Relative Strengths of Organic Acids and Bases Background knowledge for today: Proton based definition of an acid is? Recall what does it mean for an acid.

Relative Strengths of Organic Acids and Bases

Background knowledge for today:

Proton based definition of an acid is?

Recall what does it mean for an acid or base to be “strong”?

Recall that all acids and bases are in a constant equilibrium and the “one-sided-ness of that equilibrium in

measured in what numeric terms?1

New Vocabulary:

Electrophilic/Electrophiles

Nucleophilic/Nucleophiles(Req. free pair of e’s or pi bond)

Which would be a B-L acid? Base?

2

HA + H2O ↔ A- + H3Oor

HA(aq) ↔ A- (aq)+ H+ (aq)

3

HA + H2O ↔ A- + H3Oor

HA(aq) ↔ A- (aq)+ H+ (aq)

High Ka = Strong acidLow Ka = Weak acid

Low pKa (Strong Acid)High pKa (Weak Acid – Tends to reform) 4

Ka: Acid dissociation Constantp Ka Is the “negative logaritmic

version” pKa = -log10Ka

THE ACIDITY OF ORGANIC ACIDS

5

Why are these acids acidic?In each case, the same bond gets broken - the bond between the hydrogen and oxygen in an -OH group. Writing the rest of the

molecule as "X":

6

“Organic Example”

So . . . if the same bond is being broken in each case, why do these three compounds have such widely different acid

strengths?

7

The factors to consider

Two of the factors which influence the ionization of an acid are:

1. the strength of the bond being broken,

2. the stability of the ions being formed.Factor #1: In these cases, you seem to be breaking the same oxygen-hydrogen bond each time, and so you might expect the strengths to

be similar.

However: The most important factor is the nature of the ions formed. You always get a hydroxonium ion - so that's constant - but the

nature of the anion (the negative ion) varies markedly from case to case.

8

Ethanoic acid / Acetic Acid

9

The Start: The “expected”: The Actual:

This leads to a delocalized pi system over the whole of the -COO- group

Conclusion:

The more you can spread charge around, the more stable an ion becomes.

In this case, if you delocalize the negative charge over several atoms. This makes your

anion unusually stable and so you are less likely to re-form the ethanoic acid.

Meaning: It favors the formation of ions: it dissociates more: it produces more H ions: it

becomes a strong acid10

Phenol

11

Phenol

12

But here delocalization spreads this charge over the whole over only the oxygen. (Again because oxygen is more electronegative than carbon) This means: the charge is less distributed: less stable: reverts back to it’s original state more readily: produces less H+: And thus is a weaker acid

In our last example: delocalization spreads charge over the whole of the COO- group. (Because oxygen is more electronegative than carbon) You can think of most of the charge being shared between the two oxygen.

Now Try:

Ethanol

CH3CH2OH

13

Ethanol

14

This has nothing at all going for it. There is no way of delocalizing the negative charge. That intense negative charge will be highly attractive towards hydrogen ions, and so the ethanol will instantly re-form.Since ethanol is very poor at losing hydrogen ions, it is hardly acidic at all.

Deeper into the rabbit hole:

15

Variations in acid strengths

Summary Thus Far: We now know the strength is largely a factor of the ion stability via the electron delocalization/distribution.

Now: Certain attachments can either strengthen or weaken an acid.

Two Categories:•Attachments that withdraw e’s from the O/COO by pulling towards themselves (NO2, Halogens)

•Less charge gathered around O/COO•Makes the anion/conjegate base more stable•Favors formation of ions•Stronger acid results

•Attachments that induce e’s towards the O/COO by pushing away from themselves. (Alkyl groups)

•More charge gathered around O/COO•Makes the anion/conjegate base less stable•Favors formation of original reactant•Weaker acid results

Consider:

16

Variations in acid strengths

In fact, the carboxylic acids have widely different acidities. One obvious difference is between methanoic acid, HCOOH, and the other simple

carboxylic acids: pKa

HCOOH 3.75 CH3COOH 4.76 CH3CH2COOH 4.87 CH3CH2CH2COOH 4.82

Consider:

17

Variations in acid strengths

As the next table shows, the more chlorines you can attach the better:

carboxylic acids: pKa

CH3COOH 4.76 CH2ClCOOH 2.86 CHCl2COOH 1.29 CCl3COOH 0.65

Consider:

18

Variations in acid strengths

Fluorine is even better (Why?), also note the increase over Cl is not as much as you might expect:

carboxylic acids: pKa

CH2FCOOH 2.66 CH2ClCOOH 2.86 CH2BrCOOH 2.90 CH2ICOOH 3.17

Consider:

19

Variations in acid strengths

Finally, notice that the effect falls off quite quickly as the attached halogen gets further away from the -COO- end. Here is what happens if you move a chlorine atom along the chain in butanoic acid.

carboxylic acids: pKa

CH3CH2CH2COOH 4.82 CH3CH2CHClCOOH 2.84 CH3CHClCH2COOH 4.06 CH2ClCH2CH2COOH 4.52

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Figure 19.8How common substituents

affect the reactivity of abenzene ring towards

electrophiles and the acidity ofsubstituted benzoic acids

Other Push/Pull groups and how they can affect reactivity/acidity…Don’t forget to also consider a Benzene ring.

21

Substituted Benzoic Acids

Recall that substituents on a benzene ring either donate or withdraw electron density, depending on the balance of their inductive and resonance effects. These same effects also determine the acidity of substituted benzoic acids.

[1] Electron-donor groups destabilize a conjugate base, making an acid less acidic—The conjugate base is destabilized because electron density is being donated to a negatively charged carboxylate anion.

22

[2] Electron-withdrawing groups stabilize a conjugate base, making an acid more acidic. The conjugate base is stabilized because electron density is removed from the negatively charged carboxylate anion.

23

• Note that although resonance stabilization of the conjugate base is important in determining acidity, the absolute number of resonance structures alone is not what is important!

Figure 19.7Summary: The relationship

between acidity and conjugatebase stability for acetic acid,

phenol, and ethanol

Summary

Would these factors also effect organic bases?

All of the compounds we are concerned with are derived from

ammonia and so we'll start by looking at the reason for its basic properties.

24

• Just like with the acids, we’ll use the definition of a base as "a substance which combines with hydrogen ions (protons)".

• Strength in a base will be how easily they take hydrogen ions from water molecules.

Ammonia as a weak base

25

As with acids…Two of the factors which influence the strength of a base are:

•the ease with which the lone pair picks up a hydrogen ion,

•the stability of the ions being formed.

26

•The nitrogen is more negative in methylamine than in ammonia, and so it picks up a hydrogen ion more readily.•The ion formed from methylamine is more stable than the one formed from ammonia, and so is less likely to shed the hydrogen ion again.

27

28

Carboxylic Acids and the Acidity of the O—H Bond

• Carboxylic acids are compounds containing a carboxy group (COOH).

• The structure of carboxylic acids is often abbreviated as RCOOH or RCO2H, but keep in mind that the central carbon atom of the functional group is doubly bonded to one oxygen atom and singly bonded to another.

Structure and Bonding

29

• The C—O single bond of a carboxylic acid is shorter than the C—O bond of an alcohol.

• This can be explained by looking at the hybridization of the respective carbon atoms.

Structure and Bonding

• Because oxygen is more electronegative than either carbon or hydrogen, the C—O and O—H bonds are polar.

30

• Carboxylic acids exhibit dipole-dipole interactions because they have polar C—O and O—H bonds.

• They also exhibit intermolecular hydrogen bonding.• Carboxylic acids often exist as dimers held together by

two intermolecular hydrogen bonds.

Physical Properties

Figure 19.3Two molecules of acetic acid(CH3COOH) held together by

two hydrogen bonds

31

Reactions of Carboxylic Acids

The most important reactive feature of a carboxylic acid is its polar O—H bond, which is readily cleaved with base.

32

• The nonbonded electron pairs on oxygen create electron-rich sites that can be protonated by strong acids (H—A).

• Protonation occurs at the carbonyl oxygen because the resulting conjugate acid is resonance stabilized (Possibility [1]).

• The product of protonation at the OH group (Possibility [2]) cannot be resonance stabilized.

33

• The polar C—O bonds make the carboxy carbon electrophilic. Thus, carboxylic acids react with nucleophiles.

• Nucleophilic attack occurs at an sp2 hybridized carbon atom, so it results in the cleavage of the bond as well.

34

Carboxylic Acids—Strong Organic BrØnsted-Lowry Acids

• Carboxylic acids are strong organic acids, and as such, readily react with BrØnsted-Lowry bases to form carboxylate anions.

35

• An acid can be deprotonated by a base that has a conjugate acid with a higher pKa.

• Because the pKa values of many carboxylic acids are ~5, bases that have conjugate acids with pKa values higher than 5 are strong enough to deprotonate them.

36

37

• Carboxylic acids are relatively strong acids because deprotonation forms a resonance-stabilized conjugate base—a carboxylate anion.

• The acetate anion has two C—O bonds of equal length (1.27 Å) and intermediate between the length of a C—O single bond (1.36 Å) and C=O (1.21 Å).

38

• Resonance stabilization accounts for why carboxylic acids are more acidic than other compounds with O—H bonds—namely alcohols and phenols.

• To understand the relative acidity of ethanol, phenol and acetic acid, we must compare the stability of their conjugate bases and use the following rule:

- Anything that stabilizes a conjugate base A:¯ makes the starting acid H—A more acidic.

39

• Ethoxide, the conjugate base of ethanol, bears a negative charge on the O atom, but there are no additional factors to further stabilize the anion. Because ethoxide is less stable than acetate, ethanol is a weaker acid than acetic acid.

• Phenoxide, the conjugate base of phenol, is more stable than ethoxide, but less stable than acetate because acetate has two electronegative O atoms upon which to delocalize the negative charge, whereas phenoxide has only one.

40

• Note that although resonance stabilization of the conjugate base is important in determining acidity, the absolute number of resonance structures alone is not what is important!

Figure 19.7Summary: The relationship

between acidity and conjugatebase stability for acetic acid,

phenol, and ethanol

41

The Inductive Effect in Aliphatic Carboxylic Acids

42

43

Substituted Benzoic Acids

Recall that substituents on a benzene ring either donate or withdraw electron density, depending on the balance of their inductive and resonance effects. These same effects also determine the acidity of substituted benzoic acids.

[1] Electron-donor groups destabilize a conjugate base, making an acid less acidic—The conjugate base is destabilized because electron density is being donated to a negatively charged carboxylate anion.

44

[2] Electron-withdrawing groups stabilize a conjugate base, making an acid more acidic. The conjugate base is stabilized because electron density is removed from the negatively charged carboxylate anion.

45

Figure 19.8How common substituents

affect the reactivity of abenzene ring towards

electrophiles and the acidity ofsubstituted benzoic acids

1) Give the IUPAC name for each compound.

a) O

OH

3,3’-dimethylhexanoic acid

b)

O

OH

Cl 4-chloropetanoic acid

c)O

OH2,4-diethylhexanoic acid

O

OH

d)

4-isopropyl-6,8-dimethylnonanic acid

2) Draw the structure corresponding to the IUPAC name.

a) 2-bromobutanoic acid O

OH

Br

b) 2,3-dimethylpentanoic acid O

OH

c) 3,3’,4-trimethylheptanoic acid

O

OH

d) 2-secbutyl-4,4’-diethylnonanoic acidO

OH

e) 3,4-diethylcyclohexanecarboxylic acid

O

OH

f) 1-isopropylcyclobutanecarboxylic acid

O

OH

5) Give the IUPAC name for each metal salt.

a) C6H5CO2-+Li

c) (CH3)2CHCO2-+K

b) HCO2-+Na

Lithium benzoate

Sodium methanoate

Potassium 2-methylpropanoate

d) (CH3CH2)2CHCH2CHBrCH2CH2CO2-+Na

Sodium 4-bromo-6-ethyloctanoate

7) Explain how you would distinguish between these compounds using IR spectroscopy.

H3CH2CH2CH2C

O

OH H3CH2CH2CH2C

O

OCH3 O

OH

2 peaks

C=O 1710

-OH 2500-3500

1 peak

C=O 1700

1 peak

-OH 3200-3600

8) Propose a compound with the formula, C4H8O2 and the following date:

0.95 (triplet, 3H)

1.65 (multiplet, 2H)

2.30 (triplet, 2H)

11.8 (singlet, 1H)

O

OH

11)Identify the starting material in each reaction.

a) O

OHNa2Cr2O7

H2SO4, H2OOH

b) O3

H2O

2

O

OH

H3C CH3

c)

O2N

O

OHKMnO4O2N

d)CrO3

H2SO4, H2O

O

OH

O

OH

OH

19.13) Which of the following bases are strong enough to deprotonate CH3COOH?

a) F-

b) (CH3)3CO-

c) CH3-

d) NH2-

e) Cl-

pka of CH3COOH is 4.8.

b) has a pka of 18

c) has a pka of 50

d) has a pka of 38

pka of conjugate acid must be greater than that of the carboxylic acid being deprotonated.

19.14) Rank the labeled protons in order of increasing acidity.

OHc

O

H OHbHa

Ha<Hb<Hc

The more stable the conjugate base the more acidic the proton.

19.15) Match each pka value with each carboxylic acid.(3.2, 4.9 and 0.2)

a) CH3CH2COOH

b) CF3COOH

c) ICH2COOH

4.9

0.2

3.2

Electron withdrawing groups make acids more acidic.

19.16) Why is formic acid more acidic than acetic acid?

HOH

O

OH

O

The methyl group is electron donating and stabilizes the acid while destabilizing the conjugate base thus making it less acidic.

19.17) Rank the compounds within each group in order of decreasing acidity.

a) CH3COOH, HSCH2COOH, HOCH2COOH

3 2 1

b) ICH2COOH, I2CHCOOH, ICH2CH2COOh

2 1 3

19.18) Rank each group of compounds in order of decreasing acidity.

a)CO2H CO2H

Cl

CO2H

2 1 3

b)CO2H CO2H CO2H

H3CO

O

2 1 3

19.19) Is the following compound more or less acidic than phenol?

HO

OH

R

The more electron donating groups present, the less acidic a compound is. This compound has an additional hydroxy and alkyl group, both electron donating. So it is less acidic.

19.22) Comparing CF3SO3H and CH3SO3H, which has the weaker conjugate base? Which conjugate base is the better leaving group? Which of these acids has the higher pka?

CF3SO3H is the weaker conjugate base.

CF3SO3H is the better leaving group because it is the weaker conjugate base.

CH3SO3H, with the electron donating methyl group, has the higher pka and is thus a weaker acid.

20.1) What type of orbitals make up the indicated bonds? And in what orbitals do the lone pairs on the oxygen lie? O

a b

c

a. sp3-sp2

b. sp2-sp2, p-p

c. sp3-sp2

The lone pairs lie in sp2 hybridized orbitals.

20.2) Which compounds undergo nucleophilic addition and which substitution?

a) O

H3CH2CH2C

O

Cl

b)

c)

H3C

O

OCH3

d)

O

H

addition substitution

substitution addition

20.3) Which compound in each pair is more reactive toward nucleuphilic attack?

a)

H3CH2CH2C

O

H H3CH2CH2C

O

H3C(H3C)HC

O

CH2CH3H3CH2C

O

b)

O

OCH3H3CH2C

O

Cl

c)

O

NHCH3

O

OCH3

d)

H3CH2CH2C

O

H

H3CH2C

O

H3CH2C

O

Cl

O

OCH3

20.4) What alcohol is formed when each compound is treated with NaBH4 in MeOH?

a)

H3CH2CH2C

O

H

NaBH4

MeOH H3CH2CH2C

OH

H

H

b) O

NaBH4

MeOH

OH

c)

O

NaBH4

MeOH

OH

20.5) What aldehyde or ketone is needed to synthesize each alcohol by metal hydride reduction?

a)OH O

b) OH O

c)OH O

20.6) Why can’t 1-methylcyclohexanol be prepared from a carbonyl by reduction?

OH

Tertiary alcohols can not be made by reduction of a carbonyl because there are no hydrogens on the carbon with the -OH.

20.7) Draw the products of the following reactions?

a) OH

O

NaBH4

MeOH

b) OH

c) O

H2 (1 equiv.)

Pd-C

O

d) O

H2 (excess)

Pd-C

OH

O

LiAl4

H2O

e) O

NaBH4 (excess)

MeOH

OH

f) O

NaBD4

MeOH

OHD

20.8) Draw the products when the following compounds are treated with NaBH4 in MeOH.

a) O

NaBH4

MeOH

HO H

+

OHH

b) O NaBH4

MeOHOH

c)(H3C)3C O

NaBH4

MeOH(H3C)3C OH

+

(H3C)3C OH

20.9) What reagent is needed to carry out the reaction below?

O

Cl

HO

Cl

H

Two reagents are needed to carry out this reaction. First, the (S)-CBS reagent to produce the R-enantiomer. Followed by H2O to protonate the alcohol.

20.10) Draw a stepwise mechanism for the following reaction. O

Cl

LiAlH4

H2O OH

O

Cl

H3Al H

O

Cl

H

+ AlH3

O

H

+ Cl

H3Al H + AlH3

O

H

O

H

H

OH

H

H

H OH

+ OH

20.11) Draw an acid chloride and an ester that can be used to produce each product.

a) CH2OH

b)OH

O

Cl

O

OCH3

O

Cl

O

OCH3

H3CO

OHc) H3CO

O

Cl

H3CO

O

OCH3

20.12) Draw the products of LiAlH4 reduction of each compound.

a)

OH

O

OH

b)

NH2

O

NH2

c)O

N(CH3)2 N(CH3)2

d)

NH

O

NH

20.13) What amide will form each of the following amines when treated with LiAlH4?

a) NH2

NH2

O

b)N N

O

c)

NH N

H

O

20.14) Predict the products of these compounds when treated with the following reagents.

a) O O

OCH3

LiAlH4

H2O

NaBH4

MeOH

OH

OH

OH O

OCH3

b)

OH

OO

H3CO

LiAlH4

H2O

NaBH4

MeOH

HO OH

No reaction

c)LiAlH4

H2O

NaBH4

MeOH

H3CO OH3CO OH

H3CO OH

20.15) Predict the products in the following reactions.

a) OH Ag2O

NH4OH

Na2Cr2O7

H2SO4, H2O OH

O

No Reaction

b)Ag2O

NH4OH

Na2Cr2O7

H2SO4, H2O

OH O OH O

OH

O O

OH

20.16) Predict the products of the compound below when reacted with each reagent.

HO

OH

O

a)NaBH4

MeOH

HO

OH

OH

b) LiAlH4

H2O

HO

OH

OH

HO

OH

O

c)PCC

O

O

O

d) Ag2O

NH4OH

HO

OH O

OH

e) CrO3

H2SO4, H2O

HO

O O

OH

O

20.17) Write out the rea tions needed to convert CH3CH2Br to each of the following reagents.

a) H3CH2C Li

H3CH2C Br + 2 Li H3CH2C Li + Li Br

b) H3CH2C MgBr

H3CH2C Br + Mg H3CH2C MgBr

c) H3CH2C CuLi

H3CH2C Br + 2 Li H3CH2C Li + Li Br

H3CH2C Li + CuI

CH2CH3

LiCu

+ Li I2

H3CH2C

20.18) 1-octyne reacts readily with NaH, forming a gas that bubbles out of the reaction mixture. 1-octyne also reacts with CH3MgBr and a different gas is produced. Write out balanced equations for each reaction.

HC CCH2CH2CH2CH2CH2CH3 + NaH

NaC CCH2CH2CH2CH2CH2CH3 + H2

HC CCH2CH2CH2CH2CH2CH3 + CH3MgBr

BrMgC CCH2CH2CH2CH2CH2CH3 + CH4

20.19) Draw the product of the following reactions.

a) Li

+ H2O + LiOH

b)MgBr + H2O + HOMgBr

c)MgBr + H2O

+ HOMgBr

d)

LiC CCH2CH3 + H2O HC CCH2CH3+ LiOH

20.20) Draw the product formed when each compound is treated with C6H5MgBr followed by H2O.

a)

H

O

H

H

OH

H

b)

H3CH2C

O

CH2CH3

CH2CH3

OH

CH2CH3

c)

H3CH2C

O

H

CH2CH3

OH

H

d)

O

OH

20.21)Draw the products of each reaction.

a)

O

H3CH2CH2C Li

H2O

HO CH2CH2CH3

+ LiOH

b)

H

O

H

Li

H2OH

HO

H

+ LiOH

c)O

C6H5Li

H2O

OH

+ LiOH

d)

CNaH2C O

H2OC

H2C

OH

+ NaOH

20.22) Draw the products (including stereochemistry) of the following reactions.

a)

H3C

O

H

H3CH2C MgBr

H2O

H OH H OH

+

b)OH3CH2C Li

H2OOH

CH2CH3

OH

CH2CH3+

20.23) What Grignard and carbonyl are needed to prepare each alcohol?

a)OH O

H

+ H3C MgBr

b) OHO

H H

+

MgBr

c)

OHO

+ H3CH2C MgBr

orO

|+

MgBr

d)OH

O

+ H3C Br

or

MgBr+

O

20.24) Tertiary alcohols with three different R groups on the carbon attached to the OH can be prepared in three different ways using the Grignard reagent. Show them.

a)H3C

OH

CH2CH3

CH2CH2CH3

O

CH2CH3

H3CH2CH2C

H3C

O

CH2CH2CH3

H3C

O

CH2CH3

+ H3C MgBr

H3CH2C MgBr+

+ H3CH2CH2C MgBr

b) OH

O

+ H3C MgBr

O+

MgBr

O

+

MgBr

c) OH

O

O

O

+ MgBr

+

MgBr

+ H3CH2C MgBr

20.25) Show the steps for the following reaction.

HO O HO

OH

CH2CH2CH2CH3

HO OTBDMS-Cl

N

HN

TBDMSO O

BrMg CH2CH2CH2CH3

H2O

TBDMSO

OH

CH2CH2CH2CH3FN(CH2CH2CH2CH3)4

H2O

HO

OH

CH2CH2CH2CH3