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Chem 121 Winter 2016: Section 03,
Sample Problems
Aromatics
Problems, some slightly modified, from Chemistry - The Central Science, 3rd edition.
28.5 (a) Which of the following structures is aromatic? (b) What rule(s) is/are
broken by the structures you have identified as not aromatic?
(a)
Not aromatic. Compound has 8 -electrons therefore it does not
satisfy the [4n + 2] rule and is not planar.
(b)
Aromatic. Compound is planar and satisfies the [4n + 2] rule (n = 3).
All atoms in the ring have an unhybridised p orbital.
(c)
Not aromatic. The compound has 8 electrons therefore does not
satisfy the [4n + 2] rule. Looks conjugated, but this compound is not
planar.
(d)
Not aromatic. Compound satisfies the [4n + 2] rule but is not planar.
One of the carbons in the ring does not have an unhybridised p
orbital so therefore the molecule is not conjugated.
28.7
(a) Name another compound that has the same empirical formula as
benzene. (b) Briefly explain the difference between an empirical formula and a
molecular formula. (c) Why do different compounds with the same empirical
formula not all have identical chemical properties?
(a) Acetylene (ethyne), like benzene, has the empirical formula CH.
(b) The molecular formula shows the actual number of atoms of each type
in the molecule. The empirical formula shows the lowest whole number
ratio of atoms of each type. The two are sometimes the same, but if they
1
are this does not tell us anything about the two species, other than that the
ratio of the different types of atom is identical in the species.
(c) Different compounds that share the same empirical formula will
generally have different molecular formulae (when they do not, the
compounds are isomers of each other). Since these compounds are
necessarily different materials they are bound to have different properties.
28.9
What are the characteristic hybrid orbitals employed by carbon in (a) an
alkane, (b) an alkene, and (c) an aromatic compound?
(a) sp3 hybrids; (b) sp2 hybrids; (c) sp2 hybrids.
28.14
(a) Use the concept of resonance to explain why all six C-C bonds in
benzene are of equal length. (b) Those bonds are shorter than typical C-C
bonds, but longer than typical C=C bonds; why?
(a) The structure of benzene can be drawn as two different, but equivalent
resonance structures. In one structure half the C-C bonds are single and
half the C=C bonds are double. The same is true in the second structure,
but the positions of the single and double bonds are reversed. The
resonance structure is a blend of the two individual structures, so each
bond is effectively partly single and partly double, and therefore all of the
same length.
(b) Because each bond is an equal blend of single and double bond
character.
28.16
(a) What is the difference between a localised -bond and a delocalised bond? (b) How can you determine from a molecular formula whether a
molecule or ion will exhibit delocalised bonding? (c) Draw the structure of
the ethanoate ion (CH3CO2-).
Is the -bond in this ion localised or
delocalised?
(a) A localised -bond is confined between two atoms (generally, but not
always, carbon atoms). By contrast, a delocalised -bond, which can arise
if a molecule contains alternating single and double bond, is spread out
over the set of atoms that contains the alternating single and double
bonds. The -electrons in the delocalised bond are free to move from one
end of this “conjugated” system to the other end, and are therefore
“delocalised”.
(b) A pattern of alternating single and double bonds is required.
(c) The ethanoate ion is the anion of ethanoic acid. The extra electron in
the –COO- group is delocalised across the two oxygens and the carbon
atom. This delocalisation helps to stabilise the ion.
28.20 Which of the two species given below is the more stable? Explain.
CHCH3
CH2
is more stable than
because the left-hand cation is stabilised by electron donation from the
methyl group.
28.24
We can test for the presence of a double bond in an alkene by adding a
little bromine and checking to see whether the brown colour disappears;
this test does not detect the presence of aromatic molecules. Giving
suitable equations, explain these results.
Whilst an aromatic hydrocarbon such as benzene may be written as
possessing double bonds, the bonds in benzene are actually intermediate in
character between single and double bonds due to delocalisation of the electrons around the ring. Addition across any of these ‘semi-double’ bonds
will remove the delocalisation of -electrons and the additional stability this
delocalisation affords. Hence aromatic compounds are less reactive than
alkenes and do not react with molecular bromine without a catalyst.
28.28
(a) What is the difference between a substitution and an addition reaction?
Which one is commonly observed in alkenes and which in aromatics?
(b) Using condensed structural formulae, write the balanced equation
for the reaction of 2,4-dimethyl-2-pentene with bromine.
(c) Write
a balanced equation for the reaction of chlorine with paradichlorobenzene in the presence of iron (III) chloride as a catalyst.
(a)
An addition reaction is the addition of a reagent to two atoms that
were previously connected by a multiple bond. In a substitution
reaction one atom or a group of atoms replaces another atom or
group. Alkenes typically undergo addition, while aromatic
hydrocarbons usually undergo substitution.