Chemistry 211
Equilibrium Controlled Reactions:
Acid Base Reactions and Electron Energies - 7
Conjugate Acid
CH3
CH2
CH2
CH
CH2
CH2 2 CH3 CH3
CH2:
CH2
CH
CH
CH3
CH3
CH2
CH
:O:
O
CH3 C NH2
H
H
:-
16
H
-
O:
CH3 C O : -
OH
O
H
N
O
:O:
CH3 C OH
O
N
-
O
10
9.6
:O:
: :
~ 27
NH
16
CH3 C NH
: :
: :
:
-
2
~33
-
16
: :
CH
O:
: :
CH2
16
-
OH
~ 37
CH :
19
: :
: :
CH3
CH
-
C CH2:
~43
~ 37
-
CH3
CH3 CH2 O :
:
: :
:
:
NH2
-
:O:
CH3 CH2 OH
40
NH
CH
CH3 C CH3
pKa
Conjugate Base
~ 50
41
CH3 CH2 NH
NH2
CH
-
H
CH3 CH2 NH2
CH
-
:-
H
H
CH
CH2:
CH2:
CH3
CH2
CH
: O:
: :: :
CH2
~50
Conjugate Acid
: : : :
-
CH2:
CH3
CH3
pKa
Conjugate Base
: :
: :
: :
Table E
: :
:
: :
A.
Fall 2008
4.8
B. Exploration:
1. How do the structures below relate to the structures in Table E? Provide specific reactions in Table E that relate to each structure.
S
O
CH3
CH
CH
CH
S
CH2
S
CH3 CH2 C S
CH
CH3
: :
:
: :
:
: :
: :
:
:
: :
Letter a can be related to the structure with the pKa of 33 mainly because it has the same connectivity ( 2 double bonds).
Letter b can be related to the structure with the pKa of 4.8 because they both have a double bonded oxygen and a chalogen at the end .
Letter c can be related to pKa of 27, 10 and 41 because it is a ring structure with 3 double bonds whose bonded to a chalogen .
Letter d can be related to pKa of 16 , 37, 50 , 50 the closest in the way that there is a heteroatom bonded to the last carbon and the structure is single
bonded.
The pKa of the single bonded carbon chain and a ring that single bonded are the same.
2
Acid-Base & Electron Energy- 7
2.
Based upon the data in Table E: List the ions in 1. In order of decreasing energy of their highest energy electrons: (i.e. Highest energy first and
lowest energy last) Explain how you determined your order of energies citing as many supporting examples from the data as possible.
D >C>A>B D has all single bonds and as we have said before a single bond is going to have an sp3 with the lowest s-character causing it to have the
highest energy out of the 4 structures. pKa of 50 can be used to help explain why it has the highest energy. The least amount of variability of those
structures and the ones in the table.
We got the order by comparing pkas of 50, 41, 33, and 19.
3. Resonance and “Curved Arrows”: (See Review of General Chemistry and Math # 25)
Curved arrows can be used to assist in converting one structure into another. See the example below:
H
H
:
:
:
S
H
S
:
H
a. Where does arrow a begin (base of the arrow)?
Lone pair of electrons on S.
H
:
b
H
H
:
H
a
H
H
b. Where does arrow a end (point of the arrow)?
At the bond between the C and S.
c. Compare the two structures and determine the result of the change indicated by arrow a?
The lone pair of electrons on the S forms a double bond between the C and S. A pair of electrons was moved.
d. Where does arrow b begin (base of the arrow)?
At the double bond between two Carbons in the ring (one which is bonded to another carbon and one which is bonded to the S)
e. Where does arrow b end (point of the arrow)?
At the Carbon at the end of the double bond which is bonded to another carbon and a hydrogen
Acid-Base & Electron Energy-7
3
f. Compare the two structures and determine the result of the change indicated by arrow b?
The double bond between the two carbons breaks and causes a lone pair of electrons on the carbon giving it a negative charge. A pair of
electrons was moved.
g. The total effect of the operations indicate by arrows a & b together illustrate the movement of two types of structural elements. What are those
structural elements? Explain how you reached your conclusion.
Breaking a bond and going to a lone pair and going from a lone pair to a bond. We used the illustrations to visually see the differences and then
identified the structural elements which moved.
h. What is the difference between the two original structures? Be as specific as you can. Explain how you reached your conclusion.
The second structure has a double bond from a C to the S. It has a negative charge on the C due to an extra lone pair of electrons on the C.
The second structure also has 2 double bonds between Carbons whereas the first structure has 3 double bonds between carbons. The first
structure has a negative charge and extra lone pair of electrons on the S. We reached these conclusions by looking at the two structures.
i. Note that the arrow that connects the two original structures is different from those used before. This arrow is used to relate two resonance
structures. Resonance structures are similar, but are not isomers. Based on your experience in this activity, define the relationship between
resonance structures.
j. zFrom your experience in this activity, use “curved arrows” to find new resonance structures of the original structures. Explain how you
devised your structures.
k. Now use curved arrows to find all resonance structures of the other structures in question B.1.
4
Acid-Base & Electron Energy- 7

Structures that have two or more possible resonance structures that place the HEE’s on different atoms are said to have “a delocalized HEE
system.” Based on the resonance structures you have drawn, how far can the HEE’s be delocalized in each structure in question B.1.? List the
original structures in order of the length of the structure over which the HEE’s are delocalized (length of the delocalized HEE system) based on
resonance structures.
C. Considering your experience with resonance structures in B.2.:
1. Suggest a correlation between the order of energies proposed in B. 2 and the length of the delocalized HEE systems that differ only in the number of
-bonds.
As the number of pi bonds increase the energy of the structure decreases (pka values 50, 43, 33, 41). As the length of the delocalized HEE system
increases the energy increases. The ring structure pka 41 is actually slightly shorter than that of a straight carbon chain, giving it a smaller length
and less energy. More you can lengthen the delocalized system the lower the energy.
2. What effect does the presences of an additional higher effective nuclear charge atom have on the energy of a delocalized HEE system? Explain
how the data support your conclusion.
The addition of a higher effective nuclear charged atom will decrease the energy of the delocalized HEE system. (pka 19) oxygen has a higher effective
nuclear charge than carbon and so lower energy. Increased nuclear force like O, lowers the energy of the entire system. One oxygen lowers the
energy more than the addition of another bond.
D. Based upon your theory developed in C. above, predict which proton(s) in the molecule below should be most acidic.
Acid-Base & Electron Energy-7
CH3
:
N
O
CH2
CH2
:
:
CH2
CH3
5
CH3
:O:
E. Note the differences in the pKa's of the following compounds from Table E.
Conjugate Acid
CH3
CH3
CH2
CH2
CH2
CH
CH2
Conjugate Base
CH3
CH
CH3
CH3
CH3
CH2
CH 2
CH2
CH
CH
CH
CH
H
H
CH
CH2 :
-
~50
-
~43
CH2 :
:-
H
H
CH2
CH2
pKa
40
H
CH3
CH 2
CH
CH
CH
CH2 :
-
~33
:16
H
Are these differences consistent with the theory developed in C. above? Note specifically the pKa of the last compound.
Why or Why not?
The atoms in the above table do not contain any heteroatoms. The theory doesn’t work very well with our theory because the pka of 16 has only 5
carbons and pka 40 has 7 carbons and since it has a larger number of pi bonds than 16 it should have a lower energy. The pka differences of the
structures in the table are based on the number of pi bonds. Aromatic compounds are the exception to the rule that we have previously
established. In aromatic compounds there are 6pi bonds in the molecule and these pi bonds carry around all over the ring structure putting slightly
negative charge on every atom. In pka 40 there are 7 carbons and 8 pi electrons which do not work in this way. 10 electron structures also work
in the aromatic way. The 4n+2 system works in the aromatic way, able to handle that many electrons in a continuous ring.
6
Acid-Base & Electron Energy- 7
D. Out of Class Applications for acid-Base-6
1. For each of the following compounds, circle the most acidic proton(s) and explain your choice using electron energy arguments.
O
O
2. Which of the following compounds are aromatic according to the Hückel theory?
N
H
:
H
:
H
CH3
+
N
H H
3. For each of the following acid-base reactions, predict whether the equilibrium constant should be > or < 1. Explain your conclusion using electron
energy arguments.
a.
O
O
OH
O +
+
OH
O
b.
O
C
NH
O +
O
C OH
C OH
+
NH
C O
-
4. Which of the following equilibrium controlled reactions should produce more product at equilibrium. Explain your conclusion using electron
energy arguments.
O
CH3
CH3
CH2 C
O
+
CH2 C
O
O
CH3
O
CH2 C
CH3
O CH2
+
O
-
O
CH3
CH2
-
C
+
O
O
CH3
CH2
C
O
O
CH3
CH2
C
O
+
CH3 CH2 O
-
-
C. Nomenclature of Amines
1. CGWW: Ch. 2.0 p. 33
2. Tutorials:
a.
Acid-Base & Electron Energy-7
7
References:
http://chemistry.boisestate.edu/people/richardbanks/organic/nomenclature/organicnomenclature1.htm
Developed by Richard C. Banks, Professor of Chemistry, Boise State University
Provides questions with answers
Sections
b.
Amines
http://www.sci.ouc.bc.ca/chem/nomenclature/index-2.htm
Developed by Professor Dave Woodcock, Okanagan University College,
British Columbia, Canada
(Contains many examples.)
Sections:
5. Functional Groups with Suffix and Prefix
I. Introduction: Priority Rules
II. Alkanamines (Amines)
Note: Your browser must have the chemscape chime plug-in for these pages to work. Select "1. Introduction to these pages", then click
on "Nomenclature Index - Chemscape Chime" and begin with the "How and Why". If you have problems, go to the
http://www.sci.ouc.bc.ca/chem/nomenclature/nom1.htm and click on "chemscape chime" and follow directions to download the
"chemscape chime plug-in." If you have problems contact me.
c. http://www.acdlabs.com/iupac/nomenclature
Developed by Advanced Chemistry Development Laboratories
(Gives detailed rules for nomenclature.)
Recommendations 1993
R-5 Applications to Specific Classes of Compounds
R-5.4 Amines and Imines
R-5.4.1 Primary amines
R-5.4.2 Secondary and tertiary amines
3. Applications
a. Name the following:
8
Acid-Base & Electron Energy- 7
N
NH2
N
Br
NH2
H
b. Draw structural formulas for the following compounds:
N-propyl-2,3-dimethylbutanamine
N-cyclopropyl-N-methyl-2-hexanamine
N,N-diethyl-4-methoxycyclohexanamine
3-octanamine