Organic Molecule Structures and Molecular Orbital Theory Questionnaire

1. The correct ranking of the acidity (from the strongest to weakest) of the following compounds is (4 pts)I would remove ‘e’ and if such a question were of interest I would ask it separately
A. b>a>c>d>e
B. b>d>e>a>c
C. d>e>a>b>c
D. b>e>d>a>c
E. c>b>d>e>a
Version II:
I would remove ‘e’ and if such a question were of interest I would ask it separately
A. a>b>d>c>e
B. d>c>e>a>b
C. d>e>c>a>b
D. d>a>c>e>b
E. c>e>d>b>a
Version III: (same choices and key as Version II)
I would remove ‘e’ and ‘b’ (which is very close to ‘d’!) if such a question were of interest I would
ask it separately
2. Under the same condition, which one is the correct ranking of the SN2 reaction rate for the following
electrophiles? (4 pts)
I would reduce the number of choices (electrophiles) – R-F problems are easier than others, so
consider removing one of the R-Cl/R-Br or the TsOR
A. reaction rate a>e>d>c>b
B. reaction rate a>b>d>c>e
C. reaction rate e>a>d>c>b
D. reaction rate e>a>b>d>c
E. reaction rate e>c>a>d>b
Version II
Version III
Same choices and key
3. What the is the correct order for nucleophilicity of the following anions, in DMF?
a. NH2- b. OH- c.CH3CH2O- d. F- e. Cl- f. Br- g. II would reduce the number of choices (nucleophiles) – again down to 4 choices as per the previous
question – I would also not simultaneously go down AND across the periodic chart; I would
separate such questions into two different questions
A. nucleophilicity a>b>c>d>e>f>g
B. nucleophilicity g>f>e>d>c>b>a
C. nucleophilicity b>a>c>d>e>f>g
D. nucleophilicity a>c>b>d>e>f>g
E. nucleophilicity g>f>e>d>a>b>c
Version II: change DMF to THF, NH2- to CH3Version III: change DMF to DMSO
Same answers
4. Which one of the following molecules is the best reactant among the choices for an internal
(intramolecular) SN2 reaction in DMF, when no other reagents are added? (4 pts)
A.
B.
C.
D.
E.
Version II: Change -Br to -I, in A; flip C (so NH2 on the left); change Br to Cl in E; same key.
Version III: from Version I, flip the molecules in A and B; change Br to Cl in D and E; same key.
5. Among the following bromides, which one would generate the most stable carbocation in an SN1 reaction,
immediately after the leaving of Br-? (note: consider the immediately formed carbocation only, and do not
consider potential rearrangements) (4 pts)
A.
B.
C.
D.
E.
Version II:
Change C to
, E to
, which would make C the correct choice.
Version III:
Change C to
, E to
; and E is still the correct answer.
6. Consider the following reaction:
This reaction involves carbocation rearrangement. Which one of the following best describes the driving
force of the rearrangement? (4 pts)
Are you sure this is the product? The secondary carbocation formed here normally undergoes a
hydride shift to give the tertiary carbocation. It may not matter, since the question is very wellcontained and the answer is unambiguous. I prefer the question you ask.
A. The carbocation rearranges to a new structure to better fit the trigonal planar geometry for the sp 2
hybridized carbon atom.
B. A tertiary carbocation is more stable than a secondary carbocation.
C. A Secondary carbocation is more stable than a primary carbocation.
D. There is a ring expansion process to release the [add ‘angle’?] strain.
E. There is a proton transfer process to make the final product neutral.
Version II: change I to Br, change methanol to ethanol
Version III: change I to OTs, change the methyl group (β position to the OTs) to an ethyl group.
7. Which one of the following statements is correct?
A. SN1 reactions go through a carbocation [use ‘carbanion’ to avoid confusion as to what a carbocationic
transition state might be?] transition state.
B. Stronger acids have higher pKa value than weaker acids.
C. A solvent that solvates the transition state structure of the rate determining step less than the
corresponding reactant(s) would facilitate [‘accelerate’ to avoid confusion regarding what is meant by
facilitate?] the reaction.
D. I- is a stronger nucleophile than Br- in isopropanol.
E. In an E1 reaction of an alkyl bromide, the rate determining step involves both the base and the alkyl
bromide [end at ‘base’?].
Version II:
A. SN1 reactions go through a carbocation transition state.
B. Stronger acids have lower pKa value than weaker acids.
C. A solvent that solvates the transition state structure of the rate determining step less than the
corresponding reactant(s) would facilitate the reaction.
D. I- is a weaker nucleophile than Br- in isopropanol.
E. In an E1 reaction of an alkyl bromide, the rate determining step involves both the base and the alkyl
bromide.
Version III:
A. SN1 reactions go through a carbocation transition state.
B. Stronger acids have higher pKa value than weaker acids.
C. A solvent that solvates the transition state structure of the rate determining step more than the
corresponding reactant(s) would facilitate the reaction.
D. I- is a weaker nucleophile than Br- in isopropanol.
E. In an E1 reaction of an alkyl bromide, the rate determining step involves both the base and the alkyl
bromide.
8. For the following reaction, which one is the most reasonable description of the mechanisms that lead to
the major product, minor product 1, and minor product 2, respectively? (4 pts)
A. E1, SN2, SN2
B. E1, SN1, SN1
C. E2, SN1, SN1
D. E1, SN1, SN2
E. E2, SN1, SN2
Version II:
Version III:
9. There is an intermediate formed in the following reaction. (4 pts)
The scientist who made great contributions in the direct observation of these type of reactive
intermediates was
A. George Andrew Olah
B. Alexander Mikhaylovich Zaitsev
C. George Simms Hammond
D. Adolph Wilhelm Hermann Kolbe
E. Paul von Walden
Version II:
Version III:
Same choices
10. Select the correct double arrow for the following proton transfer process in DMSO. (1 pt each, 8 pts
total. However, you get 12 pts, if you got ALL 8 WRONG, i.e. 4 pts extra bonus. Be careful though – even
if you got 7 wrong, but 1 correct, you only get 1 pt, so only do it when you are certain about all 8.)
Clever, but what is the point? No sensible person would risk it, but a lucky dummy might end up
with 12 points. Lol – I recommend making it 4 points (just like the rest) and making it two problems.
Put the harder ones together [B, D, F, G] and the easier ones together. If you need 4 more points –
give them a survey question: are you a chem major? Would you rather have 1/3, 1/3, and 1/3 lecturers
as before, etc.
a.
A.
B.
C.
D.
E.
F.
G.
b.
H.
Key: abaaabaa
Version II:
Scramble orders A-H
Flip left & right (but remember to flip the key, too!)
From Version I, Change -F to -Cl in C
Change D to
Answer is b.
Put the new one as the first one (A) so any cheating students are less likely to find out the other 7 are
essentially the same.
Version III
Scramble orders A-H
Flip left & right (but remember to flip the key, too!)
From Version I, change D from propylamine to cyclopentamine; change H from isopropanol to t-butanol.
Change F to
Answer is b.
Put the new one as the first one (A) so any cheating students are less likely to find out the other 7 are
essentially the same.
11. Which one is the major organic product of the following conversion? (4 pts)
A.
B.
C.
D.
E.
Version II: change dash to wedge, which would make A the correct answer.
Version III: add a methyl to the para- position, change 2-butanol to 2-pentanol, still use dash, so answer is
B.
12. For the following conversion:
When NaH was not added, there was no reaction at room temperature for 6 hours. Upon the addition of 1
equivalent of NaH, the conversion was completed under similar conditions.
Which two basic reaction types best describe the conversion, in the correct order that it happened?
Make this bold/italics/unmissible (4 pts)
A. oxidation, substitution.
B. substitution, elimination.
C. proton transfer, substitution.
D. substitution, proton transfer.
E. addition, substitution.
Version II: change -Br to -Cl, change the methyl ether to ethyl ether.
Version III: change NaH to tBuOK, change -Br to -Cl.
13. Which one of the following compounds can form a robust monolayer of ligands on a gold surface? (4
pts)
A.
B.
C.
D.
E.
Version II: change C to
, D to ethyl acetate.
Version III: change C to
, E to acetone, question change to “Which one of the
following compounds can be used as organic ligands on gold nanoparticles?”
Key is always the compound with thiol.
14. What is the major organic product of the following reaction? (4 pts)
A.
B.
C.
D.
E.
Version II: change the reactant to
, so the correct answer is B above.
Version III: change the reactant to
, so the correct answer is still A.
15. What is the major organic product for the following reaction? (4 pts)
A.
D.
B.
C.
E.
Version II: change the reactant to
-Br on choices C, D, and E on the dashed bond.
, so the correct answer is A, but also put the
Version III: change the reactant to
, and make the choices
16. Which one of following ethers can be efficiently prepared directly by a Williamson ether synthesis
(the reaction is expected to have high yield with no considerable side reaction)? (4 pts) Consider: Only
one of following ethers can be efficiently prepared, with no significant side reaction, by Williamson
ether synthesis using an alkyl bromide and an alkoxide. Which one is it?
A.
B.
C.
D.
E.
Version II: rotate all molecules, change D to a five-membered ring, change C to
answer key.
, same
Version III: rotate all molecules, change -Br to -Cl in D, add a methoxy group to the 4- position in A.
17. For the following reaction, whether the cis or trans reactant was used, the same trans isomer was
observed as the major product. Meanwhile, the reaction rate for the trans isomer was 670 times higher than
the cis reactant. (4 pts)
(This is a very hard problem and few are going to get it right. In the question, it will not be clear that
the ‘trans isomer’ in the second statement refers to the trans starting material. I suggest you coach
them: make a model, consider that the carbony oxygen is 5 atoms away from the leaving group, etc.
In my experience, even giving them the answer in the question will not lead many of them to the right
answer – so be careful here.)
With this information, what is the likely structure of the intermediate obtained after the rate determining
step in the reaction of the trans reactant (the top scheme) that can account for the fast reaction rate and
retention of stereochemistry?
A.
D.
B.
C.
E.
Version II: change the question to “For the following reaction, on the kinetic aspect, the reaction rate for
the trans isomer was 670 times higher than the cis reactant. Meanwhile, on the stereochemistry aspect,
whether the cis or trans reactant was used, the same trans isomer was observed as the major product.”
Version III: change the question to “The following results were published by a group of researchers from
UCLA. They performed the following reaction, and found whether the cis or trans reactant was used, the
same trans isomer was observed as the major product. Meanwhile, the reaction rate for the trans isomer
was 670 times higher than the cis reactant.” Put the cis reactant on top, trans reactant at the bottom in
ChemDraw, and change the second half of the question “(the top scheme)” to “(the bottom scheme).
18. Which one is the correct description of the rate determining step (RDS), transition state(s), and
intermediate(s), and Gibbs Free Energy change, based on the reaction coordinate? (4 pts)
(Few are going to get it right. It has too many parts and is too intricate. It is not hard, per se, but if
they get it wrong what have you learned? I like the question broken into parts. How many transition
structures and intermediates are there in this RCD? Which species is most relevant to the RDS?
Which species is most relevant to the Gibbs of the reaction?)
A. The RDS is the first step. There are 3 transition states, 2 intermediates, and the reaction is exergonic.
B. The RDS is the first step. There are 2 transition states, 3 intermediates, and the reaction is exergonic.
C. The RDS is the third step. There are 3 transition states, 2 intermediates, and the reaction is endergonic.
D. The RDS is the first step. There are 5 transition states, 2 intermediates, and the reaction is endergonic.
E. The RDS is the third step. There are 3 transition states, 4 intermediates, and the reaction is exergonic.
Version II:
A. The RDS is the second step. There are 3 transition states, 2 intermediates, and the reaction is exergonic.
B. The RDS is the first step. There are 5 transition states, 2 intermediates, and the reaction is exergonic.
C. The RDS is the second step. There are 3 transition states, 2 intermediates, and the reaction is endergonic.
D. The RDS is the second step. There are 2 transition states, 3 intermediates, and the reaction is endergonic.
E. The RDS is the first step. There are 3 transition states, 4 intermediates, and the reaction is endergonic.
Version III:
A. The RDS is the second step. There are 4 transition states, 3 intermediates, and the reaction is exergonic.
B. The RDS is the first step. There are 7 transition states, 3 intermediates, and the reaction is exergonic.
C. The RDS is the second step. There are 4 transition states, 5 intermediates, and the reaction is endergonic.
D. The RDS is the second step. There are 3 transition states, 4 intermediates, and the reaction is exergonic.
E. The RDS is the first step. There are 4 transition states, 3 intermediates, and the reaction is endergonic.
19. For the following reaction, if the starting concentration of A is x, starting concentration of B is y, and
the initial reaction rate is r, then what is the initial reaction rate, when the starting concentration of A is 2x,
starting concentration of B is 3y? (4 pts)
Isn’t it easier to ask: For the reaction shown, if the concentration of A is doubled and the
concentration of B is tripled by what factor will the reaction rate increase, if at all? (then modify the
choices slightly and remove ‘r’):
A. r B. 2r C. 3r D. 5r E. 6r
Version II: change the reaction to
The correct answer is then B. 2r
Version III: use the same reaction in Version II, and change the last part of question to “if the starting
concentration of A is 3x, starting concentration of B is 2y”, then the correct answer is C. 3r.
(Many of your questions are time-consuming and challenging. I would replace this with a simple set
of T/F statements. It takes a long time to read, to consider, and to solve, imagine a student received
this question first … they will freak out. If they get it last they will give up and guess. I love the
question, give it in recitation, but not on the exam. Please consider the value of giving questions of all
similar difficulty.)
20. Isotopic labeling is an important approach to probe reaction mechanisms or metabolic pathways. In
organic reactions, the reactants can be “labeled” by replacing their atom(s) with a specific isotope, such as
2
H (D), 13C, 15N, 18O, which typically show very similar chemical behavior with the unlabeled counterparts,
but the specific isotope, can be detected and analyzed by other techniques, such as mass-spectrometry or
nuclear magnetic resonance. For example, in the reaction converting an 18O labeled alcohol R-18OH to ROTs, the 18O will remain in the R-OTs product in consistence with the mechanism:
In this reaction 18O function just like a “normal” oxygen atom (16O).
When the R-18OTs are stored in a non-nucleophilic polar solvent such as tBuOH, the “scrambling” was
often observed, evidenced by the labeled oxygen atom moved to the S=O double bonds:
Which one of the following compounds is expected to have the highest rate of oxygen scrambling? (4 pts)
A.
B.
D.
E.
C.
21. Rank the boiling points of the following molecules, from low to high. (4 pts) (Or, better(?) “Which
of these is expected to have the highest boiling point?” – If they can name things they will be able to
google it – consider more complex structures.)
A. acdb
B. cabd
C. cbad
D. cadb
E. bcad
22. For the reaction below, two organic products were observed. (4 pts, 2 pts for the first blank, 1 pt each
for the second and third blank)
In this reaction,
is the major organic product (dropdown menu, pick A/B). If potassium tbutoxide/t-butanol instead of sodium ethoxide/ethanol is used as reagents, the ratio (yield of A)/(yield of B)
will
. (dropdown menu, pick increase/decrease) Alternatively, if keeping the original reagents, but
run the reaction at 78 °C [Is this temp supposed to be negative, i.e. -78?] instead of 50 °C, the ratio (yield
of A)/(yield of B) will
. (dropdown menu, pick increase/decrease)
B, increase, decrease.
23. Which sequences of reactions below can carry out the following conversion?
A. ace B. ae C. bc D. bce E. acd