Finishing oxidation by looking at the Baeyer-Villiger reaction and then turning our attention to reduction. Once again we will see the usual suspects with a who is who of hydride sources.

1. FUNCTIONALGROUP
123.312
INTERCONVERSIONS functional group
CHAPTER7
interconversions H H O H H
Cr O H H
R O S
OH
CHAPTER seven R O
reduction
previously we had looked at
oxidations now we turn our attention to
E
the opposite, reduction
1 2 3
need to be able to control OH
oxidation
now adding hydrogen or Text H
chemoselectivity N
removing oxygen (or other HO
R H Clheteroatom) R R N O C O
HO salmefamol OMe
O O O
R OH
R R1/H R OH H2N NH2 O O O
Cl Cl O Cl Cl vs vs
R NH2 R2
R R R OR1 Cl Cl R1 H R1 R2 R1 O
R1O OR1 O
R Cl
R R R NH2
lets approach this by example
& look at the synthesis of this
Reduction anti-asthma drug
4 5 6
starting material contains many hydrogenolysis & reductive hydrogenolysis & reductive
reducible groups amination amination
OMe OMe
O OH O OH
O O O OH H2, Pd / C, O OH H2, Pd / C, O OH
MeO H+, ketone MeO H+, ketone
MeO NaBH4 MeO N MeO N MeO
HO HO
HN HN
N N Ph Ph HO Ph Ph HO
HO HO
Ph Ph Ph Ph
O O
hydrogenation reductively OMe OMe it also reduces imine formed
sodium borohydride only by condensation of amine &
reacts with the ketone & cleaves benzyl groups (as seen
earlier) but... ketone in a process called
not the ester reductive amination
7 8 9

2. ester reduction look at various reagents
many different forms of
reduction in organic synthesis
OMe OMe
O OH OH OH
LiAlH4
MeO
HN HN
HO HO
finally lithium aluminium
hydride reduces the ester
©status frustration@flickr ©golbog@flickr
10 11 12
Reduction of Aldehydes/ketones mechanism of borohydride mechanism of borohydride
reduction reduction
H Na H Et H Na H Et
Et H O O Et H O O
NaBH4 O H B H O H B H
O or LiAlH4 H OH H R1 R2 H R1 R2
solvent, reductant & cation I doubt mechanism is
are all important concerted but all these
R1 R2 R1 R2 steps must occur
H H
Et H OH Et H OH
O B H O B H
relatively easy reduction
H R1 R2 H R1 R2
13 14 15
by-product is also a (less)
powerful reductant sodium borohydride was
developed during the war but
H
not reported until 1953
H Et
Et
B
O B H Et H
H Et
is not a reducing agent, add electron donating
it is a base (eg sodium groups & get more powerful
hydride) reductant (superhydride LiBH4) ©alifaan@flickr
16 17 18

3. Reduction of esters
NaBH4 R1
O
O
R2
LiAlH4
R1
H H
OH
H
O
R2
LiAlH4
O O O O O O O O
R1 H R1 R2 R1 Cl R1 OR2 R1 NR2
R1 H R1 R2 R1 Cl lithium aluminium hydride
reduces esters all the way
only reduces reactive c=o bonds down to alcohols reduces nearly all carbonyls
19 20 21
mechanism of lithium mechanism of lithium by-product is also a (less)
aluminium hydride reduction aluminium hydride reduction powerful reductant...
H Li Li H Li Li
O O O O
H Al H R2 R1 R2 H Al H R2 R1 R2
H R1 O O H R1 O O
H H
H H
H
O
R2
H H
H
O
R2 AlH3
Li
Al
H Li Li
Al
H Li
O
O H O O H O
R1 H Al H R1 H Al H
H
H H R1 H H
H H R1 H R1 H
H
AlH3 H AlH3 H
O H O O H O cation important;
again, note H– is remove it &
R1 not the reductant R1 ...could be termed a
R1 H R1 H reaction stops
H
H H H
H H more selective reagent!
22 23 24
Reduction of amides mechanism of lithium aluminium amine normally a poor leaving group
hydride reduction of amides
Li
H O Li
O
H Al H R2
R1 N NR22
H R1
H
O H H R2 Li
AlH3
O O
R1 N
R2
LiAlH4
R1 N
R2
H Al H
H
NR22 O
AlH3
R1 NR22
X R1 H R
2
N
R2
R1 NR22 H
R2 R2 H H R1
H
second reduction (of the
lithium aluminium
H H iminium cation) does not Amine anions often used as bases
hydride can perform R2 require metal cation as it (think LDA) they have high pka so
R1 N are poor leaving groups
this reduction is already charged
R2
25 26 27

4. the reduction of Reduction of acids Reduction of acids
Text
amides really isn’t this
simple to be honest!
X
O LiAlH4 H H O LiAlH4 O
H H H
R1 O R1 O R1 O R1 O Li
just get salt formation along
lithium aluminium hydride with the evolution of hydrogen
does not reduce acids under gas (h2)
standard conditions
28 29 30
Reduction of acids Reduction of acids
big Text
difference between
lithium aluminium
hydride & borane
O BH3 H H O BH3 O BH3 H H
H B H
H H R1 O R1 O 3 R1 O
R1 O R1 O reactive
the reduction can be
achieve with a different first step is to
‘protect’ the acid the second
reagent, borane (or equivalent of borane
diborane) does the reduction ©jnthnhys@flickr
31 32 33
importance: liAlH4 is nucleophilic importance: BH3 is electrophilic
Text
H O
H O
B R2
H Al H R2 H H R1 N
R1 !+ O
H R2
more !+ve the borane needs
carbonyl the faster it activation reacts with weaker
should react more ‘electron rich’ borane reverses
carbonyls normal
chemoselectivity
attacks electron poor carbonyls attacks electron rich carbonyls ©j heffner@flickr
34 35 36

5. borane reduction of amides chemoselectivity (enantiospecific) chemoselectivity (enantiospecific)
H
H
O B H BH2 H H H H H H
H H B O LiBH4 BH3 LiBH4 BH3
R2 O H
R1 N R1 R2
R2 N CO2H EtO2C CO2H EtO2C CO2H EtO2C CO2H EtO2C
R2 R1 N H HO OH HO OH
R2 H+ H+ H+ H+
R2
H H H H
H
H H H H
BH2 H H B
R2 2 H O O start with a single O O O O select the correct O O
R1 N R1 N R R2 enantiomer of starting reagent & we can form
R1 N material either enantiomer of
R2 R2
R2 lactone
37 38 39
partial reduction very hard theoretically it is diisobutylaluminium hydride (dibal)
Text
possible...practice can
be more problematic
O O O
R1 OR2 R1 H R1 NR22 Al
difficult to prevent full H
reduction to either
alcohol or amine
dibal’s reactivity is more
similar to bh3 than liaiH4
©0olong@flickr
40 41 42
must be cold mechanism summary starting
iBu iBu material
Al iBu iBu NR2 O O O O O
O Al
H R1 R1 R1 R2 R1 OR2 R1 NR2 R1
O H H H OH
R1 OR2 NaCNBH3 reduced slow no!reaction
DIBAL R1 OR2
O –78°C O NaBH4
LiBH4
LiAlH4
R1 OR2 R1 H O H3O O
AliBu2
BH3
NHR2 OH OH OH NR2 OH
react ester at low R1 H R1 OR2
temperature & you can the crucial H R1 R1 R1 R2 R1 R1 R1
isolate the aldehyde bit is the stability of the stable at
tetrahedral intermediate low temp product
©Rachel's flickrs@flickr
43 44 45

6. want to achieve the following problem: no reagent suitable
transformation...
nabh4 only
reduces ketone
chemoselectivity OH O
O O O O O
reduction reduction OEt
OEt OH OH
OEt
which OH
functional LiAlh4 reduces
group will everything
react? ©amortize@flickr
46 47 48
solution: Acetal protecting group solution: Acetal protecting group
O O reduction O O O reduction O
diol reacts
OEt OH OEt OH
with more reactive ketone to give
acetal then we can do the
HO HO reduction
OH H2O OH H2O
H H H H
O LiAlH4 O LiAlH4
O O O O O O O O
OEt OH OEt OH
use a protecting group
49 50 51
hydrogenation is the catalytic hydrogenation can be
addition of hydrogen chemoselective...
O H2, O
H H H H Pd / C
hydrogenation H H
C X C X
hydrogen can be added across hydrogenation normally
most double or triple bonds & requires a catalyst
can cleave some single bonds this permits exquisite
chemoselectivity
52 53 54

7. catalytic hydrogenatation reduces poison the catalyst... partial hydrogenation...
alkynes
H2, H2, Pd /
H H CaSO4, Pb H H
Pd / C H H
R1 R2 R1 R2
R1 R2 Lindlar R1 R2
catalyst
can we stop this the reaction is stereoselective giving
this blocks some of the the cis product (we’ll find out how to
reaction halfway & form reactive sites on the
an alkene? make the trans isomer later)
catalyst
©furryscaly@flickr
55 56 57
FUNCTIONALGROUP
123.312
hydrogenation can cleave single hydrogenolysis of benzyl ethers
bonds
INTERCONVERSIONS
H H
C X
H H
C X R1
O
H2,
Pd / C
H
R1
O
H
CHAPTER8
seen this before during
protection/deprotection
section hydrogenolysis can also
58
cleave alkyl halides
59
E 60
looked at substrate (R-LG)... C–C bond formation is foundation of
organic synthesis
functional group
interconversions
Nuc
R LG R Nuc
CHAPTER eight
previously we looked at
c–c bond formation the substrate & which leaving
groups were good in substitution
reactions
now look at c-based nucleophiles ©Ricketts Fish@flickr
61 62 63

8. Need... Need... ...a carbanion
Cnuc Celec
virtually all C-C bond forming
reactions come down to this...(not
necessarily charged species but
certainly polarised
C C
64
Cnuc Celec C C
we are now going to
concentrate on
carbanions C–
65
C do not really exist
(just ask an inorganic
chemist) just behave
like this
66
...a carbanion 1. organometallics
C
metal
1. organometallics M R MX
R X or
R M
insert a metal into a
carbon halide bond
two ways to make
carbanions
67 68 69
behave as... grignard reagents made from organolithium reagents made
halides from chlorides (normally)
R
Mg 2 x Li R Li
R Br R MgBr R Cl
LiCl
but more much more reactive
complex insertion occurs via single electron than grignard reagents
R X Sol
transfer (SET) process and the final structure in solution depends
Mg Mg structure is solvent dependent on r. normally oligomeric (dimer,
Sol X R trimer etc)
©jurvetson@flickr
70 71 72

9. methyl lithium is a tetramer in the many other
solid state (& solution) organometallic
reagents...
Cl
R Zn R Zr
2. deprotonation
R
Cu Li
R R all with their own
Copyright: Ben Mills (2007)
pros & cons
73 74 75
2. deprotonation
base
R H R H base
this can be considered but, how easy is it
is removal of a proton
to give a carbanion but reality more complex & we to selectively
are invariably forming an
organometallic reagent
76
remove a proton
©smeerch@flickr
? 77
©dhammza@flickr
revision 78
we can use pKa to assess ease of we can use pKa to assess ease of
deprotonation... deprotonation...
low pKa
H
O
H X H H X
H
O
H
R H
conjugate
base R H base more acidic proton
base
conjugate acid
acid base
easier to form
remember, pKa indicates where
the equilibrium lies or how easy it is to
with deprotonations to
form carbanions
we are effectively looking at
(carb)anion
remove the proton from the acid removing the proton from the
conjugate acid
79 80 81

10. conjugate conjugate http://www2.lsdiv.harvard.edu/labs/evans/
acid pKa value acid pKa value
base base
CH4 CH3 49 OH O 17
O O O O
NH3 NH2 36 11
OEt
H2C CH2 H2C CH 36 OEt
O O
H H H 26 O O
O O 9
23
EtO CH3 EtO CH2 O O
O O 4.8
19 OH O best pKa
H3C CH3 H3C CH2 O O table I
S OH S O -0.6 know
19 O O
OH O
HCl Cl -7
82 83 84
You do not need to learn these values But, you do need to understand electron withdrawing groups
what factors effect them... stabilise anion & make H+ more acidic
H3C O F3C O
H H
pKa = 15.9 pKa = 12.4
F3C O F3C O
H H
CF3 F3C CF3
pKa = 9.3 pKa = 5.4
this is the inductive effect, electrons
©Graham Johnson, Graham Johnson Medical Media, Boulder, Colorado
being pulled through ! bonds
85 86 87
delocalisation stabilises anion remember, many groups involved in
resonance...
you all love the other
effect that stabilises a H N
carbanion... H H R C N
R C
pKa = 51
O
resonance O O O
R N
O
O R N
H H H H O
pKa = 28.3
O O O O
the ability to spread the charge
over more atoms causes a large degree of anion R S R S
stabilisation & hence the big differences observed Ph Ph
©stefan linecker@Flickr in pKa
88 89 90

11. we’ll concentrate on... enolates
O
©alfred sim@flickr
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