Organic Acids and Bases
This document has been written to provide you with an overview of the fundamental concepts
of organic acid-base chemistry. Many organic compounds have not been included such as the
basicity of pyrrole or the acidity of hydrogens of carbonyl compounds. These topics are
generally covered in the second and third organic chemistry courses.
1.0 Review of acids and bases 1
2.0 Trends in acid-base strengths 3
2.1 Shifting of the acid-base equilibrium 3
2.2 Electronegativity 4
2.3 Size 5
2.4 Resonance Stabilization 7
2.5 The proximity of electron-withdrawing and electron-donating groups: Inductive effects9
2.6 Hybridization 13
3.0 Base strength of Amines and Amides 14
3.1 Amines 14
3.2 Amides 17
4.0 Some exceptions to the general trends of acidity and basicity 17
Dr. Steven Forsey 1 Organic Acids and Bases
1.0 Review of acids and bases
Acid-base reactions are the first step in understanding why reactions between organic molecules occur.
This topic is studied in depth because acid-base reactions are so fundamental, that the theory will be
encountered in almost every topic in organic chemistry and when looking at the broadest sense of Lewis
acids and bases, most organic reactions are acid-base reactions. In this section you will learn why an acid
or base is stronger or weaker than another. You will not be asked to memorize the dissociation constants
but understand the trends that we see in acids and bases. This is a fundamental concept that will help you
understand Organic Chemistry. This section can be difficult for some students because there is no set
order to memorize, you must understand the concepts, look at the molecules of concern and understand
the differences in the molecule and then make your decision.
Many organic reactions involve proton (H ) transfer or electron pair transfer. Thus, it will be helpful to
review the Bronsted-Lowry and Lewis theories of acids and bases. In the Bronsted-Lowry Theory an acid is
a proton donor and a base is a proton acceptor. In the Lewis Theory an acid is an electron pair acceptor
and a base is an electron pair donor.
In the Bronsted-Lowry theory, an acid must contain an ionizable H atom and a base must contain a lone
pair of electrons onto which a proton can bind. For this reason, an acid is often represented in this theory
by the general formula HA and a base is represented by :B. An acid HA will produce H O ion3 in aqueous
solution and a base :B will produce OH ions because of the following ionization reactions.
HA(aq) + H O(l) H O (aq) + A (aq) [H 3 ] [A ]
2 3 K a
+ − [HB ] [OH ]
B(aq) + H O2l) HB (aq) + OH (aq) K b
The equilibrium constants for the ionization reactions for HA and B: are called “ionization constants” and
are denoted as K oa K .bThe strength of an acid (or a base) is determined ultimately by the value of K (ora
Kb). A strong acid (or base) has a large ionization constant and a weak acid (or base) has a small ionization
constant. Because the values of K ana K canbspan an enormous range of values (10 to 10 ), we often
use pK and pK vabues instead. The following relationships are helpful for converting K values into pK
values and vice versa.
pK log K or K 10 (K = Kaor K b
The expression Ka 10 shows that the larger the pK value, the smaller the K vaaue and the weaker
the acid. Similarly, the larger the pKbvalue, the smaller the K balue and the weaker the base. Strong
acids have negative pK values and show a strong tendency to lose a proton; strong bases have negative
pK balues and show a strong tendency to gain a proton. We can classify acids or bases as being very
strong, moderately strong, weak or very weak according to the extent to which the ionization reactions
At the end of this document there is a table of pK values for some organic and inorganic acids. Notice
that some protonated organic compounds are very strong acids, such as protonated, aldehydes and
alcohols, and show a strong tendency to lose a proton while other protonated forms, such as protonated
amines, are weak acids and show a small tendency to lose a proton. Similarly, some deprotanated organic
compounds, such as deprotonated alkanes and alcohols, are very strong bases and are easily protonated
while other deprotonated forms, such as deprotonated carboxylic acids, are weak bases. It will be
Dr. Steven Forsey 2 extremely helpful for you to know these approximate pK valuas. In general, protonated carbonyl
compounds and protonated alcohols are strong acids with pK values less than zero; carboxylic acids have
pK aalues around 5; protonated amines have pK valuea around 10; alcohols have pK values aaound 15;
and alkanes, alkenes and alkynes greater than 41.
2.0 Trends in acid-base strengths
2.1 Shifting of the acid-base equilibrium
Why do acids have different acidities? As mentioned previously a strong acid has a weak conjugate base.
Thus an acid is strong if its conjugate base is stable or not reactive. If the conjugate base is strong and not
stable it will react with a proton and shift the equilibrium to the left. Most organic reactions are not
performed in water and we are observing the acid-base reactions between different organic compounds.
However, the pK values can be use to determine the reactivity between organic acids and bases. For
example when sodium ethoxide is added to a solution containing methylamine an acid-base reaction may
CH 3H 2 + CH 3H 2¯ Na + CH 3H ¯ Na + + CH 3H O2
acid base conjugate base conjugate acid
Notice after the chemical reaction the base becomes the conjugate acid and acid becomes the conjugate
base. The conjugate base can now attack the conjugate acid to create the original acid-base pair.
CH 3H ¯ Na + CH C3 O2 CH 3H 2 + CH 3H O2 Na
conjugate base conjugate acid acid base
The original acid and base can then react again to regenerate the conjugate acid-base pair. This is the
physical process that is occurring in solution described by the equilibrium equation and when you look at
this acid-base reaction closely you will notice that there is a tug of war between two bases for the acidic
CH 3H 2 + CH 3H 2¯ Na CH 3H ¯ Na + CH C3 OH2
acid base base acid
Who wins the fight for the proton? The strongest or most reactive base does. Which side does the
equilibrium shift to? It will shift to generate the weaker or less reactive acid-base pair. In this acid-base
reaction the equilibrium is shifted to the left because the sodium methylamide ion is a much stronger
base than the sodium ethoxide ion. Remember the smaller the pK or pK value the stronger the acid or
CH 3H 2 + CH 3H O2 Na CH 3H ¯ Na + CH 3H O2
pK a40 pK b -2 pKb= -26 pKa= 16
Thus when you are comparing the strength of different acids using pK valaes you are looking at the ability
of an acid to donate a proton to another base. If the conjugate base of the acid you are concerned with is
stronger than the other bases in the system the proton will not be removed from the acid, as in the case
with methylamine and sodium ethoxide. To remove a proton from methylamine you would have to use a
stronger base such as a compound with a negatively changed carbon, for example CH CH ¯. 3 2
Acid-base reactions are equilibriums and you must look at both sides of the equilibrium to decide which
way the equilibrium will shift. The equilibrium will always shift to the weaker acid-base pair or the less
reactive species. One way of determining which way the equilibrium will shift is to look at the base
strengths on both sides of the equilibrium equation. The stronger base is more reactive and will win the
battle for the proton to produce the more stable less reactive and weaker base. But, what makes one
base less basic than another? To determine this you must look at the stability or reactivity of the base.
Dr. Steven Forsey 3 The stability of the base is dependent on the environment of the atom or atoms that donate the lone pair
of electrons to the acid. You must consider what atom the electrons are on. Is the atom electronegative, is
the atom small or large, what is the hybridization the atom and can the lone pair of electrons be
delocalized throughout the molecule. All of these factors must be considered.
In the example given above we know from the pK tablas that the sodium methylamide ion is a much
stronger base than the sodium ethoxide ion, or that ethanol is a stronger acid than methyl amine, but the
big question is why?
Consider the following molecules (CH ) 3 3, (CH ) 3 2, CH O-H3and HF. How could you determine their
relative acidity to each other without looking at the pK aables? First write out the equilibriums and
consider the stability of the conjugate bases. Which of the conjugates would be the weakest base (most
stable) and shift the equilibrium towards the products to a greater extent?
(CH3 3C-H (CH 3 3¯ + H
(CH3 2N-H (CH 3 2 ¯ + H
CH3O-H CH 3 ¯ + H
H-F F ¯ + H
B C N O F Notice that the charge is on different atoms and the most important difference
2.0 2.5 3.0 3.5 4.0 between the atoms, when considering the stability of an anion is their
Al Si P S Cl electronegativity (Table 2.1). Remember electronegativity is the measure of the
ability of an atom to attract electrons or charge density relative to another
1.5 1.8 2.1 2.5 3.0
Ga Ge As Se Br atom. Simply put electron density will shift to the atoms that have more
protons in their nucleus. As you go across the periodic table from left to right
1.8 2.0 2.2 2.6 2.8
the electronegativity of the atoms increase, (number of protons increases) thus
Te I you would expect the stability of the conjugate bases to increase, (CH ) C¯ <
2.1 2.5 3 3
(CH 3 2¯ < CH O3 < F¯ and the acidity to increase, (CH )3 3 < (CH )3 2 < CH OH3
2.1 Electronegativity < HF. This is indeed what we find, as you go across the same row in the periodic
of selected atom table the acidity of the compound increases.
Which of the following acid-base reactions will not occur?
Analyze:To determine if the reaction will proceed from left to right, first put lone pairs and charges on the
atoms and identify the acids and bases on both side of the reaction. Arrows have been drawn to show the
flow of electrons for the reactants side only. The red labeled electrons on the base show bond formation
between the base and hydrogen, the base is donating its electrons to the acid. The blue electrons,
illustrates the bond breaking that must occur for the base to form a new sigma bond with the hydrogen.
Dr. Steven Forsey 4 The electrons from this hydrogen bond are transferred to what becomes the conjugate base. Then
determine relative base strength. In these examples the least electronegative negatively charged atom
will be the stronger base. The stronger base will react with the stronger acid to form the weaker acid-base
Note: Na and K are counter ion in these reaction and do not participate in the reaction.
The alkoxide ion is a stronger base than the fluoride ion thus the above reaction will proceed from left to
The methyl carbanion (methanide ion) is a stronger base than the cyclopentylamide ion thus the reaction
will proceed from left to right
The 1-propanolate ion (alkoxide ion) is a weaker base than the amide ion. The reaction will not occur.
Another important factor is the size of the atom that the charge is on. A larger atom can disperse the
negative charge over a larger volume and thus increases the stability of the anion. Consider the following
acids, HF, HCl, HBr, and HI and order the acids from strongest to weakest. Again consider the equilibrium
and determine stability of the conjugate bases.
Dr. Steven Forsey 5 As you go down the periodic table the size of the atoms increases. The iodide ion is much larger than the
fluoride ion and will be much more stable with a negative charge even though fluorine is more
electronegative than iodine. The order of increasing acid strength should be HI > HBr > HCl . HF and this is
what we see. HF is a weak acid (pKa = 3.2) and HI is a very strong acid (pKa = -10)
Thus, when comparing atoms in the same column in the periodic table size has a greater stabilizing effect
than electronegativity and when comparing atoms in the same row electronegativity is more important. It
should be noted here that electronegativity is a much stronger effect than size. As seen in Table 2.2 the
pK aalues decrease much more dramatically from methane to HF than HF to HI.
- increasing electronegativity
- increasing ability to stabilize a charge on conjugate base
CH 4 NH 3 H 2 HF
50 36 15.7 3.2
7 -7 and ability to
Increasing stabilize a charge
electronegativity H 2e HBr
3.8 -9 on conjugate base
H 2e HI
Table 2.2 pK aalues selected acids
Order the following compounds from must acidic to least acidic.
1) CH 3H 2) CH 3H 3) CH3CH 3
Analyze:Draw the equilibriums for each acid and then determine the stability of the conjugate bases. In
this example there are two factors to consider, size and electronegativity.
First compare the negatively charged carbon and oxygen since they are in the same row on the periodic
table and are smaller than sulfur. Carbon is far less electronegative than oxygen thus a negatively charged
carbon will be a much stronger base than oxygen with a negative charge. Ethane will be a much weaker
acid than methanol. Now compare the negatively charged oxygen with the negatively charged sulfur. The
sulfur atom is larger than the oxygen atom and will be more stable with a negative charge. Methanethiol
will be a stronger acid because its conjugate base is more stable. The order of acid strength from
strongest to weakest is CH 3H > CH O3 > CH CH 3 2
Note: In the above equilibriums the acids are not shown reacting with a base, the acids are shown by
themselves. This is because we are not concerned with the acid reacting with a specific base, but with the
ability of an acid to donate a proton to determine relative acid strength between the acids. The
compound with the weakest conjugate base will be the most acidic. This is very important when designing
Dr. Steven Forsey 6 2.4 Resonance Stabilization
In first year chemistry courses the concept of resonance hybrids was introduced. Resonance hybrids occur
because more than one plausible Lewis structure can be written. A resonance hybrid is observed with the
conjugate base of acetic acid, the acetate ion.
Notice, that the arrow between the resonance hybrids is a double headed arrow and not an equilibrium
arrow. This arrow indicates that the actual structure is a hybrid of the Lewis structures and the charge is
spread over both oxygens.
Experimentally this is what is observed. Carboxylates are symmetrical with carbon bond lengths of 1.26 A
which is in between the typical carbon oxygen double (1.20 A ) and single (1.34 ) bonds in the
corresponding carboxylic acids. The negative charge is spread equally over both oxygens or we can say
that the charge is delocalized. This resonance delocalizing of the negative charge plays a critical role in
base stability. When the charge is spread over two or more atoms the base becomes more stable or less
Consider three different acids methanol, phenol and acetic acid,
Again look at the stability of the conjugate bases to explain why one compound is more acidic than the
other. The oxygen of the methoxide ion bears the full negative change alone whereas the phenoxide ion
can delocalize or spread the negative charge over three carbons. The carbons are not as electronegative
as oxygen and are not as stable with a negative charge as oxygen, but as you can see from the measured
pK aalues the equilibrium is shifted more to the right for phenol verifying that the phenoxide ion is more
stable or a weaker base than the methoxide ion.
Dr. Steven Forsey 7 A common misconception held by students is that if a compound has more resonance structures than
another it will always be a weaker base. This is an incorrect statement. For example, acetic acid is more
acidic than phenol even though there are only two resonance structures. The reason for this is that
oxygen is more electronegative than carbon and resonance onto two electronegative oxygens is more
stabilizing than resonance onto three carbons. Thus the acetate ion is a weaker base than the phenoxide
ion and the equilibrium for acetic acid is shifter farther to the right making acetic acid a stronger acid than
A fun way of understanding this concept is the hot potato analogy. If you hold onto a hot potato by
yourself you will get burnt because the heat coming form the potato is localized in your hand. If you can
share the hot potato with other people beside you, your hands will not have time to reach the full
temperature of the hot potato. You are distributing the heat coming from the potato to other people and
you will not get burnt. This is like the methoxide ion and the phenoxide ion. The methoxide ion does not
have other atoms to share the charge with. The charge is localized on a single oxygen which makes it
unstable or reactive, whereas the phenoxide ion has three carbon atoms to share the charge with and
thus is more stable or less basic than the methoxide ion. Can we now take the analogy further with the
acetate ion? The acetate ion is more stable or less basic than the phenoxide ion even though there are
only two resonance structures. In this case we could say that the oxygens are wearing oven mitts
compared to the carbon atoms. The oxygen atoms are more electronegative than carbon and can better
stabilize the charge.
It must be noted the acetic acid and phenol are weak acids and methanol is a very weak acid but when we
are thinking about acid-base reactions between organic molecules we need to know their relative
strengths to determine if a reaction will proceed.
Which way is the equilibrium shifted in the following acid-base reaction?
Analyze: To determine which way the equilibrium is shifted first put lone pairs and charges onto the
atoms and identify the acids and bases on both side of the reaction. Arrows have been drawn to show the
flow of electrons for the reactants. The base donates a pair of electrons and the acid accepts the
electrons. Then determine which base is the strongest or the weakest.
When comparing the conjugate bases of the acids we do not have to consider electronegativity or size
because the negative charge is on the same atom, nitrogen. However, the conjugate base of aniline is
resonance stabilized, i.e. the negative change is delocalized through the aromatic system making it much
more stable than the ethylamide ion.
Dr. Steven Forsey 8 The equilibrium will shift to the right.
2.5 The proximity of electron-withdrawing and electron-donating groups: Inductive effects.
Atoms bonded to each other with different electroneg