Weekly Review Lecture Answers Review 1 Chem S‐20ab 1) Draw a complete Lewis structure for each molecule below, including all lone pairs. Then draw the best alternate resonance structure for the molecule, and decide which structure is more stable. a. OCN‐ The one on the right is more stable because it puts the negative charge on the more electronegative oxygen. b. NNO The one on the right is more stable because it puts the negative charge on the more electronegative oxygen. c.
The one on the left is more stable because there are no formal charges. Review 1 Weekly Review Lecture Answers Chem S‐20ab 2) Identify the functional groups in the molecule below, and give hybridizations for the circled atoms. How do you think one would go about naming this monstrosity? If you had to name this compound you would scratch your head, then give up. The actual name is (1R,11R,18S,20S,21S,22S)‐12‐oxa‐8.17‐diazaheptacyclo[15.5.01,8.02,7.015,20] tetracosa‐2,4,6,14‐tetraene‐9‐
one. Or you could just use the common name: strychnine. 3) Provide the best systematic name for the following molecule: (E)‐4‐tert‐butyl‐3‐ethyl‐2,5‐dimethyl‐3‐octene 4) Draw the structure of (Z)‐6‐tert‐butyl‐4‐isopropyl‐3‐methyl‐3‐nonene. 5) Consider the conformations obtained for the following molecule by rotation around the C‐C bond: a) This molecule is expected to have a single lowest‐energy conformation. Using a Newman projection, show the lowest‐energy conformation of this molecule. Be sure to view the molecule along the C—C bond. Weekly Review Lecture Answers Review 1 Chem S‐20ab b) This molecule is also expected to have a single highest‐energy conformation. Using a Newman projection, show the highest‐energy conformation of this molecule. c) Using the table of interaction energies given on the right, calculate the energy difference between the highest‐ and lowest‐energy conformers of this molecule. 2.8 + 1.0 + 1.0 = 4.8 kcal/mol 6) Consider the following two molecules: A:
H3C
N
B:
H3C
O
a. For each molecule, provide a complete Molecular Orbital diagram showing the energies of the
valence orbitals in the molecule. Label each orbital with an appropriate label ( or
nonbonding), and fill each orbital with the appropriate number of electrons.
Weekly Review Lecture Answers Review 1 Chem S‐20ab b. Using curved arrows, show the result of an interaction between these two molecules. Draw in the
product that would result from this single step.
7) For each of the molecules shown below, consider HOMO = O l.p. LUMO = *CO O
CH3
H3C
O
HOMO = O l.p. (from the O on top – due to resonance) LUMO = *CO Weekly Review Lecture Answers Review 1 H2O HOMO = O l.p. LUMO = *OH E+
O
H
E
H
With electrophile or acid (for acid, E+ = H+)
Chem S‐20ab O
H
H
HOMO = CH LUMO = empty p on C This molecule would never act as an electron donor, so we don’t need to consider its reaction with an electrophile or an acid. Review 1 Weekly Review Lecture Answers Chem S‐20ab HOMO = CC LUMO = *CC This molecule is unlikely to act as an electrophile. The only cases we’ve seen where *CC acts as an acceptor orbital are in the formation of a bromonium ion or mercuronium ion. Note that in these cases, CC is acting as a donor simultaneously, so the alkene as a whole is acting as both donor and acceptor. The alkene purely as an acceptor is not something we’ll be seeing. 8) Each of the following shows one step in a reaction mechanism. Draw the product(s) that would result from the indicated curved arrows. Review 1 Weekly Review Lecture Answers Chem S‐20ab 9) Each of the following shows a single reaction step. Show the curved arrows required to get from the reactants to the products. 10) In each of the following molecules, determine which of the shown protons is more acidic. Justify your choice. Weekly Review Lecture Answers Review 1 Chem S‐20ab 11) The following compound, called a sulfonamide, undergoes an acid‐base reaction as shown below. Draw the structures of the conjugate acid and base for the reaction, and calculate the value of the equilibrium constant for the acid‐base reaction. O O
H
S
N
H
O O
S
NH
Keq = 1019
+
NH2
+ NH3
pKa ~ 35
pKa ~16
Note: The sulfur does not really have pi bonds to oxygen. This double bond notation is
really just a shorthand we use to represent sulfur/oxygen compounds. An alternative, but
more cumbersome notation, would be to represent the molecule as shown below.
O O
S
2+ NH2
12) Provide a complete curved‐arrow mechanism for the following transformation. For each step, identify the frontier molecular orbitals involved. C-C
attacks H+ 1s orbital
EtOH, H+
O
EtOH
donates to nC (the empty
nonbonding p-orbital on the
carbocation)
O
H
nO donates to
*O-H
C-C
nO (the O lone pair) donates to nC
(the empty nonbonding p-orbital
on the carbocation)
EtOH
Weekly Review Lecture Answers Review 1 Chem S‐20ab 13) Synthesize the following molecules starting from any hydrocarbons with 4 or fewer carbons. 14) Predict the product when the starting material below is subjected to the conditions shown. Explain the differences. H
2O
,H+
HO
H3C CH3
1. Hg(OAc)2
2. NaBH4
H3C
1.
2. B H
H 3
2O
2,
O
H–
Br 2, H2O
OH
Acid catalyzed hydration goes through a carbocation, which rearranges from secondary to tertiary to become more stable. Oxymercuration‐reduction gives Markovnikov addition but avoids rearrangement because it doesn’t go through a carbocation. OH
Hydroboration‐oxidation gives anti‐Mark. Addition. OH Br
Weekly Review Lecture Answers Review 1 Chem S‐20ab 15) Alkenes are known to react with CH3SH by a radical mechanism in the presence of peroxides (ROOR). a. Propose a mechanism for the following reaction. RO
OR
MeS
S
CH3
H
RO
MeS
MeS
H
SMe
b. Why is the other isomer, shown below, not formed in the reaction? CH3
CH3
S
The mechanism to make this isomer would require formation of a primary radical, which is much less stable than the tertiary radical in the mechanism above.