Post on 19-Jan-2020
Chemistry 262
Winter 2018
Final Exam Practice
The following practice contains 44 problems and mimics Monday’s 150 point final,
which will also contain 44 problems valued at 3.5 points each – with approximately ½
of the problems drawn directly from the quarter’s homework. It will also contain 2
unknown analysis problems valued at 6 points/question. Apologies for the repeats
Name_____________________________________________________
CLASSIFICATION OF ALKYL HALIDES, ALCOHOLS, THIOLS, ETHERS, AND SULFIDES
1. Neopentyl alcohol is classified as a
a. 1o alcohol
b. 2o alcohol
c. 3o alcohol
d. 4o alcohol
e. None of the above
Really no other option since there are no 4o alcohols (yes, it does contain a 4o C)
NOMENCLATURE OF ALKYL HALIDES, ALCOHOLS, THIOLS, ETHERS, AND SULFIDES
2. What is the correct IUPAC name for the following compound?
HO OH a. 1,2-Butanediol
b. Isopropanol
c. 1-Propanol
d. 1,2-Propanediol Common name is propylene glycol
e. Propylene glycol
HOMOGENEOUS INTERMOLECULAR INTERACTIONS – BOILING POINTS AND MELTING POINTS
3. Which of the following would have the greatest boiling point?
a. Benzene
b. Cyclohexene
c. Hexanoic acid
d. Ethylene glycol
e. Sodium acetate Ion-ion interactions NaAc MP = 324 oC, BP = 881 oC
HOMOGENEOUS INTERMOLECULAR INTERACTIONS – HYDROGEN BONDING
4. Which of the following compounds has the lowest boiling point?
a. I
b. II
c. III
d. IV
e. V
The one with no hydrogen bonding capacity
HETEROGENEOUS INTERMOLECULAR INTERACTIONS – SOLUTIONS AND SOLUBILITY
5. In which of the following solvents would anhydrous MgSO4 be completely insoluble?
a. Water
b. Tetrahydrofuran
c. Dimethyl formamide
d. Dimethyl sulfoxide
e. Hexane
Hopefully you recognized the correct answer from all the hard work you have put into
the lab. Ion-ion interactions are much stronger than ion-dispersion force interactions
HETEROGENEOUS INTERMOLECULAR INTERACTIONS – SOLUTIONS AND SOLUBILITY
6. Which of the following compounds would have the greatest solubility in water?
a. Diethyl ether
b. Methyl propyl ether
c. 1-Butanol
d. 1,2-Butanediol
e. 4-Mercapto-1-butanol
2:1 C:O for option (d). Although polarizable, and very weakly acidic, sulfur does not
hydrogen bond, and so corresponding thiols have significantly lower water solubility.
For example, mercapotethane (ethanethiol) has a molar mass of 62.1, a boiling point of
35 oC, and a water solubility of 0.7 % @ 20 oC. Controlling for mass, n-propanol has
MM = 60.1, BP = 97-98 oC, and at 3:1 C to O is completely miscible with water
I II III IV V
OOH
OH CH2OHOH
NUCLEOPHILIC SUBSTITUTION AND -ELIMINATION REACTIONS – REACTION RATES
7. Consider the reaction of 2-iodopentane with acetate. Assuming no other changes to
the system, what effect on the rate would result from simultaneously doubling the
concentrations of both 2-iodopentane and acetate?
I
ONa
O OO
NaI
a. No effect
b. The rate would double (2x)
c. The rate would triple (3x)
d. The rate would increase 4 fold (4x)
e. The rate would increase 6 fold (6x)
SN2: Substitution Nucleophilic Bimolecular, or both reactants participate in the rate
determining step
NUCLEOPHILIC SUBSTITUTION AND -ELIMINATION REACTIONS – TRANSITION STATE THEORY
8. Which of the following is true for any nucleophilic substitution reaction?
a. H > 0
b. H < 0
c. G‡ > 0
d. G > 0
e. G < 0
From ground state to transition state is necessarily an energy demanding process; i.e.
with nothing to impede proceeding from reactants to products, all favorable reactions
would [essentially] instantly reach equilibrium
NUCLEOPHILIC SUBSTITUTION AND -ELIMINATION REACTIONS – THE SN2 REACTION
9. When 0.10 mol of ICH2CH2CH2CH2Cl reacts with 0.10 mol of NaOCH3 at 40 oC, the
major product formed is
a. CH3OCH2CH2CH2CH2Cl
b. CH3OCH2CH2CH2CH2I
c. CH3OCH2CH2CH2CH2OCH3
d. CH2=CHCH2CH2Cl
e. CH2=CHCH2CH2I
Iodide is a better leaving group. Note these are primary halides which resist elimination
NUCLEOPHILIC SUBSTITUTION AND -ELIMINATION REACTIONS – STEREOCHEMISTRY
10. What would be the major product(s) of the following reaction?
CH2Cl
H CH3
OH
CH2OH
H3C H
CH2Cl
HO CH3
I
III
II
IV
CH2Cl
H OH
CH2OH
H CH3
SN2
a. I
b. II
c. III
d. IV
e. More than one of the above
Recall the absolute configuration does not change if the chiral center is not involved in the
reaction
NUCLEOPHILIC SUBSTITUTION AND -ELIMINATION REACTIONS – CRITICAL FACTORS
11. Which of the following is the strongest nucleophile in an aprotic solvent such as
dimethylsulfoxide?
a. I
b. Br
c. Cl
d. F
e. All are equally nucleophilic in this environment
Nucleophilicity follows basicity in polar aprotic solvents
NUCLEOPHILIC SUBSTITUTION AND -ELIMINATION REACTIONS – E2 REGIOCHEMISTRY
12. What is the principal product of the following reaction sequence?
a.
b.
c.
d.
e.
None of the above represents a significant product
Recall t-BuOK is a bulky base which has difficulty acting as a nucleophile. It also
gives the anti-Zaitsev product, but that is of no consequence in this case. Step 2
saturates the terminal double bond
NUCLEOPHILIC SUBSTITUTION AND -ELIMINATION REACTIONS – E2 STEREOCHEMISTRY
13. What is major product of the following reaction?
a.
b.
c.
Br
1. t-BuOK, t-BuOH
2. H2, Pd/C, EtOH
Ot-Bu
Br
EtOH, EtONa
heat
OEt
d.
e.
In order to adopt the concerted antiperiplaner configuration between the hydrogen
removed and the bromide as a leaving group, the methyls have to be on the same side
of the Neuman projection
NUCLEOPHILIC SUBSTITUTION AND -ELIMINATION REACTIONS – FACTORS INFLUENCING SN1 AND E1 REACTIONS
14. Which statement(s) is/are true of acid-catalyzed alcohol dehydrations?
a. Protonation of the alcohol is a fast step
b. Formation of a carbocation from the protonated alcohol is a slow step
c. Rearrangements of less stable carbocations to more stable carbocations are
common
d. Loss of a proton by the carbocation is a fast step
e. All of the above
A blast from the past, and well worth knowing
NUCLEOPHILIC SUBSTITUTION AND -ELIMINATION REACTIONS – GENERAL PRODUCT PREDICTIONS
15. What are the principal products formed when 2-bromo-2-methylpentane reacts
with sodium propynide?
BrC: Na
I
HC CH
IV
II
V
HC CH
III
?
a. I
b. II
c. III
d. IV
e. V
With a pKa = 25, terminal alkynes are strong bases. Zaitsev’s rule in action
NUCLEOPHILIC SUBSTITUTION AND -ELIMINATION REACTIONS – GENERAL PRODUCT PREDICTIONS
16. What would be the major product outcome for the following reaction?
HH3C
H Br
HH3C
H CN
CNH3C
H Br
I II
III IV
CNH3C
H H
HH3C
NC Br
C
25oC
a. I
b. II
c. III
d. IV
e. Equal amounts of I and III
Unlike a terminal alkyne, HCN has a pKa = 9.2, which makes the conjugate base
relatively weak. Coupled with the room temperature conditions, this is a
straightforward SN2 reaction
ORGANOMETALLIC COMPOUNDS – FORMATION OF GRIGNARD AND ORGANOLITHIUM REAGENTS
17. Which of the following is the correct order of capacity towards Grignard formation,
from easiest to hardest?
RBr RCl RI RF
I II III IV
a. I, II, III, IV
b. III, I, II, IV
c. IV, II, I, III
d. III, I, IV, II
e. I, IV, III, II
ORGANOMETALLIC COMPOUNDS – REACTIONS OF GRIGNARD AND ORGANOLITHIUM REAGENTS
18. Which of the following reactions will not produce a primary alcohol?
a.
(CH3)2CHCH2CH2CHO
1. LiAlH4/Et2O
2. H3O+
b.
c.
d.
e.
These are all very good reactions to know. We haven’t covered (e), but that is not
necessary to solve the problem since you know (from lab) NaBH4 will reduce ketones to
2o alcohols
CH2
O
CH2
(CH3)2CHCH2CH2MgBr
1.
2. H3O+
C6H5CCH3
O 1. NaBH4
2. H3O+
CH3CH2CH2CH2Li1. H2C=O
2. H3O+
CH3(CH2)5COOH1. LiAlH4, Et2O
2. H3O+
CARBENES, CARBENOIDS, AND THE SIMMONS-SMITH REACTION
19. What is the major product for the following reaction
a.
b.
c.
d.
e.
More than one of the above
This is a concerted reaction, just in case that comes up in the near future
ALCOHOL AND THIOL ACID BASE CONSIDERATIONS
20. Which of the following has the lowest pKa value in aqueous solution?
a. Mercaptoethane
b. Phenol
c. Phenyl mercaptan
d. tert-butyl alcohol
e. 1,2-ethanediol
I am very pleased with myself on this question, since it is an additive effect, much like
one finds for NMR shift values. Thioethane has pKa = 10.6, phenol pKa = 10,
thiophenol pKa = 6.6 (it would be lower, but the orbital overlap with the system of
the benzene ring is not as good with S as it is with O)
I2CH2
Zn(Cu)
I I
I
DEHYDRATION OF ALCOHOLS
21. Carbocations are frequently intermediates in acidic reactions of alkenes, alcohols, etc.
Which of the following is a common carbocation property?
a. Rearrangement to a more stable carbocation
b. Loss of a proton to form an alkene
c. Combination with a nucleophile
d. Reaction with an alkene to form a larger carbocation
e. All of the above
Again a blast from the past and well worth knowing all of these options
REACTION OF ALCOHOLS WITH HYDROGEN HALIDES
22. Which of the following could be used to synthesize 2-chlorobutane?
a. CH3CH2CHCH2 + Cl2 (aq)
b. CH3CH2CHOHCH3 + HCl
c. CH3CH2CCH + HCl
d. CH3CH2CCH + Cl2
e. None of the above
Cannot rearrange to a more stable carbocation in this case
BASE COMPATABLE ALCOHOL-DERIVED LEAVING GROUPS
23. What would be the principal product outcome of the following reaction sequence?
OH
H
CH3
H
pCH3C6H4SO2Cl
basetosylate
NaBr
ethanol
?
H
H
CH3
Br
Br
H
CH3
H
H
Br
CH3
Br
H
H
CH3
SO2C6H4CH3
I II III IV
a. I
b. II
c. III SN2 reaction between bromide and the tosylate with inversion
d. IV
e. An equimolar ratio of II & III
MODERN METHODS FOR CONVERSION OF ALCOHOLS TO ALKYL HALIDES
24. Which reagent(s) would transform propyl alcohol into propyl bromide?
a. Conc. HBr/
b. PBr3
c. NaBr (aq)/
d. Two of the above
e. All of the above
With regard to option (d), hydroxide is a much worse leaving group than bromide
ORGANIC REDOX REACTIONS
25. Which of the following is the correct order of decreasing level of oxidation?
HCOOH CH3OHCH2Cl2
I II III
a. I, II, III
a. III, II, I
b. II, III, I
c. II, I, III
d. I, III, II
Yes, it is true, CCl4 is at the same level of oxidation as CO2, will not burn, and has been
used in [toxic] fire extinguishers
OXIDATION OF ALCOHOLS
26. CrO3 in H2SO4/H2O will fail to give a positive test with which of these compounds?
a. CH3CH2CH2CH2OH
b.
c. (CH3)3COH
d.
e. All give a positive test with Jone’s reagent
A nice reminder to look for hydrogens on the carbon being oxidized…for alcohols – KMnO4
turns tetrasubstituted alkenes into 2 ketones
CH3CHCH2CH3
OH
CH3CH2CH2C O
H
OCTET EXPANSION AND OXIDATION OF THIOLS
27. Which of the following would oxidize a thiol to a sulfonic acid?
a. KMnO4
b. HNO3
c. HCl
d. Two of the above
e. All of the above
HCl is not an oxidizing acid – do a formal analysis of the oxidation number on Cl vs N
or Mn and you will see this
SYNTHESIS OF ALCOHOLS
28. Which product(s) are the expected outcome of the following reaction sequence?
a. I
b. II
c. III
d. IV
e. V
Anti-Markovnikov alkylborane formation as a concerted syn addition
CH3
CH3
OH
CH3
OH
CH3H3C CH2OH
OOH
+
enantiomer
+
enantiomer
+
enantiomer
I II III IV V
1. BH3-THF
2. H2O2, NaOH?
SYNTHESIS OF ETHERS AND SULFIDES
29. What is the major product for the following reaction?
a.
b.
c.
d.
e. More than one of the above
Variation of the halohydrin reaction, following Markovnikov’s rule. A very nice way to
form complex ethers that may then be transformed at the adjacent carbon
SYNTHESIS OF EPOXIDES
30. If cis-2-butene is treated with meta-chloroperbenzoic acid, what is the final product?
O
H3C CH3
H H
O
H3C H
H CH3
O
H3C H
H3C H
O
H3CH2C H
H H
I II III IV
a. I
b. II
c. III
d. IV
e. None of the above
A concerted syn addition – notice the methyl groups are still on the same side (of the
Neuman projection)
Br2, CH3OH
+ enantiomerOH
Br
+ enantiomer
OCH3
Br
+ enantiomer
Br
H3CO
+ enantiomer
OCH3
Br
EPOXIDES – NUCLEOPHILIC SUBSTITUTION REACTIONS
31. Which of the following is the principal product for the following reaction?
a. CH3CH2CH2
18OH
b.
c.
d.
e.
This is just how we figure these mechanisms out. In an acidic medium the nucleophile
isn’t as hungry, and approaches the C with the most carbocation character
EPOXIDES – NUCLEOPHILIC SUBSTITUTION REACTIONS
32. Which of the following is the principal product for the following reaction?
O
?CH3O-
CH3OH
OCH3OH
OCH3OCH3 OHOCH3 OHOH
I II III IV
H3O+
a. I
b. II
c. III
d. IV
e. Equal amounts of I & III
Methoxide is a hungry base, and goes after whatever is most exposed
CH3CH
O
CH2 HA
H218O
?
CH3CHCH3
18OH
CH3CHCH2OH
18OH
CH3CH
OH
CH2
18OH
CH3CHCH218OH
18OH
INFRARED SPECTROSCOPY
33. An optically active compound of molecular formula C7H12, reacts with cold, dilute
KMnO4 and gives a characteristic IR band around 3300 cm-1. On catalytic
hydrogenation a new optically active compound is formed, with a formula C7H16. Which
of the following compounds is a possible candidate?
II III
IV V
I
a. I
b. II
c. III
d. IV
e. V
Reaction with KMnO4 does not help us here, since all of the candidates will react.
Optical activity rules out I, III, and the absorption at 3300 cm-1 is the sp hybridized
C-H stretch, which translates to a terminal alkyne
1H-NMR SPECTROSCOPY – NUMBER OF RADIOFREQUENCY ABSORPTIONS
34. Assuming all are resolved, how many different signals would be observed in the 1H-NMR
of the following compound?
a. 1
b. 2
c. 3
d. 4
e. 5
Symmetry, symmetry, symmetry. Of course the same symmetry gives rise to 5 peaks
in the 13C NMR
1H-NMR SPECTROSCOPY – SPIN COUPLING AND CHEMICAL EQUIVALENCY
35. Predict the splitting pattern you would observe for the proton(s) at C1 of 2,3-
dimethyl-2-phenylbutane
a. Singlet
b. Doublet
c. Quartet
d. Septet
e. Octet
Really more of a nomenclature problem than anything else, with the phenyl indicated at
C2 making the problem pretty simple
1H-NMR SPECTROSCOPY – DESHIELDING AND SPIN COUPLING
36. Which of the following best represents the approximate predicted chemical shift and
coupling for the hydrogen indicated with the arrow?
a. 1.0 ppm, quartet
b. 2.4 ppm, singlet
c. 2.4 ppm, triplet
d. 2.4 ppm, quartet Notice this is decidedly upfield from an H-C-O shift
e. 3.0 ppm, quartet 13
C-NMR SPECTROSCOPY – CHEMICAL EQUIVALENCY AND 13
C VS 1H ABSORPTIONS COMPARED
37. Assuming all peaks are resolved, which of the following would give rise to 4 peaks in
both the 1H-NMR and broadband decoupled 13C-NMR spectra?
a. I 6
b. II 3 (Dig that crazy symmetry)
c. III 4
d. IV 5
e. More than one of the above
O
I II III IV
Br
13C-NMR SPECTROSCOPY – CHEMICAL EQUIVALENCY, NUMBER OF SIGNALS, AND DESHIELDING
38. In 13C-NMR spectroscopy, which of the following functional groups gives rise to
the signal furthest downfield?
a. –CN
b.
This should be money
c.
d.
e.
–CC–
MASS SPECTROMETRY – BASIC DEFINITIONS AND FRAGEMENTATION
39. Which of the following C6H14 compounds would give an MS base peak of m/z = 43?
a. CH3CH2CH2CH2CH2CH3
b. (CH3CH2)2CHCH3
c. (CH3)3CCH2CH3
d. (CH3)2CHCH(CH3)2 C3H7 = 43
e. None of these
Actually, this question was taken out of the Solomons and Fryhle test bank and is not
quite correct, since the base peak for both (c) and (d) is m/z = 43. It is reasonable to
expect m/z = 57 would dominate for 2,2-dimethylbutane (as the t-butyl cation), and this
is a dominant peak. However, methyl radical loss (note the M-15 loss for both 2,2-
dimethylbutane and 2,3-dimethyl butane on the following page) also gives rise to a tertiary
carbocation for 2,2-dimethyl butane (note the larger peak at m/z =71) which can then
undergo a loss of neutral ethylene (m/z = 28), once again leaving the 2o propyl cation.
See if you can draw this out, noting that you will need to transfer a hydrogen to the m/z
= 71 carbocation before withdrawing a pair of electrons to form the ethylene
C O
C O
C X
MASS SPECTROMETRY – BASIC DEFINITIONS AND FRAGEMENTATION
40. The data below for the isotope envelope of the molecular ion region of a halogen
containing species are consistent with the presence of what halogens for the compound
in question?
𝑀+⋅, I = 100.0 % 𝑀+⋅+ 2, I = 64.0 % 𝑀+⋅+ 4, I = 10.0 %
a. 1 Br
b. 1 Cl
c. 1 Br, 1 Cl
d. 2 Br
e. 2 Cl
Repeat from the E3 practice key: The fact that 79Br and 81Br exist 1:1 (atomic weight ≈
80) and 35Cl and 37Cl exist in a 3:1 ratio (atomic weight 35.5) is very useful diagnostically.
If 1 Cl were present M+2 would be 33 % of M; if 1 Br was present, then M and M+2
would be approximately the same size; however if 2 Br were present, there would be twice
the likelihood at least one of the bromine was 81Br and so M+2 would be twice the size of
M (and M+4). For 2 chlorine present, the possibility of one of the 2 positions being 37Cl
doubles – DON’T BE FOOLED! Since the ratio of 35Cl to 37Cl is approximately 75:25 (73:27
actual) , when the 75% peak is scaled to 100%, the 25 % peak increases to 33 % (37 %
actual), and 64 % for M+2 is a good indication 2 chlorine atoms are present.
ALKYNES – BASIC STRUCTURE AND NOMENCLATURE
41. What is the IUPAC name for the compound shown?
a. (S,E)-3-chlorohex-4-en-1-yne
b. (R,E)-3-chlorohex-2-en-5-yne
c. (S,E)-3-chlorohex-2-en-5-yne
d. (R,E)-3-chlorohex-4-en-1-yne
e. (R,E)-3-chloropent-4-en-1-yne
When determining longest chain numbering between alkenes and alkynes, treat them
equivalently and assign based on which gives the lowest numbering. The phantom rule
gives the alkyne higher priority (C,C,C vs C,C,H) and so the assignment is (S) as the Cl
is receding as shown. Note the repeat of the n in the base name
Cl
ALKYNES – ADDITION REACTIONS
42. Addition of excess HCl to 2-methyl-3-heptyne produces
a. 2-methyl-3,3-dichloroheptane
b. 2-methyl-2,3-dichloroheptane
c. 2-methyl-4,4-dichloroheptane
d. 2-methyl-3,4-dichloroheptane
e. 2-methyl-3-chloroheptene
f. More than one of the above
When adding hydrogen halides to alkynes do it twice, with the halogen ending up on the
same C owing to the carbocation stabilization by the non-bonded valence electrons of
the halogen. Follow Markovnikov’s rule initially. In this case both of the sp hybridized
C are 2o, so (a) and (c) are both formed
ALKYNES – ADDITION REACTIONS
43. What is the major product for the following reaction sequence?
a.
b.
c.
d.
e.
Lindlar’s catalyst gives cis-alkenes from alkynes, followed by anti Br2 addition
1. H2, Lindlars catalyst
2. Br2, CCl4
+ enantiomer
Br
Br
+ enantiomer
Br
Br
+ enantiomer
Br
+ enantiomerBr
Br Br
Br Br
ALKYNES – REACTIONS OF TERMINAL ALKYNES
44. Which of the following would not be a successful method for preparing 5-
methyl-1-hexyne
a.
Br1) NaNH2 (2 mol), liq NH3
2) NH4+
b.
HC CNa Cl+ c.
LiHC CH+
d.
Br
Br
1) NaNH2 (3 mol), liq NH3
2) NH4+
e.
Br
Br1) NaNH2 (3 mol), liq NH3
2) NH4+
NaNH2 is strong enough to cause the elimination of HBr twice (or once from
haloalkenes such as (a)) so long as an extra equivalent is available to drive the reaction
by acid-base reaction with the terminal alkyne formed (thus the need for a weak acid
in step (2)). The alkyl lithium species shown would certainly deprotonate acetylene,
but then you would just have an alkane