Classes 8 and 9: Predicting Acid Strength – Qualitative

Question 1

Predict the relative acidity of one acid compared to another based on the stabilities of their conjugate bases.

Answer 1

• Principle: The more stable the conjugate base, the stronger the parent acid
• Factors increasing conjugate base stability:
  
  • Delocalization/resonance
  • More electronegative atom attached to anion
  • Electron-withdrawing inductive effects
• Delocalization/resonance:
  
  • e.g. CH3COO- more stable than Cl- (resonance)
  • Therefore, CH3COOH is stronger acid than HCl
• Electronegativity:
  
  • e.g. F- more stable than Cl-
  • Therefore, HF is stronger acid than HCl
• Inductive effects:
  
  • e.g. Cl3CCOOH > Cl2CHCOOH > ClCH2COOH > CH3COOH
  • More Cl atoms, stronger -I effect, more stable anion
• Compare trends in anion stability:
  
  • More stable anion = stronger conjugate base = stronger parent acid

So in summary, analyze and compare factors like charge delocalization, electronegativity, and inductive effects on the conjugate base anions to determine the stronger acid.

Question 2

Predict which side of an acid-base equilibrium will be favored based on which side has the weaker conjugate acid-base pair.

Answer 2

• Principle: Equilibrium favors the side with the weaker conjugate acid-base pair
• Comparing conjugate acid strengths:
  
  • Weaker acid has more stable conjugate base
  • More stable conjugate base from resonance, electronegativity, inductive effects
• For equilibrium: HA + BOH ⇌ A- + B+H2O
  
  • If HA is weaker acid than B+H2O, equilibrium favors products (A- and B+H2O)
  • If B+H2O is weaker acid than HA, equilibrium favors reactants (HA and BOH)
• Comparing conjugate base strengths:
  
  • Weaker base has more stable conjugate acid
  • For equilibrium above, compare A- and BOH base strengths
• So predict equilibrium position by:
  1) Identify conjugate acid-base pairs
  2) Determine which is the weaker acid/base pair
  3) Equilibrium favors side with weaker conjugate pair
• Factors like resonance, electronegativity, inductive effects determine conjugate strengths

So in essence, analyze the relative strengths of the conjugate acid-base pairs to predict which side of the equilibrium is favored.

Question 3

Use curved arrows to show how one resonance structure is converted into another.

Answer 3

• Resonance structures represent the same molecule/ion
• They differ only in position of electrons (π bonds, charges)
• Using curved arrows:
• Arrows track movement of electrons
• From the initial π bond or lone pair
• To the new position in the resonance form
• General steps:
  1) Identify π bonds and formal charges in each structure
  2) Use curved arrow(s) starting from:
  
  • π bond that needs to relocate
  • Lone pair that needs to relocate
  • Negative formal charge (excess electrons)
    3) Arrow points to new position in resonance form
• Example: Nitromethane anionO O-
  || |
  N-C ←→ N+=C
  | ||
  H O
• Curved arrow starts at N-C π bond
• Moves to N+=C resonance position

So in summary:
- Curved arrows represent π and lone pair relocations
- Start at initial π bond or lone pair position
- Arrow points to new location in resonance form

Question 4

Use the following to predict stability:
Element effects
Resonance effects
Inductive effects
Hybridization effects

Answer 4

• Element effects:
• More electronegative elements stabilize negative charges
• e.g. F- more stable than Cl-, Br-, I-
• Resonance effects:
• Delocalization of charge stabilizes species
• More resonance structures, more stabilization
• e.g. Carboxylate anion (R-COO-) stabilized by resonance
• Inductive effects:
• Electron-withdrawing groups stabilize adjacent negative charges
• Electron-donating groups destabilize adjacent negative charges
• e.g. F3C- more stable than CH3- due to -I effect
• Hybridization effects:
• Species with more s-character are more stable
• Higher s-character = lower electrostatic repulsion
• sp > sp2 > sp3 hybridization stability
• Analyzing stability:
  1) Identify resonance possibilities
  2) Consider inductive effects of substituents
  3) Note electronegative element effects
  4) Evaluate hybridization state
  5) Most stable has best resonance, -I effect, electronegative elements, higher s-character

So in essence, evaluate and compare resonance, inductive, element, and hybridization effects to predict and rationalize the relative stability of charged and resonance species.

Question 5

Predict the relative acidity of one acid compared to another based on the stabilities of their conjugate bases.

Answer 5

Acid strength: Look at stability of the conjugate base to determine the strength of the acid (if more stable then the acid)
Base strength: if the conjugate acid is also stable then more likely to be favored with the addition of a proton

Question 6

Predict which side of an acid-base equilibrium will be favored based on which side has the weaker conjugate acid-base pair

Answer 6

If the conjugate base is very very stable, then the reverse reaction will functionally not happen
Whichever is more stable will be the reaction that is favored

Question 7

Use the following to predict stability:

Answer 7

Element effects
Size: proton bound to an element farther down the periodic table = more acidic (covalent bonds weaker due to an increase in radius and the bad overlap of different bonds)
HI > HCl
RSH > ROH (R=generic hydrocarbon group)
Electronegativity: More EN = more acidic (stabilize neg charge left behind)
R-OH&raquo_space; R-CH
Basticity is the opposite
size matters more (if extreme)
MOST IMPORTANT INDICATOR
Resonance effects
If the conjugate base (lone pair) is involved in resonance then the proton will be more acidic because of the spread of the negative charge throughout the atom.
Inductive effect
More electronegative elements near the atom = more acidic
DECREASES with distance
The electronegative atom will pull electrons from the other side, which creates a partial positive and spreads out the effect of the negative charge
Hybridization effect
More s character = more acidic
sp>sp2>sp3

Question 8

What is meant by “base stability”? Is a stable base more or less reactive with an acid than an
unstable base?

Answer 8

A stable base is less reactive towards accepting a proton in a Bronsted-Lowry acid-base
reaction. It is lower in free energy compared to a more reactive base, which more readily
accepts a proton

Question 9

2. Consider the trend in the table and provide evidence and reasoning for the trend.

Answer 9

Evidence: As you move across the periodic table, the conjugate bases bear the negative charge
on increasingly electronegative atoms.
Reasoning: A more electronegative atom with a negative charge pulls electron density towards
itself, stabilizing the negative charge and therefore making the base more stable and less
reactive with a proton. This drives Ka eq. to the right. (As Ka increases, pKa decreases)

Question 10

Without looking up pKa data, let’s consider H2O vs. H2S.

Answer 10

a. Make an argument for why the pKa of H2O should be lower than H2S.
O is more electronegative than S and therefore is more stable with the negative charge. (See
explanation for #2)
b. Make an argument for why the pKa of H2O should be higher than H2S.
S is larger than O and it is more polarizable. The negative charge is distributed over a bigger
volume, stabilizing the negative charge on sulfur and decreasing the pKa of H2S compared to
H2O.
c. Now look up the actual pKa of H2S and compare it to that of H2O in the table on the first page.
Which argument fit with the actual data?
pKa of H2O is approximately 15 and pKa of H2S is approximately 7. Argument b) fits the data.
When comparing charges on atoms of different size, the polarizability of the atom is more
important in determining pKa.
d. Based on 3 a-c, why are HCl, HBr, and HI strong acids but HF is weak?
The first three are larger and more
polarizable than F-
. The effect is so great that
their conjugate acids are strong and these ions are spectator ions in an aqueous solution.

Question 11

Write the reaction between methanol and the acetate anion. Is the acetate anion a strong enough base
to deprotonate the methanol such that the products of the reaction will be favored at equilibrium?
Explain

Answer 11

The pKa of methanol is approximately 15, the pKa of acetic acid is approximately 5. The
reactants will be favored at equilibrium as the equilibrium reaction favors the weaker
conjugate acid/base pair.

Question 12

We have seen in the table in Part I that a proton attached to a carbon atom is typically
a very weak acid. However, the pKa of acetone (shown) is 20. Explain this drastic
difference.

Answer 12

The conjugate base of acetone has a resonance structure with the negative charge on oxygen.
Because it is more electronegative, oxygen is better able to support the negative charge.

Question 13

The conjugate base of acetone has a resonance structure with the negative charge on oxygen.
Because it is more electronegative, oxygen is better able to support the negative charge.

Answer 13

No, HCCH has the lowest pKa (most acidic), but the highest bond dissociation energy.

Question 14

Draw the conjugate bases of each of the compounds. Which conjugate base is least reactive with a
proton? Explain.

Answer 14

HCCis the least reactive because the lone pair is in an sp orbital. The sp orbital is 50% scharacter and, since s is lower in energy than p,
the lone pair is in a lower energy orbital (and
located closer to the positively charged nucleus)
than the conjugate bases of ethene or ethane

Question 15

Rank them in order of increasing basicity,
where the most basic is 1 and the least basic
is 3. Explain

Answer 15

The lone pair in the sp3 hybridized orbital is in the highest energy orbital (sp2
and sp are
both lower in energy) so it is easier to donate the electron pair to a proton.

Question 16

Main concepts:

Answer 16

Predicting Acid Strength:
a. The stability of an acid’s conjugate base determines the acid’s strength.
b. More stable conjugate bases result in weaker acids.
Element Effects:
a. More electronegative atoms stabilize the negative charge on conjugate bases, making them weaker bases and stronger acids.
b. Polarizability of atoms also plays a role in determining acid strength.
Resonance Effects:
a. Resonance stabilizes conjugate bases, leading to weaker acids.
b. Drawing resonance structures and using curved arrows to show movement of electrons.
Inductive Effects:
a. Electronegative atoms inductively withdraw electron density, stabilizing conjugate bases and increasing acid strength.
Hybridization Effects:
a. Conjugate bases with lone pairs in lower energy (more s-character) orbitals are more stable and less basic.
b. Relationship between hybridization, orbital energy, and basicity/acidity.
Acidity is not directly correlated with bond energy between the proton and the atom it is bound to.
Understanding the equilibrium between acids and conjugate bases, and predicting the favored side based on the weaker conjugate acid-base pair.
Drawing conjugate bases and acids, and understanding their relative stabilities and reactivities.
Use curved arrows to show how one resonance structure is converted into another when analyzing resonance effects.
For element effects, look at the periodic trends in electronegativity and polarizability to predict which conjugate base will be more stable.
When comparing charges on atoms of different sizes, polarizability is more important than electronegativity in determining acid strength (e.g., H2O vs H2S example).
For resonance effects, look for resonance structures that can delocalize the negative charge, especially onto more electronegative atoms like oxygen.
For inductive effects, identify the atom(s) three atoms away from the conjugate base and consider their electronegativity to determine how they withdraw or release electron density.
For hybridization effects, determine the hybridization of the atom bearing the lone pair in the conjugate base. More s-character means lower energy and more stable lone pair.
Draw out the conjugate bases and acids to visualize and compare their stabilities based on resonance, inductive effects, etc.
Consider the pKa values and use the understanding that weaker conjugate acid-base pairs will be favored at equilibrium.
Compare bond lengths and bond energies, but recognize they don’t directly correlate with acidity trends.

Question 17

Rank the following four carboxylic acids from lowest pKa (1) to highest pKa (4). Briefly explain
your ranking

Answer 17

Acetic acid has the highest pKa (rank #4; weakest acid). It has no electronegative atoms in proximity
to the ionizable CH3CO2H proton so will have the least stable conjugate base.
Don’t conflate your knowledge of the HX acids (X = F, Cl, Br, or I; HF = weakest acid and HI =
strongest) with the important inductive effect involved in this question. The conjugate base is placing
the negative charge on oxygen in each acid. F (electronegativity = 4.0) will have the strongest
inductive effect. Br (electronegativity = 2.8) will have the weakest inductive effect.

Question 18

Draw the Lewis structure for H3PO4. If you look up the pKa for H3PO4, you find three very
different values listed as pKa1 = 2.14, pKa2 = 7.20, and pKa3 = 12.37. This may seem surprising
at first as your Lewis structures shows three identical OH groups attached to P=O. Explain why
there are three different pKa values listed.

Answer 18

Removal of the first proton of H3PO4 will produce a negatively charged molecule. Removal of the
second proton will now be more difficult (i.e., higher pKa) as there is already a full negative charge in
the molecule. We use conjugate base stability to assess acidity. Removal of the second proton will
require formation of a molecule with two negative charges. Removal of the third will be most difficult
as the conjugate base has a -3 charge

True or False? The correct ranking of Ka values for H3PO4 is Ka1 > Ka2 > Ka3.

True. pKa = -log[Ka] The proton with the lowest pKa has the highest Ka.

Question 19

5. True or False? Kw changes with temperature.

Answer 19

True. KW is an equilibrium constant. We know that equilibrium constants change with temperature

The autoionization
reaction of water is endothermic.

heat + 2 H2O <-> H3O+ + –OH

If we increase the temperature, the KW will increase. The equilibrium shifts toward the products.
K = [products/reactants]

Question 20

One way to think about basicity of a lone pair is to consider the stability of the lone pair. For example,
a chloride anion’s lone pair is much more stable than the hydroxide anion’s lone pair. We can do the
same for the two nitrogens of quinine.
The circled (more basic) N lone pair resides in an sp3-hybridized atomic orbital. The other nitrogen
lone pair resides in an sp2-hybridized atomic orbital. Electrons in hybrid atomic orbitals with a greater
percentage of s-character are more stable than electrons in orbitals with less s-character.

Question 21

Ethers do not react with bases, even very strong ones. Though there is an electronegative oxygen, it
has no protons attached (as in an alcohol or carboxylic acid). The conjugate base of the ether should
be a bit more stable than an alkane’s conjugate base due to an inductive effect from the oxygen.

Question 22

other notes

Answer 22

Drawing Resonance Structures and Using Curved Arrows: Identifying resonance forms vs constitutional isomers. Using curved arrow notation to show movement of pi electrons and formation of new resonance structures. Drawing all important resonance contributors for conjugate bases.

Predicting Relative Acidities: Considering resonance stabilization of conjugate bases - more resonance delocalization leads to stronger acids. Analyzing inductive effects of electronegative atoms - electronegative atoms near the ionizable proton stabilize the conjugate base. Accounting for hybridization - more s-character in the lone pair orbital increases basicity/acidity. Analyzing 3D geometry and conjugation effects on resonance stabilization.

Polyprotic Acid Behavior: Successive pKa values arise from increasing charge on the conjugate base after each deprotonation, making further deprotonation more difficult.

Curved Arrows for Acid-Base Mechanisms: Using curved arrows to show proton transfer mechanisms. Predicting favored direction based on relative stability of conjugate acid-base pairs.

Factors Affecting Acidity: Resonance, inductive effects, intermolecular interactions/solvation, geometry/conjugation all impact acidity trends. Some functional groups like ethers cannot be deprotonated.

Temperature Dependence of Kw: Kw increases with temperature for the endothermic autoionization of water.

Comparing Hybridization: Differences in carbon hybridization in cyclic vs acyclic compounds, impacting properties.

Identifying Basic Sites: Considering orbital hybridization to determine most basic sites.

Resonance Effects: Analyzing resonance contributions of groups like nitro on acidity.