CH341/CH540, Exam 4 Key, Fall 2009
This exam consists of several parts of unequal difficulty. Look over the whole exam right now, and start
on something you know well. If you don’t know something, skip it, and come back to it later. Use your
Keep Your Eyes On Your Own Paper!
Page 2. _______________ (24) Page 5. _______________ (14)
Page 3. _______________ (24) Page 6. _______________ (12)
Page 4. _______________ (16) Page 7. _______________ (10+)
Total _______________ (100)
A Partial Periodic Table of the Elements
1A IIA IIIB IVB VB VIB VII VIII IB IIB IIIA IVA VA VIA VII VIII
B A A
3 4 5 6 7 8 9 10
Li Be B C N O F Ne
6.94 9.01 10.8 12.0 14.0 16.0 19.0 20.2
11 12 13 14 15 16 17 18
Na Mg Al Si P S Cl Ar
23.0 24.3 27.0 28.1 31.0 32.1 35.5 40.0
19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36
K Ca Ga Ge As Se Br Kr
39.1 40.1 69.7 72.6 74.9 79.0 79.9 83.8
37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54
1. Give the major organic product of each of the following reactions. If no reaction will occur, write
“No Reaction”. If a mixture of ortho and para products would form, draw only the para.
a. NO2 HO 3S NO2
H 3C H3C
CH Br2 CH
CH3 FeBr3 CH3
c. O H3C O
H3C O Cl C
1. KMnO4, H2O,
e. O CH3
H 3C CH3
2. Fill in the needed reagents and conditions over the arrows. An arrow may require more than one step.
O H2N N
CH3 H+ (cat.) CH3
1. NaCN 2. H2SO4 or C N
3. Give IUPAC names for the following structures.
3. Give reasonable arrow-pushing mechanisms for the following reactions. Show all important
+ H Cl
Cl Cl AlCl3 Cl
H2N H2N H2N
Cl AlCl 3
+ H Cl
O H+ (cat.)
+ H OCH3 OCH3
O O +
O H O H
4. Multiple Choice: Choose the best answer for each of the following questions.
1. In electrophilic substitution reactions, a -Br group is:
A. Meta directing and activating
D B. Meta directing and deactivating
_________ C. Ortho-para directing and activating
D. Ortho-para directing and deactivating
2. The compound to the right is (was): NO2
A. A crown ether
B B. An explosive invented by Alfred Nobel
_________ C. An antimicrobial compound synthesized by Dr. Hathaway O2N NO2
D. An oxidizing agent
3. Which reaction is reversible?
A. Nitration of benzene
C B. Bromination of benzene
_________ C. Acetal formation
D. Friedel-Crafts Acylation
4. Which benzene ring is the least activated towards electrophilic substitution?
A. B. C. D.
_________ F O H3C N
H3C O -
5. Which TWO statements are true?
A. All ortho-para directors are activating groups.
B. All activating groups are ortho-para directing.
B, E C. All ortho-para directors are deactivating groups.
_________ D. All deactivating groups are ortho-para directors.
E. All meta directors are deactivating groups.
F. All deactivating groups are meta-directors.
6. Which statement is true about benzene?
A. It is less stable than a hypothetical cyclohexatriene.
D B. It is an equilibrium mixture of two similar structures.
_________ C. It is the smallest aromatic compound.
D. It is shaped like a regular hexagon.
Draw the O
acetaldehyde. H3C C H
5. Show how you would synthesize the following products from the indicated starting materials, plus
any other reagents.
Product Starting Material
H3C C CH2CH3 H3C C H
CrO3, H2SO4, H2O
H3C C CH2CH3 H3C CH CH2CH3
H3C C Cl
6. What are the four “rules” for an aromatic compound?
2. Planar (flat) or nearly so
3. Every ring atom has a p-orbital
4. 4N + 2 pi-electrons in the system of p-orbitals, (N = 0, 1, 2, 3…)
Classify the following structures as aromatic, anti-aromatic, or non-aromatic. Briefly explain your
reasoning. Don’t merely say, “It follows the rules,” or something similar.
Cyclic, flat, p-orbital on each ring atom, but only Cyclic, flat, p-orbital on each ring atom, and 6
12 electrons in the p-orbitals, so it can’t be electrons in the p-orbitals, so it is aromatic.
aromatic. 12 is a 4N number, so it is anti-
Bonus! If 1,4-di-tert-butylbenzene is treated with 2-chloro-2-methylpropane and AlCl3, and heated, the
major product is 1,3,5-tri-tert-butylbenzene. Explain how this forms, and why.
CH3 CH3 CH3
C Cl C CH3 C
CH3 + H Cl
H3C AlCl3 H3C CH3
C C C
H3C H3C CH3
CH3 CH3 CH3
The t-butyl group adds ortho to one of the t-butyl groups on the ring. Two t-butyl groups next to each
other causes a lot of steric hindrance, which isn’t good.
C H CH3
+ H Cl CH3
H3C CH3 H3C CH3
C C C C
H3C CH3 H3C CH3
CH3 CH3 CH3 CH3
H adds to the ring, and a t-butyl group comes off as a cation.
+ C H
C C H3C CH3
H3C CH3 C C
CH3 CH3 H3C CH3
The t-butyl group adds meta to minimize the steric hindrance.