Chapter 15 HOMEWORK—Benzene and Aromaticity
SHORT ANSWER
Drawing Instructions: Draw structures corresponding to the given names.
1.
Draw: m-fluoronitrobenzene
ANS:
2.
Draw: p-bromoaniline
ANS:
3.
Draw: o-chlorophenol
ANS:
4.
Draw: 3,5-dimethylbenzoic acid
ANS:
1
Benzene and Aromaticity
5.
Draw: p-chlorobenzaldehyde
ANS:
6.
Draw: 1-phenyl-3-methylpentane
ANS:
IUPAC Naming Instructions: Provide proper IUPAC names.
7.
Name:
ANS:
m-nitrotoluene or 1-methyl-3-nitrobenzene
8.
Name:
ANS:
(1-methylethyl) benzene or isopropylbenzene or cumene
2
Chapter 15
9.
Name:
ANS:
1-((1S, 3R)-3-methylcyclohexyl)benzene
10.
Name:
ANS:
1-methyl-2,4-benzenediamine or 2,4-diaminotoluene
11.
Name:
ANS:
(E)-2,3-diphenyl-2-hexene
Exhibit 15-1
For each molecule below, predict whether the molecule would be expected to show aromatic
character or not. Explain your answer in each case.
12.
3
Benzene and Aromaticity
ANS:
To meet the criterion of a cyclic, planar, conjugated molecule, the oxygen and the nitrogen bearing
the hydrogen must both be sp2-hybridized. If this occurs, both heteroatoms contribute two
electrons each to the pi system. This results in a total of eight pi electrons, which is not aromatic
according to the Hückel (4n + 2) Rule.
13.
ANS:
The carbocation is sp2-hybridized, so if the oxygen is also sp2-hybridized then each atom in the
cycle has a conjugated p-orbital. Each double bond contributes two pi electrons and the oxygen
contributes two pi electrons, for a total of six pi electrons in the conjugated system. Therefore, this
species is predicted to show aromatic character.
14.
ANS:
This compound, [16]annulene, appears to be a cyclic, planar, conjugated molecule with 16 pi
electrons. However, 16 is not a Hückel number, so the compound is probably not planar, and is not
aromatic.
15.
ANS:
This compound has a pyridine-like nitrogen, which is sp2-hybridized and has its lone pair of
electrons in the plane of the ring. If oxygen is also sp2-hybridized, then it contributes two pi
electrons to the cyclic, conjugated pi system for a total of six π-electrons. Therefore, this molecule
is predicted to show aromatic character.
4
Chapter 15
Exhibit 15-2
Answer the following question(s) concerning sulfathiazole, below.
16.
Refer to Exhibit 15-2. What is the hybridization of the nitrogen atom in sulfathiazole?
ANS:
The hybridization of the nitrogen atom is sp2 in sulfathiazole.
17.
Refer to Exhibit 15-2. Assuming that the sulfur atom is sp2-hybridized, how many πelectrons are there in the sulfathiazole ring?
ANS:
There are six π-electrons in the sulfathiazole ring.
18.
Refer to Exhibit 15-2. What reactivity do you predict for sulfathiazole?
ANS:
Sulfathiazole is predicted to have aromatic reactivity since it is cyclic, planar, conjugated and has
six π-electrons.
PTS: 1
Exhibit 15-3
Refer to the data below to answer the following question(s).
The 1H NMR spectrum of [14]annulene at −60°C shows two signals, one at 0 ppm and one at 7.6
ppm, with an area ratio of 5:2.
19.
Refer to Exhibit 15-3. Is [14]annulene aromatic?
ANS:
[14]annulene contains a cyclic, conjugated pi system with 14 pi electrons (14 = 4n + 2; n = 3). It is
difficult to predict, however, if the molecule is planar without making a model. The 1H NMR
signal at 7.6 ppm is consistent with the presence of a ring current, which is characteristic of
aromatic molecules. Thus, [14]annulene is aromatic.
5
Benzene and Aromaticity
20.
Refer to Exhibit 15-3. Draw in the protons on the [14]annulene skeleton provided below
which are responsible for the 1H NMR signal at 0 ppm.
ANS:
21.
Refer to Exhibit 15-3. Explain why these protons have absorption at such high fields.
ANS:
When an aromatic ring is oriented perpendicularly to a strong magnetic field the induced ring
current deshields protons outside the ring and shields protons inside the ring. Thus, the protons
inside the ring absorb at a field higher than normal (0 ppm versus 4.5−6.5 ppm).
Exhibit 15-4
Consider the data below to answer the following question(s).
6
Chapter 15
22.
Refer to Exhibit 15-4. Conjugate bases QB and ZB are both resonance stabilized. Draw the
indicated number of resonance forms for QB and ZB.
ANS:
23.
Refer to Exhibit 15-4. Which of the compounds above, Q or Z, would you predict to be most
acidic? Explain your answer.
ANS:
The conjugate bases of Compounds Q and Z are both resonance stabilized. Let's examine all
resonance structures for both conjugate bases:
7
Benzene and Aromaticity
For the conjugate base QB the negative charge is delocalized into the ring. This generates a
cyclopentadienyl anion that is aromatic. For the conjugate base ZB, the negative charge is
delocalized by resonance but no aromatic ions result. Therefore, Compound Q should be the most
acidic, since its anion is the most stable.
24.
Most alkyl halides are nonpolar covalent compounds and, therefore, are soluble in nonpolar
solvents and insoluble in water. Cycloheptatrienyl bromide is an unusual alkyl halide in that
it is insoluble in nonpolar solvents, but is readily soluble in water! This behavior is
consistent with cycloheptatrienyl bromide being an ionic compound. Why does
cycloheptatrienyl bromide exist as an ionic compound? Explain by comparing the covalent
structure to the ionic structure.
ANS:
In the covalent structure, the carbon bonded to the bromine is sp3-hybridized, so the ring cannot be
aromatic. In the ionic structure, the carbocation is sp2-hybridized, so the structure is a cyclic,
conjugated, six π-electron system and aromatic.
MULTIPLE CHOICE
25.
Consider the following compound. Atoms other than carbon and hydrogen are labeled.
Which of the following is not valid?
a. parent compound in naming, benzene.
b. 1,4-disubstitution is present.
c. para prefix applicable
d. classified as aromatic
ANS: A
26. How many monosubstituted products would you expect for the bromination of naphthalene?
a. 2
c. 6
b. 4
d. 8
ANS: A
8