Textbook: Bruice 6th Ed. (2010)

Chapter 17: Carbonyl Compounds I: Reactions of Carboxylic Acids and Carboxylic Derivatives

Practice Problems (No matching mendel sets were found.)

Short link: mendelset.com/chapters/1116

Printer-friendly version

Individual Problems

Individual Problems

Problem # 669

In your own words, what is the major difference in the addition of a Grignard reagent to an oxidation state III carbonyl (ester/acid chloride) versus an oxidation state II carbonyl? (aldehyde/ketone)

Solution

Oxidation state III carbonyls (esters, acid chlorides) contain a built-in leaving group (such as ^-OR or ^-Cl) and so undergo nucleophilic acyl substitution reactions. The product is another carbonyl.

Oxidation state II carbonyls (aldehydes and ketones) do not contain a built-in leaving group and so undergo nucleophilc acyl addition reactions (instead of substitution). The product is an alcohol.

Because Grignards react with all carbonyls- esters and aldehydes/ketones- esters and acid chlorides will react twice with Grignards: once in a Nuc Acyl Sub mechanism to form a ketone, which will then react with another equivalent of Grignard in a Nuc Acyl Add mechanism to form an alcohol.

Problem Details

MendelSet practice problem # 669 submitted by Matt on July 18, 2011.

Subject: Organic Chemistry

Subject Topic(s): Grignard and Masked Carbanions (alcohols from carbonyls), Nucleophilic Acyl Addition (Aldehydes & Ketones, Imines/Enamines, Acetals/Ketals), Nucleophilic Acyl Substitution (Carboxylic Acid Derivatives)

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

Keyword(s): nucleophilic acyl substitution, nucleophilic acyl addition

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 # 735

Show how each ketone below can be prepared from sodium cyanide and either ethylene or propene.

You may also use methyl Grignard and ethylene oxide.

Solution

Nitriles react with Grignard reagents to form ketones, so the carbonyl carbon on each product must have came from a nitrile.

You also have the restriction that the longest carbon chain building must be three carbons.

a) Working backwards, the left side must have come from a nitrile, so the left side came from a Grignard.

How can we make each starting from propene?

To make the Grignard, prepare propyl bromide from propene via anti-Markovnikov HBr addition (HBr/peroxides), and then add Mg⁰/ether.

To make the nitrile, add sodium cyanide (NaCN) to propyl bromide.

Then add the nitrile and Grignard together, which forms the imine intermediate, which will hydrolyze to the ketone after you add water (H₃O⁺).

b) This problem is harder because you're limited to starting materials with three carbons or less, so we have to make three "cuts" instead of two, like in a).

Working backwards again, prepare propyl Grignard from propene as described in a), and then add ethylene oxide to form 1-pentanol. Form the 6-carbon nitrile by first converting the alcohol to a bromide using PBr₃, and then adding NaCN. Finally, add methyl Grignard followed by H₃O⁺ to form the desired ketone.

Problem Details

MendelSet practice problem # 735 submitted by Matt on July 25, 2011.

Subject: Organic Chemistry

Subject Topic(s): Nitriles

Question Type: Synthesis (show how to prepare X from Y..)

Keyword(s): carboxylic acids, nitriles

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 # 670

Draw out the mechanism for the addition of excess phenyl Grignard to the carbonyl compound below.

Solution

This carbonyl has two leaving groups attached to it- each of those oxygens can take part in a nucleophilic acyl substitution reaction and form a new carbonyl product.

First the Grignard attacks the oxidation state IV carbonyl carbon (4 oxygen bonds, so oxidation state 4). The carbonyl itself will act as a leaving group and form a tetrahedral intermediate. But tetrahedral intermediates don't last if there are any leaving groups attached to the carbon, so the ^-O will "come back down again", kick off an oxygen leaving group, and reform the carbonyl.

Then a second equivalent of Grignard will attack that carbonyl (an ester), and we will do another nucleophilic acyl substitution reaction to form yet another carbonyl.

Finally, the third carbonyl doesn't have any leaving groups built in (it's a ketone), so when the third equivalent of Grignard attacks it, it will do a nucleophilic acyl addition reaction, and the product will be an alcohol.

Notice that as the reaction progresses the oxidation state of the carbonyl carbon (number of oxygen bonds attached to it) goes down form 4 to 3 to 2 and then to 1.

Problem Details

MendelSet practice problem # 670 submitted by Matt on July 18, 2011.

Subject: Organic Chemistry

Subject Topic(s): Grignard and Masked Carbanions (alcohols from carbonyls), Nucleophilic Acyl Addition (Aldehydes & Ketones, Imines/Enamines, Acetals/Ketals), Nucleophilic Acyl Substitution (Carboxylic Acid Derivatives)

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

Keyword(s): nucleophilic acyl substitution, nucleophilic acyl addition

Problem # 726

Show two esters that would yield the two alcohols below after treatment with lithium aluminum hydride.

Solution

Adding LiAlH₄ to an ester yields two alcohols. The carbonyl "side" becomes one alcohol, and the ester "side" becomes the other alcohol.

So there are two esters that would yield these two alcohols after reduction with LiAlH₄.

Problem Details

MendelSet practice problem # 726 submitted by Matt on July 24, 2011.

Subject: Organic Chemistry

Subject Topic(s): Nucleophilic Acyl Substitution (Carboxylic Acid Derivatives)

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

Keyword(s): nucleophilic acyl substitution

Problem # 727

Show how the ester below can be prepared from propene.

Solution

Esters are prepared from a carboyxlic acid and an alcohol (Fischer esterification).

So the problem becomes, "how can you prepare this ester from an alochol and a carboxylic acid, each containing three carbons? (propene)"

The "pair" of carboxylic acid and alcohol to make this ester is propanoic acid and propanol.

From propene, to make propanol, just do an anti-Markovnikov H₂O addition using borane (BH₃ or B₂H₆) then hydrogen peroxide (H₂O₂).

To make propanoic acid, oxidize propanol to propanoic acid using Jones reagent (aqueous chromium).

You can also convert propanoic acid to its acid chloride using SOCl₂, and then add propanol.

Problem Details

MendelSet practice problem # 727 submitted by Matt on July 24, 2011.

Subject: Organic Chemistry

Subject Topic(s): Nucleophilic Acyl Substitution (Carboxylic Acid Derivatives)

Keyword(s): nucleophilic acyl substitution, Fischer esterification

Problem # 728

The acyl group is a protecting group for amines. Amines can be acylated using acetic anhydride, and deacylated with base.

Propose a mechanism for each reaction.

Solution

These reactions are both nucleophilic acyl substitution under basic conditions. See problem 705 for the mechanism in detail.

Problem Details

MendelSet practice problem # 728 submitted by Matt on July 24, 2011.

Subject: Organic Chemistry

Subject Topic(s): Nucleophilic Acyl Substitution (Carboxylic Acid Derivatives), Amines (Basicity, Hoffman Elimination, Gabriel Amine Synthesis)

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

Keyword(s): nucleophilic acyl substitution

Problem # 729

The ester below was dissolved in a solution of water, a small amount of which was isotopically labeled with O-18, denoted with an asterisk.

After a few hours, some isotopically labeled oxygen was found in the ester. Where was it found in the ester? Can you explain why?

Solution

To answer this problem, you must be familiar with the nucleophilic acyl substitution mechanism. (see problem 725)

Water reacts with an ester to form a carboxylic acid. This is what happens here.

If there is only trace amounts of water, it's possible that the water reacts with the ester to form a carboxylic acid, and then goes back to reform the ester. But, in this process, leaves an isotopically labeled oxygen in the carbonyl position.

Problem Details

MendelSet practice problem # 729 submitted by Matt on July 24, 2011.

Subject: Organic Chemistry

Subject Topic(s): Nucleophilic Acyl Substitution (Carboxylic Acid Derivatives)

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

Keyword(s): nucleophilic acyl substitution, Fischer esterification

Problem # 730

N,N-dimethylformamide (DMF) is shown below. Based on its structure, you might expect to see only one -CH₃ signal in the ¹H NMR spectrum. But instead DMF shows two different -CH₃ signals. Explain.

Solution

DMF appears to have two identical methyl groups. Since these six protons are all equivalent, its ¹H NMR should only show one methyl signal (singlet, 6H).

So why is that the real life the ¹H NMR of DMF shows two methyl signals? (two singlets with integration of 3H).

Because DMF is an amide.Recall that the "real" structure of molecule is the a mixture of its resonance forms. DMF doesn't look like either of the two resonance forms below. In real life, its somewhere in between.

For most carboxylic acid derivatives (such as esters), the resonance form is only a minor contributor and so the real "picture" looks very close to the carbonyl Lewis structure.

But for amides, its resonance form is fairly stable (it's common for nitrogen atoms to be positively charged), and so is a major resonance contributor.

In an amide, the bond between the carbonyl carbon and the nitrogen atom has a high degree of double bond character. (This also explains why it's harder to rotate the C-N "single bond" than you would expect from its Lewis structure- it's sort of like a double bond).

Because the C-N "single bond" is closer to a double bond, the two methyls are not equivalent. One methyl is cis, and the other is trans, and so they show two signals in the ¹H NMR.

Problem Details

MendelSet practice problem # 730 submitted by Matt on July 24, 2011.

Subject: Organic Chemistry

Subject Topic(s): Nuclear Magnetic Resonance Spectroscopy (NMR), Nucleophilic Acyl Substitution (Carboxylic Acid Derivatives), Amino Acids, Peptides, and Proteins

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

Keyword(s): proton NMR, amides

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.