Organic chemistry section test (4)

OH– adds to the carbonyl carbon in the second molecule, generating a new chiral center.. The reaction proceeds as follows: the oxygen atom on the alcohol acts as a nucleophile and attack

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Organic Chemistry Subject Test 4

1 Rank the following structures in order of increasing

basicity:

N

H H

I

analine

N

H H

IV

p -nitro-analine

N

H

H

H

II

ammonia

N

III

trimethylamine

NO2

A I, IV, II, III

B IV, I, II, III

C II, III, IV, I

D III, II, I, IV

E III, IV, I, II

2 Which of the following structures properly represents a zwitterion?

A

NH2

H

H COOH

B

NH3

H

H COOH

C

NH3

H

H COO

D

NH2

H

H COO

E None of the above

3

CH3 OH + CH 3 COH

O

?

A CH3OH

B CH3OCH3

C

CH3 OCCH3

O

D

CH3 CCH 3

O

E CH3CH2OH

4 What is the major product of the following reaction?

NO2

Cl2 AlCl3

A

Cl

NO2

Cl

B

Cl

NO2

C

NO2

Cl Cl

D

Cl

NO2

E

NO2

Cl

5 What will be the favored product of the following reaction?

CH3 C

CH3

CH3

O K

+

O

H

S

O

O

CH3

A

O

H

C

CH3

CH3

CH3

B

C

O

O

CH3

CH3

S

O

O

CH3

D

E

O

H

C

CH3

CH3

CH3

6 What reagents are necessary for the following reaction

to occur?

CH3CH2MgBr + ? ∅ CH3(CH2)3OH

A CH3CH2OH

B CO2, HCl

C

CH3 CH O, H2 O

D

H 2 C

O

CH2 , H 2 O

E None of the above

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

NO2

N(CH )

HNO 3

H 2 SO4

3 2

N(CH3 ) 2

NO2

O 2 N

NO2

N(CH3 ) 2

O 2 N

O 2 N

NO2

N(CH3 ) 2

NO2 N(CH3 ) 2

NO2

O 2 N

NO2

N(CH3 ) 2

NO2

N(CH3 ) 2 NO2

O 2 N

N(CH3 ) 2

NO2

NO2

N(CH3 ) 2

NO2

A

B

C

D

E None of the above

8 At the isoelectric pH of a certain amino acid solution, the amino acid may exist as a zwitterion This is

A a positively charged ion

B a negatively charged ion

C either a positively or a negatively charged ion

D an ion that carries both a positive and a negative charge

E an ion without a constant charge

9 Which is an example of anti-Markovnikov addition?

A CH3CH=CH2 + HBr  ROOR    , ∆ → CH3CH2CH2Br

B (CH3)2C=CH2 + HCl ∅ (CH3)3CCl

C (CH3)2C=CH2 + Cl2/H2O ∅

OH

(CH 3 ) 2 CCH 2 Cl

D

+ Br2 anti

Br H

E

+ Br2 syn

Br Br

10

H 3 C

Br2

The product(s) of the above reaction is (are)

A

Br Br

H 3 C CH3

B

Br H

H 3 C Br

and

Br

H 3 C

H Br

C

H

H 3 C

Br

and

Br

H 3 C

H

D

Br

H

H 3 C Br

and

Br

H 3 C

H Br

E

Br

H

H 3 C

Br

and

Br

H 3 C

H Br

11

OCH 3

OCH 3

C

O

C 4 H 9

H 2 N NH2 KOH

P

The product P is

A

OCH 3

OCH 3

C HO

C 4 H 9 H

B

H

H

C

O

C 4 H 9

C

H

OH

C

O

C 4 H 9

D

OCH 3

OCH 3

C 5 H 11

E

OCH 3

12 What is the most effective reducing agent for ethene?

A KMnO4

B H3O+/H2O

C H2/Pt

D Fe/HCl

E HNO3/H2SO4

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

A

B

NO2

OH

C

Cl

OH

D

E

OH

NO2

14 An aldol condensation involves the base-catalyzed reaction between

A an ether and an alcohol

B an alcohol and an alcohol

C an aldehyde and an alcohol

D a ketone and an alcohol

E an aldehyde and an aldehyde

15 When placed in a basic solution, the ketone:

C C

O

CH3

H

undergoes racemization, while the ketone of a similar

structure:

C C

O

CH3

C 2 H 5

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

16 In the light-activated chlorination of methane, the

chain-initiating step is

A Cl2 uv

 →  2Cl•

B CH4 + Cl2 uv

 →  CH3Cl + HCl

C •CH3 + Cl2 uv

 →  CH3Cl + Cl•

D Cl• + CH4 uv

 →  •CH3 + HCl

E Cl• + Cl• uv

 →  Cl2

17 Which of the following compounds would be most reactive towards chlorination of the aromatic ring?

A

NH2

B

NO2

C

D

SO3 H

E

Br

18 Which of the following is the most reactive as a hydride donor?

A H2NNH2

B CH3MgBr

C CH3+

D LiAlH4

E All are equally reactive

19 Which compound below will undergo oxidation without the cleavage of any carbon-carbon bond?

A t-butyl alcohol

B acetone

C acetaldehyde

D ethyl methyl ketone

E None of the above

Questions 20-24 refer to the following choices:

A ortho/para-directing with activation

B ortho/para-directing with deactivation

C meta-directing with activation

D meta-directing with deactivation

E neither directing nor activating

What is the effect of each of the following substituents

on electrophilic aromatic substitution?

20 —F

21 —NO3

22 —OCH3

23

O

24 —SO3H

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

C(CH3 ) 3

Br2

h v

C(CH3 ) 3

Br

C(CH3 ) 3

Br

C(CH3 ) 2 Br

C(CH3 ) 3 Br

Br

C(CH3 ) 3 E.

D.

C.

A.

B.

26 IR spectroscopy would be most useful in distinguishing

between which of the following pairs of compounds?

CH3 CH2 CH2 CCH 2 CH3

O

and CH3 C CH2 CH2 CH3

O

CH3 CH CH2 CH2 CH3

CH3

and CH3 CH2 CH2 CH2 CH2 CH3

CHO

NO2

CHO

NO2 and

CH3 CH2 O CH2 CH2 CH3 and CH3 CH2 CH2 CH2 O CH3

CH3 CH2 CH2 CH2 O CH3 and CH3 CH2 CH2 CH2 CH2 OH

A.

B.

C.

D.

E.

27 What is the major product of the reaction below?

H 3 O +

1

2

B.

A.

C.

D.

E.

OCH 2 CH3

Br OH

CH2 CH3

OH

O

28 Which of the structures below corresponds to

para-nitrobenzenesulfonic acid?

E

D.

C.

B.

A.

SO3 H

SO3 H

SO3 H

H 2 N

SO3 H

O 2 N

SH

O 2 N

29 Which of the following is a productive propagation step

in the free radical bromination of ethane?

A CH3CH2 + Br ∅ CH3CH2Br

B CH3CH3 + Br ∅ CH3CH2Br + H

C CH3CH3 + Br ∅ CH3CH2.+ HBr

D CH3CH3 + Br ∅ CH3Br + CH3

E CH3CH2 + CH3CH2 ∅ CH3CH2CH2CH3

30 Which of the following is a major organic product of

the reaction below?

OCH 3

AlCl3

CH3 COCl

A

Cl

OCH 3

C

O

H 3 C

C

H 3 C

O

OCH 3 OAlCl 2

OCH 3 Cl Cl

E.

D.

C.

B.

STOP! END OF TEST.

ORGANIC CHEMISTRY SUBJECT TEST 4

ANSWER KEY

1 B

2 C

3 C

4 B

5 B

6 D

7 B

8 D

9 A

10 B

11 D

12 C

13 E

14 E

15 D

16 A

17 A

18 D

19 C

20 B

21 D

22 A

23 A

24 D

25 D

26 E

27 B

28 E

29 C

30 B

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

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

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

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

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+)

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

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.)