Content check — Organic Chemistry, Chapter 15: Benzene and Aromaticity

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

Chapter 15: Benzene and Aromaticity

12 problems · 26 figures

15-1 §15.1 source ↗

Tell whether the following compounds are ortho-, meta-, or para-disubstituted:

(a)

(b)

(c)

Answer (3 parts)

(a)

Meta

(b)

Para

(c)

Ortho

15-2 §15.1 source ↗

Give IUPAC names for the following compounds:

(a)

(b)

(c)

(d)

(e)

(f)

Answer (6 parts)

(a)

m-Bromochlorobenzene

(b)

(3-Methylbutyl)benzene

(c)

p-Bromoaniline

(d)

2,5-Dichlorotoluene

(e)

1-Ethyl-2,4-dinitrobenzene

(f)

1,2,3,5-Tetramethylbenzene

15-3 §15.1 source ↗

Draw structures corresponding to the following IUPAC names:

(a)

p-Bromochlorobenzene

(b)

p-Bromotoluene

(c)

m-Chloroaniline

(d)

1-Chloro-3,5-dimethylbenzene

Answer (4 parts)

(a)

(b)

(c)

(d)

15-4 §15.2 source ↗

Pyridine is a flat, hexagonal molecule with bond angles of 120°. It undergoes substitution rather than addition and generally behaves like benzene. Draw a picture of the π orbitals of pyridine to explain its properties. Check your answer by looking ahead to Section 15.5.

Answer

Pyridine has an aromatic sextet of electrons.

15-5 §15.3 source ↗

To be aromatic, a molecule must have 4n + 2 π electrons and must have a planar, monocyclic system of conjugation. Cyclodecapentaene fulfills one of these criteria but not the other and has resisted all attempts at synthesis. Explain.

Answer

Cyclodecapentaene is not flat because of steric interactions.

15-6 §15.4 source ↗

Draw the five resonance structures of the cyclopentadienyl anion. Are all carbon–carbon bonds equivalent? How many absorption lines would you expect to see in the ¹H NMR and ¹³C NMR spectra of the anion?

Answer

All C – C bonds are equivalent; one resonance line in both ¹H and ¹³C NMR spectra.

15-7 §15.4 source ↗

Cyclooctatetraene readily reacts with potassium metal to form the stable cyclooctatetraene dianion, C₈H₈^2–. Why do you suppose this reaction occurs so easily? What geometry do you expect for the cyclooctatetraene dianion?

Answer

The cyclooctatetraenyl dianion is aromatic (ten π electrons) and flat.

15-8 §15.4 source ↗

The relative energy levels of the five π molecular orbitals of the cyclopentadienyl system are similar to those in benzene. That is, there is a single lowest-energy MO, above which the orbitals come in degenerate pairs. Draw a diagram like that in Figure 15.6, and tell which of the five orbitals are occupied in the cation, radical, and anion.

Six p atomic orbitals each with an unpaired electron become three bonding molecular orbitals, each having two electrons, and three antibonding molecular orbitals which are empty. — Figure 15.6 in the book ↗

Answer

15-9 §15.5 source ↗

Draw an orbital picture of furan to show how the molecule is aromatic.

Answer

15-10 §15.5 source ↗

Thiamin, or vitamin B₁, contains a positively charged five-membered nitrogen–sulfur heterocycle called a thiazolium ring. Explain why the thiazolium ring is aromatic.

Answer

The thiazolium ring has six π electrons.

15-11 §15.6 source ↗

Azulene, a beautiful blue hydrocarbon, is an isomer of naphthalene. Is azulene aromatic? Draw a second resonance form of azulene in addition to that shown.

Answer

Yes, it's aromatic.

15-12 §15.6 source ↗

How many electrons does each of the four nitrogen atoms in purine contribute to the aromatic π system?

Answer

The three nitrogens in double bonds each contribute one; the remaining nitrogen contributes two.

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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.