivity and Control for Organic Synthesis 110
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.
Reactivity and Control for Organic Synthesis 111
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
Reactivity and Control for Organic Synthesis 112
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
Reactivity and Control for Organic Synthesis 113
“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
Reactivity and Control for Organic Synthesis 114
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
increasing nucleophilicity
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
Reactivity and Control for Organic Synthesis 115
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)
Reactivity and Control for Organic Synthesis 116
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
Reactivity and Control for Organic Synthesis 117
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
Reactivity and Control for Organic Synthesis 118
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
Reactivity and Control for Organic Synthesis 119
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
Reactivity and Control for Organic Synthesis 120
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
Reactivity and Control for Organic Synthesis 121
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
increasing electrophilicity
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
Reactivity and Control for Organic Synthesis 122
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
Reactivity and Control for Organic Synthesis 123
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
Reactivity and Control for Organic Synthesis 124
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
Reactivity and Control for Organic Synthesis 125
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
Reactivity and Control for Organic Synthesis 126
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.
Reactivity and Control for Organic Synthesis 127
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
Reactivity and Control for Organic Synthesis 128
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
Reactivity and Control for Organic Synthesis 129
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
Reactivity and Control for Organic Synthesis 130
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
Reactivity and Control for Organic Synthesis 131
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
Reactivity and Control for Organic Synthesis 132
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
Reactivity and Control for Organic Synthesis 133
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
Reactivity and Control for Organic Synthesis 134
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
Reactivity and Control for Organic Synthesis 135
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
Carboxylic acids and related groups
mul%ple types of delocalisa%on are possible
Reactivity and Control for Organic Synthesis 136
Carboxylic acids and related groups
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
increasing electrophilicity
R Me
O
R OMe
O
R OH
O
R NR2
O
R Cl
O
R
O
O
O
R
Reactivity and Control for Organic Synthesis 137
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
Reactivity and Control for Organic Synthesis 138
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
Reactivity and Control for Organic Synthesis 139
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
Reactivity and Control for Organic Synthesis 140
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
Reactivity and Control for Organic Synthesis 141
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
Reactivity and Control for Organic Synthesis 142
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
Reactivity and Control for Organic Synthesis 143
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
Reactivity and Control for Organic Synthesis 144
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
Reactivity and Control for Organic Synthesis 145
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
Reactivity and Control for Organic Synthesis 146
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
Reactivity and Control for Organic Synthesis 147
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
Reactivity and Control for Organic Synthesis 148
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
Reactivity and Control for Organic Synthesis 149
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
Reactivity and Control for Organic Synthesis 150
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
Reactivity and Control for Organic Synthesis 151
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%
Reactivity and Control for Organic Synthesis 152
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
Reactivity and Control for Organic Synthesis 153
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;
Reactivity and Control for Organic Synthesis 154
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
Reactivity and Control for Organic Synthesis 155
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
Reactivity and Control for Organic Synthesis 156
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 =
Reactivity and Control for Organic Synthesis 157
-‐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
Reactivity and Control for Organic Synthesis 158
-‐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
Reactivity and Control for Organic Synthesis 159
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
Reactivity and Control for Organic Synthesis 160
“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
Reactivity and Control for Organic Synthesis 161
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?
Reactivity and Control for Organic Synthesis 162
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
Reactivity and Control for Organic Synthesis 163
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
Reactivity and Control for Organic Synthesis 164
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
Reactivity and Control for Organic Synthesis 165
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
Reactivity and Control for Organic Synthesis 166
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
+ + +
Reactivity and Control for Organic Synthesis 167
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
Reactivity and Control for Organic Synthesis 168
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
Reactivity and Control for Organic Synthesis 169
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
Reactivity and Control for Organic Synthesis 170
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