1. Explain two reasons why conjugated systems are more stable than isolated systems.
Conjugated systems have more stable, stronger single bonds. See classnotes for the two
explanations of hybridization and mmolecular orbitals. Hybridization, sp2-sp2 overlap, forms
stronger sigma bonds and the lowest energy, most stable molecular orbital gives the “single
bonds” double bond character.
2. Explain why the π electrons are able to travel throughout the entire conjugated system.
The lowest energy, most stable molecular orbital has complete p-orbital overlap allowing the π
electrons to move completely around the ring.
3. Which of the following compounds has a higher λmax?
O
O
The first compound - the more conjugated the system, the higher the wavelength for maximum
absorption in the UV region.
4. Of the two isomeric hexadienols, 5-methyl-2, 4-hexadien-1-ol and 2-methyl-3, 5hexadien-2-ol, one has a UV maximum absorption peak at 223 nm and the other at 236 nm.
Draw the structures of each and assign the absorption bands to the correct compounds.
OH
OH
The first molecule has the higher wavelength, 236 nm, because it is more substituted with more
alkyl groups.
5. Draw the possible products of the 1,2- and 1,4-addition of HCl to the following:
Cl
HCl
Cl
Label the kinetic and thermodynamic product.
The first is the kinetic (1,2 addn) and the second is the thermodynamic (1,4 addn).
Explain why the kinetic product is easier to form.
Forms through the most stable carbocation intermediate.
Explain why the thermodynamic product is more stable.
Formed because it is the most stable alkene product.
6. What is the one requirement for the diene of a Diels-Alder reaction?
Must be able to become the s-cis conformation.
7. Explain how the monobromination experiment of benzene assisted in the assignment of
the structure of benzene, to which the molecular formula of C6H6 had been assigned.
What key information did this experiment provide?
A single product formed because all the H atoms in benzene are all perfectly symmetrical.
8. Which of the following species should be aromatic by Huckel’s Rule?
N
N
O
The second and third. Both the first and the fourth have a tetrahedral carbon in their ring
structure. The third compound requires the lone pair on N to participate in the resonance.
9. Why is 4n+2 the perfect number of pi electrons?
Completely fills the bonding molecular orbitals.
10. Two positions are indicated on the substituted naphthalene molecule shown below. On
which position will the electrophilic reaction more readily occur? Why?
OH
A
B
Position B is the more nucleophilic position due to the relative relationship to the hydroxyl group
on the aromatic ring system. There are six resonance forms for this compound, five of which
have negative charges in the ring system, including one where the negative charge is at position
B. Be sure you can draw each of them.
11. Explain why the first molecule shown below has a larger dipole moment than the
second molecule.
O
O
Ph
Ph
Ph
Ph
The first molecule is aromatic, if the carbonyl pi electron density shifts entirely towards the
oxygen, leaving a positive charge on the ring system. With the shift and formation of the anion
and cation, the molecule becomes more polar.
12. Explain why a nitro group is a meta-director?
The resonance forms for nitrobenzene place full positive charges on the ortho and para positions.
This leaves the meta positions more electron-rich to react with E+. Be sure you can draw the
other resonance forms:
O
N
O
O
N
O
two more forms
13. Why can pyridine be used as a base but not pyrrole?
Pyridine does not need the lone pair on N to be an aromatic system but pyrrole will need to use
the lone pair to have enough π electrons to be aromatic. Once the lone pair becomes part of the
π system, the lone pair electrons cannot act to react with protons (acids).
14. p-Nitrophenol has a pKa of 7.14 while the pKa of m-nitrophenol is 8.39. Why?
The placement of the nitro group on the phenol can either help stabilize the anion or not. The
para postion allows the nitro group to participate in further stabilizing the anion of phenol, thus
the para-nitrophenol is more acidic.