Transcript Lectures 34-35

H
=
H
H
O
C
C
C
C
O
C
C
H
C
O
H
=
C
H3C
O
aspirin
CHEMISTRY 2000
Topic #4: Organic Chemistry
Spring 2008
Dr. Susan Lait
Organic Acids (Carboxylic Acids)


When you hear the term “organic acid”, it’s generally referring to
a carboxylic acid.
Carboxylic acids are readily deprotonated by bases such as NaOH
..O ..
..O ..
e.g.
H .. H
- .. H
C .. H
C .. - +
+ .. O
O
.
..
..
H3C
acid

H3C
O
..
base
O
...
conjugate base conjugate acid
The reason that this reaction is favoured in the forward direction
is that the products are more stable than the reactants. In
particular, the conjugate base (acetate; CH3CO2-) is much more
stable than the original base (hydroxide, OH-). This makes
hydroxide a stronger base than acetate. It also makes acetic acid
(pKa=4.7) a stronger acid than water (pKa=14).
2
Organic Acids (Carboxylic Acids)

The strength of an acid is dependent on the stability of its
conjugate base:


The conjugate base of HCl (a strong acid; pKa=-7) is Cl(a stable anion and very weak base)
The conjugate base of H2O (a weak acid; pKa=14) is OH(less stable than Cl- and a relatively strong base)
The strength of an acid can also be said to be inversely related
to the strength of its conjugate base (and vice versa).

Why is CH3CO2- more stable than OH-?
3
Organic Acids (Carboxylic Acids)

Is the reaction below product-favoured or reactant-favoured?
..O ..
C
H3C

..O ..
..
O
..
H
+
H
..
O
..
H
C
H3C
...O
...
+
H
+
..
H
O
H
Since this reaction is equivalent to the acid dissociation equation
(as long as the solution is sufficiently dilute that the XH2O  1)
CH3CO2H(aq)
CH3CO2-(aq)
+
H+(aq)
its equilibrium constant is the Ka for CH3CO2H:

Recall that we can relate the Ka and pKa for an acid via the
following equation:
An acid is stronger if it has a large Ka and a small pKa.
4
Organic Acids (Carboxylic Acids)

We can increase the strength of an acid by adding electronwithdrawing groups, further stabilizing its conjugate base.
e.g. To increase the acidity of acetic acid, replace one or more
hydrogen atoms of the methyl group with halogens:
H
C
C
H

..
O
..
H
H
C
C
H
..O ..
..O ..
..O ..
..O ..
..
O
..
H
F
C
C
H
..
O
..
H
F
C
C
F
..
O
..
H
F
F
F
pKa = 4.74
pKa = 2.66
pKa = 1.24
pKa = 0.23
H
This stabilization through  bonds is called an inductive effect.


Inductive effects are strongest when close to the acidic hydrogen.
(CF3CH2CH2CH2CO2H is not significantly more acidic than
CH3CH2CH2CH2CO2H)
We saw inductive effects in CHEM 1000 when we looked at the
strength of the oxoacids (e.g. HClO2 vs. HClO3 vs. HClO4)
5
Organic Acids (Others)

Carboxylic acids are among the most acidic organic molecules;
however, a number of other functional groups contain acidic
hydrogen atoms:


Alcohols are typically about as acidic as water. Most have pKa
values of ~15-18. (Inductive effects can lower these values to ~12)
Phenols are typically more acidic than alcohols. Most have pKa
values of ~8-10. (The pKa can go as low as 1 with the right groups
attached to the aromatic ring)



Thiols (R-SH) are also typically more acidic than alcohols. Most
have pKa values of ~10-12. (Also subject to inductive effects)
Amines are not acidic in water, but can be deprotonated by some
*very* strong bases in unreactive solvents like alkanes or ethers.
Their pKa values are typically ~35-40!
Sulfur and nitrogen are both less electronegative than oxygen.
Why are thiols more acidic than alcohols, but amines less acidic?
6
Organic Acids (Others)

Phenol is “just an alcohol with a benzene ring”. Why is phenol
more acidic than most non-aromatic alcohols?
7
Calculating pH of an Acidic Solution

We know that a strong acid (pKa < 0) dissociates fully in water
because it is a stronger acid than H3O+ (pKa = 0) so reacts fully
with H2O to generate H3O+ and its conjugate base. This effect is
known as solvent leveling:


No acid stronger than the conjugate acid of the solvent can exist in
any solution.
No base stronger than the conjugate base of the solvent can exist in
any solution. (Hydroxide isn’t the strongest base – not by a long
shot! It’s just the strongest base that can exist in water.)

Weak acids aren’t subject to solvent leveling, so we must consider
what percentage of acid molecules have dissociated.


If we have a relatively concentrated solution of a relatively weak
acid, it is likely that the percent dissociation of the acid is small and
that the actual concentration of acid is relatively close to the nominal
concentration (i.e. concentration listed on the bottle or in the
question).
As a solution becomes more dilute, this assumption becomes less fair.
ALWAYS CHECK YOUR ASSUMPTIONS AFTER CALCULATING!!!
8
Calculating pH of an Acidic Solution

Consider a 0.32 M solution of phenol (pKa=9.95) at 25.00 C.
9
Calculating pH of an Acidic Solution

Consider a 4.210-5 M solution of CH3CO2H (pKa=4.74) at 25.00 C.
10
Distribution Curves

We can also work backwards, using the Ka equation to work out
the percent dissociation of an acid at a given pH. The pH tells
us the activity (and therefore the concentration) of H+(aq) so, as
long as we know the Ka value for the acid, we’re set.
aH  a A
Ka 
a HA



so
K a aA
[A  ]


aH
a HA [HA]


Calculate the percent dissociation of acetic acid (pKa=4.74) at
pH 4.00.
11
Distribution Curves

If we repeat this calculation at a large number of different pH
values, we generate what is known as a distribution curve:
acetate (CH3CO2-)
acetic acid (CH3CO2H)
Note that the two curves cross at the pKa of the acid!
The pKa of an acid is the pH at which it is exactly 50% dissociated.
12
Distribution Curves

It is easy to see why an acid will be 50% dissociated at its pKa if
we take the negative logarithm of the Ka equation on page 11:
 aH   a A 
 logK a    log
 aHA



 a A 

 logK a    logaH    log
 aHA 

 aA 
pK a  pH  log
 aHA



 aA 
pH  pK a  log
 aHA



This is the Henderson-Hasselbach equation. We can see
that if aA-/aHA = 1 then log(aA-/aHA) = 0 and pH = pKa. As the
acid’s percent dissociation rises, so does the pH and vice versa.13
Distribution Curves for Polyprotic Acids

If an acid has multiple acidic protons whose pKa values are
different by several units, the distribution curve will look like
several monoprotic distribution curves superimposed:
If the pKa values are too close, the distribution curve will become more
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complex.
Organic Bases (Amines)


When you hear the term “organic acid”, it’s generally referring
to an amine.
Amines are readily protonated by acids:
e.g.
H
..
N
R
R
base

R
+
..+
H
O
H
acid
N
H
R
+
R
R
conjugate acid
+
H
..
O
..
H
conjugate base
The basicity of amines is due to the lone pair on the nitrogen
atom which makes all amines (R = H, alkyl or combination)
Lewis bases.
15
Organic Bases (Amines)


We have seen one other functional group containing nitrogen.
Why is an amine considered a good base, but an amide is not?
The nitrogen atom of an amide is so weakly basic that the
oxygen of the carbonyl group will be protonated over it!
16
Quantifying Basicity

We can measure a base’s strength using either of two measures:

We can refer to the Ka (or pKa) of the conjugate acid. Since the
strength of a base is inversely related to the strength of its conjugate
acid, we can deduce the strength of a base from the strength of its
conjugate acid.
H
N
H
H
+
H
N
H
pKa = 9.3
H3C
+
H
H
pKa = 10.6
Smaller pKa = Stronger Conjugate Acid = Weaker Base
17
Quantifying Basicity

Alternatively, we can refer to the Kb (or pKb) of the base itself.
is the equilibrium constant for reaction of the base with water:
H
..
N
R
This
R
R
H
+
..
O
..
H
N
R
+
R
+
R
-.. O..
..
H
Kb is inversely proportional to Ka for the conjugate acid:
K a K b  Kw
where Kw = 10-14 under standard conditions.
..
..
N
N
H
H
H
pKb = 4.7
H3C
H
H
pKb = 3.4
Larger Kb = Smaller pKb = Stronger Base
18
Calculating pH of a Basic Solution

We can calculate the pH of a solution of weak base by:



Using the Kb expression and nominal solution concentration to
calculate aOH- then
Using the Kw expression (Kw = aH+· aOH-) to calculate aH+ then
Using aH+ to calculate pH.
Recall that, at 25 C, Kw = 10-14.

The pKa of aniline (C6H5NH2) is 28 and the pKa of C6H5NH3+ is 4.
What are the Kb and pKb values for aniline?
19
Calculating pH of a Basic Solution

Calculate the pH of a 0.71 M aqueous solution of aniline at 25 C.
20
Amino Acids: Acid and Base

Some textbooks present the structure of an amino acid as follows
..O ..
H ..
N
C .. H
H
C
O
..
R
H
in order to show that the name comes from having an amine
group and a carboxylic acid group.

This is not how an amino acid actually exists in biological systems
(or other pH ~7 solution). Instead, it exists as the zwitterion:
..O ..
H
H
N
+
H
C
C
R
.. .-
O
.. .
H
21
Amino Acids: Acid and Base

Why is this?



The pKa for the –CO2H proton is typically ~2 (slightly lower than the
standard 3-5 range for carboxylic acids).
The pKa for the conjugate acid of the amine group (i.e. for the –NH3+
proton) is typically ~9-10.
Use this data to sketch a distribution curve for the amino acid,
labeling with the major species within each pH range.
22
Beyond Water: Implications of Acid-Base Chemistry


It’s easy to forget that not all chemistry is done in aqueous
solution, but the pKa scale doesn’t just go from 0 to 14! Often
times, the solvent for a reaction is chosen based (at least in part)
on acid-base considerations…
Consider the following reaction:
H
C
H
H
.. . O.
+
..
H
C
H
H
.. .
..I .
H
C
H
H
..
O
..
H
C
H
H
+
.. ..I .. ..
What would happen if we tried to perform it in aqueous solution?
23
Beyond Water: Implications of Acid-Base Chemistry

Solvent leveling can be useful, though. Sodium methoxide
(NaOCH3) can be prepared via a couple of different methods,
both of which use methanol as the solvent:


We saw in CHEM 1000 that adding Na(s) to H2O gives NaOH and H2(g)
Similarly, adding Na(s) to CH3OH gives NaOCH3 and H2(g):
Adding a very strong base (e.g. a source of H- such as NaH) to
CH3OH will also produce NaOCH3 along with the conjugate acid of
the base:
24