Textbook: Wade 7th Ed. (2009)

Chapter 20: Carboxylic Acids

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Individual Problems

Individual Problems

Problem # 721

Rank each of the four compounds below in order of decreasing acidity (1 = most acidic).

Solution

The general rule is that the pK₁ or a dicarboxylic acid is lower (more acidic) than the pK_a of a regular carboxylic acid, and the pK₂ is a little higher (less acidic).

Why is this? Because carboxylic acids are electron withdrawing groups (EWG). They stabilize the conjugate base of an acid, making the acid more acidic. So Both D and A should be more acidic than C. Because A has its carboxylic acid groups closer together, the EWG effect is stronger, and it will be the most acidic.

B will be the least acidic of the group. Why? Because a carboxylate group has a negative charge, which is electron donating and destablizing to the conjugate base (is a molecule is already -1, it doesn't help to become -2.)

So the overall order is:

(strongest acid) A > D > C > B (weakest acid)

Problem Details

MendelSet practice problem # 721 submitted by Matt on July 24, 2011.

Subject: Organic Chemistry

Subject Topic(s): Carboxylic Acids (Inductive Effects, Esterification)

Question Type: Concept (explain why this is so..)

Keyword(s): acid strength, carboxylic acids

Problem # 724

Use curved arrows to show the formation of the tetrahedral intermediate of a Fischer esterification reaction (shown below). There are three steps in total.

Solution

This is a good mechanism to know (nucleophilic acyl substitution). If you don't understand it, see problems 706 and 708.

Problem Details

MendelSet practice problem # 724 submitted by Matt on July 24, 2011.

Subject: Organic Chemistry

Subject Topic(s): Carboxylic Acids (Inductive Effects, Esterification), Nucleophilic Acyl Substitution (Carboxylic Acid Derivatives)

Question Type: Mechanism (show a mechanism using curved arrows..)

Keyword(s): nucleophilic acyl substitution, Fischer esterification, carboxylic acids

Problem # 708

The overall mechanism for Fischer esterification is shown below. This isn't a real mechanism, just an outline.

Methanol (the nucleophile) attacks the carbonyl carbon, forming a tetrahedral intermediate, which then loses a water to reform the carbonyl. This mechanism is called nucleophilic acyl substitution.

Use curved arrows to draw a full mechanism for this reaction. I've included structures for you to use as a guide.

This reaction takes place under acidic conditions, so the mechanism you draw will be similar to those in problem 706.

Solution

Acidic mechanisms only appear complicated because they contain several proton transfer steps.

Nucleophilic acyl substitution mechanisms have only three real steps- the "up, down, and kick."

First, the nucleophile attacks the carbonyl carbon, forming a tetrahedral intermediate (the "up").

Then the carbonyl reforms (the "down") and a leaving group leaves (the "kick").

Problem Details

MendelSet practice problem # 708 submitted by Matt on July 22, 2011.

Subject: Organic Chemistry

Subject Topic(s): Carboxylic Acids (Inductive Effects, Esterification), Nucleophilic Acyl Substitution (Carboxylic Acid Derivatives)

Question Type: Mechanism (show a mechanism using curved arrows..)

Keyword(s): nucleophilic acyl substitution, Fischer esterification, up-down-kick carbonyl mechanism

Problem # 717

Rank each of the eight compounds A through H below in order of decreasing acidity (1 = most acidic).

Don't get intimidated! What are the differences between these compounds?

Consider electron withdrawing groups (EWG), resonance, hybridization, and the functional group of the acidic proton.

Solution

Carboxylic acids are much more acidic than alcohols (because of resonance in their conjugate bases), so A and D are the least acidic of the group. The conjugate base of D has resonance, so D is more acidic than A:

(everything) > D > A

Electronegative atoms increase acid strength (by stabilizing the conjugate base), and sp² carbons are more electronegative than sp³ carbons (due to higher s-character). So B (all sp³ carbons) will be the least acidic of the carboxylic acids, and H will be more acidic than B (two sp² carbons):

(everything) > H > B > D > A

Of the remaining 4 benzoic acids, C has the most EWG (two nitros) and so is the most acidic. F has no EWG and so is the least acidic. Both E and G have only one EWG and so will be similar in acidity, with E slightly more acidic than G because the EWG affect is strongest in the ortho position (closer to the action). So the overal order is:

(strongest acid) C > E > G > F > H > B > D > A (weakest acid)

Problem Details

MendelSet practice problem # 717 submitted by Matt on July 23, 2011.

Subject: Organic Chemistry

Subject Topic(s): Carboxylic Acids (Inductive Effects, Esterification)

Question Type: Concept (explain why this is so..)

Keyword(s): acid strength, carboxylic acids

Problem # 720

Base your answers to the three problems below on your knowledge of electron donating groups and electron withdrawing groups (EDG and EWG).

a) Based on the pK_a's listed below, Is formic acid more or less acidic than acetic acid? Propose an explanation why.

b) If acetic acid were added to a pH = 4.7 buffer solution, what percentage of it would be in its acetate (conjugate base) form?

c) Methoxy (-OCH₃) is usually considered an EDG. But based on the pK_a of methoxy acetic acid, do you think this is always the case? Explain.

Solution

a) The pK_a of formic acid is less than that of acetic acid, so formic acid is the stronger acid.

The pK_a difference is about ~1, and pK_a is logorithmic, so formic acid is ~10x as acidic as acetic acid.

Why? Because formate (formic acid's conjugate base) is more stable than acetate (acetic acid's conjugte base).

Why? Probably because of EDG. Alkyl groups such as methyl as weakly electron donating. Carboxylate is already negatively charged, and adding an EDG (the methyl group in acetate) makes it less stable, so acetate is a strong base, and acetic acid is the weaker acid. Formate has an hydrogen, which is neither EDG nor EWG, so doesn't have this problem.

b) The Henderson–Hasselbalch equal is pH = pK_a + log (A^-/HA). In this case, A^- is ^-OAc (acetate) and HA is HOAc (acetic acid). So plugging in the numbers:

pH = pK_a + log (^-OAc/HOAc)

4.7 = 4.7 + log (^-OAc/HOAc)

0 = log (^-OAc/HOAc),

so (^-OAc/HOAc) =1, and ^-OAc = HOAc.

So the ratio of acid to conjugate base (acetic acid to acetate) would be about 1:1.

The math above illustrates a rule you might have learned in general chemistry- when pH = pK_a, the concentration of acid equals the concentration of conjugate base.

c) The pKa of methoxy acetic acid (~3.6) is less than that of formic acid (~3.8), which means methoxy acetic acid is the strong acid, so methoxy acetate must be more stable than formate. This implies that methoxy must be somewhat electron withdrawing (an EWG).

This might be suprising because during the aromatic reactions chapter(s) (EAS) methoxy is considered an EDG. Oxygen is electron donating in terms of resonance. But remember that oxygen is also very electronegative. So in this case, it seems that the inductive EWG effect (electronegativity) outweighs the EDG effect.

Problem Details

MendelSet practice problem # 720 submitted by Matt on July 24, 2011.

Subject: Organic Chemistry

Subject Topic(s): Carboxylic Acids (Inductive Effects, Esterification)

Question Type: Concept (explain why this is so..)

Keyword(s): acid strength, carboxylic acids

Problem # 725

A chemist carried out a Fischer esterification using methanol that was isotopically labeled with ¹⁸O (indicated with an asterisk).

Which one of the esters below (A-D) was formed?

Solution

To answer this problem, you must be familiar with the nucleophilic acyl substitution mechanism.

In this mechanism, the nucleophile (methanol) becomes the -OCH₃ group in the ester.

At least 80% of second semester organic chemistry is two mechnanisms: nucleophilic acyl addition and nucleophilic acyl substitution. It's worth your time to become familiar with these mechanisms. See problems 705 (basic conditions), 706 (acidic conditions), 707, and 708.

Problem Details

MendelSet practice problem # 725 submitted by Matt on July 24, 2011.

Subject: Organic Chemistry

Subject Topic(s): Carboxylic Acids (Inductive Effects, Esterification), Nucleophilic Acyl Substitution (Carboxylic Acid Derivatives)

Question Type: Mechanism (show a mechanism using curved arrows..)

Keyword(s): nucleophilic acyl substitution, Fischer esterification, carboxylic acids

Problem # 723

Show how to prepare each compound from vinyl benzene.

How would you prepare the methyl ester of each compound?

Solution

Vinyl benzene is a common starting material and intermediate in second semester organic chemistry synthesis problems, so you should be familiar how to make vinyl benzene from benzene (see problem 722).

The trick to many synthesis problem is to realize how many carbons need to be added, as that will determine which route you use.

a) Adding zero carbons: alcohol then oxidation

All you need is an alcohol added to the last carbon, which you can oxidize to a carboxylic acid. So from the alkene do an anti-Markovnikov addition of H₂O via hydroboration, followed by oxidation using Jones reagent (aqueous chromium, Cr⁶⁺ in acid).

b) Adding one carbon: nitrile (cyanide, CN)

When you add one carbon that screams "cyanide!" The most common way to add CN is using NaCN with an alkyl halide in an S_N2 reaction.

Because you want to add the extra carbon (CN) to the end of the chain, you want to do an anti-Markovnikov addition using HBr and perioxides. Then add NaCN to form the nitrile.

To convert a nitrile to a carboxylic acid, hydrolyze under acidic conditions (add water with acid, H₃O⁺).

c) Adding two carbons: epoxide

When you need to add two carbons, ethylene oxide is the way to go. But ethylene oxide is an electrophile, so first you must transform vinyl benzene into a nucleophile, specifically, a Grignard reagent.

We want the Grignard part (MgBr) on the end, so do an anti-Markovnikov addition with HBr/peroxides followed by Mg⁰/ether. Then add the epoxide which adds two carbons with an alcohol on the end. Finally, oxidize using Jones reagent.