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Acid/Base stuff Beauchamp 1

y:\files\classes\Organic Chemistry Tool Chest\Acid,Base,Tautomers\314 acid-base list, answers newer.DOC

Important preliminary questions before looking at specific acids and bases. 1. What is an acid? What is a base? What are the Bronsted and Lewis definitions of acids and bases?

Arrhenius definitions (before 1900) – water is emphasized acids: substances which increase the hydronium ion concentration in water, [H3O+] bases: substances which increase the hydroxide ion concentration in water, [HO--] Bronsted definitions (1924) – proton transfer is emphasized acid: a proton donor (no reference to the solvent) base: a proton acceptor (no reference to the solvent) Lewis definitions (1924) – electron pair transfer is emphasized acids: substances which accept a pair of electrons bases: substances which donate a pair of electrons

2. Write an equation using water as the base with generic acid, H-A. Use curved arrows to show how the reaction occurs between an acid and a base (water)? Always push electrons with your arrows! Use full-headed arrows for two electron movement (in acid/base chemistry), and half-headed arrows for one electron movement (in free radical reactions). Remember how you used arrows in resonance. Practice this skill at every opportunity. This is the organic way of looking at reactions. It can be qualitative (which side does the equilibrium favor?) or quantitative (what is the value of Keq?). Water is the reference base in reactions with various acids. Arrow pushing shows how the reactions work.

A H O A OHH

H H

HKa

pKa

This is the freshman chemistry way of looking at acid/base chemistry. Symbols are mainly used for quantitative numerical calculations. We won't write H+ by itself. We will always attach the proton to some pair of electrons.

C2H3O2H H

C2H3O2Ka =

]

C2H3O2H[

[

]H+

C2H3O2

[ ]][

3. Write an equilibrium expression for the reaction of acid ionization in water.

Keq =

(A )(H3O )(HA)(H2O)

4. How does Ka differ from Keq? What is the Ka and what does it tell us about an acid? What magnitude is Ka for a

strong acid? What magnitude is Ka for a weak acid?

Ka = Keq (H2O) =

bigger number (>1) = strong acid (up to 1020)

smaller number (<1) = weak acid (low as 10-50)

= #

We almost always use Ka instead of Keq in acidity tables. They only differ by the concentration of water, which is essentially constant ( 55.6 M). An acid with a large Ka value (Ka > 1) is called a strong acid (up to 1020). An acid with a small Ka value (Ka < 1) is called a weak acid (as low as 10-50). Acid strengths span a remarkable range of 1070!

(A )(H3O ) (HA)



5. What is the pKa and what does it mean? (strong acids = ?)(weak acids = ?) What’s an order of magnitude? pKa is another way to look at Ka It's a little confusing because it is the negative log of the Ka, (the negative of the tens exponent). Every power of 10 is an order of magnitude (103 is an order of magnitude larger than 102 and 10-4 is two orders of magnitude smaller than 10-2). (pKa = negative number for strong acids and pKa = positive number for weak acids)

pKa = - log (Ka)

A very useful way to think of pKa is as (G)x(1.4) kcal/mole. Very approximately: pKa = (G)x(1.4) kcal/mole

G = -2.3RT (log Ka) = 2.3RT (-log Ka) = (1.4)(pKa) kcal/mole pKa

Ka = 10-pKa

R = 2 cal/mole-KR = 8.3 joule/mole-K

6. Can you think of a better base that could be used in water (but similar looking)? What are the limits of basicity in

water? Using a less acidic solvent can allow for more basic environments. Some solvents are essentially nonacidic and can tolerate very strong bases. This is often necessary in organic chemistry.

Hydroxide is a stronger (better) base than water, although water is still most likely the solvent. On cannot go higher in pH than the pKa of water, where 50% of the water would be ionized to hydroxide and no longer liquid.

A H O A OHH H

pKa = 16Ka = 10-16

pKa = variesKa = varies

7. Write an equation with water as the acid with generic base, B:. We won’t consider Kb or pKb.

OOH HH B BHWater is the reference acid in reactions with various bases. Arrow pushing shows how the reactions work.

Kb

pKb

pKa = 16

Ka = 10-16

pKa = varies

Ka = varies

8. Can you think of a better acid that you could use in water (but similar looking)? What are the limits of acidity in

water? Using a less basic solvent can allow for more acidic environments. This is sometimes necessary in organic chemistry.

Hydronium ion is a stronger (better) acid than water, although water is still most likely the solvent. On cannot go lower in pH than the pKa of H3O+, where 50% of the water would be ionized to H3O+ and no longer liquid.

OHB BHOH

H

H

pKa = -2

Ka = 10+2

pKa = varies

Ka = varies

9. How does one draw an energy diagram (PE vs POR) for strong acid ionization equation? How does one draw an

energy diagram (PE vs POR) for weak acid ionization?

B AH AHB

stronger acid & base

AH

APE

POR

stronger acid

HB A+ -

weaker acid & base

TS

The equilibrium shifts towards the weaker conjugate acid and base (away from the stronger acid and base).



The equilibrium shifts towards the weaker conjugate acid and base (away from the stronger acid and base).

YH Y

stronger acid & baseYH

Y

B HB

PE

POR

weaker acid

HB Y+ -

weaker acid & base

TS

10. What makes an acid stronger? What makes an acid weaker? In our course, we present two reasons: a.

inductive effects (related to electronegativity) and b. resonance effects through 2p orbitals. These are mainly used to explain stabilities of the ‘less stable’ conjugate bases. Do stronger acids have more stable or less stable conjugate bases (more stable)? What about weak acids (less stable bases)? At our peril, we ignore salvation effects in our presentation and focus mainly on the stability of the conjugate bases.

AH

APE

stronger acid

HB A+ -

TS As the conjugate base, A: gets more stable, the acid, HA, gets stronger.

As the conjugate base, Y: gets less stable, the acid, HY, gets weaker.

YH

YPE

POR

weaker acid

HB Y+ -

TS

POR

11. Consider the base, instead of the acid. What makes a base stronger (less stable)? What makes a base weaker

(more stable)? Turn acidity around to evaluate basicity. Electron donating ability is related to the reasons provided for relative acidities in question 10? We often use available pKa tables of acidities to determine relative basicities of the conjugate bases from their inverse relationships with one another. (The stronger acid pairs with the stronger base and the weaker acid pairs with the weaker base.)

OH

H O H H

stronger acid stronger base weaker acidweaker base

pKa = 16

Ka = 10-16

pKa = 37

Ka = 10-37

Keq =

G = (pKa1 - pKa2) x 1.4 = (16 - 37) x 1.4 = (-21) x 1.4 = -29 kcal/mole

Ka1 10-16

Ka2 10-37= = 10+21

equilibrium lies completely to the right

H3CCH2

HH H H

stronger acidstronger baseweaker acid

weaker base

pKa = 50

Ka = 10-50pKa = 37

Ka = 10-37

Keq =

G = (pKa1 - pKa2) x 1.4 = (50 - 37) x 1.4 = (13) x 1.4 = +18 kcal/mole

Ka1 10-50

Ka2 10-37= = 10-13

equilibrium lies completely to the left

H3CCH2



Having a negative charge (or lone pair of electrons) on a more electronegative atom makes it more stable (F is more stable than HO is more stable than H2N is more stable than H3C ). More delocalized electrons are more stable than less delocalized electrons. This delocalization could be due to the size of the atoms: I > Br > Cl > F or due to resonance.

F is more stable than H3C

FH

FPE weaker

acid

HB F+ -

TS

POR HH3C

PEmuch

weaker acid

HB CH3

+ -

TS

POR

H3C

F

Iis more stable than

because F has a higher Zeff (+7 > +4) making it more electronegative than C. Both are similar size second row elements.

because iodide is a larger anion and its electrons are more delocalizedthan fluoride's electrons, while both have the same Zeff of +7.

Delocalization of electrons in organic chemistry usually refers to "resonance".

O

OO

is more stable than

O

O

pKa = +3G = +4

G = +70pKa = +50

delocalizedcharge

(resonance)localizedcharge

12. Many examples follow, providing opportunities to use organic logic of points 10 and 11.

Generic acid/base equilibrium equation, the organic way, with curved arrows. Keq and G can be estimated for a proton transfer reaction involving two generic acids, as shown below.

Ka

pKa

A1 H A2 A1 A2H

Keq = (:A2 )(HA1 )

(HA2)(:A1 )

acid = proton donorbase = proton acceptor

acid = electron pair acceptorbase = electron pair donor

=Ka1

Ka2

G = (pKa) x 1.4 = (pKa1 - pKa2) x 1.4

Bronsted definitionsLewis definitions

An estimate of the equilibrium constant, Keq, can be calculated by dividing the Ka of the acid on the left (reactant) by the Ka of the acid on the right (product). An estimate of G for the reaction can be calculated by subtracting the pKa of the acid on the right from the pKa of the acid on the left, and multiplying by 1.4 kcal/mole.

G = (1.4)(pKa) kcal/mole pKa

OH

H

H A OH

H

AH

Acid ionization reactions use full headed arrows to show two electron movement. Water is the reference base in usual Ka and pKa tables.

You should be able to match pKa values with their acids in each group below and explain the differences. You should be able to draw an arrow-pushing mechanism with general base, B:- for any of the acids, H-A. Include resonance structures whenever appropriate. If there was a reaction shown between any two conjugate acids and bases, you should be able to qualitatively and quantitatively indicate which side of the equilibrium is favored, and what an approximate G is for the reaction.



C H

HpKa = 50

N H

H

pKa = 35O H

H

pKa = 16

F H pKa = 3HH

H1

What is Keq and G for the following reactions?

C H

H

H

H

N H

H

HO

H

HOstronger acid

stronger baseweaker acid

weaker base

pKa = 16

Ka = 10-16

pKa = 37

Ka = 10-37

Keq =

G = (pKa1 - pKa2) x 1.4 = (16 - 37) x 1.4 = (-21) x 1.4 = -29 kcal/mole

Ka1 10-16

Ka2 10-37= = 10+21

equilibrium lies completely to the right

C

H

H

H

N H

H

H

weaker acid

pKa = 50Ka = 10-50

stronger acid

pKa = 16Ka = 10-16

stronger baseweaker base

equilibrium lies completely to the left

Keq =Ka1 10-50

Ka2 10-37= = 10-13

G = (pKa1 - pKa2) x 1.4 = (50 - 37) x 1.4 = (13) x 1.4 = +18 kcal/mole

N H

H

N H

H

H

H3C

H2C

CH

H3C

H2C

NH

H3C

H2C

OH

HHH

pKa = 16

2

pKa = 37

pKa = 50

pKa = NA

H3C

H2C

F

O H O H O HpKa = 11I Br Cl O HF

not stable

3pKa = 8.7

pKa = 7.5

pKa = NA

OH

OHCl

OHCl

Cl

OHCl

ClCl pKa's = 16

4pKa's = 14.3

pKa's = 12.2

pKa's = 12.8

C H

HpKa 32*

C Cl

H

pKa = 25C H

H

pKa 40*

pKa = 50HH

H5 H Cl

Cl C H

Cl

Cl

Cl

* = my estimate

H3CO

H

H3C

H2C

OH

H3CCH

OH

H3CC

OH pKa's = 17

6 pKa's = 19

pKa's = 15.8

pKa's = 15.5

CH3 CH3H3C



CH

H

H

7. What is the expected order of stability of these reactive organic intermediates? (most stable = 1)

CR

H

HC

R

R

HC

R

R

R

a. free radicals

CH

H

HC

R

H

HC

R

R

HC

R

R

R

b. carbocations

CH

H

HC

R

H

HC

R

R

HC

R

R

R

c. carbanions

relative energies(most stable = 0)

12 kcal/mole 5 2 0


70 kcal/mole3515 0


? kcal/mole

Cl H Br H

pKa = -9I H

pKa = -10F H pKa = 3

pKa = -78

S H

H pKa = 7

Se H

H

pKa = 4Te H

H

pKa = 3O H

H

pKa = 16

9

OHO

O

O

pKa = 10

HH

10

pKa = 4

pKa = 16

H3CC

O

CH

H3CC

O

NH

H3CC

O

OH

HHH

pKa = 20

11

pKa = 15

pKa = 5

H2C

H

C

O

OH

H2C

CH3

C

O

OH

H2C

I

C

O

OH

H2C

Br

C

O

OH

H2C

Cl

C

O

OH

H2C

F

C

O

OH

Ref. pKa = 4.7other pKa's = 4.9, 3.2, 2.90, 2.85, 2.59

12

CH2

C

O

OH pKa = 1.3

13

pKa = 2.8

pKa = 5H3C

C

O

OH Cl

CHC

O

OHCl

CC

O

OHCl

Cl ClCl

pKa = 0.7



pKa = 4.5

14

pKa = 2.8

pKa = 4.0

O

OH

pKa = 4.8O

OH

O

OH

O

OH

Cl

Cl

Cl

C

H

H

H

N

H

HO

H

C

H

H

H

N

H

H

O

H

pKa's = 10, 18, 28, 37, 41, 50

15

H

C

O

O H

H3C

C

O

O H

F

C

O

O H

C

C

O

O H

O2N

C

O

O H

N

Ref. pKa = 4.2

16

pKa = 3.5pKa = 3.6pKa = 3.9pKa = 4.3

N

H

H

C

N

N H

H

N

H

HCNpKa = 23.6

pKa = 24.8pKa = 28

17

H H H H H H

F3C CF3

H H

O O OO O19

pKa = 9

pKa = 5

pKa = 20

pKa = 50

OH

O2N

OH

N

O

O

pKa's = 8.4, 7.1, 10.0, 10.2

OH

OH

20

OH

OH

Cl

pKa's = 85, 9.0, 9.4, 10.0

OH

Cl

OH

21

Cl



H2C

C

C

O

OH

O OH

H2C

C

C

O

OH

O OH3C

C

O

OH

pKa = 2.85 pKa = 5.70

pKa = 4.7

22

H

H

H

H

H

HpKa = 50 pKa = 42

pKa = 15*23

* also aromatic

C C

H

H

H

H

H

H

C C

H

H

H

H

CH C H

pKa = 50

pKa = 44 pKa = 25

24

Electrons in 2s orbitals are held tighter than electrons in 2p orbitals. Orbitals which have a greater %s character are more electronegative than lesser %s character. Therefore the electronegativity of hybrid orbitals is: sp (50% s) > sp2 (33% s) > sp3 (25% s). Greater electronegativity is better able to stabilize the negative charge in the conjugate base, so sp C-H bonds are the most acidic of these hydrocarbons, then sp2 C-H and lastly sp3 C-H (lowest acidity of any acid in our course).

C N

H

H

H

H

H

H

C N

H

H

H

H

CH N H

pKa = 9

pKa = 5pKa = -10

25

Bascity

Use full headed arrows to show two electron movement. Water is the reference acid in usual Kb tables. We won't use pKb values. Instead, we will compare pKa values and judge bases to be stronger when their acids are weaker and bases to be weaker when their acids are stronger. In the examples that follow pair up each base with the pKa value of its conjugate acid.

OH

H

O

H

B B H

judge the strength of the base from the weakness of the acid

use this acid's pKa value to judge the electron donating power of B: to the reference acid, H2O.



Here are a few qualitative examples. Where is the most basic site in each molecule below? Can an order of basicity be explained for some or for all of the bases (approximate pKa’s of some of the conjugate acids are provided)?

H3CC

O

CH3 H3CC

O

NH

H3CC

O

OCH3

H

N

N

H

NC

O

NH

H

NC

N

NH

H

R

H

HH

H

N

O

H3CC

O

OH

pKa -7 pKa -7 pKa -6 pKa -0.5

pKa 0.2 pKa 5 (guess) pKa 7 pKa 13.6

What is the order of basicity among the following molecules of each group (1 = most basic)? Explain your reasoning. Match the given pKa values with the conjugate acids of the indicated bases. Write arrow-pushing mechanisms with general acid, H-A to illustrate the reactions. Include resonance structures whenever appropriate. Where is the most basic site in each molecule? Explain your reasoning using arguments of inductive effects (electronegativity), resonance effects (electron delocalization) or both. For any reaction between two conjugate acids and bases, you should be able to qualitatively and quantitatively indicate which side of the equilibrium is favored, and what is an approximate G is for the reaction.

H3CC

O

NH

H3CC

O

OH

H

H3CC

N

NH

H

R

pKa's of the bases' conjugate acids = -7, 0, 7

1

C N H

H

H

H

H

C N

H

H H

CH N pKa's of the bases' conjugate acids = 9, 5, -10

2

N

N

H

NC

N

NH

H

H

H

N

R

pKa's of the bases' conjugate acids = 5, 7, 13

3

OO

O

O


4



pKa's of the bases' conjugate acids = 50, 20, 9, 5

H HH

F3C CF3

H

OO O O O5

H3CC

O

CH2H3CC

O

NH

H3CC

O

O


6

C C H

H

H

H

H

C C

H

H H

CH C pKa's of the bases' conjugate acids = 50, 44, 25

7


HHH

8

Acid/Base arrow pushing worksheet

1. These proton transfer reactions are the first step in multistep mechanisms to be studied later in the course. Supply the necessary curved arrows, lone pairs of electrons and/or formal charge to show how the first step each reaction proceeds. Except for the first reaction, they are all simple proton transfer reactions generating a carbanion. Generally, there is some stabilizing feature that allows a carbanion to form via acid/base chemistry, such as inductive and/or resonance effects. In working the problem below, show any important resonance structures or identify the inductive effect that makes the reaction possible. a.

N H

CH2CH2CH2CH3

Li

NLi

CH2CH2CH2CH3H

Formation of lithium diisopropyl amide (LDA) using butyl lithium.

n-butyl lithium is commercially available

Ka = 10-37

Ka = 10-50

b.

N

R

R Li

LDA

CC

H

HH

O

Rketone

-78 oC resonance

Ka = 10-37

R2NH

Ka = 10-20



c.

N

R

R Li

LDA

CC

H

HH

O

O

ester

-78 oCresonanceR

Ka = 10-37

R2NH

Ka = 10-25

d.

CC

H

HH

O

Ntertiary amide

R

R

N

R

R Li

LDA

-78 oC

resonance

Ka = 10-37

R2NH

Ka = 10-30

e.

CC N

nitrile

N

R

R Li

LDA

-78 oC resonance

H

H

HKa = 10-30

Ka = 10-37

R2NH

f.

CC

C

O

CH3

sodium hydride

-78 oC

resonance

O

H3C

H HNa

Hresonance

1,3-dicarbonyl

Ka = 10-9 Ka = 10-37

H-H

g.

N

R

R Na

sodium amide

-33 oCCH C R

terminal alkyneterminal acetylide ammoniaKa = 10-25

Ka = 10-35

R2NH

h.

n-butyl lithiumdithiane

-78 oCH2C

H2C

CH2

CH3

Li

S

SH

HCH2CH2CH2CH3H

Ka = 10-50

Ka 10-35

possibleresonancewith sulfurd orbitals?

i.



S C

n-butyl lithiumsulfur ylidPh = phenyl

-78 oC

possibleresonancewith sulfurd orbitals?

H

H

HH2C

H2C

CH2

CH3

LiPh

Ph CH2CH2CH2CH3H

Ka = 10-50Ka 10-35

j.

P C

n-butyl lithium

phosphorous ylid (Ph = phenyl)

-78 oC

possibleresonance

with phosphorousd orbitals?

H

H

H

H2C

H2C

CH2

CH3

LiPh

Ph

Ph

Br

Ka 10-35

CH2CH2CH2CH3H

Ka = 10-50

2. Lone pair donors to very strong acid

All of the following functional groups react with protic acid as the first step of a reaction studied in organic chemistry. Often subsequent chemistry occurs after that initial step and you will study most of those reactions later in the course. Show how they react in the first step by including all lone pairs, curved arrows to show electron movement and formal charge.

OH H

Acid/Base arrow pushing worksheet

lone pair donors lone pair acceptor

OH S

O

O

OHKeq = ?

pKa's = -10, -2equilibrium

NH H

H

OH H

H

Keq = ?

pKa's = -2, 9ammonia

equilibrium

NR H

H1o amine

Keq = ?

pKa's = -2, 10equilibrium

OH H

H

NR H

R2o amine

Keq = ?

pKa's = -2, 10equilibrium

OH H

H

equilibriumNR R

R3o amine

Keq = ?

pKa's = -2, 10

OH H

H



OR H additionalchemistrypossiblealcohol

Keq = ?

pKa's = -2, -3

OH H

H

additionalchemistrypossible

OR RKeq = ?

pKa's = -2, -3ether

OH H

H


O

epoxide (ether)

Keq = ?

pKa's = -2, -3

OH H

H


Keq = ?

pKa's = -2, -7

C

O

R Haldehyde

OH H

H

resonance(2)


Keq = ?

pKa's = -2, -7

C

O

R R'ketone

OH H

H

resonance(2)


CR N

nitrile

Keq = ?

pKa's = -10, -10

OH SO3H

sulfuric acid

resonance(2)


carboxylic acid

C

O

R OH

Keq = ?

pKa's = -2, -6

OH H

H

resonance(3)


C

O

R OR'

Keq = ?

pKa's = -2, -6ester

OH H

H

resonance(3)


amide

C

O

R NH2

Keq = ?

pKa's = -2, 0

OH H

H

resonance(3)



additionalchemistrypossibleC

N

R NH2

H

Keq = ?

pKa's = -2, 12

OH H

H

resonance(3)



Carbon-carbon pi bonds as weak electron pair donors to very strong acid


Keq = ?

pKa's = -10, -10

OH SO3H

sulfuric acidC

CH2

H

Ralkene


CCH2

NR2

Renamine

Keq = ?

pKa's = -2, 5

OH H

H

resonance(2)


CCH2

O

Renol ether

Keq = ?

pKa's = -2, -7

R

OH H

H

resonance(2)


C C HR

alkyne

Keq = ?

pKa's = -10, -10

OH SO3H

sulfuric acid


resonance(3)

Keq = ?

pKa's = -10, -10

E+ = electrophile(Lewis acid = electron pair acceptor)

C

CC

C

CC

E

aromatic

H

H

H

H

H

H

At this point we are mainly interested in understanding acid/base proton transfers, curved arrow pushing, formal charge, recognizing resonance structures and using the logic arguments of inductive effects and resonance effects to explain relative stabilities of acids and bases. If you can do these things, you are well on your way to understanding organic chemistry and biochemistry. The Tautomer Game – an arrow-pushing training reaction in acid and in base, forward and reverse Tautomers are isomers that differ by the location of a proton and a pi bond. To be official tautomers, a heteroatom or atoms (different than carbon, often oxygen or nitrogen or both) is part of the system. In the simplest case, there are at two isomers in equilibrium with one another (there may be many, many more tautomers possible in more complex systems). The tautomers are interchangeable by 1. proton transfer, 2. resonance intermediates and 3. proton transfer. The “keto” isomer, has a heteroatom in a pi bond and in the “enol” tautomer has two carbons forming a pi bond. This simple pattern can occur in an infinite number of systems, from very simple to very complex. A possible approach to figuring out what to do in keto/enol tautomer problems is shown below. Slightly more complex tautomer relationships are shown in the second example. They are mainly more complicated mainly because there are more than two tautomers and interchanges may require one or more simple tautomer interconversions.



C

H H

CC

O

C

H H

CC

O

H

HH

H

H

HH

acid orbase

Use H3O /H2O or HO /H2O to accomplish the given transformations. Every transformation will always follow a 3 step sequence.

H1. proton transfer (in acid = proton on) (in base = proton off)2. resonance delocalized intermediates 3. proton transfer (in acid = proton off) (in base = proton on)

keto tautomer enol tautomer

in acid

best acid best base

in base

H3O H2O

H2O HO

Tautomers in acid (simplest examples)

a. “keto” “enol”

C

H H

CC

O

H

HH

H

O

H

H H

OH H

C

H H

CC

O

H

HH

H

resonance

H

C

H H

CC

O

H

HH

H

H OH H

C

H H

CC

O

H

HH

H

enol tautomer

keto tautomer in acid = proton on in acid = proton off

b. “enol” “keto”

C

H H

CC

O

H

HH

H

O

H

H H

OH H C

H H

CC

O

H

HH

H

resonance

H

C

H H

CC

O

H

HH

H

H OH H

C

H H

CC

O

H

HH

H

enol tautomer keto tautomerin acid = proton on in acid = proton off

Tautomers in base (simplest examples)

a. “keto” “enol”

C

H H

CC

O

H

HH

HOH H C

H H

CC

O

HH

H

resonance

OH H

C

H H

CC

O

H

HH

H

enol tautomerketo tautomerin base = proton off in base = proton on

O H

C

H H

CC

O

H

HH

enolate

b. “enol” “keto”

C

H H

CC

O

H

HH

HOH H C

H H

CC

O

HH

H

resonance

OH H

C

H H

CC

O

H

HH

H

enol tautomer keto tautomer

O H

C

H H

CC

O

H

HH

enolatein base = proton off in base = proton on



A slightly more complicated keto/enol tautomer problem – keto/enol with an additional pi bond

O

H

H

O

H

H

O

H

H

O

H

H

O

H

H

H

H

H

H

H

H

HH

H

HHH

H

H

H

H

H H

H

H

H

H

H H

H

H H

1

3

24

5

base

also possible

1

3

H H

H

base base

base

base

The tautomer interconversions shown above are possible in one step in base because of shared resonance intermediates. The total number of tautomer changes required to change any tautomer into any other tautomer are shown below for base (on the left) and acid (on the right). The number of tautomer changes in parentheses was worked out in my head, not on paper, so there may be some wrong estimates.

Number of tautomer changes to transform one tautomer into another in base.

Number of tautomer changes to transform one tautomer into another in acid.

1 2 3 4 5

(1x) (1x) (1x) (2x)

2 1 3 4 5

(1x) (1x) (2x) (2x)

3 1 2 4 5

(1x) (1x) (2x) (1x)

4 1 2 3 5

(1x) (2x) (2x) (3x)

5 1 2 3 4

(2x) (2x) (1x) (3x)

1 2 3 4 5

(1x) (2x) (1x) (1x)

2 1 3 4 5

(1x) (1x) (1x) (1x)

3 1 2 4 5

(2x) (1x) (2x) (1x)

4 1 2 3 5

(1x) (1x) (2x) (2x)

5 1 2 3 4

(2x) (1x) (1x) (2x)

1. Circling the protons that change is always a good idea, because you know these are going to have to move.

However, these may not show every tautomer change because sometimes a necessary change is reversed in a later step. The circled protons have to be moved, either taken off (with the best base available) or put on (with the best acid available) and there is always resaonance delocalization in the intermediate. Best acid in H3O

+/H2O is H3O+, best base in H3O

+/H2O is H2O Best base in H2O/HO-- is HO-- and the best acid in H2O/HO-- is H2O

2. Always work from a "keto" (CH-C=O or CH-C=N-) part or "enol" (C=C-OH or C=C-NH-) part of the molecule. Do not use isolated pi bonds (C=C) to initiate change in the structure. With an allowed change an



isolated pi bond may become conjugated with a “keto” or “enol” part of another tautomer. Any keto or enol part will be the better base or the better acid, as is indicated because it will form a resonance stabilized intermediate with the oxygen (or nitrogen) assisting in the resonance structures.

Not possible in in base in one keto/enol cycle.

Changing tautomer 2 into tautomer 4 is possible in base by first converting tautomer 2 into tautomer 1 and then changing tautomer 1 into tautomer 4. On the other hand tautomer 2 could be converted to tautomer 4 in a single tautomer change in acid.

OH

H

H

H

H

H

H

H2

OH

H

H

H

H

H

H

H

4

OH

H

H

H

H

H

H

H2

O

H

H

H

H

H

H

H

H

1

OH

H

H

H

H

H

H

H

4This is an isolated C=C bond. Don't begin here in acid or base.

This is an isolated C=C bond. Don't begin here in acid or base.

This is an isolated C=C bond. Don't begin here in acid or base.

3. If in acid, use the strongest acid (H3O

+ in our examples) to put on a "gained" proton first and take off a "lost" proton second, with a weak base (usually the solvent = H2O in our examples).

4. If in base, use the strongest base (HO- in our examples) to take off a "lost" proton first and put on a "gained" proton second with a weak acid (usually the solvent = H2O in our examples).

5. In all tautomer mechanisms there will be resonance structures in the intermediate formed. The intermediate structure will show the way to all other reasonable tautomers from that intermediate. You may have to repeat the tautomer process once, twice, etc. until you accomplish an overall indicated transformation. Counting the number of protons lost or the number of protons gained will give you an indication of how many times you may have repeat the tautomerization process. This may not always match however because sometimes a tautomer sequence is reversed and hidden from the overall change indicated (See rule 2.).



Acids and Bases and Tautomerism The following transformations can be done in base or acid. Intermediate resonance structures lead to stable structures.

O

H

H

O

H

H

OH

H

H

H

H

H

H

H

H

H

H

H

H

H

H

H

H

H

H H

1

3

2

acid orbase

acid orbase

acid orbase

Use generic acid, H-A, or generic base, B:to accomplish the given transformations.For every transformation there will beresonance delocalized intermediates that lead toward the path desired.

This is the most thermodynamically favored keto/enol structure because it retains the C=O and has conjugatedpi bonds.

1AH

1B

2

2

2AH

2B

1

1

3AH

3B

1

1

1AH

1B

3

3

2AH

2B

3

3

3AH

3B

2

2

OH

H

H

OH

H

H

H

H

H

H

H H H

H

H H45

Additional tautomeric structures. Additional tautomeric interconversions (40 different problems).



Arrow-Pushing Practice – Fill in missing formal charge, lone pairs and curved arrows.

O

H

H

O

H

H

O

H

H

O

H

H

O

H

H

H

H

H

H

H

H

H

H

H

H

H

H

H

H

H

H

H H

H

H

H

H H H

H

H H

1

32

45

H

HH

O H

O

H

H

H

H

H

H

H

OH H

O

H

H

H

H

H

H

H

AC OH H

B

OH H

D

Steps in acid for each tautomeric change:1. proton transfer (proton off, best base = HO )2. resonance intermediates3. proton transfer (proton on, best acid = H2O)

H

O

A

B

O

H

H

H

H

H

H

H

O

H

H

H

H

H

H

H

O

H

H

H

H

H

H

H

D

OH

O

H

H

H

H

H

H H

O

H

H

H

H

H

H H

C

E

EF

F

OH H

G

G

O

H

H

O

H

H

O

H

H

O

H

H

O

H

H

H

H

H

H

H

H

H

H

H

H

H

H

H

H

H

H

H H

H

H

H

H

H H

H

H H

1

3

2

4

5

H

H

H

OH

H

H

O

H

H

H

H

H

H

H

H

H

OH H

O

H

H

H

H

H

H

H

H

H

AA

OH H

OH

H

H

B

B

O

H

H

H

H

H

H

H

H H

O

H

H

H

H

H

H

H

H H

O

H

H

H

H

H

H

H H

H

OH H

C

OH H

D

CD

Steps in acid for each tautomeric change:1. proton transfer (proton on, best acid = H3O+)2. resonance intermediates3. proton transfer (proton off, best base = H2O)

Remember: each tautomer has the same overall formal charge and the same total number of pi bonds.




Possible Key for arrow-pushing in “tautomer” problems

O

H

H

O

H

H

O

H

H

O

H

H

O

H

H

H

H

H

H

H

H

H

H

H

H

H

H

H

H

H

H

H H

H

H

H

H H H

H

H H

1

32

45

H

HH

O H

O

H

H

H

H

H

H

H

OH H

O

H

H

H

H

H

H

H

AC OH H

B

OH H

D

Steps in acid for each tautomeric change:1. proton transfer (proton off, best base = HO )2. resonance intermediates3. proton transfer (proton on, best acid = H2O)

H

O

A

B

O

H

H

H

H

H

H

H

O

H

H

H

H

H

H

H

O

H

H

H

H

H

H

H

D

OH

O

H

H

H

H

H

H H

O

H

H

H

H

H

H H

C

E

EF

F

OH H

G

G

O

H

H

O

H

H

O

H

H

O

H

H

O

H

H

H

H

H

H

H

H

H

H

H

H

H

H

H

H

H

H

H H

H

H

H

H

H H

H

H H

1

3

2

4

5

H

H

H

OH

H

H

O

H

H

H

H

H

H

H

H

H

OH H

O

H

H

H

H

H

H

H

H

H

AA

OH H

OH

H

H

B

B

O

H

H

H

H

H

H

H

H H

O

H

H

H

H

H

H

H

H H

O

H

H

H

H

H

H

H H

H

OH H

C

OH H

D

CD

Steps in acid for each tautomeric change:1. proton transfer (proton on, best acid = H3O+)2. resonance intermediates3. proton transfer (proton off, best base = H2O)





Ka/pKa table for a variety of acid types The sign and magnitude of an acid’s pKa represents the approximate energy change to form the conjugate base from the acid with water as the general base. Remember a difference of 2 pKa units is the same as the difference between a 6’ person and a 600’ person, a pretty obvious difference.

A pKa table provides us with immediate access to an acid’s proton donating ability and indirectly to its conjugate base electron donating ability. You can decide from the values in the pKa table whether an acid is strong or weak and its relative acidity (or basicity) compared to other acids (or bases) in the table. If it is weak (most of them are, pKa > 1 to very large), you can evaluate approximately how large an energy input is necessary to form the conjugate base. Remember, water is the reference base for all of the listed Ka’s of the acids even though as a solvent for many acids, it is meaningless.

pKa Table for a Variety of Acids – Approximately equal to Gacid ionization (in kcal/mole = (1.4)x(pKa))

Carbon Acids – There is a fair amount of uncertainty in the higher pKa values.

CHO2N NO2

H

pKa = 4

CHOHC CHO

H

pKa = 5

CHRO2C NO2

H

pKa = 6

CH

H

pKa = 9

O O

CH

N

pKa = 9

CHR NO2

H

pKa = 10

CN

pKa = 11

H

CN

H = acidic hydrogen atom

CH

HpKa = 11

O O

ORCH

RO2S SO2R

H

pKa = 13

CHRO2C CO2R

H

pKa = 13 pKa = 15

CHR

HpKa = 16

O

H

pKa = 20

H

HCH

R

H

pKa = 19

O

H

H H

CH2

S

HpKa = 23 pKa = 24 pKa = 25

CHR

H

pKa = 25

O

OR

OO

R

RC

CH

C

Cl

Cl

Cl

H

pKa = 30

CN

R

HpKa = 31

C

Ph

Ph

Ph

H CH2

S

HpKa = 32

R

O

pKa = 23

H H

pKa = 34

C

Ph

Ph

H

H

S

S

H

pKa = 23 pKa = 40

C

H

Ph

H

HCH2

H

pKa = 43

H

pKa = 43

H2C

HC

H

pKa = 44

H

HpKa = 46 pKa = 50 - 60

C

R

R

R

H

Oxygen Acids

H = acidic hydrogen atom

pKa = -3

RO

H

S

O

O

O

H

HO

pKa = -10

Cl

O

O

O

H

O

R

O

R

Haldehydes, ketonesesters, acids pKa = -8 to -6amides pKa = 0 pKa = -1

S

O

O

O

H

R

pKa = -3

R

RO

H

pKa = -3

H

HO

H

pKa = -2

H

ON

O

OH

pKa = -1

RC

O

OH

pKa = +5

RC

O

OO

pKa = +8

HO

H

pKa = +10

HO

OH

pKa = +11.6

H3CO

H

pKa = +15.5

HO

H

pKa = +15.7

RO

H

pKa = +16-19

HO

pKa = +25



C

O

OH

pKa = +4.8

N O

H

pKa = +1 pKa = +5

N

R

R

O

H

R

Compare the following groups.

C

O

OH C

O

OH C

O

OH

Cl

Cl

ClpKa = +4.5 pKa = +4.0 pKa = +2.8

H3CC

O

OH

pKa = +4.7

H2CC

O

OH

Cl pKa = +2.9

CHC

O

OH

Cl pKa = +1.3

ClC

C

O

OH

Cl pKa = +0.7

Cl

ClH2C

C

O

OH

FpKa = +2.6

H2CC

O

OH

Cl pKa = +2.9

H2CC

O

OH

Br pKa = +3.0

H2CC

O

OH

I pKa = +3.1

Nitrogen Acids

CR

NH

N

Ph

Ph

HPh

pKa = -10 pKa = -5

N

Ph

H

HPh

pKa = +1

N

H

H

HPh

pKa = +5

N H

pKa = +5

HO

NH

pKa = +6

HH

NN

H

H

pKa = +7

HN

NH

pKa = +8

HH

HpKa = +10

N

H

H

HH

pKa = +9.2

N H

pKa = +17

N

R

R

HR

pKa = +9-11

N

O

O

HC

H2N

H2N

N

H

H

pKa = +13

H2NC

O

NH

H

H3CC

O

NH

H

pKa = +14 pKa = +15

HN

H

pKa = +35

RN

H

pKa = +37

H R

Other Miscellaneous Acids

ON

H

pKa = +13.7

Compare the following groups.

S

H

HR S HH

pKa = +7pKa = -5

S HPh

pKa = +8

S HR

pKa = +10 pKa = +0

P

H

H

HH

pKa = +9

N

H

H

HH

pKa = +8.1

H

H2NN

H

HH

H

Compare the following groups.F5SbF H pKa = -20

FSO3 H pKa = -15

F4B H pKa = -15

O3ClO H pKa = -10

I H pKa = -10

Br H pKa = -9

Cl H pKa = -7

F H pKa = +3

H2PO4 H pKa = +2.1

HPO4 H pKa = +7.2

PO4 H pKa = +12.4-2

O2NO H pKa = -1

ONO H pKa = +3

HO2CO H pKa = +6.4

O2CO H pKa = +10.3 ClO H pKa = +7.5

O3ClO H pKa = -10

O3ClO H pKa = -1

O3ClO H pKa = +2

HTe H pKa = 3

HSe H pKa = 4

HS H pKa = 7

HO H pKa = 16

ClO H pKa = +7.5

BrO H pKa = +8.7

IO H pKa = +11

HO3SO H pKa = -3 HO2SO H pKa = +2

O2SO H pKa = +7O3SO H pKa = +2

HSe H pKa = +4

HS H pKa = +7

HOO H pKa = +12

HO H pKa = +16

HO H pKa = +16



Table of acidities of some phenols

O

H

SO

S

S o = ortho m = meta p = para

-Cl 8.48 9.02 9.38-Br 8.42 9.11 9.34-OH 9.98 9.44 9.96-CH3 10.48 10.08 10.19-NO2 7.23 8.35 7.14-CHO 6.79 8.00 7.66

Table of acidities of some very strong acids (= 100% ionization in water)

HF / SbF5 pKa -20FSO3H pKa -15

HF / BF3 pKa -15

HF / BF3 pKa -15

HI pKa -10

HClO4 pKa -10

HBr pKa -9

HCl pKa -7

SR

H

H pKa -7

CR N H pKa -10

S

O

O

OH pKa -2

SR

R

HpKa -5

OR

H

H pKa -3

OR

R

H pKa -3

OH

H

H pKa -2

H2MnO4pKa -1

H2CrO4 pKa -1

HNO3pKa -1

HClO3pKa -1

HH

H pKa -8

CR

O

H

H

CR

O

OH

H

CR

O

R

H

CR

O

OR

H

CR

O

NR

H

R

CN

H

CN

NC

NC CN

N

O

O

N

H

H

H

NO2

NO2

R N

O

O H

pKa -8

pKa -6

pKa -7

pKa -7

pKa 0

pKa -10

pKa -10

pKa -11

H2SO4 pKa -3

CF3SO3H pKa -14

The above pKa tables dramatically demonstrate how much Bronsted acids can vary in strength. The magnitude of the numbers is really beyond our comprehension. The strongest acid in the table has a pKa 10+20, while the weakest acid has a Ka 10-50. That’s 70 orders of magnitude! What does 1070 mean? Even so, we will only use two simple arguments to rationalize the differences in acidity (…and basicity). We will only use two reasons for these large differences: 1. inductive effects (based on relative electronegativity) and 2. charge delocalization effects (usually based on resonance through 2p orbitals). We will not emphasize steric effects, hydrogen bonding or solvation effects, which can also modify relative acidities, sometimes greatly.

Because we use water as our reference base, the differences in Ka values of all the various acids are mainly due to the differences in energy between each acid (HA) and its conjugate base (A:-) in the table. We can focus our attention on just the factors that raise or lower these two components. Quite often one of these components is neutral and one of them is charged. The two most common possibilities are shown below; the first being much more common to us. Each reaction is drawn as though G is positive (a weak acid), though there are examples of both that are strong acids (negative G). Of the two components in each equation below (conjugate acid and conjugate base), the one that is charged usually has the larger effect on Gionization in comparisons with other acids.



A3H

More variations in the energies of A: because of the excess negative charge. Resonance and inductive effects will generally have a larger effect on the conjugate base, because it is charged. Differences seen here will be more important in explaining differences in relative acidities between different acids.

A2

A1

A3

A2HA1H

A3H

More variation in H-A-H because of the excess positive charge. Resonance and inductive effects will generally have a larger effect on the charged component.

A2HA1H

A3H

A2H

A1H

AH A

Equation 1 - The acid is neutral, conjugate base is charged. We will see more reactions like this than equation 2.

AH

Equation 2 - The acid is charged, conjugate base is neutral. The most common situation like this is when "A" is a positively charged nitrogen atom.

AH

H

H

H

H

H (sol)H (sol)

neutral

anionic

neutral

cationic

G = (1.4)(pKa)

G = (1.4)(pKa)

General Examples

The factors that stabilize the charged component have the larger effect on the acidity of an acid. Resonance and inductive effects are key concepts to your understanding of acidity, and much more in organic chemistry and biochemistry. Learn these concepts well.

Specific Examples

Examples

HO

H pKa = +16

OH

HN

H

pKa = +35

NH

H

H

Inductive / electronegative effect affects the stability of the anionic conjugate base more than the neutral acid. Negative charge on oxygen is more stable than negative charge on nitrogen.

Resonance / delocalization effect affects the stability of the cationic acid more than the neutral conjugate base.

H3NH

H2NC

NH2

NH

pKa = +13

H

pKa = +9

NH3 H2NC

NH2

N

H

A larger pKa G means a weaker acid. In the first example this is due to a less stable conjugate base (negative charge on nitrogeninstead of oxygen). In the second example this is due to a more stable acid (delocalization of charge / resonance).

H = lost protonweaker acid

HA4 is more stable.

stronger acidHA3 is less stable.

stronger acid weaker acid

:A1 is more stable.

:A2 is less stable.

resonance

arrhenius definitions (before 1900) – water is...

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