Organic Chemistry I & II
Organic Chemistry I & II

Organic Chemistry I & II

Lead Author(s): Steven Forsey

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Organic Chemistry I & II is designed for instructors who want an active, dynamic, and understandable approach to support their own efforts in the classroom. This ever-evolving textbook includes auto-graded questions, videos and approachable language in order to make difficult concepts easier to understand and implement.

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Pricing

Average price of textbook across most common format

Top Hat

Steven Forsey, “Organic Chemistry”, Only one edition needed

Up to 40-60% more affordable

Lifetime access on any device

McGraw-Hill

Carey & Giuliano, “Organic Chemistry”, 10th Edition

$219

Hardcover print text only

Wiley

Solomons et al., “Organic Chemistry”, 12th Edition

$301

Hardcover print text only

Wiley

David R. Klein, “Organic Chemistry”, 3rd Edition

$301

Hardcover print text only

Always up-to-date content, constantly revised by community of professors

Constantly revised and updated by a community of professors with the latest content

Top Hat

Steven Forsey, “Organic Chemistry”, Only one edition needed

McGraw-Hill

Carey & Giuliano, “Organic Chemistry”, 10th Edition

Wiley

Solomons et al., “Organic Chemistry”, 12th Edition

Wiley

David R. Klein, “Organic Chemistry”, 3rd Edition

In-book Interactivity

Includes embedded multi-media files and integrated software to enhance visual presentation of concepts directly in textbook

Top Hat

Steven Forsey, “Organic Chemistry”, Only one edition needed

McGraw-Hill

Carey & Giuliano, “Organic Chemistry”, 10th Edition

Wiley

Solomons et al., “Organic Chemistry”, 12th Edition

Wiley

David R. Klein, “Organic Chemistry”, 3rd Edition

Customizable

Ability to revise, adjust and adapt content to meet needs of course and instructor

Top Hat

Steven Forsey, “Organic Chemistry”, Only one edition needed

McGraw-Hill

Carey & Giuliano, “Organic Chemistry”, 10th Edition

Wiley

Solomons et al., “Organic Chemistry”, 12th Edition

Wiley

David R. Klein, “Organic Chemistry”, 3rd Edition

All-in-one Platform

Access to additional questions, test banks, and slides available within one platform

Top Hat

Steven Forsey, “Organic Chemistry”, Only one edition needed

McGraw-Hill

Carey & Giuliano, “Organic Chemistry”, 10th Edition

Wiley

Solomons et al., “Organic Chemistry”, 12th Edition

Wiley

David R. Klein, “Organic Chemistry”, 3rd Edition

About this textbook

Lead Authors

Dr. Steven Forsey, Ph.D.University of Waterloo

Steven Forsey is currently a Professor at University of Waterloo, teaching a variety of organic chemistry courses to Chemistry, Science, Chemical Engineering, Nanotechnology and distance education students. He received his Ph.D. (2004) for Synthetic Organic Chemistry from University of Waterloo, Ontario. He is a recipient of the Excellence of Science Teaching Award and has acted as the Teaching Fellow for the Department of Chemistry since 2016.

Contributing Authors

Felix NgassaGrand Valley State University

Neil GargUCLA

Jennifer ChaytorSaginaw Valley State University

Greg DomskiAugustana College

Christian E. MaduCollin Community College

Christopher NicholsonUniversity of West Florida

Franklin OwEast Los Angeles College, UCLA

Robert S. PhillipsUniversity of Georgia

Grigoriy SeredaUniversity of South Dakota

Simon E. LopezUniversity of Florida

Brannon McCulloughNorthern Arizona University

Jason JonesKennesaw State University

José BoquinAugustana College

Stephanie BrouetSaginaw Valley State University

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Chapter 4: Acids and Bases

We encounter acids and bases in our everyday life all the time. Two very common examples are acetic acid (vinegar) and citric acid which is found in all citrus fruits. [1]


Contents

Learning Objectives

  • Identify acids and bases and distinguish between strong and weak acids and bases.
  • Understand the effect of electronegativity, size, hybridization and resonance has on the stability of bases and how this relates to both base and acid strength.
  • Understand dynamic equilibrium and why the equilibrium is shifted towards the weaker acid-base pair.
  • Evaluate a series of acids or bases and predict their relative acid or base strength.

4.0 Introduction

This module will provide you with an overview of the fundamental concepts of organic acid–base chemistry. Heteroaromatic organic compounds (like pyrroles) and acidity trends in α hydrogens of carbonyl compounds have not been included.

4.1 Acid–base Equilibrium Reactions

Acid–base reactions are presented mostly in terms of base strength and not in terms of the hydrogen bond strength. Acid–base reactions are presented this way because many organic reactions involve a base attacking another molecule. The attacking molecule can either donate its electrons to a hydrogen through an acid–base reaction, or act as a nucleophile and attack an electrophilic center. Many reactions will involve both mechanisms, and it is up to you to be able to figure out which mechanism dominates. Thus, you must thoroughly understand how bases influence the equilibrium in acid–base reactions.

Brønsted–Lowry acids and bases are proton transfer reactions:

Brønsted–Lowry acid: A substance that can donate a proton (H+)

Brønsted-Lowry base: A substance that can accept a proton (H+)

Example: Acetic acid and sodium hydroxide. Note: Try to get into the habit of using arrows to show the flow of electrons. Arrows show bond formation and bond breaking. This will help you in future topics. A double-headed arrow shows the movement of two electrons. Electrons have been added to sigma bonds in the diagrams to help you follow the flow of electrons.

4.01.png
​Figure 4.1. Proton transfer reaction between acetic acid and sodium hydroxide.

Lewis acid: A substance that can accept a pair of electrons

Lewis base: A substance that can donate a pair of electrons

Lewis acid–base theory expands the species that can accept electrons beyond that of a proton to include any electron-deficient atom. Lewis bases and Brønsted–Lowry bases have essentially the same definition since a Brønsted base must donate an electron-pair to accept a proton.

Example: Water and trifluoroborane, and hydrogen thiolate ion and bromomethane

Figure 4.2a. Lewis acid and base reactions between water and trifluoroborane.​


​Figure 4.2b. Lewis acid and base reactions between thiolate and bromomethane.

Most organic reactions are Lewis acid–Lewis base reactions. However, before discussing these reactions you must understand proton transfer reactions, since these concepts will be used in describing organic reactions between molecules (nucleophiles and electrophiles).

Presented below is a review of acid–base equations and will not be discussed in detail. 

Figure 4.3. Review of Bronsted-Lowry acids and bases.

Equilibrium values are calculated by dividing the concentration of the products by that of the reactants. The acidity of a solution is determined by measuring the concentration of hydronium ions in solution with a pH meter. Thus, a solution is acidic if the proton is not bonded to the acid and the more the equilibrium is shifted to the products the more acidic the solution and the larger the Ka value (or the smaller the pKa) will be.

The strength of an acid depends on its ability to donate its proton to another base. In the following example, the base is water.

Figure 4.4. Strong acids have weak conjugate bases and the concentration of the conjugate base and hydronium ions that are measured by the pH meter is large. Thus, the acidic proton is mostly bonded to water not the organic acid. ​


4.2 What Makes one Acid Stronger than the Other?

Look closely at the acid–base reaction equilibrium and you will notice that there is an acid–base pair on each side of the equilibrium. It is the base on each side of the reaction that is donating its electrons to form new sigma bonds.

Figure 4.5 Generic acid-base equilibrium reaction and expression. The stronger the acid the more the equilibrium is shifted to the right. This produces a larger Ka or smaller the pKa value

Also, note that the base on one side of the equilibrium becomes the conjugate acid on the other side. Likewise, the acid on one side of the equilibrium becomes the conjugate base. This is important. The only difference between the two sides of the equilibrium is to which molecule the proton is bonded. If the proton is bonded to the acid, the acid is weak because it does not readily donate its proton. If the proton is bonded to water to form the hydronium ion, then the acid (HA) is strong because it has donated its proton to the base (H2O). Thus, there is a tug-of-war between the two bases for the proton; in this case, between the water and the conjugate base of the acid.

The equilibrium will shift to form the weaker acid–base pair, which means the equilibrium will shift to form the stronger acid-hydrogen bond and the weaker base.

Consider the following acid–base reaction that does not involve water. You add sodium ethanethiolate to methanol. What are the acid–base reactions and which way is the equilibrium shifted?

The first possible acid–base reaction is:

Figure 4.6. Acid-base reaction between methanol and sodium ethanethiolate.

After the above acid–base reaction occurs, 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. In the reaction below, the counter ion, Na+, has been removed from the equation since it is not involved in the reaction.

Figure 4.7. Acid-base reaction between methoxide and ethanethiol.

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 as described by the equilibrium equation. The two bases are in a tug-of-war for the proton.

Figure 4.8 ​

Who wins the fight for the proton? The strongest or most reactive base does. It will form the strongest bond with the proton. 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 sodium methoxide is a much stronger base (smaller pKb values).

When you are comparing the strengths of different acids using pKa values, you are looking at the ability of an acid to donate a proton to another base. If the conjugate base of the acid is stronger than the other bases in the system, then the proton will not be removed from the acid, as in the case with methanol and sodium ethanethiolate. To remove a proton from methanol you would have to use a stronger base, such as CH3CNH¯.

Click on the video for a visual explanation:

Before going on, make sure you can do the following questions. These questions ask you to identify acids and bases and determine acidities and equilibria using pKa and pKb values.


Q4.1 - Level 1

On each side of the equilibrium arrow there is a Bronsted-Lowry acid-base pair. Identify each compound as an acid or a base.

question description
Premise
Response
1

1)

A

Acid

2

2)

B

Base

3

3)

C

Acid

4

4)

D

Base


Q4.2 - Level 1

On each side of the equilibrium arrow there is a Lewis acid-base pair. Identify each compound as an acid or a base.

question description
Premise
Response
1

1)

A

Acid

2

2)

B

Acid

3

3)

C

Base

4

4)

D

Base


Q4.3 - Level 1

Identify the Lewis acid-base pair on the left side of the equilibrium.

question description
Premise
Response
1

1)

A

Acid

2

2)

B

Base


Q4.4 - Level 2

Using the pKa_a values, is the equilibrium shifted to the left or right?

question description
A

Right

B

Left


Q4.5 - Level 2

Using the pKa_a values, is the equilibrium shifted to the left or right?

question description
A

Right

B

Left


Q4.6 - Level 2

Using the pKb_b values, is the equilibrium shifted to the left or right?

question description
A

Right

B

Left


Q4.7 - Level 2

Using the pKb_b values, is the equilibrium shifted to the left or right?

question description
A

Right

B

Left


4.3 Factors that Affect the Acid–Base Equilibrium

Acid–base reactions exist in a state of equilibrium: 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, more reactive base will win the battle for the proton to produce the more stable, less reactive weaker base. What makes one base less basic than another? To determine this, you must look at the stability or reactivity of the base. The reactivity of a base is dependent on the atom or atoms that are donating the lone pair of electrons to the acid. Is the atom electronegative, is the atom small or large, what is the hybridization of the atom, and can the lone pair of electrons be delocalized throughout the molecule? All of these factors must be considered.

4.3.1 Electronegativity

Electronegativity is the measure of the ability of an atom to attract electrons or charge density relative to another atom. The more electronegative an atom is, the greater its ability to stabilize a negative charge.

Example: Order the following molecules (CH3)3CH, (CH3)2NH, CH3OH, and HF from the strongest to weakest acid.

How could you determine their relative acidity to each other without looking at the tables? First, write out the equilibria and consider the stability of the conjugate bases.

Figure 4.9. Determining relative acidity by considering the stability of the conjugate bases.

Which of the conjugate bases would be the weakest or the most stable? Look at the electronegativity values on the periodic table. The stability of the negatively charged conjugate bases follows the electronegativity of the atom. The fluoride ion is the most electronegative and is the most stable conjugate base, followed by the alkoxide ion, amide ion and, finally, the carbanion. Now, think about the equilibrium:

Figure 4.9b. ​

The weaker the conjugate base the more the equilibrium is shifted to the right and the stronger the acid (HA). Thus, the order from strongest to weakest acid is as follows.

4.19.png
Figure 4.10. pKa's of acids examined from strongest to weakest

The theory is consistent with the pKa values.

When going across the periodic table, electronegativity is an important factor when determining the strength of the conjugate base.


Q4.8 - Level 1

Which of the following acid-base reactions favors the reactants?

question description
A

a

B

b

C

c

Watch the following video to see the explanation to Q4.8.


4.3.2 Size of the Atom

When determining conjugate base stability of anions in the same column of the periodic table, the size of the atom matters more than its electronegativity.

A larger atom can disperse the negative charge over a larger volume, which increases the stability of the conjugate base.

Example: Arrange the acids HF, HCl, HBr, and HI from strongest to weakest.

First, write out the equilibriums and consider the stability of the conjugate bases.

Figure 4.11. Determining relative acidity by considering the stability of the conjugate bases for hydrogen halides.

Since the size of atoms increases as you go down the periodic table and anion stability increases with increasing volume, the iodide anion will be the most stable and the fluoride ion will be the least stable.

Now, think about the equilibrium.

Figure 4.12. Equilibrium reaction of generic acid HA.

The weaker the conjugate base, the more the equilibrium is shifted to the right and the stronger the acid (HA). Thus, the order from strongest to weakest acid is

Figure 4.13. pKa’s of hydrogen halide acids examined from strongest to weakest.

The theory is consistent with the pKa values.

The same holds true for column 16 of the periodic table.

4.24.png
Figure 4.14. pKa’s of column 16 acids from strongest to weakest.

4.3.3 What is More Important: Electronegativity or Size?

When comparing atoms in the same column in the periodic table, size has a greater stabilizing effect than electronegativity because the charge is spread over a larger volume. When comparing atoms in the same row electronegativity is more important. It should be noted here that electronegativity has a much stronger effect than size; as seen in the table, the values decrease much more dramatically from methane to HF than from HF to HI.

Figure 4.15. Trends in pKa values – electronegativity vs size.​


Q4.9 - Level 2

Which hydrogen is the most acidic in the following molecule?

question description
A

1)

B

2)

C

3)

Watch the following video to see the explanation to Q4.9.


Q4.10 - Level 2

Order the following compounds from most acidic to least acidic.

A

CH3_3CH3_3

B

CH3_3SH

C

CH3_3OH

Watch the following video to see the explanation to Q4.10.



4.3.4 Inductive Effects

Induction: when electronegative atoms pull electron density toward themselves and away from adjacent atoms through sigma bonds. This phenomenon creates partial charges on molecules.

Figure 4.16. Partial charges of fluoroethane.

What effect do electron withdrawing groups have on the acidity of molecules?

Electronegative atoms and electron withdrawing groups increase the stability of the conjugate base by drawing electron density away from the negatively charged region of the molecule.

Trend 1: Because induction is transmitted through sigma bonds, the closer the electronegative atom is to the acidic atom or group, the more acidic the molecule. The electron withdrawing effect helps stabilize and decrease the basicity of the conjugate base and shifts the equilibrium to the products.

Figure 4.17. Effect of induction on acidity demonstrated by the pKa’s of chlorobutanoic acid constitutional isomers.

Trend 2: Because induction is transmitted through sigma bonds, increasing the electronegativity of the atom adjacent to the negatively charged region of the conjugate base decreases its basicity and increases the acidity of the acid.

Figure 4.18. effect of increased electronegativity on pKa.

Trend 3: Because induction is transmitted through sigma bonds, increasing the number of electron withdrawing groups increases the acidity by decreasing the basicity of the conjugate base.

Figure 4.19. The additive effect of multiple electron withdrawing groups on pKa.


Q4.11 - Level 1

Order the following compounds from most acidic to least acidic.

question description
A

4)

B

2)

C

3)

D

1)

Watch the following video to see the explanation to Q4.11.



4.3.4.1 Inductive Effects: The Nitro Group

Another important electron withdrawing group is the nitro group. The nitro group is neutral. However, the nitrogen is positively charged and a negative charge resonates onto the two oxygens. It is a highly electron withdrawing group and increases the stability of the conjugate base by drawing electron density away from the negatively charged region of the molecule.

Figure 4.20. Resonance structures of nitro group.

For example, nitroethanol is significantly more acidic than fluoroethanol.

Figure 4.21. Relative acidity of ethanol, fluorethanol and nitroethanol.

4.3.5 Solvation Effects

In water, the negatively charged conjugate base is also stabilized by solvation. The solvent helps to disperse the charge. If the conjugate base is poorly solvated by water due to unfavorable interaction with the solvent, then the equilibrium shifts to the reactants making the compound less acidic. (Note: the solvent is not always water; therefore, different solvents will affect the equilibrium in different ways. However, for now let us only consider water as the solvent.)

This trend is observed in the following carboxylic acids. As the carboxylic acid functional group becomes more sterically hindered by adjacent alkyl groups, the acidity of the compound decreases.

Figure 4.22. Solvation effect and steric hindrance for simple carboxylic acids.

Example: Formic acid is much more acidic than 1,1-dimethyl propanoic acid.

Figure 4.23. Solvation stabilization effect on formic acid vs. Pivalic acid.


Q4.12 - Level 3

Order the following acids from most acidic to least acidic.

question description
A

4)

B

3)

C

2)

D

1)

Watch the following video to see the explanation to Q4.12.


A notable exception to the trend given above is the acidity of methanol and water (methanol pKa = 15.7). This is due to the entropy and enthalpy of solvation. For more details, consult a resource such as Modern Physical Organic Chemistry by E. V. Anslyn and D. A. Dougherty, University Science Books, Sausalito, California.

4.3.6 Delocalization of Charge (Resonance)

Delocalization of a charge onto different atoms through resonance helps stabilize charged molecules. Thus, resonance stabilization will make a conjugate base more stable, less reactive, and weaker. This will shift the equilibrium to the right and make the acid stronger.

We can see this when we compare the weak acids, methanol and acetic acid.

Figure 4.24. The effect of resonance on pKa, a comparison of methanol and acetic acid.

We can also see this with the resonance of carbanions.

Figure 4.25. The effect of resonance on pKa, a comparison of propane, prop-1-ene, and toluene.

A fun way of understanding this concept is the hot potato analogy. If you hold onto a hot potato by yourself, then you will get burned because the heat coming from the potato is localized on your hand. If you can share the hot potato with other people beside you, then 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 burned. This is like the methoxide ion and the acetate 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 and a strong base, whereas the acetate ion has two oxygen atoms to share the charge with and, thus, is more stable and less basic than the methoxide ion.

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 it has only two resonance structures compared to four for phenol. 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 shifted farther to the right, making acetic acid a stronger acid than phenol. Thinking about the hot potato analogy and the acetate ion, we could say that the oxygens are wearing oven mitts compared with the bare-handed carbon atoms, and are less affected by the heat and better at stabilizing the hot potato (negative charge) than the carbons; therefore, fewer atoms are needed to stabilize the negative charge .

Figure 4.26. The number of resonance contributors vs. the electronegativity of atoms bearing the charge. The pKa of phenol vs. acetic acid.


Q4.13 - Level 2

Which of the hydrogens shown in the following molecule would be most acidic?

question description
A

1)

B

2)

Watch the following video to see the explanation to Q4.13.



4.3.7 Delocalization of a Charge onto the Nitro Group

Nitro groups, inductively, are electron withdrawing groups that can also resonance stabilize the conjugate base. For example, the pKa of phenol decreases when an electron withdrawing group such as chlorine is added (see below). The pKa decreases even more when the strong electron withdrawing nitro group is added to the third position on the benzene. When the nitro group is added to the fourth position, the negative charge of the conjugate base can now resonate onto the nitro group. This extra resonance structure further increases the stability of the conjugate base and decreases the pKa by approximately 3 pKa units more than phenol.

Figure 4.27. Delocalization of a charge on to the Nitro group is stabilized by an additional resonance structure.


4.4 Hybridization

Increasing the s character of an orbital decreases the average energy of the orbital.

Consider the following very weak acids.

4.42.png
Figure 4.28. The effect of S character on pKa, 2-methylpropane vs. ethene vs. ethyne.


As the electrons in the hybridized orbital get closer to the nucleus, the average energy of the electrons decreases and they become more stable and less reactive. Thus, as the hybridization of the conjugate base changes from sp3 to sp2 to sp, the basicity of the conjugate base decreases, and the acidity of the acid increases.

Q4.14 - Level 2

Does the following reaction favor the reactants or products?

question description
A

Reactants

B

Products

Watch the following video to see the explanation to Q4.14.


4.5 Amines are Less Acidic than Alkynes

Alkynes and amines are very weak acids and comparing their acidities it is initially surprising to discover that: amines are less acidic than alkynes.

4.44.png
Figure 4.29. The relative acidity of amines, 2-methylpropane vs. ethane vs. ammonia vs. ethyne.

The relative acidity of amines, 2-methylpropane vs. ethene vs. ammonia vs. ethyne.

For carbon species, acid strength increases as the s character of the carbanion increases (i.e., sp > sp2 > sp3). This is because the negative charge is stabilized to a greater extent by carbons having more s character since the electrons are closer to the nucleus and are lower in energy.

Now, introduce amines into the above trend. Since nitrogen is more electronegative than carbon, the conjugate base should be more stable; indeed, that is what we see. An sp3 nitrogen is a stronger acid than an sp3 carbon. This is also true for an sp3 nitrogen versus an sp2 carbon. However, when we compare an sp3 nitrogen and an sp carbon, the sp hybridization on the carbon has a greater stabilizing effect than the difference in electronegativity of the sp3 nitrogen.

This is not something that you or I could have figured out by just looking at the molecule. This had to be determined experimentally first. In fact, NaNH2 is often used to deprotonate a triple bond.

Figure 4.30. Proton transfer reaction of acetylene with sodium amide. Sodium counter-ion omitted for clarity. Lone pairs are coloured and shown in sigma bonds to help show the flow of electrons in the mechanism.


 Review questions:

Q4.15 - Level 2

Rank the following compounds from strongest to weakest acid.

question description
A

1)

B

3)

C

4)

D

2)


Q4.16 - Level 1

Rank the following carboxylic acids from strongest to weakest acid.

question description
A

1)

B

3)

C

2)


Q4.17 - Level 1

Rank the following carboxylic acids from strongest to weakest acid.

question description
A

3)

B

2)

C

1)


Q4.18 - Level 2

Rank the following compounds from strongest to weakest acid.

question description
A

2)

B

1)

C

3)


Q4.19 - Level 2

Rank the following compounds from strongest to weakest acid.

question description
A

3)

B

4)

C

2)

D

1)


Q4.20 - Level 3

Rank the following compounds from strongest to weakest acid.

question description
A

3)

B

1)

C

2)


Q4.21 - Level 1

Which of the two indicated hydrogens would be more acidic?

question description
A

1)

B

2)


Q4.22 - Level 1

Rank the following compounds from strongest to weakest acid.

question description
A

1)

B

2)

C

3)


Q4.23 - Level 3

Rank the following compounds from strongest to weakest acid.

question description
A

2)

B

3)

C

1)

D

4)