Section Handout Solutions: Module 2
Chemistry 20
Spring 2016
1. In each of the following molecules, determine which of the shown protons is more acidic. Explain your
reasoning in one sentence.
O
More acidic
(a)
Atom effect: S is larger than O, and
thus better able to stabilize the
negative charge in the conjugate base.
SH
HO
O
Resonance effect: Removing the
indicated proton results in a conjugate
base that is stabilized by resonance.
N
(b)
H
More acidic
N
N
H
H
H
(c)
Orbital effect: An sp atom is more
acidic than an sp3 atom. (This beats the
atom effect in this case, as you know
from your knowledge of pKa values.)
N
H
H
More acidic
O
(d)
OH
HO
Br
O
Inductive effect: The indicated proton is
more acidic because it is closer to the Br
atom, which withdraws electron density
from it via the sigma bond framework.
More acidic
Resonance effect: Removing the indicated
proton results in a conjugate base in which
the negative charge is shared between an
N atom and an O atom (rather than being
localized only on an O atom).
O
(e)
O
N
H
H
More acidic
O
O
O
N
O
H
N
H
Page 1 of 6
Section Handout Solutions: Module 2
Chemistry 20
Spring 2016
2. Using your knowledge of pKa values, calculate the equilibrium constant for the following acid-base reaction.
O
Keq = ?
OH
+
+
HS
pKa = 10
H2S
pKa = 7
pKa = pKa(product) _ pKa(reactant) = 7 _ 10 = _ 3
Keq = 10
pKa
= 10-3
Page 2 of 6
Section Handout Solutions: Module 2
Chemistry 20
Spring 2016
3. For each pair of bases, briefly explain why the base on the left is stronger than the base on the right.
(a)
O
Basicity measures reactivity of a lone pair. The less
reactive the lone pair, the weaker the base.
O
N
stronger base
While both structures have 3 additional resonance
structures that place the anion on 3 carbons, the compound
on the right has the extra resonance stabilization imparted
by the nitro group. This lessens the overall reactivity of the
anion and makes it the less basic compound.
O
O
O
O
N
O
O
N
O
O
O
O
(b)
H3C
O
This is the key extra
resonance structure that
the compound on the
left does not have.
NH2
H3C
First point: The amide basic site is at the oxygen, not the
nitrogen. The most basic site of the molecule leads to the
weakest conjugate acid.
CH3
stronger base
O
OH
OH
versus
H3C
NH3
H3C
Nitrogen protonation gives no further
resonance delocalization of the charge.
This conjugate acid is more reactive.
NH2
H3C
NH2
Oxygen protonation allows for resonance delocalization of
the charge. This conjugate acid is more stable than the
option on the left, so this is the correct conjugate acid.
Second point: The amide is more basic than the ketone for the reasons discussed already: the
protonated amide will have a resonance delocalized charge while the protonated ketone will not. The
conjugate acid of amide is more stable (less reactive) than the conjugate acid of the ketone.
OH
H3C
OH
OH
NH2
H3C
NH2
The stronger base leads to the weaker conjugage acid.
This charge is delocalized.
H3C
CH3
This charge is not delocalized.
Page 3 of 6
Section Handout Solutions: Module 2
Chemistry 20
Spring 2016
4. Consider the following two species:
A:
H
C
O
B:
C
(a) For each species, provide a complete molecular orbital diagram showing the energies of all the orbitals in the
molecule. Label each orbital with an appropriate label ( , *, , *, or n), and fill in each orbital with the appropriate
number of electrons. Identify the HOMO and LUMO of each molecule by labeling them on the diagram.
A:
H
C
B:
C
O
H
H
H
Energy
H
Energy
*C–H
*C–H
*C–C
*C–C
*C=C
LUMO
*C–O
LUMO
Clp
HOMO
Nlp
HOMO
C=C
C–H
C–H
C–C
C–C
C–O
(b) When these two molecules react, the HOMO of one molecule will react with the LUMO of the other. Identify which
HOMO will react with which LUMO.
The HOMO of A (Clp) will react with the LUMO of B ( *C–O).
(c) Provide the curved arrows for this single-step reaction, and draw the product that would result.
O
O
H
C
C
H
C
C
Page 4 of 6
Section Handout Solutions: Module 2
Chemistry 20
Spring 2016
5. The following species react in a two-step reaction. For each step:
(a)
(b)
(c)
(d)
Identify the HOMO and LUMO of each species.
Determine which species will be the nucleophile and which will be the electrophile.
Show how the species would react using curved arrows
Predict the product.
+
Step 1
H
+
HOMO:
C=C
HOMO:
LUMO:
*C=C
LUMO:
nucleophile
C–H
LUMO: Cvacant p
electrophile
Cl
H
Cllp
*H–Cl
electrophile
+
HOMO:
Cl
Step 2
Cl
HOMO:
Cl
Cllp
LUMO: none
(all orbitals
are filled)
nucleophile
Page 5 of 6
6. Based on the structures given, draw curved arrows to account for each step of the following reactions. Be
aware that you may need to draw in lone pairs if they are not shown. Also identify the donor and acceptor
orbitals for each step of the reaction.
O
H
+
OH
O
O
H
H
+
OH
O
H
Donor:
Olp
*O–H
Acceptor:
Donor:
Acceptor:
Olp
*C=O
HO
HO
H
O O
+
H
O
H
OH
HO
H O O H
OH
Donor:
Olp
*O–H
Acceptor:
Olp
Donor:
*O–H
Acceptor:
H
HO
OH
+
O
HO
Donor:
Acceptor:
O
O
H
OH
Olp
Donor:
*C–O
Acceptor:
Olp
*C=O
H
O
O
+
H
O
O
OH
HO
O
H
OH
Donor:
Acceptor:
Olp
*O–H
Page 6 of 6