Module 3 Practice Set: Acidity and Electron Pushing Arrows

Ranking Acid Strength Sample Question:

Rank the strengths of the following acids from 1-4 with 1 being the strongest acid and 4 being the weakest.

Answer Key:

The acidic hydrogen in each molecule is highlighted red.

The first property we need to look at is the length of the bond to the acidic hydrogen, which is effected by the number of electron shells the atom bonded to each acidic hydrogen has. Bromine has 4 electron shells, chlorine has 3, and oxygen has 2.

1) HBr is the strongest acid because the acidic H is bonded to an atom with 4 electron shells. All of the other acidic hydrogens are bonded to atoms with less than 4 electron shells. This makes the H-Br bond the longest and therefore the weakest, so that H is the easiest to lose which makes it the strongest acid.

2) HCl is the second strongest acid because the acidic H is bonded to an atom with 3 electron shells. All of the remaining acidic hydrogens are bonded to oxygen atoms with 2 electron shells. This makes the H-Cl bond longer and therefore weaker than the H-O bonds, so that H is the easiest remaining H to lose which makes it the second strongest acid.

The remaining two molecules have acidic hydrogens bonded oxygen atoms which both have 2 electron shells, so we need to check the next property to break the tie.

The second property we need to look at is the electronegativity of the atom the acidic hydrogen is bonded to. The two remaining acidic hydrogens are both bonded to oxygen atoms, so this does not give us any information either.

The third property we need to look at is the number of highly electronegative atoms in the molecule as a whole. The fourth molecule has a double bonded oxygen next to the O-H bond that the third molecule does not have.

3) The carboxylic acid is the third strongest acid because it has an additional oxygen in the molecule whereas EtOH does not. The additional oxygen atom creates an additional dipole away from the acidic hydrogen, which makes that H-O bond weaker than the H-O bond in EtOH. Because the H-O bond in the carboxylic acid is weaker, it is easier to lose that H which makes it more acidic than the H in EtOH. This makes it the third strongest acid overall.

4) EtOH is the weakest acid. Oxygen is tied for the lowest number of electron shells and the molecule does not have the additional oxygen atom that the carboxylic acid does. This means that the dipole towards the oxygen is the smallest, so the bond to the acidic hydrogen is the strongest. Therefore, that H is the hardest to lose which makes it the weakest acid.

 

Identifying pKa Values Sample Question:

*Please note that the pKa values in this answer key are approximates. Any pKa table can be used for this worksheet, however different pKa tables will give slightly different values for each hydrogen. The values in this answer key are assigned from the pKa table shown below.

Assign pKa values for the highlighted hydrogens in the following molecule.

Answer Key:

The red hydrogen has a pKa of ~15. It is bonded directly to an oxygen, so we need to look for an O-H bond in the pKa table. The adjacent group is an sp3 hybridized carbon which will be represented by an R group in the table. The representation of the group we’re looking for in the pKa table is shown in the following image:

The blue hydrogen has a pKa of ~45. It is bonded directly to a carbon, so we need to look for a C-H bond in the pKa table. The adjacent groups are both sp3 hybridized carbons that will be represented by one R group in the pKa table. The representation of the group we’re looking for in the pKa table is shown in the following image:

 

Electron Pushing Arrows Sample Question:

For the transformation listed below, draw the missing electron pushing arrows necessary to account for the chemical changes or resonance structures.

Answer Key:

In reaction below, the electrons that move are highlighted green and/or blue to help illustrate the change occurring.

In the product, the nitrogen gained 1 hydrogen, lost 1 lone pair, and has a formal positive charge. The chlorine in the product lost its hydrogen, gained 1 lone pair, and has a formal negative charge. This tells us that a lone pair on the nitrogen must have formed a bond to the hydrogen in HCl.

An arrow from the lone pair on the nitrogen is drawn to the hydrogen in HCl. The arrow has to start at the lone pair and not the nitrogen atom because electron pushing arrows always have to start at a pair of electrons. The formal positive charge in the products comes from the fact that the nitrogen donated 1 electron from its lone pair to hydrogen in order to form the bond. Therefore, it has 1 less electron than it started with which causes its formal charge to increase by 1.

At the same time, the hydrogen in HCl lets go of its bond and both electrons in that bond go to the chlorine. The arrow starts at the sigma bond and is drawn to the chlorine to show both electrons in the bond moving. The formal negative charge in the products comes from the fact that the chlorine gained 1 electron from the sigma bond that became a lone pair. Therefore, it has 1 more electron than it started with which causes its formal charge to decrease by 1.

 

Acid/Base Reactions Sample Question:

*Please note that the pKa values in this answer key are approximates. Any pKa table can be used for this worksheet, however different pKa tables will give slightly different values for each hydrogen. The values in this answer key are assigned from the pKa table shown below.

Instructions:

For each of the following problems, complete the following steps:

1. Identify the Acid:
   • Determine which molecule in the given reaction is acting as the acid.
2. Draw Electron Pushing Arrows and Generate Products:
   • Use electron pushing arrows to show the flow of electrons during the acid/base reaction.
   • Draw the products of the reaction, indicating any changes in the molecular structure due to the movement of electrons.
3. Identify pKa Values:
   • Using a pKa table, find the pKa values for the acids on both sides of the equation.
4. Determine the Favored Side of the Equation:
   • Based on the pKa values, determine which side of the equilibrium is favored. Recall that equilibrium favors the formation of the weaker acid (the acid with the higher pKa value).

 

Answer Key:

H3O+ will be the acid in the reactants. We know this because H3O+ is formally positive so it must be the acid. This tells us that the oxygen atom will lose 1 hydrogen and nitrogen will gain 1 hydrogen. The products will therefore be H2O and NH4+.

An arrow is drawn from a lone pair on the nitrogen atom directly to a hydrogen in H3O+. The arrow has to start at the lone pair and not the nitrogen atom because electron pushing arrows always have to start at a pair of electrons. The formal positive charge in the products comes from the fact that the nitrogen lost 1 electron from the lone pair that became a sigma bond. Therefore, it has 1 less electron than it started with which causes its formal charge to increase by 1.

At the same time, the acidic hydrogen lets go of its bond and both electrons in the bond go to the oxygen. The arrow starts at the sigma bond and is drawn to the oxygen atom to show both electrons in the bond moving. The formal charge goes to zero in the products because the oxygen gained 1 electron from the sigma bond that became a lone pair. Therefore, it has 1 more electron than it started with which causes its formal charge to decrease by 1.

In order to determine which side of the equation is favored, we first need to find the acid on each side of the equation. H3O+ is the acid in the reactants, and NH4+ is the acid in the products for the reverse reaction. We know that NH4+ is the acid in the products due to its formal positive charge. The acidic hydrogens are highlighted blue and green in the answer shown above.

Now we need to determine which acid is weaker. The pKa of H3O+ is ~0, and the pKa of NH4+ is ~10. The pKa values tell us that NH4+ is a weaker acid than H3O+, so the products are favored in this reaction. Because the difference in pKa values is greater than 5, the reaction has an irreversible arrow.

 

Boiling Point Sample Question:

Rank the relative boiling points from 1-4 with 1 being the highest boiling point and 4 being the lowest boiling point.

Answer Key:

EtOH has the highest boiling point. It has hydrogen bonding, and the hydrogen bonding present is stronger than the hydrogen bonding in EtNH2 due to oxygen being significantly more electronegative than nitrogen. This results in larger dipoles and partial charges, which results in stronger intermolecular forces. Because it has the strongest intermolecular forces out of all the molecules, it requires the most energy to boil.

EtNH2 has the second highest boiling point. It has hydrogen bonding, and although the hydrogen bonding is weaker than the hydrogen bonding in EtOH,  all of the other molecules do not have any hydrogen bonding. Therefore, it has the second strongest intermolecular forces so it requires the second highest amount of energy to boil.

C5H12 has the third highest boiling point. It does not have any hydrogen bonding, however it is a longer carbon chain than C3H8 so there are more LDFs present. More LDFs mean it has stronger intermolecular forces than C3H8, so it requires more energy to boil.

C3H8 has the lowest boiling point. It does not have any polar bonds and has the smallest surface area. Therefore, it has the weakest intermolecular forces so it requires the least amount of energy to boil.

 

 

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