Alcohols : Page 1

Alcohols Oxidation and Reduction

1 Nomenclature Notation: Recall the following notation (primary, secondary etc.)

• IUPAC naming priority, alcohol > alkene ~ alkyne > halide (more oxidized functional groups have higher priority) • suffix: -ol • MULTIPLE FUNCTIONAL GROUPS: highest priority functional group suffix (e.g. -ol for alcohol) at the end • number to give the -OH the lowest number, and if this is the same numbering from both "ends", THEN (and only then), number to give the first alkene the lowest number • number to give the first substituent the lowest number ONLY IF ALL ELSE IS EQUAL • with ONE -OH group (monoalcohol) names as alkanOL (e.g. 1-penatol) • HOWEVER, with TWO -OH groups (diol) named as alkanEdiol (e.g. 1,3-pentanEdiol)

• LAST EXAMPLE: the -OH group is #4 numbering from EITHER end, therefore it is numbered from the left end to give the first alkene the lowest number - but ONLY because the alcohol is #4 from either end, in all cases number to give the alcohol the lowest number unless there is a special case such as this one • note the use of number directly before the functional group in the examples above, used when we have multiple functional groups • WHEN WE HAVE MULTIPLE FUNCTIONAL GROUPS, they are included in the name in the order: first -ene (if there is one), then -yne (if there is one) followed by -ol. In addition to IUPAC names, many organic structures have "common" names, that are often historical, but are still widely used, more than the IUPAC names. Some Common Alcohols with Common Names:

• Those with the * symbol you NEED TO KNOW!

RCOH R

CRCOH

R' R''R'

OH1o, primary 2o, secondary 3o, tertiary

H H H

HO1 2

3 45

(2S,5)-dimethylhept-(4E)-en-1-ol

HO 12

34

56

7

6-methyl-3-propyl-2-heptanol

sterochemistryignored

67

HO

sterochemistryNOT ignored

12 3

4 5 6 73,3,6-trimethylhepta-1,5-dien-(4R)-ol

OHH4 1

3 (C,C,H)H2 (C,C,C)

(R)*=

H *(S)

sterochemistryNOT ignored

A because DIENE

OHHOOH

glycerol

HO

OHethylene glycol

OH

iso-propanol

OH

phenol

CH3 OH

methanolCH3 OH

ethanol* **

Alcohols : Page 2

2 Alcohol Acidity, Return to Substituent Effects • Alcohols are weak acids, the -OH bonds are similar to those in water Example: Simplest alcohol methanol

• The conjugate base anion of water is the hydroxide anion • The conjugate base anion of an alcohol is generically an alkoxide anion, i.e. the conjugate base anion of methanol is the methoxide anion, the conjugate base of ethanol is the ethoxide anion etc. • In general we will find that the acidity of alcohols is determined by the energy of the electrons in the conjugate base anion, the lower the electron energy in the base, the weaker the base, the stronger acid is the alcohol • For Example: Simple Resonance effects significantly influence alcohol Bronsted acidity

• The energy of the non-bonding electrons in the conjugate base anion of phenol are lower compared to cyclohexanol due to resonance delocalization/stabilization, phenol is the stronger acid, has the smaller pKa

2.1 Substituent Effects: Important General Concept • There are THREE main kinds of substituent effects, INDUCTIVE effect substituents, ALKYL substituent effects and RESONANCE effect substituents • Substituent effects primarily influence the ENERGIES of the ELECTRONS that are involved in Bronsted acidity, and therefore influence this simple chemical reaction in ways that we are already familiar with • A complicating factor in Bronsted acidity is that because liberation of a proton generates ions (the proton and the conjugate base anion), we will also sometimes need to take into account ionic solvation effects 1) Recall the Inductive Substituent Effect: • Inductive Effect: This is the polarization of electrons in sigma-Bonds due to electronegative elements • We previously saw this effect as being responsible for generating BOND DIPOLE MOMENTS

• The polarization of electrons occurs because the electrons are STABILIZED by proximity to electronegative elements • Electronegative elements are electronegative because they have more concentrated positive charges on the nucleus that are not highly shielded by electrons, which lowers the energy (increases the stability) of electrons not only on the electronegative atom but also on adjacent atoms that are connected via sigma-bonds.

H2O

CH3OH

pKa~15.5

~15.5

HO

CH3O

H2O

H2O

+ H3O+

+ H3O+

hydroxide

methoxideconjugate base anion

RO = generically alkoxide anion

methanol

OH

OH

H2O

H2O

O

O

~19.0

~9.0

pKa

OOO

cyclohexanol

phenol

+ H3O+

lower energy electrons, weaker base

higher energy electrons, stronger base

lower pKastronger acid

HO

Hδ δ

δ

Alcohols : Page 3

• The inductive effect normally stabilizes the conjugate base alkoxide anion • The inductive effect decreases rapidly with increasing distance between the electronegative element and the electrons of the base anion, here is an example from earlier in the course

2) Alkyl groups as substituents: • Alkyl groups stabilize cations by HYPERCONJUGATION, and destabilized anions due to electron repulsion • Alkyl groups stabilize carbocations by hyperconjugation, a form of resonance, hyperconjugation delocalizes electrons and charge, lowers the total energy of the electrons in the cation

• Extra methyl groups weakly destabilize the non-bonding electrons in alcohol conjugate base anions

• The electron donation effect is actually pretty weak in this case, AND, probably more important is that the extra methyl groups also decrease solvation of the conjugate base anions, lowering the propensity of the alcohols to ionize in water, decreasing their acidity

CH3OH~15.5 CH3OH2O

+ H3O+

.CF3 CH2OH~12.4 H2O + H3O+

CF3 CH2O

pKa

inductive effect

C OH

HC

F

F

F

weaker basestronger acid

F

O

F

O

+ H+

+ H+

stronger acid MORE stable anion STRONGER inductive effect

weaker acidF

OH

F

OH LESS stable WEAKER inductive effectδ

δδ

δ-H+

-H+

CH

H

H

methyl primary secondary tertiary

CH

H

RC

R

H

RC

R

R

R

increasing hyperconjugation/stability, INCREASING stability

CH

HC

H

HH

hyperconjugation - a form of resonancedelocalizes charge, stabilizes cations

CH

H

C

H

HH

CH

H

H

methyl primary secondary tertiary

CH

H

RC

R

H

RC

R

R

R

INCREASING electron repulsion, DECREASING stability

CH

H

C

H

HH

electron repulsion - raises non-bonding electron energy, destabilizes anions

electron repulsiondestabilization

H3CCH

H3COH~19.0

H3CCH

H3CO

H2O+ H3O+

Me- donating groups weakly

destabilize anion and LOWER

solvation of the anion

CH3 CH2OH~15.9 CH3 CH2OH2O + H3O+

H2O

CH3OH

pKa~15.5

~15.5

HO

CH3O

H2O

H2O

+ H3O+

+ H3O+

Alcohols : Page 4

3) Alkyl (electron donating) substituents (on a pi-system):

• Simple Alkyl substituents weakly DONATE electron density into the ring as a result of HYPERCONJUGATION • The conjugate base anion is thus weakly destabilized by the methyl group. Alkyl groups are WEAKLY donating, because hyperconjugation is a much less effective form of electron donation compared to conventional resonance (below), because the donated electrons are already in a strong sigma bond. 2.2 Substituent Effects on Pi-Systems : Resonance and Inductive Effects (Resonance wins !) • There are THREE main kinds of substituent effects, INDUCTIVE effect substituents, ALKYL substituent effects and RESONANCE effect substituents 1) Resonance Donation into pi-Systems: • Other substituents can stabilize positive charges and destabilize negative charges on pi-systems such as benzene rings via the inductive AND resonance electron donation and electron repulsion effects • Such effects are often larger than alkyl group substituent effects • Substituents can be classed as ELECTRON DONATING or ELECTRON WITHDRAWING when attached to pi-systems, depending upon whether they have inductive or resonance effects • The resonance in a substituent usually wins over the inductive effect when both are in effect! 2) Stronger Electron Donating substituents on a pi-system: • Minor resonance contributors show how substituents DONATE electrons into a pi-system, e.g. benzene ring:

• The -NMe2 substituents withdraws electron density via the inductive effect, BUT, the inductive effect is overwhelmed by the resonance donating effect, -NMe2 is overall ELECTRON DONATING • Electron donating substituents DECREASE the Bronsted acidity of phenols

• The conjugate base pi-anion is resonance DESTABILIZED by the electron DONATING -NMe2 group. The resonance donating effect is stronger that any inductive stabilization by the electronegative nitrogen 3) Electron Withdrawing substituents on a pi-system: • Minor resonance contributors show how substituents WITHDRAW electrons from a pi-system, e.g. benzene ring:

sp2

HO O O OO

-H +~9.0

pKa

O O

MeMe

O

C

O

Me

-H +HO

Me

~10.0

π-anion weakly DESTABILIZED by electron repulsion

H HH

donatingsubstituent donating

substituent

weakeracid

NMe2 NMe2 NMe2 NMe2NMe2inductive effect

WITHDRAWS electrons from ringresonance effect DONATES electron

density into the ring - WINS!

BUT.....

donating

O O

NMe2NMe2

O

NMe2

O

NMe2

-H +HO

NMe2pKa ~9.0donating destabilizing

HO

weakeracid

pKa ~11.0

Alcohols : Page 5

• The -CHO substituent withdraws electron density via the inductive effect, AND via the resonance effect illustrated by the minor resonance contributors, -CHO is overall ELECTRON WITHDRAWING • Electron withdrawing substituents INCREASE the Bronsted acidity of phenols

• The conjugate base pi-anion is resonance STABILIZED by the electron WITHDRAWING -CHO group • The -CHO group is electron withdrawing on a pi-system, electron withdrawal occurs by both resonance and inductive effects 4) Position of Substituents on a pi-system: • The pKa of unsubstituted phenol is ca. 9.0 • Consider 4-nitrophenol:

• The conjugate base anion of 4-nitrophenol is directly stabilized by the inductive AND resonance effect of the nitro (-NO2) substituent, the formal negative charge is delocalized onto the carbon that the -nitro group is attached, the negative charge is further resonance stabilized by the nitro group, the base anion is stabilized and the acid is thus stronger. • The pKa of unsubstituted phenol is ca. 9.0 • Consider 3-nitrophenol:

• The conjugate base anion is NOT DIRECTLY stabilized by the nitro (-NO2) substituent because the formal negative change is not delocalized onto the carbon that the substituent is attached to, there is no resonance stabilization although there is still some inductive stabilization of the conjugate base anion, 3-nitrophenol is more acidic than phenol but less acidic than 4-nitrophenol.

C

inductive effectWITHDRAWS electrons from ring

resonance effect WITHDRAWS electron density from the ring

AND....

withdrawingOHδ+ C

OHC

OHC

OHC

OH

HO

CHO

O O

CHOCHO

O

C

O

CHO

-H +

OH

O

COHwithdrawing

strongeracid

pKa ~6.0

resonance withdrawal

pKa ~9.0

HO

OO OOHO

-H +

N

~7.0

strongeracid

OOwithdrawing

N N NNO OOO O OOO

O

NOO

pKa

OO OOHO-H +

~8.0

withdrawing

N O

O

N N N NO

O

O

O

O

O

O

O

pKaweakeracid

nitro

Alcohols : Page 6

Summary of Electron Withdrawing/Donating Substituents WHEN ATTACHED TO PI-BONDING SYSTEMS • donating and withdrawing ability when bonded to a pi-system measured relative to hydrogen

• distinguishing the D- and W- groups is easier than it looks (no memorization!!) • the donating groups have non-bonding electrons or electrons in pi-bonds that can be used to DONATE to the attached pi-system • just about every other substituent is withdrawing due to the presence of electronegative elements, W- groups do NOT have non-bonding electrons on the atoms that is connected to the pi-system 3 Oxidation/Reduction: Definition • the General Chemistry Definition of oxidation and reduction is the addition and subtraction of electrons • Counting electrons in ORGANIC structures is difficult because the electrons are mainly shared in covalent bonds • There are formal ways of counting electrons in bonds that require learning rules, so we will skip this • There is also a slightly less formal way that we WILL use, that focuses attention on the ATOMS that are involved in the bonds that are made and are broken for the organic structure in question • When a new bond is made (for example) from a carbon atom to a MORE ELECTRONEGATIVE atom (usually oxygen), that is OXIDATION (the more electronegative element "takes" an electron from the carbon) • When a new bond is made (for example) from carbon atom to a LESS ELECTRONEGATIVE atom (usually hydrogen), that is REDUCTION (the more electronegative element "gives" an electron to the carbon) Oxidation: Making NEW BONDS to more electronegative elements, usually oxygen Oxidation: Breaking OLD BONDS to (removing or replacing) less electronegative elements, usually hydrogen Reduction: Making NEW BONDS to less electronegative elements, usually hydrogen Reduction: Breaking OLD BONDS to (removing or replacing) more electronegative elements, usually oxygen Attention is focused on the atoms that are involved in the BONDS THAT ARE MADE and BROKEN, other atoms are not important Examples:

• for the starting organic structure, the new bonds are to both to the less electronegative hydrogen (2 x H added), thus, overall REDUCTION of the organic structure

• for the starting organic structure, the new bonds are to both to the more electronegative bromine (2 x Br added), thus, overall OXIDATION of the organic structure

NR2

OH

NH2

OR

NH C

O

R

O C

O

RCH

R

CH2

C

O

R

C

O

RO

C

O

R2N

CN

HO3S

O2N

R4N+

H

increasing electron donating abilityincreasing electron withdrawing ability

F3C

neither donating or withdrawing non-bonding electrons

hyperconjugation

-W -D

δ+δ-

these substituents STABILIZE a negative charge on a benzene ring

these substituents DESTABILIZE a negative charge on a benzene ring

H2reduction(alkene reduced) Pd/C

H Hnew bond to Hnew bond to H

Br Br Br Broxidation(alkene oxidized)

new bond to Brnew bond to Br

Alcohols : Page 7

• for the starting organic structure, bonds are broken to less electronegative H atoms (2 x H are removed), thus, overall OXIDATION of the organic structure

• for the starting organic structure, the new bonds are to the more electronegative bromine AND to the less electronegative H (1 x Br and 1 x H added), thus, overall NEITHER oxidation or reduction of the organic structure

• for the starting organic structure, the new bonds are to the more electronegative oxygen AND to the less electronegative H (1 x O and 1 x H added), thus, overall NEITHER oxidation or reduction of the organic structure: NOTE the fact that a H atom is also added to the structure via the -OH group is irrelevant, it is only the atoms that are involved in forming bonds to the organic structure that are important. 4 Preparation of Alcohols 4.1 Review of Reactions We Have Already Seen Recall:

• There are condition where catalytic hydrogenation (reduction) of C=C bonds can be performed without reducing a C=O bond (see the last example above), but this s a bit specialized for this course 4.2 Hydride Reduction of the Carbonyl Group Catalytic hydrogenation can actually get quite subtle, with different catalysts, one functional group can be reduced in the presence of another, e.g.:

CrO3

HOH O

H2SO4oxidation

(alcohol oxidized)

break bond to H

break bond to H

H Br H Brnew bond to Brnew bond to H

neitheroxidized or

reduced

H3O+ H Oneitheroxidized or

reduced

new bond to Onew bond to H

H

This H is added but it is NOT BONDED to any of the original

atoms of the alkene, it is therefore irrelevant to oxidation or reduction

1. Hg(OAc)2 / H2O

HO(±)

1. BH3 . THF

HO

H

H

Anti-MarkovnikovSyn-addition

Pd/CH2

O OH

MarkovnikovAnti-addition

2. -OH/H2O2

2. NaBH4

adds 2 H to BOTH C=C and C=O bonds

H

H H

Raney NickelH2

Oadds 2 H to BOTH C=C and C=O bondsO

H

H

Alcohols : Page 8

To prepare an alcohol, however, we need to do selective reduction of ONLY the C=O bond, HOW?

• A Lewis base (e.g. hydride anion) tends not to react with another Lewis base, and so does not react with the alkene, but DOES react with the carbonyl (C=O) group, which can act as a Lewis acid Some (new) reagents:

In Principle:

• Both hydride anion and the alkene are nucleophiles (Lewis bases), thus no reaction there • A hydride anion (H-) and a proton (H+) are equivalent to 2 hydrogen atoms (2 H.) In Practice: • NaH is too reactive and too strong a Bronsted base (less selective), NaH will usually deprotonate an aldehyde/ketone rather than add to the C=O bond, LiAlH4 or NaBH4 used instead

• The BH4– anion is less reactive than H– because the electrons are in a bond, therefore lower in energy • Overall, BH4– supplies H–, EtOH supplies H+. Together H– and H+ make H2.

O OH???

CC O+

LB LA "end" LB "end"HLB

X

(±)addition (of 2 x H)

reductionH

Na Sodium Hydride (NaH)

Li Lithium Aluminum Hydride (LiAlH4)

Sodium Borohydride (NaBH4)

very reactive

"masked hydrides"less reactive, more useful

HHAl HHH

NaHB HHH

decreasing electron pair energy

Al larger, electrons in weaker Al-H bond

B smaller, electrons in stronger B-H bond

electrons not in bond, very reactive

least reactive, most selective

OH X O

H

H

OHH

reduction accomplished

LA

LB H

(±)

H + H = 2 H

O NaBH4EtOH

OH

HB HHH O

HH O

Et

LB

LA

LA/BA

LB/BB

(±) reductionH

addition

Na

Alcohols : Page 9

Example: (stereochemistry ignored)

Why does the NaBH4 reduce the ketone and not the ester, and the LiAlH4 reduce both? • The less reactive NaBH4 reduces aldehydes and ketones but not esters • The more reactive LiAlH4 also reduces esters (and acids) • The (H3Al-H)– bond is weaker than the (H3B–H)– bond, and so is more reactive • Esters and acids are less reactive than aldehydes and ketones due to better resonance stabilization

• minor resonance structures Emphasize the Lewis acid character and PARTIAL POSITIVE CHARGE on the carbonyl carbon in a ketone, a LB will react FASTER with an aldehyde and ketone • minor resonance structures DEemphasize the Lewis acid character and MUCH SMALLER PARTIAL POSITIVE CHARGE on the carbonyl carbon in an ester, a LB will react SLOWER with an ester Alternatively:

We can consider that C=O to be a simple pi-system (the same way that a benzene ring is a larger pi-system), and the ester has a strong donating group attached to the carbon of the C=O, which decreases its reactivity towards a Lewis base/nucleophile, aldehydes/ketones have only weak donating groups attached to the carbon of the C=O group, they are more reactive. • To reduce the less reactive esters, the more reactive LiAlH4 is required More on LiAlH4: • The AlH4– ion will react violently with water and alcohols, so the proton has to be added in a second ACID WORKUP step, hence the notation: 1. LiAlH4 ... 2. H3O+

• in this second acid workup step, just enough dilute acid is used to "complete" the reaction • the protonation is essentially instantaneous, i.e. this is NOT the same as acid catalyzed addition of water to an alkene (for example), which requires higher concentrations of acid, a lot of time and usually some heat

NaBH4

EtOH

1. LiAlH4

2. H3O+

OO

O

OHO

O

OH

HO

more reactive less reactive

O

O

OOO

OLBLB

larger partial charge smaller partial charge

R RCO

ROCO

LBLBweak

donating STRONGdonating

simple π-system

less reactivemore reactive

OR

O

HAl HHH

ORO

HH

O

HAl HHH

HO

HH

OH

HLB

LALA

LB

OH

HH

LB/BB

LA/BA

addition elimination

ORleaving group

favored by entropy

+

OR

+

OH

HH HOR

+

BY-PRODUCT

Alcohols : Page 10

• this is our first example of an addition/elimination mechanism, we will see this again later..... • note REMOVAL of the -OR group of the ester in the elimination step • elimination occurs because the -OR is a reasonable leaving group (but not great!), AND elimination is favored by entropy, AND, occurs because the species that is eliminating already has a negative charge on oxygen • NOTE that the -OR leaving group will get protonated in the acid step to form usually a simple alcohol (HOR) that is often not included in the products since it is not part of the main organic structure, it is a BY-PRODUCT • A negatively charged oxygen CAN BE a leaving group, but only if the reaction STARTS with high chemical potential reactants that provide the energy to allow an oxygen anion to leave, for example a strong Lewis base that is an anion, e.g. the aluminum hydride anion (AlH4-) Examples: (stereochemistry ignored)

• NaBH4 reacts ONLY with the aldehyde, the ester is less reactive and the alkene is also a Lewis base • LiAlH4 reacts with the aldehyde AND the carboxylic acid (the acid reaction proceeds via addition/elimination) • The H3O+ in the LiAlH4 reaction does NOT react with the alkene, because in this context, H3O+ means an "acid workup step", which in turn means "add just enough dilute aqueous acid to protonate the negatively charge oxygen atoms". When acid catalyzes water addition to the C=C bond of an alkene the acid concentrations are high, the reaction time is long and the temperature has to be high, the context defines the meaning of H3O+. 5 Oxidation Reactions of Alcohols Here are a couple of oxidation reactions

• These reactions can be CONTROLLED by appropriate choice of reagent/conditions New Cr(VI) Reagent #1:

• the reagent is sodium dichromate and sulfuric acid dissolved in water, this generates chromic acid "in situ" Example with a SECONDARY Alcohol • oxidation to form a KETONE

H

OOMeO

1. LiAlH42. H3O+

NaBH4

EtOH

OHOMeO

OHHO(+ MeOH)

this came from the -OMe part of the ester, often

not included in the reaction productsBY-PRODUCT

R CH2 OH R CO

HR C

O

OHoxidation further oxidationalcohol

add 1 oxygen atomremove 2 H atoms

aldehyde carboxylic acid

Na2Cr2O7 + H2SO4 HO CrO

OOH + Na+ –HSO4

chromic acid

H2O

sodium dichromatesolvent

HO CrO

OOH2 O Cr

O

OOH

HCR

R'H

chromate ester

+ H+

RC OHR'H

HO CrO

OOH

– H+

O CrO

OOHC

RR'

HH

HO

don't have to know!!!

ketone productNa2Cr2O7/H2SO4/H2O

CR

R'O

oxidation

Alcohols : Page 11

• mechanisms involving oxidation/reduction of metals do not follow the usual patterns of Lewis acid/base Bronsted acid/base reactions, AND they tend to be specialized and specific, for this reason you do not have to know this mechanism Example with a PRIMARY Alcohol: • oxidation to form a carboxylic acid THE overall process is as follows:

• note that formation of the aldehyde from the alcohol is oxidation (removes 2 H atoms) • note that formation of the acid from the hydrate is oxidation (removes 2 H atoms) • note that formation of the hydrate from the aldehyde is NEITHER oxidation or reduction Details: FIRST, formation of the aldehyde via the same mechanism as above

Details: SECOND, conversion of the aldehyde into the hydrate in the presence of water and an acid catalyst, you DO NEED TO KNOW THIS MECHANISM. Remember, the reaction conditions involve sulfuric acid in water, the next step is simply acid catalyzed addition of water to the aldehyde

Final Detail: THIRD, conversion of the hydrate (a geminal di-alcohol) into a carboxylic acid, via the same mechanism as before, the chromic acid removes 2 hydrogen atoms from the hydrate, which forms the carboxylic acid, the mechanism by which chromic acid does this is the same as formation of the aldehyde from the primary alcohol, you don't need to know this and it is not shown again here

aldehyde

HC HROH H2O

CRO

HCROH

HOH

Na2Cr2O7H2SO4 H3O+

H2O

Na2Cr2O7H2SO4 CR

O

OH

hydrate acidoxidation oxidation

HO CrO

OOH2 O Cr

O

OOH

HCH

RH

+ H+

HO CrO

OOH

– H+

HH O

O CrO

OOHC

HR

H

HC HROH H2O

CRO

HNa2Cr2O7/H2SO4

same as before, don't need to know!!

OH

H

HH

HO

H2O

Na2Cr2O7/H2SO4CRO

H

CROH

OHH

CR

O

H

H

H

H O

CROH

HO

HH

hydrate

need toKNOW THISmechanism!!

Alcohols : Page 12

• the hydrate gets oxidized to a carboxylic acid because it is now a (di) alcohol Example with a TERTIARY Alcohol:

• tertiary alcohols cannot be oxidized, the necessary hydrogen atom is missing

• for the same reason, ketones cannot be oxidized, the necessary hydrogen atom is missing New Cr(VI) Reagent #2:

• CH2Cl2 is the SOLVENT, therefore there is NO WATER here, so any aldehydes that are formed cannot make a hydrate, so further oxidation to a carboxylic acid will not occur PCC with a PRIMARY Alcohol:

OHC HROH H2O

CRO

OHNa2Cr2O7/H2SO4

-H2

R

C OHR1R2

O CrO

OOHC

RR1

R2

no hydrogens to eliminate, 3° alcohols can not be oxidized!

Na2Cr2O7/H2SO4/H2O

HO CrO

OOH

X

X

X

R HCOH

H1°

-H2

RCO

H RCOH

HOH

-H2

RCOH

O

R HCOH

R2°

-H2

RCO

R

no hydrogens to eliminate, can't oxidize a ketone

R RCOH

R3°

no hydrogens to eliminate, can't oxidize

CrO3 + HCl + Cl CrO

OO

pyridinium chlorochromate (PCC)N

pyridine

NHCH2Cl2

HC OHRH

HO CrO

OCl O Cr

O

OOH

HCH

RH

O CrO

OOHC

HRH

N

OCH

R

no water, no hydrate formationaldehyde stable product!

–H+

PCC

CH2Cl2

don't need to know!!!

Alcohols : Page 13

Summary of Oxidation Reactions

Examples:

6 Substitution Reactions of Alcohols (SN1 and SN2) : Water, and other species as Good Leaving Groups 6.1 Recall: SN1 and SN2 Reactions of Alkyl Halides Substitution reactions of alkyl halides (SN1 and SN2)

• Substitution via SN2 is a very important substitution mechanism

R CH2 OH

PCC/CH2Cl2Na2Cr2O7

H2SO4/H2O

1° alcohol

R CH OH 2° alcoholR'

R C OH 3° alcoholR'

R''

Carboxylic Acid Aldehyde

Ketone Ketone

no reaction! no reaction!

R C H aldehydeO

Carboxylic Acid no reaction!via the hydrate

H

O OH

PCC/CH2Cl2

HO

O C OHO

H

O C HO

Na2Cr2O7H2SO4/H2O acid

aldehyde

acid

no reaction here

HO OH

HO O

HO OPCC/CH2Cl2

Na2Cr2O7H2SO4/H2O

Br Na OH OH SN2

1° halide, fast SN2

Br heatCH3OH

OCH3

3° halide, weak Nu and weak base, polar protic and heat, must be

+E1 would also occursubstitution

SN1

makes an alcohol!

Alcohols : Page 14

• We will try to avoid SN1 if we can since it involves a carbocation intermediate that can rearrange and/or eliminate, although there will be some cases where it will not be possible to avoid SN1. 6.2 Substitution of Alcohols (SN1 and SN2) : Formation of Alkyl Halides Consider the following substitution reaction, is it possible?

• It doesn't work! in fact, this reaction goes in the OPPOSITE direction • –OH is a good nucleophile and will substitute for a halide, which is a good leaving group (think about standard SN2 reactions) Reactions with Haloacids: HCl, HBr, HI, etc.

• this works because WATER IS AN EXCELLENT LEAVING GROUP • This is a REALLY IMPORTANT principle, if good leaving groups have low energy electrons in the anions, then leaving groups that are neutral, stable small molecules that don’t even have a charge are even better! • NOTE the tertiary (3°) alcohol and water being an excellent leaving group, substitution by SN1, SN1 is unavoidable here because the carbon is tertiary Compare

• The alcohol is now on a primary carbon, therefore only SN2 is possible (it is not possible to form a carbocation on a primary carbon) • This reaction goes even though the bromide anion is a poor nucleophile because water is such a good leaving group • These previous two mechanisms are IMPORTANT, we will see mechanistic step such as these several times However, the SN1 mechanism still prevails for 2° and 3° alcohols, thus, there is still the usual problem with the cation intermediates such as elimination and rearrangements • SN2 mechanism for 1° alcohols, but still, the halide anions are poor nucleophiles, something better is needed Compare: Reaction with PBr3 (phosphorous tribromide) • An alternate way of converting an –OH group into a good leaving group

XR CH2 OH + OH

X

R CH2 XSN2

LB

LA not a very good leaving group

OH

H Br

OH2

water, good leaving group

BrBr

LB/BB

LA/BALB

LA

+ OH2

SN1 SN13°3°

OH

H Br

water, good leaving groupLB/BB

LA/BA

+ OH2SN21°

1°OH21°

BrBr

R C OH P BrR'

H

Br

Br

R C OR'

H HPBr

Br

Br

R C HR'

Br

SN2HO P

Br

Br+

good leaving groupdon't have to know....

+

1° + 2° alcohol bromide

Alcohols : Page 15

• In the first step the leaving group is bromide, in the second step it is the neutral and stable HOPBr2 • An SN2 mechanism works well with both 2° and 1° alcohols due to the good leaving group, this is a good reaction • The mechanism is not quite as straightforward as standard SN1 and SN2, and it a bit specific for PBr3, therefore you don’t need to know the details of this mechanism • The only problem is 3° halides, which still don't work well for steric reasons (recall, no SN2 at 3° centers!) (Another issue is the stereochemistry of the reaction, if it is SN2 there should be inversion at a chiral center, but the stereochemical outcome of this reaction is less reliable than other SN2 reactions, therefore, we will ignore it!) Compare: Reaction with SOCl2 (thionyl chloride): • Another way of converting an –OH group into a good leaving group

• Here we have addition/elimination, followed by deprotonation, followed by an SN2 reaction to make an alkyl chloride • In the SN2 reaction the leaving groups are Cl– (a good leaving group), and SO2 (a very good leaving group) • An SN2 mechanism works well with both 2° and 1° alcohols due to the good leaving group, this is a good reaction • The mechanism is not quite as straightforward as standard SN1 and SN2, and it a bit specific for SOCl2, therefore you don’t need to know the details of this mechanism • The only problem is 3° halides, which still don't work well for steric reasons (recall, no SN2 at 3° centers!) (As with PBr3, another issue is the stereochemistry of the reaction, if it is SN2 there should be inversion at a chiral center, but the stereochemical outcome of this reaction is less reliable than other SN2 reactions, therefore, we will ignore it again!)

Reaction Summary

• The mechanisms for PBr3 and SOCl2 are a bit complicated, you don’t need to know these, but you DO NEED TO KNOW the mechanisms of the the HCl and HBr reactions, even when they are not the preferred reagents 6.3 Substitution of Alcohols : Formation of Tosylate Esters (better leaving group) Why are Tosylates Useful? They are yet ANOTHER even MORE versatile way of converting –OH from a poor leaving group into a good leaving group • This is what we have just learned

R C OHR'

HCl Cl

SO

R C OR'

H HSCl

Cl

O R C OR'

HH

SCl

O

Cl

R C OR'

HS

Cl

OCl

R C HR'

ClSO2 + Cl +

elimination

SN2

addition+

good leaving groups

1° + 2° alcohol

chloride

don't need to know...

Preferred Reagents BromideChloride

1° alcohol

2° alcohol

1°, 2°, 3° alcohol

SOCl2 PBr3

SOCl2 PBr3

HCl HBr

"SN2""SN2"

SN1 & SN2

use THESEfor 1° & 2° alcohols

R OH2

Nu

R Nu + H2O Good leaving group, worksif Nu can tolerate acid.

Alcohols : Page 16

• This is OK, except that is rarely useful beyond HCl, HBr, H2SO4, etc. because you need to protonate the -OH first, THEN add the nucleophile (which is a Lewis base and often a reasonable Bronsted base) and hope that the nucleophile doesn't just get protonated, hmmmm, something better is really needed…. • This reaction looks a lot more complicated (but it isn’t) and it fixes this problem

• The tosylate anion is highly resonance stabilized, it is low energy non-bonding electrons, which is why it is such a good leaving group

Where do tosylate (esters) come from?

• you don't need to know the mechanism, but you DO need to know that you need pyridine to remove the proton some clarification on notation and structure…..

Examples: Useful reactions:

SO

OCH3O R Nu

tosylate anion, excellent leaving group, works under lots of conditions, this is why tosylates are useful!

R

Nu

+ SO

OCH3O

tosylate (ester)

SO

OCH3O S

O

OCH3O S

O

OCH3O

tosylate anion is resonance stabilized

R O

tosyl chloride

SO

OCH3Cl

+

SO

OCH3OR

H

alcohol

SO

OCH3O

H

R

tosylate (ester)R–OH Ts–Cl R–OTs

pyridine

N=

N

pyridineorganic

base

NH+

don't need to know

"Tosyl" group = -TsSO

OCH3

(para-toluenesulfonate)

"Tosylate" group = OTsSO

OCH3O

HO TsO

Na+ –Br

Na+ –CNNH3

Na+–OCH3

Br

CNH3N

H3CO

weak nucleophile

weak nucleophile

pyridineTsCl

Alcohols : Page 17

7 Elimination Reactions of Alcohols (E1 and E2) 7.1 Recall: E1 and E2 Reactions of Alkyl Halides Elimination reactions of alkyl halides (E1 and E2)

• Elimination via E2 is a very important elimination mechanism • We will try to avoid E1 if we can since it involves a carbocation intermediate that can rearrange and/or substitute, although there will be some cases where it will not be possible to avoid E1. 7.2 Elimination of Alcohols (E1 and E2) : Formation of Alkenes • By now we KNOW that –OH can be a good leaving group if it is protonated, water is a good leaving group • This works for substitution (SN1 and SN2), and we should not be surprised if we can also do eliminations • Again, the reactions start by protonating the -OH of an alcohol to form a good leaving group, and then standard E1 and/or E2 mechanisms after that Example Problem: Give the mechanism for the following elimination reaction.

• Sulfuric acid protonates the -OH in a Bronsted acid/base reaction to convert the -OH into a good leaving group • Water is such a good leaving group that the elimination is almost always E1 with 3° and 2° alcohols • Water is such a good leaving group that E1 is occurs even at a secondary carbon to make a secondary cation • carbocation intermediates mean rearrangements! (that hasn't changed, of course)

Br

BrNa+ –OMe

Br

t-BuO– +K

Sayetzeff

Sayetzeff

2°

3°

bulky base, avoids SN2

3°

bulky base

E2 elimination

E2 elimination

nonbulky base

HK+ –O-t-Bu

t-BuO

Anti-Sayetzeff - HofmannE2 elimination

Br

MeOH3°

weak base

Sayetzeff AND SUBSTITUIONE1 elimination

heat

OHconc. H2SO4

heat

OH2 H

Sayetzeff major

LA/BALB/BB

LA/BA LB/BB

standard E1 elimination (including rearrangement)

no water!!

E1

H O S O HO

O

O S O HO

O

FIRSTconvert poor leaving group

to good leaving group

+ H2O very goodleaving group

Alcohols : Page 18

• the conjugate base anion of the sulfuric acid, the bisulfate anion, is the most likely base to deprotonate the carbocation intermediate, thus regenerating the acid catalyst • The alkene formed will be the Sayetzeff (Zaitsev), there are no stereochemical constraints in the E1 mechanism and the most stable alkene will form. Why does the alcohol make an alkene + water when previously we learned that water + alkene gives an alcohol? • THIS is what we learned previously

• The addition reaction "goes" because the weaker pi-bond is converted into a stronger sigma-bond • The reagents/conditions have a LARGE quantity of water and a SMALL quantity of sulfuric acid • THIS is what we now learned

• The reagents have ZERO water and a HIGH concentration of sulfuric acid (opposite of previous reaction) • The elimination reaction "goes" because the water is highly solvated in the concentrated sulfuric acid • note a special kind of SOLVENT EFFECT here! In an aqueous medium, acid catalyzes water ADDITION to the alkene to make an alcohol. In conc. sulfuric acid medium, the acid helps to REMOVE water from an alcohol to make an alkene (the sulfuric acid DEHYDRATES the alcohol) • Alternate reagents and conditions are H2SO4/P2O5, and others…. Example: Primary (1°) Alcohols: E2 elimination (with rearrangement…)

• With a primary alcohol the mechanism must be E2, formation of a primary carbocation CAN'T occur

C CH2SO4 (cat.)/heat

addition of H2OH2O H

C COH

WATER

SMALL quantity of acid

HC C

OHC C

conc. H2SO4

heat+

NO water!

elimination

HIGH acid concentration (100%)

STRONG IMFH-bonding

OH

HH O S O H

O

O

H O S O HO

O

OH

OH2

HLB/BB

LB/BB

LA/BA

LA/BAHH

LB/BB

LB/BB

conc. H2SO4

heat

E2 elimination

LA/BA

rearranged stable alkene formed

same as from E1!

protonation/deprotonation

LA/BAH O S O H

O

O

O S O HO

O H O S O HO

O

O S O HO

O

E2 elimination for 1° alcohols, NO E1!

FIRSTconvert poor leaving group

to good leaving group

Alcohols : Page 19

• BUT, even though the elimination does not involve a rearrangement, the final alkene product is usually the same one that would have been formed via an E1 reaction due to protonation followed by deprotonation (isomerization) of the primary alkene into a final more stable product Look AGAIN at the second part of the mechanism, the rearrangement

• This effectively converts a less stable less substituted alkene into a more stable more substituted alkene, this is why this ISOMERIZATION reaction "goes" • To solve the mechanism problem, ADD SOME hydrogen atoms back to the line-angle structure, the H atoms tell you exactly where you need to protonate and deprotonate • in the presence of acid, PROTONATION will occur first, followed by deprotonation • a less substituted/less stable alkene is converted into a more substituted/more stable alkene • this is a REARRANGEMENT, the acid is only the catalyst (no atoms are overall added or subtracted) • In a strong acid, especially with heat, protonation and deprotonation can OFTEN occur, and if this can result in formation of a more stable alkene, then the more stable alkene will form, and you should always include this step when doing acid catalyzed dehydrations of alcohols The final product is the SAME MOST SUBSTITUTED ALKENE, whether the mechanism is E1 followed by cation rearrangement (2° and 3° alcohols) or E2 followed by protonation/deprotonation (1° alcohols) Examples:

7.3 A Different E1 Elimination: The Pinacol Rearrangement Mechanism The Pinacol Rearrangement: (another kind of alcohol dehydration)

• Let's treat this as a mechanism problem, how to solve it and what basic principles can we use to guide us? • Look carefully at the reagents/conditions (in this case, acid in water)

HH

rearranged stable alkene formed

H O S O HO

O

OSOHO

OH

H

H HH

H

H H HH

H

OH conc. H2SO4

heat

OHconc. H2SO4

heat

1° ALCOHOLE2 followed by rearrangement

2° ALCOHOLE1 carbocation rearrangement

OH2 H H

PARTIAL mechanism

CCOH

CH3

CH3CH3

H3COH

CCCH3

CH3CH3

H3COH2SO4

H2O+ H2O

removal of water

Alcohols : Page 20

• Look for differences in start and end structures • Need to remove –H and –OH and do an alkyl shift • Acid catalyzed, therefore protonate first, no anionic intermediates in presence of acid The Mechanism:

• protonate to make a good leaving group (H2O) • usual carbocation rearrangement to make a more stable (resonance stabilized) cation intermediate • Deprotonation at the end regenerates the acid catalyst Example Mechanism Problem: A "Hidden" Pinacol rearrangement

CCOH

CH3

CH3CH3

H3COH

OH

HH

CCOH2

CH3

CH3CH3

H3COH

H2Oeliminated

CCCH3

CH3

CH3

H3COH

CC

CH3

CH3H3C

OH

CH3alkylshift

CCCH3

CH3H3CO

CH3

H

CCCH3

CH3CH3

H3CO

O HH

OH

HO

HOO

OHCl / H2O

OH

HH

OH2

HO HO H

H

O H

Alcohols : Page 21

6 Alcohols : Summary of Reactions Do NOT start studying by trying to memorize the reactions here! Work as many problems as you can, with this list of reactions in front of you if necessary, so that you can get through as many problems as you can without getting stuck on eth reagents/conditions, and so that you can learn and practice solving reaction problems. Use this list AFTER you have worked all of the problems, and just before an exam. By then you will have learned a lot of the reagents/conditions just by using them and you will only have to memorize what you haven't learned yet. Then do the following: • Cover the entire page of reagents/conditions with a long vertical strip of paper, see if you can write down the reagents/conditions for each reaction, check to see which you get correct, if COMPLETELY correct, circle Y, if incorrect or even slightly incorrect, circle N. In this way you keep track of what you know and what you don't know. • Keep coming back to this list and so the same thing only for those reactions you circled N, until all are circled Y. • Knowing the reagents/conditions on this page is INSUFFICIENT to do well on an exam since you will ALSO need to recognize how to use and solve reaction problems in different contexts, this page ONLY helps you to learn the reagents/conditions that you have not YET learned by working problems.

CH2OH CO

HPCC

CH3(CH2)8CH2OH CH3(CH2)8CO2HNa2Cr2O7

OHSOCl2 Cl

OOH PBr3

OBr

OH ClHCl

HBrOH Br

CH2Cl2

H2SO4 / H2O

O H2

Pd/C

OH

O

O OH

2. H3O+

(±)

O OHNaBH4 (±)

1. Excess LiAlH4Y / N

Y / N

Y / N

Y / N

Y / N

Y / N

Y / N

Y / N

Y / N

LiAlH4 also does this reactionEtOH

NaBH4 does NOT do this reaction

specifying CH2Cl2 solvent is optional

specifying the H2O solvent is optional

SOCl2 does NOT work for 3° alcohols

PBr3 does NOT work for 3° alcohols

and many other SN2 reactions of tosylates

OH CH3SO

OCl CH3S

O

OO

CH3SO

OO CNNa+ –CN

(TsCl)

(R-OTs)

(R-OTs)

Y / N

Y / N

OH OY / N

PCC can also be used to do this reaction

Na2Cr2O7H2SO4 / H2O

(±)