CHEM 2312

ORGANIC CHEMISTRY II

SECOND HOUR EXAM

Answer Key


July 13, 1998

Dr. Hanson

1. For the following compounds, provide a name where a structure is given or provide a structure where a name is given. IUPAC names are always acceptable, common trivial names are also acceptable. (12 points)

a)


Answer: 1,3-dinitrobenzene


b)


Answer: 2-Bromo-5-fluorotoluene


c) aniline

Answer: C6H5-NH2

d) o-chlorobenzoic acid

Answer: Cl-C6H4-CO2H with the chloro and acid substituents 1,2




2. For the compounds provided, use Huckelís Rule to decide if they are aromatic, antiaromatic, or nonaromatic. Assume that all rings are planar unless clearly shown as non-planar. (15 points)

a)


Answer: antiaromatic

b)


Answer: aromatic

c)


Answer: antiaromatic

d)


Answer: aromatic

e)


Answer: nonaromatic




3. Draw all resonance structures for the ion shown below. (Two different ways of drawing this ion are provided for clarity.) Is this compound aromatic by Huckelís Rule?. (8 points)


Answer: there are five major resonance structures, with the negative charge on each of the five carbons of the ring in turn. The cyclopentadienide anion has 6 electrons and is therefore aromatic by Huckelís rule.



4. Sketch a molecular orbital diagram for benzene. The diagram should include the relative energies of the molecular orbitals and also sketches of the molecular orbitals themselves, indicating the contributions of the individual atomic orbitals to each molecular orbital. (12 points)


Answer: The molecular orbital diagram should look like a hexagon (by Frostís Rule): one lowest energy bonding orbital; two orbitals of equal energy lying higher but still bonding; two orbitals of equal energy lying still higher and antibonding; and a single antibonding orbital at highest energy. The nodal structure of these orbitals is as follows: no nodes in the low energy bonding orbital; one node in the two equal energy bonding orbital (one has a node that is ìverticalî and one a node that is ìhorizontalî); two nodes in the two equal energy antibonding orbitals (one set of nodes looks like a ì+î the other like an ìxî); three nodes in the highest energy antibonding orbital.




5. Complete the following reaction sequences, providing starting materials, products, or reagents as required. (25 points)

a)


Answer: 3-bromobenzonitrile (the CN group is a meta director!!!)

b)


Answer: 4-fluorobenzoic acid

c)


Answer: Reagent 1 = butanoyl chloride and AlCl3; Reagent 2 = Zn and HCl.

d)


Answer: Toluene

e)


Answer: phenol




6. Provide a detailed, step-by-step mechanism for the following reaction. Show all steps! Draw all resonance structures for the key intermediate, to explain the orientational selectivity (or regiochemistry) of the reaction. (16 points)


Answer: This is a standard electrophilic aromatic substitution. In the presence of sulfuric acid, nitric acid forms the nitronium ion (NO2+) by protonation and loss of water. The nitronium ion adds as an electrophile to the anisole to form an arenium ion. Addition is favored at the para position shown here (also at the ortho positions) due to the extra resonance structure that can be drawn which delocalizes the positive charge onto the oxygen of the methoxy group. The arenium ion loses a proton to regenerate an aromatic structure, in this case 4-nitroanisole.



7. Propose syntheses for the following compounds, beginning with benzene and using any necessary organic or inorganic reagents. (12 points)

a)


Answer: The first step is nitration (HNO3/H2SO4); the second is electrophilic chlorination (Cl2/FeCl3) which is oriented meta by the nitro group; the third is reduction of the nitro to an aniline (Sn/HCl); the fourth diazotization (NaNO2/HCl); and finally reaction with KI.



b)


Answer: The first step is Friedel-Crafts alkylation (CH3Cl/AlCl3); the second and last is sulfonation (SO3/H2SO4).