Chem 2500 Final Exam 2003, December 10th, 9:00 am to 12:00 am
You are permitted the use of a model kit and note cards which were prepared by hand, by yourself. You may
also use a periodic table. A periodic table and a table of pKas appears at the end of this paper. Answer all
questions in this booklet. Print and sign your name below. Your signature indicates that you agree not to
divulge or discuss the contents of this exam in any way until after Jan. 20th., 2004.
.
Name:
Signature:
Question One (6 marks)
Give systematic names for the following compounds. R/S configurational symbols are not necessary.
i
ii
iii
CH3
O
O
O
F
C
O CH 3
NH 2
OH
H3 C
CH3
i)
ii)
iii)
Question Two (6 marks)
For each of the following molecules, label each chiral center as R or S and double bonds as E or Z as is
appropriate.
O
S
O
HOH 2 C
H
H R
S
R
CN
H
CHO
R
OH
CH 2 O H
E
H 3 CH 2 C
CH 2 CH 2 Br
CH 2 O H
Que
stion Three (3 marks)
You wish to prepare an enolate anion of acetone (right side) for use as a nucleophile.
Give an example of an appropriate reagent to accomplish this, and explain how you made your choice.
You need to recognize that a base is required to effect this process:
O
O
C
H3 C
CH 3
B:
HB
C
H3 C
CH 2
Your base must be strong enough to remove a proton from acetone. In Brønsted A/B reaction of this type, the
reaction favours the side of the weakest acid. You therefore want HB to be a weaker acid than acetone. The
pKa of acetone is 20 so your base must be the conjugate of something with a higher pKa. H- or H2N- would be
suitable choices.
Question Four (3 marks)
For the following two species,
i) draw them showing their correct structure showing all bonds with as many atoms in the plane of the page as
possible. All non-bonding electrons are shown.
ii) Indicate the hybridization of all non-hydrogen atoms.
H2 C
H
sp
sp 2 C
C
H
H2 C
CH
C
NH
H
H
sp 2 C
H
C
sp
N
H
Question Five (8 marks)
The following preparation was just published in the Journal of Organic Chemistry. Classify each step in the
reaction as an addition, elimination, substitution, reduction, oxidation, pericyclic or rearrangement process.
Br
CH 3
CH 3
Br 2 , Fe, CCl4
H3C
CH 2 Br
NBS, CCl 4
H3C
Substitution
Br
Substitution
Br H 2 C
Br
Br
Ca CO 3 ,
Dio xa n e / H 2O
Substitution
Br
Br
CH( OCH 3 ) 2
(CH 3 O) 2 HC
Br
CHO
PCC
CH 3 OH/ H+
CH 2 OH
OHC
Addition
Br
HOH2 C
Oxidation
Br
Br
1 ) n -Bu Li
2 ) DM F
3 ) H2O
Substitution
C HO
CH 2 OH
CH(OCH 3 ) 2
CH( OCH 3 ) 2
Na BH4 , CH 3 OH
(CH 3 O) 2 HC
CH 2 OH
CH 3
N
O
( CH 3O ) 2 HC
Reduction
CHO
O
N CH 3
O
CH 3 O H / H +
Substitution
O
O
O
OCH 3
LDA
O
CH 3 O
O
N
H3C
O
Pericyclic
O
O
Elimination
O
Question Six (4 marks)
What is the stereochemical relationship between the following molecules: are they diastereomers, enantiomers,
constitutional isomers or identical?
i) enantiomers
ii) diastereomers
iii) diastereomers
iv) enantiomers
i)
ii)
F
H
F
H
H
H
F
F
H
H
OH
CH3
HO
H
CH 3
H3 C
H
H3 C
H
H
CHO
H
OH
H
OH
H
OH
HO
H HO
HO H2 C
H
CHO
HO
H
CH2 O H
iii)
iv)
H
H
OH
OH
Question Seven (4 marks)
i)
Is the equilibrium constant for the following conformational equilibrium constant greater or less than 1?
Explain your reasoning.
H
OH
OH
H
Less than one. The more stable conformer is the one with the substituent in the equatorial position that
minimizes 1,3-diaxial interactions.
ii)
The equilibrium constant varies depending on the solvent. In polar solvents like methanol, Keq is less
than non-polar solvents like hexane. Explain.
Polar solvents will cluster around the polar OH group to form H bonding interactions in solvents like
methanol. These interactions are stabilizing. There will be more room for solvent molecules to cluster
around the OH group when it is equatorial. Such interactions are not as important in non-polar solvents,
so there will not be such a distinction between the two conformations.
Question Eight (7 marks)
Diazomethane is a useful, if somewhat dangerous, compound.
i)
Add formal charges to complete the resonance structures below
.
H2 C
N
H2C
N
N
N
ii)
Which of these two resonance structures contributes more to the character of diazomethane.
Explain.
The second. Neither structure has any electron deficient atoms and both have the same charges, so the better
structure will be that with the negative charge on the more electronegative atom.
iii) Diazomethane reacts with carboxylic acids to give esters. The first step is a Brønsted acid/base reaction in
which diazomethane is protonated on carbon and the second is an extremely fast SN2 process.
Why does diazomethane protonate on carbon?
Being less electronegative, C anions are more basic than N anions.
iv) Give a mechanism for the two step reaction of a formic acid with diazomethane.
O
O
C
H
H
C
O
H2C
N
N
H
CH 3
O
+ N2
O
C
H
O
H3 C
N
N
v) Give two reasons why the SN2 reaction is fast compared to a comparable reaction with 1-bromopropane.
Methyl substrates react more quickly than 1° substrates and N2 is a better leaving group.
Question Nine (6 marks)
Give a mechanism for the following acid catalyzed reaction.
H+
O
We did not cover this reaction.
Question Ten (4 marks)
Of the following pairs of reactions, which one (if either) will be faster? Explain your reasoning and show the
structure of the product of the faster reaction.
1)
HO
Br
OH
H 2O
Br
OH
Same rate. 3° substrate, so SN1, the rate of which does not depend on the nucleophile.
2)
Cl
CH3 S
SCH 3
Cl
CH3 S
SCH 3
1° substrates so SN2. Same nucleophile. The second is faster because the first substrate suffers from betabranching.
Question Eleven (6 marks)
Benzaldehydes can be prepared by the solvolysis reaction of dibromomethylarenes as shown below. This
reaction is first order and works best for benzylic substrates. Give a mechanism that is consistent with this
information and explain your reasoning.
H
O
Br
C
H2 O
Br
C
H
If it’s first order and works best for benzylic substrates, then it is likely to be an SN1 process.
H
O
Br
C
H2O
C
Br
H
H
O
Br
H
H
H
H
O
C
O
H
H
C
Br
Br
H
H
H
H
O
C
H
H
O
H
O
H
C
Br
H
H
O
O
H
H
H
O
C
H
O
H
C
H
Question Twelve (5 marks)
The allylic substrate shown below is a good substrate for both SN1 and SN2 reactions.
i) Why?
Br
Allyl Bromide
It is 1° so is it’s good for SN2. It forms a resonance stabilized carbocation upon ionization, so it’s good for
SN1.
ii) Suggest an experiment that would enable you to determine which of the two mechanisms operates faster for
this kind of substrate.
There were lots of answers to this question. The best would be to use a chiral substrate and see what the
balance of inversion to racemization would be.
Br
H D
If the only product isolated has the opposite absolute configuration, then this indicates that SN2 is much
faster than SN1. If complete racemization occurs, then that would indicate that SN1 is much
faster.Question Thirteen (4 marks)
Give the structure, showing the correct stereochemistry, of the starting material required to prepare only the Zalkene shown via an E2 reaction. Explain your reasoning.
?
H3 C
H
Ph
Ph
KOt -Bu
Didn’t cover this this term.
Question Fourteen (5 marks)
i)
On one of the two diagrams below (whichever is most appropriate) sketch the * (anti-bonding) orbital
of ethylene.
H
H
C
H
ii)
H
H
C
C
C
H
H
H
Platinum forms coordination compounds with ethylene in which the Pt-C bonding is the result of overlap
of a filled d-orbital on Pt with the * orbital of ethylene. In the diagram below, all of the Pt—Cl bonds
lie in one plane (that of the page). Given this information, which of the following structures show the
best orientation of the alkene in the complex? As part of your answer, show how the overlap between
the d- and *orbitals occurs by adding to the Pt d orbital drawing below.
HH
Pt
C
C
H H
H H
Cl
Cl
Cl
H
Only the middle possibility
allowsHfor the necessary in-phase overlap of the orbitals
involved. H
H
C
C
Cl
P
t
Cl
Pt
Cl
Pt
C
C
C
C
H
H
Cl
Cl
Cl
H
H
H H
Question Fifteen (12 marks)
Predict the organic product(s) of the following reactions. If more than one organic product is formed, show
them all and indicate, if possible, which is/are the major products. Be careful to indicate the relevant
stereochemical results clearly (using dashed and wedge bonds where appropriate).
i)
ii)
Br
OCH 3
OCH 3
CH 3 OH
Br
Br
Br
With this nucleophile, only SN1 is possible, so only the benzylic site reacts. Note that the two products formed
are diastereomers, not enantiomers.
iii)
iv)
H
H
H
Ph
CO 2 H
H 2 , P t/ C
Ph
C
C
CO 2 H
H
H
H
v)
Na (s )
CH 3 CH 2 O H
Br
CH 3 CH 2 O
Na
OCH 2 CH 3
vi)
OH
Na N 3
Ra ce m ic.
O
N3
vii)
O
OH
CH 3 OH , H 2 SO4
H 3 CO