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