216 S09 Exam #1 - Page 2.
Name ___key_____________________
I. (9 points) Silica gel thin layer chromatography (TLC) is often used to monitor the progress of an organic
reaction. For the following ester hydrolysis reaction, a solvent system is selected to give the starting
material an Rf value of about 0.5. (1) Provide in the box below the structure of the expected product 2. (2)
Fill in the spots that would be expected when the reaction is 50% compete and 100% complete, each after
acidic work-up. Make sure to assign each spot you draw to the corresponding compound number (1, 2, or
3). Consider only the compounds that can be visualized as a spot on TLC upon exposure of each solventdried plate to a 254 nm-UV lamp.
O
O
H2SO4 (cat)
H2O
Δ (heat)
H
H
H
H
O
O
H
O
H
H
HO
2
1
+ HO
3
H
3
100% completion
50% completion
solvent
front
1
On each TLC plate, a student has placed
a sample of the starting material (1) as a
reference on the left of the plate, a spot
of the reaction mixture on the right, and
a co-spot in the center of each. Cospotting is where some of 1 and some of
the reaction mixture are spotted together
in order to make better comparisons.
2
3
1 co-spot
1 co-spot
reaction mixture
reaction mixture
II. (8 points) What would be the expected order of Rf values for each of the following sets of organic
compounds when subjected to TLC analysis on silica gel? For each set, circle the compound that is expected
to give the higher Rf value. Assume dichloromethane is used as the developing solvent for the TLC analysis.
H3C
OH
(1)
O
OCH3
OCH3
A
CH3
(3)
CH3
NH2
(2)
B
(4)
O
OH
A
B
OH
OH
A
B
O
A
B
216 S09 Exam #1 - Page 3
Name_________key_______________
III. (15 points) N-Benzylaniline (6) can be prepared by a reaction of benzyl chloride (4) and aniline (5)
in the presence of aqueous sodium bicarbonate (NaHCO3).
Cl
4
NH2
+
+
NaHCO3
H 2O
90 °C
+ NaCl + H2O + CO2
N
H
6
5
The procedure calls for 372 g of aniline (5), 127 g of benzyl chloride (4), 105 g of NaHCO3, and 100 mL
of water. A typical reaction produces 160 g of pure benzylaniline (6). Use the following atomic
weights: H, 1; C, 12; N, 14; Na, 23; Cl, 35.5.
(1) (2 points) What is the molecular weight of benzylaniline (6)?
Mol formula of 6 is C13H13N . So, the molecular weight is: 183 (g/mol).
(2) (3 points) What is the molar ratio of aniline (5): benzyl chloride (4) used in this reaction? Show your
work.
4: 127 / 126.5 = 1.004 mol
5: 372 / 93 = 4.0
Therefore, the molar ration of 5 : 4 is 4 / 1.004 = 4
(3) (2 points) What is the liming reagent in this reaction?
Benzyl chloride (4): 1.004 mol; aniline (5): 4.0 mol; NaHCO3: 1.25 mol
Therefore, the limiting reagent is benzyl chloride (4).
(4) (3 points) What is the theoretical yield of benzylaniline (6)? Show your work.
1 mol x 183 g / mol = 183 g
Theoretical yield is 183 g.
(5) (3 points) What is the percentage yield of benzylaniline (6) if 160 g is produced? Show your work.
(160 g / 183 g) x 100 = 87.4
Yield: 87.4%
(6) (2 points) What volume of benzyl chloride (density 1.10 g/mL) is used in this reaction? Show your
work.
127 g / (1.10 g/mL) = 115.5 mL
Volume of benzyl chloride used: 115.5 mL
216 S09 Exam #1 - Page 4
Name ______key__________________
IV. (8 points) The pKa values of the conjugate acids of aniline and 4-cyanoaniline are determined to be,
respectively, 4.60 and 1.57. Explain in the box below why the latter is considerably more acidic than the
former using pertinent resonance structures of both the conjugate acid of 4-cyanoaniline and 4cyanoaniline itself.
N C
NH2
4-cyanoaniline
4-cyanoaniline
For the conjugate acid of 4-cyanoaniline:
N C
NH3
N C
For 4-cyanoaniline:
NH3
N C
NH2
N C
NH2
The NH2 lone-pair electrons
are delocalized into the CN group,
i.e., more stable than C6H5NH2.
This makes the NH2 nitrogen less
basic. Therefore, the conjugate
acid of 4-cyanoaniline is more
acidic than that of aniline.
charge-charge
repulsion, making the
conjugate acid less stable than
C6H5NH3+. Therefore, this
conjugate acid is more acidic
than C6H5NH3+
V. (10 points) For the following reaction, provide in the boxes below the structures of the expected
products and a step-by-step mechanism through the use of the curved arrow convention.
O
NH3
O
+
benzoic anhydride
O
H
N
O
CH2Cl2
room
temperature
+
HO
O
2
2
Reaction mechanism:
O
NH2
O
NH2
O
intermediate str: 1 pt each
each set of mechanistic
arrows: 1 pt each
O
O
O
O
O
H
H
N
O
O
H
N
+
O
OH
6
216 S09 Exam #1 – Page 5
Name ______key_________________
VI. (10 points) Treatment of aldehyde ester 7 with 1-mole equivalent of NaBH4 in tetrahydrofurandimethyl sulfoxide at room temperature results in, upon aqueous acidic work up, the formation of
compound 8 (C9H10O3). Compound 8 shows strong IR peaks at 3400 (broad) and 1735 cm-1. Draw in
the box below the structure of compound 8 and provide the step-by-step mechanism for its formation
using the curved arrow convention.
O
O
1. NaBH4
O
2.
(protonation)
H
7
O
OH
H3O+
O
8
3
Mechanism for the formation of 8:
O
O
O
O
7
H
H
H B
Na
intermediate str: 1 pt each
each set of mechanistic
arrows: 1 pt each
O
O
H
H H
H
H
H
O
O
H
H
O
O
O
O
O
OH
O
O
8
H
7
VII. (8 points) For each of the following pairs of compounds, match the expected IR wavenumbers for
the C=O bond stretching vibration absorption to the wavenumbers given.
(1) 1700 or 1660 cm-1
(2) 1685 or 1655 cm-1
O
O
SPh
1700 cm-1
N(CH3)2
1660 cm-1
(3) 1760 or 1710 cm-1
O
O
1685 cm-1
N(CH3)2
1655 cm-1
(4) 1715 or 1695 cm-1
O
O
OCH3
O
O
O
1710 cm-1
1760 cm-1
1695 cm-1
1715 cm-1
216 S09 Exam #1 – Page 8
Name ___key_____________________
VIII. (continued)
CH3
A
H
HO
H3C
H
N
CH3
O
B
OH
I
HO
OCH3
C
OH
NH2
J
NH2
K
O
D
OH
H
N
CH3
E
O
L
OCH3
O
CH3
F
OH
M
N
H
O
O
G
CH3
H3C
O
OH
N
O