Organic chemistry section test (4)

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Organic Chemistry Subject Test 4

2.
1.

Rank the following structures in order of increasing
basicity:
H

Which of the following structures properly represents a
zwitterion?
A.

H

NH2

N

H

N
H

COOH
H

H

B.

H

NH3
I
analine

II
ammonia

H

COOH
H

H

C.

H
N

NH3

N

CH3

H

CH3

H

CH3
III
trimethylamine

A.
B.
C.
D.
E.

I, IV, II, III
IV, I, II, III
II, III, IV, I
III, II, I, IV
III, IV, I, II

COO

D.

NO 2

IV
p-nitro-analine

NH2
H

COO
H

E. None of the above
3.

O
CH3OH + CH3COH

?

A. CH3OH
B. CH3OCH3
C.

O
CH3OCCH3
D.

O
CH3CCH3
E. CH3CH2OH

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4.

What is the major product of the following reaction?

NO2

5.

What will be the favored product of the following
reaction?

O

Cl2

CH3

O S

AlCl 3

+

O

A.

H

NO2
Cl

CH3 C

CH3

OK

CH3

A.

CH3
O

Cl

C

CH3

CH3

B.

Cl

H
B.

NO2
C.

C.

O

NO2

O

S
O

Cl

Cl

CH3

O

D.

CH3

Cl

C

CH3

CH3

NO2
D.

E.

NO2

E.

CH3
O

C

CH3

CH3
Cl

H

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6.

What reagents are necessary for the following reaction
to occur?

7. For the following reaction, what will be the two major
products?

N(CH3 )2

CH3CH2MgBr + ? ∅ CH3(CH2)3OH
A. CH3CH2OH
B. CO2, HCl
C.

CH3CH

HNO3
H2SO4
NO2

O, H2O

D.

A.

O
H2C

N(CH
3)2

CH2 , H2O

N(CH
3)2

O2N

E. None of the above

NO2

NO2 O2N
B.

N(CH
3)2

N(CH
3)2

O2N
NO2

NO2
NO2
C.

N(CH
3)2

N(CH
3)2

O2N

NO2
NO2
N(CH
3)2

D.

O2N

NO2
N(CH
3)2

NO2

NO2
NO2

E.

8.

None of the above

At the isoelectric pH of a certain amino acid solution,
the amino acid may exist as a zwitterion. This is
A.
B.
C.
D.

a positively charged ion.
a negatively charged ion.
either a positively or a negatively charged ion.
an ion that carries both a positive and a negative
charge.
E. an ion without a constant charge.

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9.

ROOR
,∆
A. CH3CH=CH2 + HBr → CH3CH2CH2Br
B. (CH3)2C=CH2 + HCl ∅ (CH3)3CCl

OH
(CH3) 2CCH2Cl
C. (CH3)2C=CH2 + Cl2/H2O ∅

H3C

Br

H Br

E.

Br

H
Br

D.

+ Br2

and H3C Br

H Br

Which is an example of anti-Markovnikov addition?

and

H3C

Br

anti

addition

H

Br

H3C

H
Br

Br H
E.

+ Br2

syn
addition

H

H

Br Br
10.

Br2
CCl4
H3C
The product(s) of the above reaction is (are)

A.

Br Br
H3C

CH3

B.

and H3C Br

Br H
H3C

Br

Br H

C.

Br

H

and
Br
H3C

H
H3C

D.
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12. What is the most effective reducing agent for ethene?
11.
OCH3
H2 N

NH 2 KOH

O
C

P

A.
B.
C.
D.
E.

KMnO4
H3O+/H2O
H2/Pt
Fe/HCl
HNO3/H2SO4

OCH3

C4H9

13. Which of the following substituted phenols is the most
acidic?

The product P is

A.

A.
OCH3

CH3 O

OH

B.
HO

C
C 4H9

OCH3

H

OH

B.

NO2

H

C.

OH

O
C

H

C4 H9

Cl
D.

C.
H

H2N
E.

O
C

OH

OH

C4H 9

D.

NO2

OCH 3

C5 H11

OCH 3

E.

14. An aldol condensation involves the base-catalyzed
reaction between
A.
B.
C.
D.
E.

OCH3

NH2

OH

an ether and an alcohol.
an alcohol and an alcohol.
an aldehyde and an alcohol.
a ketone and an alcohol.
an aldehyde and an aldehyde.

NH2

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15. When placed in a basic solution, the ketone:

CH3
C

C

O

H

undergoes racemization, while the ketone of a similar
structure:

17. Which of the following compounds would be most

reactive towards chlorination of the aromatic ring?
A.

NH2

B.

NO2

CH3
C

C

O

C2H5

C.

does not. What could account for this observation?
A. Alkyl groups are electron-releasing.
B. The second compound forms a carbanion in basic
solution, while the first one does not.
C. The base catalyzes carbocation rearrangement.
D. There is no a-hydrogen in the second compound.
E. OH– adds to the carbonyl carbon in the second
molecule, generating a new chiral center.

D.

SO3H

E.

Br

16. In the light-activated chlorination of methane, the
chain-initiating step is

uv

→ 2Cl•
A. Cl2 

uv

B. CH4 + Cl2 
→ CH3Cl + HCl

uv

C. •CH3 + Cl2 
→ CH3Cl + Cl•

uv

D. Cl• + CH4 
→ •CH3 + HCl

uv

E. Cl• + Cl• 
→ Cl2

18. Which of the following is the most reactive as a
hydride donor?
A.
B.
C.
D.
E.

H2NNH2
CH3MgBr
CH3+
LiAlH4
All are equally reactive

19. Which compound below will undergo oxidation
without the cleavage of any carbon-carbon bond?
A.
B.
C.
D.
E.

t-butyl alcohol
acetone
acetaldehyde

ethyl methyl ketone
None of the above

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Questions 20-24 refer to the following choices:
A.
B.
C.
D.
E.

25. What is the major organic product of the reaction below?

ortho/para-directing with activation
ortho/para-directing with deactivation
meta-directing with activation
meta-directing with deactivation
neither directing nor activating

C(CH 3) 3
Br 2
hv

What is the effect of each of the following substituents
on electrophilic aromatic substitution?

C(CH 3) 3

20. —F
A.

21. —NO3

Br

22. —OCH3
23.

C(CH 3) 3

NH

C
O

CH3
B.

24. —SO3H

Br
Br

C(CH 3) 2

C.

Br

C(CH 3) 3

D.

E.

Br

C(CH 3) 3

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26. IR spectroscopy would be most useful in distinguishing
between which of the following pairs of compounds?
A.

O

O

CH 3CH 2CH 2CCH 2CH 3

and

28. Which of the structures below corresponds to paranitrobenzenesulfonic acid?
SO 3H
A.
NO 2

CH 3C CH2CH 2CH 3

SO 3H

B.

B.

CH 3
and

CH 3CH CH 2CH 2CH3

NO 2

CH 3CH 2CH 2CH 2CH2CH 3

C.

NO 2

SH

C.
O 2N

and

SO 3H

NO 2
CHO

CHO

D. CH 3CH 2OCH 2CH 2CH 3

and

CH 3CH 2CH 2CH 2OCH 3

E. CH 3CH 2CH 2CH 2OCH 3

and

CH 3CH 2CH 2CH 2CH2OH

D.
H 2N
SO 3 H

E.

O 2N

27. What is the major product of the reaction below?

O

1. CH 3CH 2MgBr
2.

+

H 3O

A. CH3CH2

C.

Br

D.
OH

B.

.

.

+ Br ∅ CH3CH2Br

.
.

CH CH + Br. ∅ CH CH .+ HBr
CH CH + Br. ∅ CH Br + CH .
CH CH . + CH CH . ∅ CH CH CH CH

B. CH3CH3 + Br ∅ CH3CH2Br + H

OCH 2CH 3
A.

29. Which of the following is a productive propagation step
in the free radical bromination of ethane?

E.

3

3

3

3

3

2

3

2

3

3

2

3

3

2

2

3

CH 2CH 3
Br

OCH 2CH 3

C.

OH

D.

E.

O

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30. Which of the following is a major organic product of
the reaction below?
OCH 3

CH 3COCl

AlCl 3
Cl

OCCH 3

A.

CH 3
OCH 3
H 3C

B.

C
O
OAlCl 2

C.

O
C
D.

OCH 3

H 3C

OCH 3
E.

Cl

Cl

STOP! END OF TEST.

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ORGANIC CHEMISTRY SUBJECT TEST 4

ANSWER KEY
1. B

7. B

13. E

19. C

25. D

2. C

8. D

14. E

20. B

26. E

3. C

9. A

15. D

21. D

27. B

4. B

10. B

16. A

22. A

28. E

5. B

11. D

17. A

23. A

29. C

6. D

12. C

18. D

24. D

30. B

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EXPLANATIONS
1.

B
The basicity of amines is dependent upon the stability of the unshared electron pair. Electron-withdrawing groups
decrease basicity, while electron-donating groups increase basicity. The benzene ring stabilizes the electron pair through
resonance, while methyl groups are electron-donating. Using ammonia as a reference point, it is clear that the two aniline
derivatives (compounds I and IV) must be less basic than ammonia (compound II) which is itself less basic than
trimethylamine (compound III). p-Nitroaniline would be even less basic that aniline because the nitro group which is
electron-withdrawing would add to the electron-withdrawing characteristics of the benzene ring. Thus, the order of
increasing basicity is IV, I, II, and III, or choice B.

2.

C
The definition of a zwitterion is a species that has both a positive and a negative charge. Choice C, which represents
the amino acid glycine at neutral pH, is indeed a zwitterion. Choice A is incorrect because it is not even an ion! Choice B is
simply a cation, while choice D is an anion.

3.

C
Alcohols and carboxylic acids will condense to form esters, a reaction known, appropriately, as esterification. This
information in itself is enough to answer the question, since only choice C is an ester. The reaction proceeds as follows: the
oxygen atom on the alcohol acts as a nucleophile and attacks the carbonyl carbon, which becomes sp3 hybridized with four
groups attached. The carbonyl double bond is restored as the —OH group originally on the carboxylic acid becomes
protonated and leaves as a water molecule.

4.

B
Nitro groups are strongly deactivating meta directors for electrophilic aromatic substitutions. The chloro group will
attach at the meta position. However, halides are also deactivating, so the presence of two deactivating species (nitro and
chloro) makes a third substitution unlikely. In addition, halides are ortho/para directors, so their influence would be in
conflict with that of the nitro group.

5.

B
The p-toluenesulfonate group, or tosylate group, is an excellent leaving group. The molecule will therefore be
expected to undergo either a nucleophilic substitution or an elimination reaction. The other reactant, a tert-butoxide ion, is a
strong and bulky base which tends to favor an elimination reaction. The extraction of a proton from the cyclohexane and the
departure of the leaving group will thus lead to the formation of cyclohexene.

6.

D
The Grignard reagent (ethyl magnesium bromide) is a potent nucleophile. Choice A, ethanol, is an unlikely target for a
nucleophile, as it does not have a good leaving group in the absence of a proton source. Carbon dioxide would be a good
target for the Grignard reagent, but would only result in a three carbon species, whereas the target product has four carbons.
Choice C will give a four carbon alcohol, but it will yield 2-butanol instead of 1-butanol as desired: the ethyl group from
the Grignard reagent will add to the carbonyl group. However, when the Grignard reagent is added to the reagents in choice
D, the nucleophile will add to one of the carbons of the epoxide. The bond between that carbon and the oxygen atom would
break, thus opening up the ring. The other carbon will retain the oxygen, which will become a primary alcohol when water
is added.

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7.

B
The —N(CH3)2 group is a strongly activating ortho/para director for electrophilic additions. While it is true that nitro
groups are deactivating and meta directing, the effect of the dimethylamino group overpowers that of the nitro group. This
will lead to nitration in the ortho/para positions relative to the —N(CH3)2 substituent, which are shown in choice B. The
second product shown in choice C is also an ortho-substituted product, but compared to the para-substituted product in
choice B, it will not be as favored because of steric effects.

8.

D
Zwitterions are neutral molecules that carry both a positive and a negative charge. Many amino acids are zwitterions
in neutral aqueous solutions because they have a positively charged group (—NH3+) and a negatively charged group (—
COO–). At other pHs, one of the groups may change their protonation state, thus losing the charge it carries, resulting in the
amino acid acquiring a net positive or negative charge. Choices A and B are incorrect because they are cations and anions,
respectively. Choice C is actually the definition of an ion, while choice E describes multivalent ions, such as iron (Fe2+,
Fe3+).

9.

A
It is perhaps best to first understand what Markovnikov addition is before addressing the case of anti-Markovnikov
addition. In the addition of a hydrogen halide (or any nonsymmetrical reagent) to a double bond, the reaction often
proceeds through an ionic mechanism: the bond between the two parts of the molecule breaks heterolytically as the positive
part acts as an electrophile and adds to one of the carbon atoms, leading to a carbocation intermediate. The other, negative
portion of the molecule then add to the other carbon atom. In the case where the substituent groups on the two carbon atoms
are different, Markovnikov’s rule predicts which portion of the molecule will add to which carbon: the addition will

proceed such that the most stable (most highly substituted) carbocation is formed as the intermediate. For a hydrogen
halide, then, the hydrogen (as a proton) will add to the less substituted carbon atom, so that the resulting positive charge
will reside on the more heavily substituted one. The halide ion will then add to this more highly substituted carbon. With
HBr (but not any other kind of hydrogen halide!) in the presence of peroxides (the conditions in choice A), however, it has
been observed that the addition product is different from what is predicted by Markovnikov’s rule: the bromine ends up on
the less substituted carbon. This is known as anti-Markovnikov addition. This phenomenon arises because in the presence
of peroxides, the addition proceed through a radical mechanism rather than the usual ionic one: the peroxide ROOR cleaves
upon heating to generate two alkoxyl radicals which extract the hydrogen atom from HBr. The neutral bromine atom (not
bromide ion!) then attacks the less substituted (and hence less sterically hindered) carbon atom to produce the more stable
(more highly substituted) radical intermediate. Addition is complete with the extraction of hydrogen atom by the radical
intermediate.
Choice B is incorrect because it depicts a normal Markovnikov addition: the proton adds to the terminal carbon,
leading to a stable tertiary carbocation intermediate, to which the chloride ion then adds. Choice C represents halohydrin
formation: the first step is identical to halogen addition, i.e. the formation of a cyclic carbocation. The next step, however,
has water as the attacking nucleophile that opens up the ring. Subsequent deprotonation then yields the halohydrin. The
water molecule attacks the more highly substituted carbon atom preferentially. Choices D and E are incorrect because in the
case of a symmetrical addition reagent (molecular bromine here) the Markovnikov vs. anti-Markovnikov designation is
irrelevant.

10.

B
The bromination of alkene proceeds through a cyclic bromonium intermediate, forming a three-membered ring
consisting of bromine and the two carbon atoms involved in the double bond. The remaining bromide ion, acting as a
nucleophile, will add in an orientation anti to the first one as it opens up the ring. Two possible products (enantiomers of
each other) will form. (See diagram on the next page.)

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Choice A is incorrect: where did the extra methyl group come from? Choices C and E are both incorrect because in
the absence of radical-generating agents, the bromine will add only to the unsaturated carbons involved in the double bond.
Choice D is incorrect because the two bromine atoms will add to adjacent carbons rather than the same carbon.

Br
C

Br

H3C Br

+

Br

C

C

H3C

C

H3C

Br

H

Br+

Br

Br

OR
C

H

C

H3C

H3C

Br

Br-

11.

D
This is an example of the Wolff-Kishner reduction of a ketone to an alkane (conversion of —C=O groups into —CH2
groups), often seen following a Friedel-Crafts acylation reaction. This is a synthetically useful pathway because a
straightforward Friedel-Crafts alkylation may lead to the rearrangement of the alkyl group one is trying to add. Another
way to reduce the acyl group to an alkyl group is by the Clemmensen reduction which, unlike the Wolff-Kishner reduction,
takes place under acidic conditions.

Choice A is a partially reduced compound that would result if we had used hydride donors like LiAlH4 or NaBH4
instead. With the conditions and reagents in this question, however, the oxygen is completely removed. Choices B and C
are incorrect because the reaction conditions are insufficient to break the relatively stable Ar–O bonds (requires HI or HBr
at high temperatures). Choice E is incorrect for the same reasons as B and C, and also because hydrazine is not a reagent
that is used to donate amine groups to a benzene ring.

12.

C
Hydrogenation of alkenes in the presence of a metal catalyst (Pt, Ni, or Pd) is a standard reaction that every student
should be well familiar with. Choice A is incorrect because KMnO4 is a strong oxidizing agent that will cleave a double
bond by oxidation or produce a geminal diol, depending upon reaction conditions. Choice B lists the conditions for acidcatalyzed hydration: it will lead to the addition of a hydrogen atom on one side but a hydroxyl group on the other. Neither
choice D nor choice E is capable of addition of hydrogen atoms: in particular, you should recognize the reagents in choice
E as those for the nitration of aromatic compounds in electrophilic aromatic substitution reactions.

13.

E
To stabilize the negative charge of the phenolate ion (the resultant anion when a phenol acts as an acid and
deprotonates), an electron-withdrawing group is required. Both methoxy and amine groups are electron-donating, so
choices A and D can be eliminated. Chlorine is electron-withdrawing, but only weakly so compared to the nitro group, so
choice C can also be eliminated. This leaves either the m-nitrophenol or o-nitrophenol (choices B and E respectively).
Nitro-substituted aromatic rings are meta directing. This means that the electron density will be higher in the m position

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than in the o/p position. Since we are looking for low electron density (to stabilize the negative charge) at the hydroxyl

group position, choice E would be more acidic than choice B.

14.

E
Aldol condensations occur between aldehydes, one acting as a carbanion nucleophile and the other as an electrophile.
The end product is a b-hydroxyaldehyde, i.e. it has both an aldehyde and an alcohol functionality, hence the name aldol
condensation. The base catalyzes the reaction by increasing the rate of tautomerization of the aldehyde into the enolate ion.
The enolate ion is a nucleophile and can attach to the carbonyl carbon of the other aldehyde molecule. The reaction is
illustrated below for the case of two formaldehyde molecules:

O-

O
CH2

CH

CH2

O
-:CH2

CH

CH

H
H3C

-:OH

CH
O
OH O

O

CH3CHCH
2CH

CH3CHCH
2CH
: OH

OH

Note that the β-hydroxyaldehyde product is rather unstable and can undergo dehydration to give an unsaturated
compound in which a carbon-carbon double bond is conjugated with the carbonyl bond.

15.

D
The hydrogen atom attached to the carbon atom next to the carbonyl carbon (the α-hydrogen, the carbon to which it is
attached being the α-carbon) is relatively acidic because the resulting anion upon deprotonation can be stabilized via
resonance by the carbonyl oxygen. The resulting carbanion can then be reprotonated from above or below the plane of the
ion, giving a racemic mixture. Since there is no α-hydrogen in the second compound, it will not racemize in a basic
solution. While the statement in choice A is correct, it has nothing to do with the phenomenon we are trying to account for.
Choice B is false: none of the hydrogens in the second compound is expected to be acidic enough to be plucked off,
whereas the first molecule definitely yields a carbanion upon deprotonation. Choice C is incorrect because no carbocation is

formed; besides, rearrangement would change the constitutional structure of the molecule, instead of its chiral properties.
Choice E is incorrect for two reasons. First, both compounds could conceivably be attacked by OH- groups. Second, attack
of the carbonyl by OH- would result in a gem diol, which would not be chiral, as is stated.

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16.

A
There are three steps in free radical reactions. The first is the initiation, where a bond is cleaved homolytically to
produce free radicals. This is depicted in choice A: the UV radiation breaks the Cl-Cl bond to give two neutral chlorine
atoms, which are radicals because they each possess an unpaired electron. Chain propagation steps are steps in which one
free radical is used to generate another. This is shown in choices C and D. Finally, chain termination steps describe
reactions where two free radicals combine to form a molecule, as in choice E. These steps “soak up” free radicals, and
eventually will stop the chain reaction. Choice B does not involve any radicals, and can be thought of as just a double
displacement reaction.

17.

A
Amine groups are electron-donating, and thus will activate electrophilic substitutions to the benzene ring. Choices B,
D, and E are all incorrect because the substituents to the benzene rings shown are all electron-withdrawing, and thus
deactivating. Choice C is incorrect because electron-donating species increase the rate of electrophilic aromatic substitution
over and above that of the straight benzene ring.

18.

D
This question may be very easily answered if one remembers the name of LiAlH4 (or LAH) to be lithium aluminum
hydride. It is a powerful hydride donor and is commonly used as a reducing agent. Choice A, hydrazine, although a
reducing agent, is not a hydride donor. Choice B is a Grignard reagent and can be considered a carbanion donor, but not a
hydride donor per se. Choice C is incorrect because a cation is much more likely to grab hydrides than to donate them.

19 .

C
This question can be answered without knowledge of any specific organic reactions, as long as one is able to identify
the structures of the molecules and knows what oxidation involves. In the context of organic chemistry, oxidation is most
conveniently thought of as increasing the number of bonds to oxygen. The carbon atoms in molecules A–D all have four
bonds already (either four single bonds or two single and one double bond). To oxidize them further would require the
formation of another carbon-oxygen bond (either forming a new C-O single bond or turning a C-O bond to a double, i.e.
carbonyl bond), which can only be formed at the expense of cleaving another (non C-O) bond. In choices A, B, and D, all
the non-C-O bonds are carbon-carbon bonds, and so oxidation would involve the cleavage of one such bond. Aldehydes,
however, possess a C-H bond which can be cleaved as a hydroxyl group attaches to the carbonyl carbon, yielding a
carboxylic acid.

20. B 21. D 22. A 23. A 24. D
In electrophilic aromatic substitution reactions, the rate-determining step is the formation of the positively-charged
arenium ion intermediate. Substituents that are electron-donating would stabilize this intermediate, thereby increasing the
rate of reaction (activating the aromatic substrate). Resonance structures also show that in such cases, further substitution
at the o/p positions (relative to the electron-donating group) are favored in general. Examples of such groups include the
methoxy group (#22), the amide group (#23), the hydroxyl group, and, to a more modest extent, alkyl groups. On the other
hand, electron-withdrawing groups destabilize the intermediate, deactivating the aromatic substrate. Substitutions that are
forced to occur by imposing more rigorous reaction conditions would take place preferentially in the meta position, because
it is there that their unfavorable effects are least felt in general. Examples of such groups include the nitro group (#21), the
sulfonic acid group (#24), the cyano group, and the carboxyl group. Halides (such as fluorine in #20) are weakly electronwithdrawing and are ortho/para-directing. As such, they are the exception to the general rule which states that all electronwithdrawing species are deactivating and meta-directing. The reason for this is that their high electronegativity exerts an
inductively electron-withdrawing effect (hence deactivating the ring), yet they can stabilize positive charges at the o/p

positions by resonance. The resonance structure involves it forming a double bond to the phenyl ring and bearing a formal
positive charge. It is therefore not a significant resonance structure and is unable to overcome the deactivating effect caused
by induction.

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25.

D
This reaction is an example of free radical halogenation. In the initiation step, UV light is used to homolytically split
the bromine into two bromine free radicals. During the propagation steps a bromine radical will attack the alkane,
specifically it will remove a hydrogen from the carbon atom that can form the most stable fee radical (the most substituted
carbon atom). The alkyl radical will then attack another Br2 molecule, resulting in the addition of Br to the radical carbon
and the formation of another bromine radical. The carbon attached to the ring, at the center of the t-butyl group, has no
hydrogens to give, so the bromine cannot add there. The carbon atom to which the t-butyl group is attached is the next
most substituted (it is 3o), so that is the site of the bromine addition.

Br

Br

H3C
H3C

26.

Br

2 Br

CH3
C

C

H

H3C
H3C

CH3
C

C

Br

Br

H3C
H3C

CH3
C

C

Br

E
IR spectroscopy is useful in identifying the functional groups a molecule possesses. In order for IR to be useful in
distinguishing between molecules, they must have different functional groups. Only choice E, an ether and an alcohol, can
be differentiated (the C–O of an ether peaks at 1050-1150 cm-1, the O–H of the alcohol has a broad peak at 3100-3500 cm-1
. Choices A and B could be differentiated using H-NMR, since the have different amounts of hydrogens. Choices C and D
could also be differentiated using H-NMR, because the different locations of the electron withdrawing groups would cause
different downfield shifting of the associated hydrogens.

27.

B
This reaction is an example of nucleophilic acyl addition, with a Grignard reagent acting as the nucleophile. The
carbonyl carbon of cyclobutone (or any carbonyl carbon) is susceptible to nucleophilic attack for two reasons: (1) it has a
partial positive charge because of the electron withdrawing effects of oxygen. (2) it is trigonal planar, so there is no steric
hindrance. The alkyl group of the Grignard reagent has a partial negative charge (because it is more electronegative than
magnesium, a metal), and so adds to the carbonyl carbon. You can also think of this reaction as electrophilic addition
across a double bond, with Mg as the electrophile. The intermediate of the reaction, bromomagnesiumcylcobutoxide,
removes a proton from H3O+ and becomes an alcohol.

28.

E
Two substituents on a benzene to are para when they are three carbons away from each other. The groups in choice A
are ortho (next to one another), and the groups in choice B are meta (2 carbons away). Choices C and D are wrong because
they are not nitrobenzenesulfonic acid. Choice C is para-nitrothiolphenol and D is para-aminobenzenesulfonic acid.

29.

C
A productive propagation step is when a free radical attacks a non-radical compound, producing another free radical
that continue the reaction to form the desired product. Choices A and D are termination steps: two free radicals are
combining, decreasing the amount of free radicals that can continue the reaction. Choice B is incorrect; when bromine
radical attacks ethane, it removes a proton and becomes HBr, and the ethane becomes a radical. Choice D is a nonproductive propagation: the desired product is ethyl bromine, not methyl bromine.

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30.

B
This reaction is an example of Friedel-Crafts acylation. The AlCl3 serves as a Lewis acid catalyst, and makes the
ethanoyl chloride a stronger electrophile (a strong electrophile is necessary to disrupt the stable conjugated benzene ring).
The ethanone adds to the benzene ring para to the methoxide group, because –OR groups are ortho/para directing. The
other products of the reaction are HCl and the AlCl3 catalyst .

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