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Chemistry 261 Exam 2 Practice Fall 2017 The following practice examination contains 30 questions valued at 3 point/question unless otherwise noted. Wednesday’s exam will also contain 30 questions, with an additional 9 points available as bonus/challenge questions Name: KEY ELECTRON DEFICIENT COMPOUNDS AND LEWIS ACID-BASE DEFINITIONS 1. Which of the following is not a Lewis acid? a. AlCl 3 b. H 3 O + c. FeCl 3 d. SO 3 e. N/A; all of the above will act as Lewis acids By definition, a Lewis acid is an electron pair acceptor. G.N. Lewis’s development in 1906 (the same year Brønsted and Lowry developed the proton transfer concept) focuses on acceptors of an electron pair, is the broadest definition of acidity, and as such Bronsted acids are a subset of Lewis acids 1 . Notice that SO 3 reacts with H 2 O to generate H 2 SO 4 , a principal component of acid rain 1 For the interested student, the Lewis acid Wiki is pretty good
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Page 1: KEY Exams … · para directing, activating the benzene ring towards electrophilic aromatic substitution (EAS, chapter 16). However, when in the presence of the common catalyst for

Chemistry 261

Exam 2 Practice

Fall 2017

The following practice examination contains 30 questions valued at 3 point/question

unless otherwise noted. Wednesday’s exam will also contain 30 questions, with an

additional 9 points available as bonus/challenge questions

Name: KEY

ELECTRON DEFICIENT COMPOUNDS AND LEWIS ACID-BASE DEFINITIONS

1. Which of the following is not a Lewis acid?

a. AlCl3

b. H3O+

c. FeCl3

d. SO3

e. N/A; all of the above will act as Lewis acids

By definition, a Lewis acid is an electron pair acceptor. G.N. Lewis’s development in 1906

(the same year Brønsted and Lowry developed the proton transfer concept) focuses on

acceptors of an electron pair, is the broadest definition of acidity, and as such Bronsted

acids are a subset of Lewis acids1. Notice that SO3 reacts with H2O to generate H2SO4,

a principal component of acid rain

1 For the interested student, the Lewis acid Wiki is pretty good

Page 2: KEY Exams … · para directing, activating the benzene ring towards electrophilic aromatic substitution (EAS, chapter 16). However, when in the presence of the common catalyst for

ELECTRON DEFICIENT COMPOUNDS AND LEWIS ACID-BASE REACTIONS

2. The amino group on the weak base aniline (anilinium ion pKa = 4.87) is strongly ortho and

para directing, activating the benzene ring towards electrophilic aromatic substitution

(EAS, chapter 16). However, when in the presence of the common catalyst for

electrophilic aromatic bromination of the benzene ring, FeBr3, the reaction is strongly

inhibited, with complex I being formed. The aniline amino group has acted as

Complex I

a. A Lewis acid

b. A Lewis base

c. A Brønsted acid

d. A Brønsted acid

e. An electrophile

Sorry for the wordiness on this one, but it shows that identifying the potential for Lewis

adduct formation can be critical to assessing reaction potential. One way to make better

sense of FeBr3 as a Lewis acid is to name it – Iron(III) bromide…iron is in the +3

oxidation state and has available d orbitals for coordinate covalent bond formation

NUCLEOPHILES, ELECTROPHILES, AND LEAVING GROUPS

3. In the reaction between NaCN and propyl chloride to generate cyanopropane (formally

butanenitrile) the chloro group acts as

a. A nucleophile

b. An electrophile

c. A Lewis acid

d. A Lewis base

e. A leaving group

Nucleophiles, electrophiles, and leaving groups, oh my! Most of the reactions you will see

in this class are 2 electron processes involving a nucleophilic attacking species (I suppose it

“attacks” based on curved arrow conventions), an electrophilic species being attacked, and

a leaving group to maintain the requisite number of bonds for a noble gas electronic

configuration

Page 3: KEY Exams … · para directing, activating the benzene ring towards electrophilic aromatic substitution (EAS, chapter 16). However, when in the presence of the common catalyst for

NUCLEOPHILES, ELECTROPHILES, AND LEAVING GROUPS

4. Which of the following is the strongest electrophile?

a. CH3CO2CH3

b. CH3CO2H

c. CH3OCH3

d. CH3COCH3

e. CH3CCNa+

Clearly, you must be able to translate condensed structural formulas into full Lewis

structures: (a) is an ester (b) is a carboxylic acid (c) is an ether, and (d) is a ketone. As

a [poorly] stabilized carbanion, (e) is a strong base and nucleophile. The carbonyl is the

most polarizing, but the extra oxygen in options (a) and (b) make them more electron rich

BRØNSTED-LOWRY ACIDS AND BASES – CONJUGATE ACID AND BASE DEFINITION

5. Which of the following is the conjugate base of ammonia?

a. NH4+

b. NH3

c. NH2

d. N2H4

e. N/A; none of the above is the conjugate base of ammonia

Critical to understand this, and to understand the weaker the acid (in this case the weak

base ammonia has somehow been forced to act as an acid) the stronger the conjugate base

– the amide ion is a very strong base

BRØNSTED-LOWRY ACIDS AND BASES – CONJUGATE ACID AND BASE DEFINITION

6. Which of the following are Brønsted bases in the following equilibrium?

H2O + H2O OH + H3O+

a. H2O and OH

b. H3O+ and OH

c. H2O and H2O

d. H3O+ and H2O

e. Only OH as water cannot simultaneously be an acid and a base

Water in the forward direction, hydroxide in the reverse direction. Of course water is

amphiprotic, accepting or donating H+ depending on circumstance – this is fundamentally

the basis for the pH scale

Page 4: KEY Exams … · para directing, activating the benzene ring towards electrophilic aromatic substitution (EAS, chapter 16). However, when in the presence of the common catalyst for

FUN WITH pKa’s

7. What is the approximate pKa of hexanoic acid (commonly caproic acid)?

a. 2

b. 3

c. 4

d. 5

e. 6

Actual pKa = 4.88. Acetic acid pKa = 4.76. Without anything to distort the electron

distribution in relation to acetic acid, the pKa will not change much

FUN WITH pKa’s

8. Which of the following compounds would be deprotonated by potassium tert-butoxide?

a. Hexane

b. 1-Hexene

c. 1-Hexyne

d. (b) & (c)

e. None of the above

This question really isn’t that difficult given tBuOK is the conjugate base of an alcohol (an

alkoxide, with the negative charge on an electronegative oxygen) while the answer options

would leave a carbon anion. Options (1) or (2) sinply are not happening regardless of the

base employed, while (c) has an approximate pKa = 25 while tBuOH has a pKa = 18. A 7

unit difference is too large to affect any significant deprotonation

FUN WITH pKa’s

9. Which of the following compounds would be deprotonated by sodium hydride?

a. Hexane

b. 1-Hexene

c. 1-Hexyne

d. (b) & (c)

e. None of the above

With a pKa ≈ 35, NaH easily deprotonates terminal alkynes of pKa = 25. I suppose by

this point you have rightly concluded the free energy of activation for proton transfer

reactions is sufficiently low that if there is a favorable pKa difference, reaction will occur.

In fact, some of the very strong base reactions must be run at reduced temperature so

that things don’t get out of hand

Page 5: KEY Exams … · para directing, activating the benzene ring towards electrophilic aromatic substitution (EAS, chapter 16). However, when in the presence of the common catalyst for

FREE ENERGY AND CHEMICAL EQUILIBRIUM

10. Which of the following correctly expresses the relationship between pKa and the standard

free energy of dissociation?

a. Gao = 2.3RTpKa

b. Gao = -2.3RTpKa

c. Gao = 10pKa/2.3RT

d. Gao = 10pKa/2.3RT

e. pKa = 2.3RTGao

Math! Note the change from the more usual Ka form given the definition of p as –log; i.e.

Gao = -2.3RTlogKa. Rearranging and taking the antilog places -Ga

o/2.3RT as an

exponent. Since 2.3RT translates to 1.4 kcal/mol under standard conditions, slight

changes in Gao correspond to large scale changes in equilibrium ratios. Spending some

time with table 3.2 gives you a very nice feel for this fact

STRUCTURE-ACIDITY RELATIONSHIPS

11. Rank the bold-faced hydrogens in the following compounds from most acidic to least

acidic.

a. I > II > III > IV > V

b. III > V > II > I > IV

c. V > II > IV > III > I

d. III > I > V > II > IV

e. V > III > I > II > IV

Protonated ether (-2.5) benzoic acid with electron withdrawing group (≈3) phenol (10)

vinylic (44) and straight up alkane (50)

STRUCTURE-ACIDITY RELATIONSHIPS

I II III IV V

CF3

COOH

H

O

H H

OH

CH3

H3C

H3C

Page 6: KEY Exams … · para directing, activating the benzene ring towards electrophilic aromatic substitution (EAS, chapter 16). However, when in the presence of the common catalyst for

12. Rank the identified hydrogens from the most acidic to least acidic in the compound

shown below

a. Ha > Hb > Hc > Hd

b. Ha > Hc > Hd > Hb

c. Hc > Hd > Ha > Hb

d. Hd > Ha > Hc > Hb

e. Hc > Ha > Hd > Hb

I really like this question – imagine you were starting at pH = -1 and then the pH was

slowly raised…can you see the protons being sequentially removed? The ring structure

corresponds to the non-nuclephilic (too bulky and inductive removal of lone pair electron

density) weak base pyridine (pyridinium pKa = 5.2) which is a very useful “proton sink” for

reactions that generate acid by-products

STRUCTURE-ACIDITY RELATIONSHIPS

13. For the simple hydrides listed below, which is the correct order of decreasing pKa

values?

a. CH4 > NH3 > H2O > H2S > HBr

b. HBr > H2S > H2O > NH3 > CH4

c. HBr > H2O > NH3 > H2S > CH4

d. NH3 > H2S > CH4 > H2O > HBr

e. H2S > H2O > HBr > NH3 > CH4

Decreasing? With methane on the list? Game over! Notice H2S has a pKa = 7 (thiols

similar to phenol @ 10-12), while water pKa = 15.7 just as HCl is a much stronger acid

than HF

Page 7: KEY Exams … · para directing, activating the benzene ring towards electrophilic aromatic substitution (EAS, chapter 16). However, when in the presence of the common catalyst for

ACID-BASE REACTION PRODUCT PREDICTION – CURVED ARROW APPROACH

14. What is/are the products of the following acid-base mechanism?

a.

b.

c.

d.

e. None of the above

Pretty straightforward – nowhere else for the carbanion to resonate to, and having an

electron deficient C next to a carbanion simply isn’t tenable (option (d))

H

BrMg

+ BrMg H

MgBr

+ CH3CH3

H

MgBr

+ CH3CH3

Page 8: KEY Exams … · para directing, activating the benzene ring towards electrophilic aromatic substitution (EAS, chapter 16). However, when in the presence of the common catalyst for

ACID-BASE REACTION PRODUCT PREDICTION – CURVED ARROW APPROACH (IMPLIED)

15. What is/are the products of the following acid-base reaction?

a.

b.

c.

d.

e. N/A; There is no acid-base reaction between the reactants shown

Option (a) seems pretty tempting, but comparing the pKa values shows the alkyl amide ion

(yes, I don’t like the naming any more than you do) competes much more fiercely than the

alkynide ion for the hydrogen in question

N

H

+Na

N +

Na

H

NH

+

Na

H

H

N

NH +

Na

H

Page 9: KEY Exams … · para directing, activating the benzene ring towards electrophilic aromatic substitution (EAS, chapter 16). However, when in the presence of the common catalyst for

STRUCTURE AND BONDING IN ALKENES

16. Which of the following unknown acyclic molecules contains no rings or double bonds

a. C5H13N

b. C5H10O

c. C6H10Cl2O

d. C7H12IBr

e. More than one of the above

CnH2n+2, adding 1 for N, ignoring O, and counting halogens as hydrogen

STRUCTURE AND BONDING IN ALKENES

17. For which of the following molecules can resonance give a completely equivalent, non-

charge separated form?

a. Benzene

b. 1,3,5-Hexatriene

c. 1,3-Cyclobutadiene

d. 1,3-Butadiene

e. More than 1 of the above

Benzene is pretty obvious, 1,3-cyclobutadiene less so. If the molecule is not cyclic, then

a charge separated form must be the end result. 1,3-cyclobutadiene is the poster child

for anti-aromatic compounds, and has a very short half-life, since electrons need to be

placed in non-bonding MO’s2

DOUBLE BOND STEREOISOMERS/CIS-TRANS & E,Z NOMENCLATURE

18. Please draw the structure for (Z)-2-vinyl-2-pentenoic acid

Z for zusammen or together. Phantom rule on carbonyl oxygen gives higher priority than

phantom rule on ethenyl group

2 Much more on this later – the interested student should see Hückel’s Rule

Page 10: KEY Exams … · para directing, activating the benzene ring towards electrophilic aromatic substitution (EAS, chapter 16). However, when in the presence of the common catalyst for

DOUBLE BOND STEREOISOMERS/CIS-TRANS & E,Z NOMENCLATURE

19. Please name the following compound, adhering to IUPAC conventions

Name: 2,5-cyclohexadien-1-ol

The higher oxidation state of the alcohol dictates numbering. Alkenes are within a 6

membered ring – on need to indicate Z/cis or E/trans. Indicating the alcohol as the “1”

position is overkill in this case

UNSATURATION NUMBER OR INDEX OF HYDROGEN DEFICIENCY

20. What is the unsaturation number for the compound immediately below

a. 2

b. 3

c. 4

d. 5

e. 6

5 bonds and 1 ring

PHYSICAL PROPERTIES OF ALKENES

21. Which of the following will have the highest boiling point

a. 1-pentene

b. 1-hexene

c. cyclopentene

d. cyclohexene

e. All of the above boil within 5 oC of one another

Some questions I like so much I may just revisit them (wink). See practice quiz 2 for

further details

Page 11: KEY Exams … · para directing, activating the benzene ring towards electrophilic aromatic substitution (EAS, chapter 16). However, when in the presence of the common catalyst for

RELATIVE STABILITIES OF ALKENE ISOMERS

22. Which alkene would you expect to be most reactive toward acid-catalyzed hydration?

Hint: resonance possibilities may surprise you

a. Styrene (vinyl benzene)

b. Toluene (methyl benzene)

c. 1-Hexene

d. 2-Hexene

e. Cyclohexene

The benzylic cation is highly resonance stabilized, which gives it the approximate stability

of a 2o carbocation. Notice that the cation formed has stability more like a 3o

carbocation since it 2o to begin with (C+ bonded to benzene ring as well as the terminal C)

ADDITION OF HYDROGEN HALIDES/CARBOCATION STABILITY/CARBOCATION REARRANGEMENTS

23. Which of the following carbocations would not undergo rearrangement?

a.

CH3CHCHCH3

CH3 b.

CH3CHCCH3

CH3

CH3

c.

CH3CCH2CH3

CH3

d.

CH3CHCH2

CH3

e.

CH3CCHCH2CH3

CH3

CH3

The one that is already a 3o carbocation. (a) and (d) would undergo hydride shifts, while

(b) and (e) undergo methide shifts

Page 12: KEY Exams … · para directing, activating the benzene ring towards electrophilic aromatic substitution (EAS, chapter 16). However, when in the presence of the common catalyst for

ADDITION OF HYDROGEN HALIDES/CARBOCATION STABILITY/CARBOCATION REARRANGEMENTS

24. Treating 1-methylcyclohexene with H3O+ would produce which of the following as the

principal product(s)?

II III

IV V

I

HO HO

HO

HOOH

a. I & V

b. II

c. III & IV

d. IV

e. I, III, & V

Direct formation of a tertiary carbocation followed by the addition of water

ADDITION OF HYDROGEN HALIDES/CARBOCATION STABILITY/CARBOCATION REARRANGEMENTS

25. What is the principal product of the following reaction?

a.

b.

c.

d.

Br2, CH3OH

+ enantiomerOH

Br

+ enantiomer

OCH3

Br

+ enantiomer

Br

H3CO

Page 13: KEY Exams … · para directing, activating the benzene ring towards electrophilic aromatic substitution (EAS, chapter 16). However, when in the presence of the common catalyst for

e. More than one of the above

Variation on halohydrin formation where alcohol is the solvent, leading to an ether (as

opposed to an alcohol product). Other principles apply – since you are adding Br2, not

HBr, do not expect a rearrangement. 2o carbocation is more stable, so we expect

alcohol addition adjacent the quaternary positon

ADDITION OF HYDROGEN HALIDES/CARBOCATION STABILITY/CARBOCATION REARRANGEMENTS

26. Reacting 1-methylcyclopentene with bromine monochloride (BrCl) yields which of the

following as principal products?

I II III

Cl

Br

CH3 CH3

Br

Cl

Cl

Br

CH3

IV

CH3

Br

Cl

V

CH2Cl

Br

+

enantiomer

+

enantiomer

+

enantiomer

+

enantiomer

+

enantiomer

a. I

b. II

c. III

d. IV

e. V

Too hard? Not if you apply fundamental principles! Chlorine is more electronegative than

bromine, making the bromine end of the molecule more electrophilic. Thus, a bromonium

ion bridges between the [formerly] alkene carbons in question, with a large degree of

carbocation character at the 3o position. Chloride attacks this site of greatest

electrophilicity from “below”, leaving the chloro and bromo groups on opposite faces of

the ring

+ enantiomer

OCH3

Br

Page 14: KEY Exams … · para directing, activating the benzene ring towards electrophilic aromatic substitution (EAS, chapter 16). However, when in the presence of the common catalyst for

REACTION RATE CONSIDERATIONS

27. Which of the following correctly expresses the relationship between rate and the

standard free energy of activation?

a. log(rate) = 2.3RTGo‡

b. log(rate) = -2.3RTGo‡

c. log(rate) = -Go‡/2.3RT

d. log(rate) = Go‡/2.3RT

e. ln(rate) = -Go‡/2.3RT

Notice the negative value associated with the free energy change – larger free energy

barriers lead to larger negative exponents and much slower rates of reaction. To wit, just

as the 1.4 kcal/mol difference changes the equilibrium concentration by a factor of 10, a

1.4 kcal/mol increase in free energy of activation slows the reaction by a factor of 10

REACTION RATE CONSIDERATIONS

28. Briefly, what is the difference between the standard free energy of activation and the

standard free energy of dissociation?

The standard free energy of activation pertains to the relationship between free

energy changes and rate of reaction, whereas the standard free energy of

dissociation relates overall free energy changes to equilibrium concentrations of

reactants and products. Importantly, there are only positive values for Go‡, since

it is the difference between the transition state and reactant system free energy.

For Gao, negative free energy changes (release of free energy) between product

and reactant free energy levels are common, if not expected, with large negative

changes driving the reaction to completion, as should be evident with an overall

positive value of Gao in Ka = 10^-Ga

o/2.3RT

Page 15: KEY Exams … · para directing, activating the benzene ring towards electrophilic aromatic substitution (EAS, chapter 16). However, when in the presence of the common catalyst for

CATALYSIS/CATALYTIC HYDROGENATION OF ALKENES

29. Under usual hydrogenation conditions, how many moles of diatomic hydrogen will react with

the compound immediately below

a. 2

b. 3

c. 4

d. 5

e. 6

5 bonds and 1 ring, and the ring won’t react

CATALYSIS/CATALYTIC HYDROGENATION OF ALKENES

30. Which of the following will convert cyclohexene into cyclohexane?

a. H2/Ni/high pressure

b. H2/Pd/C

c. H2O/Ni

d. More than 1 of the above

e. All of the above

Hydrogenation of an alkene requires hydrogen (and a metal catalyst). LiAlH4 and NaBH4

provide hydride, which effectively amounts to hydrogenation upon acidic workup, but only

for more electrophilic systems like carbonyls, and NOT for alkenes. Similarly, carbonyls

are typically resistant to conditions that hydrogenate alkenes – a very useful difference


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