Content check — Organic Chemistry, Chapter 20: Carboxylic Acids and Nitriles

Every problem in the chapter beside the book’s answer, a chapter at a time as you scroll.

Chapter 20: Carboxylic Acids and Nitriles

16 problems · 17 figures

20-1 §20.1 source ↗

Give IUPAC names for the following compounds:

(a)

(b)

(c)

(d)

(e)

(f)

Answer (6 parts)

(a)

3-Methylbutanoic acid

(b)

4-Bromopentanoic acid

(c)

2-Ethylpentanoic acid

(d)

cis-4-Hexenoic acid

(e)

2,4-Dimethylpentanenitrile

(f)

cis-1,3-Cyclopentanedicarboxylic acid

20-2 §20.1 source ↗

Draw structures corresponding to the following IUPAC names:

(a)

2,3-Dimethylhexanoic acid

(b)

4-Methylpentanoic acid

(c)

trans-1,2-Cyclobutanedicarboxylic acid

(d)

o-Hydroxybenzoic acid

(e)

(9Z,12Z)-9,12-Octadecadienoic acid

(f)

2-Pentenenitrile

Answer (6 parts)

(a)

(b)

(c)

(d)

(e)

(f)

20-3 §20.2 source ↗

Assume you have a mixture of naphthalene and benzoic acid that you want to separate. How might you take advantage of the acidity of one component in the mixture to effect a separation?

Answer

Dissolve the mixture in ether, extract with aqueous NaOH, separate and acidify the aqueous layer, and extract with ether.

20-4 §20.2 source ↗

The K_a for dichloroacetic acid is 3.32 × 10^–2. Approximately what percentage of the acid is dissociated in a 0.10 M aqueous solution?

Answer

43%

20-5 §20.3 source ↗

Calculate the percentages of dissociated and undissociated forms present in the following solutions:

(a)

0.0010 M glycolic acid (HOCH₂CO₂H; pK_a = 3.83) at pH = 4.50

(b)

0.0020 M propanoic acid (pK_a = 4.87) at pH = 5.30

Answer (2 parts)

(a)

82% dissociation

(b)

73% dissociation

20-6 §20.4 source ↗

Which would you expect to be a stronger acid, the lactic acid found in tired muscles or acetic acid? Explain.

Answer

Lactic acid is stronger because of the inductive effect of the −OH group.

20-7 §20.4 source ↗

Dicarboxylic acids have two dissociation constants, one for the initial dissociation into a monoanion and one for the second dissociation into a dianion. For oxalic acid, HO₂C–CO₂H, the first ionization constant is pK_a1 = 1.2 and the second ionization constant is pK_a2 = 4.2. Why is the second carboxyl group far less acidic than the first?

Answer

The dianion is destabilized by repulsion between charges.

20-8 §20.4 source ↗

The pK_a of p-cyclopropylbenzoic acid is 4.45. Is cyclopropylbenzene likely to be more reactive or less reactive than benzene toward electrophilic bromination? Explain.

Answer

More reactive

20-9 §20.4 source ↗

Rank the following compounds in order of increasing acidity. Don’t look at a table of pK_a data to help with your answer.

(a)

Benzoic acid, p-methylbenzoic acid, p-chlorobenzoic acid

(b)

p-Nitrobenzoic acid, acetic acid, benzoic acid

Answer (2 parts)

(a)

p-Methylbenzoic acid < Benzoic acid < p-Chlorobenzoic acid

(b)

Acetic acid < Benzoic acid < p-Nitrobenzoic acid

20-10 §20.5 source ↗

How would you prepare the following carboxylic acids?

(a)

(CH₃)₃CCO₂H from (CH₃)₃CCl

(b)

CH₃CH₂CH₂CO₂H from CH₃CH₂CH₂Br

Answer (2 parts)

(a)

1. Mg; 2. CO₂; 3. H₃O⁺

(b)

1. Mg; 2. CO₂; 3. H₃O⁺ or 1. NaCN; 2. H₃O⁺

20-11 §20.6 source ↗

How might you prepare 2-phenylethanol from benzyl bromide? More than one step is needed.

Answer

1. NaCN; 2. H₃O⁺; 3. LiAlH₄

20-12 §20.6 source ↗

How might you carry out the following transformation? More than one step is needed.

Answer

1. PBr₃; 2. NaCN; 3. H₃O⁺; 4. LiAlH₄

20-13 §20.7 source ↗

How would you prepare the following carbonyl compounds from a nitrile?

(a)

(b)

Answer (2 parts)

(a)

Propanenitrile + CH₃CH₂MgBr, then H₃O⁺

(b)

p-Chlorobenzonitrile + CH₃MgBr, then H₃O⁺

20-14 §20.7 source ↗

How would you prepare 1-phenyl-2-butanone, C₆H₅CH₂COCH₂CH₃, from benzyl bromide, C₆H₅CH₂Br? More than one step is needed.

Answer

1. NaCN; 2. CH₃CH₂MgBr, then H₃O⁺

20-15 §20.8 source ↗

Cyclopentanecarboxylic acid and 4-hydroxycyclohexanone have the same formula (C₆H₁₀O₂), and both contain an –OH and a C═O group. How could you distinguish between them using IR spectroscopy?

Answer

A carboxylic acid has a very broad − OH absorption at 2500–3300 cm⁻¹.

20-16 §20.8 source ↗

How could you distinguish between the isomers cyclopentanecarboxylic acid and 4-hydroxycyclohexanone by ¹H and ¹³C NMR spectroscopy? (See Problem 20-15.)

Answer

4-Hydroxycyclohexanone: H–C–O absorption near 4 δ in the ¹H spectrum and C═O absorption near 210 δ in the ¹³C spectrum. Cyclopentanecarboxylic acid: –CO₂H absorption near 12 δ in the ¹H spectrum and –CO₂H absorption near 170 δ in the ¹³C spectrum.

Loading chapter 21…

Problems and answers are adapted from Organic Chemistry: A Tenth Edition by John McMurry, OpenStax, and are used under the CC BY-NC-SA 4.0 licence. Access for free at https://openstax.org/books/organic-chemistry/pages/1-why-this-chapter. Illustrations: Copyright Rice University, OpenStax, under CC BY-NC-SA 4.0 license. This adaptation is released under the same licence, is non-commercial, and is neither affiliated with nor endorsed by OpenStax or Rice University.