BELLARMINE COLLEGE
DEPARTMENT OF CHEMISTRY & PHYSICS
CHEM217 INTERMEDIATE CHEMISTRY II
FIRST MIDTERM EXAM ANSWER KEY
2/7/99
TIME: 50 MINUTES
PART ONE: REACTION PRODUCTS - For any seven (7) of the following reactions provide the major organic product.
(21) NOTE: there is no Q10.
The key to doing these kinds of problems is to correctly identify the substrate and reagents. Once identified, there should be
a limited number of options open to you.
1.
Due to a typographical error, the substrate given was an alcohol rather than the intended carboxylic acid. As
shown below, the alcohol is converted to an ether whilst the carboxylic acid is turned into an ester under these
reaction conditions. Interestingly, many of you changed the given alcohol into the carboxylic acid and where able
to get the required ester. A few of you stuck with the alcohol and came up with the correct ether.
(CH3)2CHCO2H
carboxylic acid
(CH3)2CHCH2OH
alcohol
SOCl 2
thionyl
chloride
SOCl 2
thionyl
chloride
(CH3)2CHCOCl
acid chloride
(CH3)2CHCH2Cl
alkyl chloride
CH 3CH 2CH 2OH
(CH3)2CHCOOCH 2CH 2CH 3
alcohol
ester
CH 3CH 2CH 2OH
(CH3)2CHCH2OCH 2CH 2CH 3
alcohol
ether
2.
H
O
shows where new bond will be
O
O
OH
+
H
HO(CH2)4CHO
This is a hydroxyaldehyde
which can be redrawn as
shown to the right.
Aldehydes react with
alcohols to form
hemiacetals. When the
alcohol and the aldehyde
are on the same molecule
the result is a cyclic
hemiacetal.
cyclic hemiacetal
3.
OH
O
O
ester
HO
1) NaOH
O
2) H+
hydrolysis
conditions
alcohol
(phenol)
carboxylic acid
4.
H+
(CH3)3CCH2CH2OH (excess) + C6H5CH2CO2H
alcohol
carboxylic acid
(CH3)3CCH2CH2O2CCH2C6H5
ester
OR
O
OH
HO
Fischer esterification excess alcohol used to
drive equilibrium to
product side.
H+
O
(+ H2O)
O
5.
via
H+
+ HOCH2CH2CH2OH
O
O
diol
O
ketone
hemiacetal OH
O
diols are quite often used to protect
aldehydes and ketones by converting
them to cyclic acetals.
cyclic acetal
HO
alcohol
6.
CH3CH2CH2Cl
+ NaOH
o
1 alkyl chloride
substrate
hydroxide
nucleophile
CH3CH2CH2OH + NaCl
o
1 alcohol
chloride
product
leaving group
This is an example of a SN1 reaction
7.
CH3COCl
+
acid chloride
OR
carboxylic acid
salt
O
O
Cl
8.
CH3CO2-Na+
-O
CH3CO2COCH3
(+ NaCl)
carboxylic acid
anhydride
O
O
O
H2O/H+/
C6H5CN
C6H5CO2H
hydrolysis
conditions
nitrile
(+ NH3)
via C6H5CONH2
carboxylic acid
amide
O
OR
O
N
C
NH2
OH
-NH3
+ H2O
9.
O
O
O
O
+
+ HO
O
HO
O
carboxylic acid
anhydride
alcohol
(phenol)
ester
carboxylic
acid
11.
OH
OCH 3
H+
+ CH3OH
(+ H2O)
OCH 3
hemiacetal
OCH 3
alcohol
acetal
12.
OH
H+
O
(+ CH3OH)
OCH 3
hemiacetal
ketone
alcohol
PART TWO: REACTION MECHANISMS - Provide a reasonable mechanism for either one of the following reactions.
(14)
13.
H
H+
O
O
O
O
O
OH2
+ HO
acetal
OH
H
H
+
O
O
H
H
O
O
H+
O
H
O
H
protonated hemiacetal
O
H+
ketone
14.
O
O
O
H2O
O
O
H+
O
H
H
ester
H
H+
OH
O
O
HO
O
OH
O
alcohol (phenol)
H
protonated tetrahedral intermediates
H
O
OH
H+
H
O
O
OH
OH
carboxylic acid
H
H
PART THREE: LE CHATELIER'S PRINCIPLE - Do each of the following questions. The equilibrium constant for the
Fischer esterification of isopropyl alcohol (2-propanol) with acetic acid (ethanoic acid) is 2.35. (15)
15. Write the chemical equation for this equilibrium.
H+
(CH3)2CHOH + CH3CO2H
2-propanol
(isopropyl
alcohol
(CH3)2CHO2CCH3 + H2O
ethanoic acid
acetic acid
2-propyl ethanoate
isopropyl acetate)
16. If the initial amounts of isopropyl alcohol and acetic acid used are 1.0 mol, how much isopropyl acetate (in mol) is
produced at equilibrium?
K = 2.35 = [ester][water]/[acid][alcohol] = x2/(1.0 - x)(1.0 - x)
=> 1.35x2 - 4.7x +2.35 = 0
and x = 2.88 or 0.61,
(2.88 makes no sense)
So, answer is 0.61 mol isopropyl acetate produced
17. If the amount of isopropyl alcohol used is increased 20-fold (no increase in amount of acetic acid used) how much ester
will be produced at equilibrium?
K = 2.35 = [ester][water]/[acid][alcohol] = x2/(1.0 - x)(20.0 - x)
 1.35x2 - 49.35x + 47 = 0, therefore x = 35.6 or 0.98, (35.6 makes no sense)
So, answer is 0.98 mol isopropyl acetate produced