Organic chemistry section test (5)

It exists predominantly in the enol form because its keto tautomer is nonaromatic.. It exists predominantly in the enol form because its keto tautomer is antiaromatic?. It exists predomi

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MCAT Subject Tests

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Organic Chemistry Subject Test 5

1 What is the major organic product of the reaction

sequence shown?

CH3 CH2 I Mg

ether

D 2 0

?

A CH3CH2OD

B CH3CH2MgD

C CH3CH2D

D CH3CH2MgI

E CH3CDO

2 What is the major organic product of the synthetic

sequence outlined below? [A]

NH3

CH3 Br

?

A

B

C

D

E

C C

H 3 C

Br

C C

H 3 C

CH3

H 3 C CH2 CH2 CH3

3 Which of the following factors usually increases the

solubility of a compound in a given solvent?

I Higher temperature

II Similar polarities

III Higher molecular weight of the compound

IV Lower density of the solvent

A I only

B I and II only

C I, II, and III only

D I, II, and IV only

E I, II, III, and IV

4 Which of the phrases below best describes the kinetics

of the dehydration of 3-methyl-3-hexanol in concentrated (85%) phosphoric acid?

A independent of concentration

B first order in alcohol, first order overall

C first order in acid, first order overall

D first order in alcohol, second order overall

E second order in alcohol, second order overall

5 A compound to be analyzed is known to be a six carbon cyclic hydrocarbon The compound is found to be inert

to bromine in water and to bromine in dichloromethane, yet it decolorizes bromine in carbon tetrachloride when a small quantity if FeBr3 is added Which of the following could be the identity of the compound?

A cyclohexane

B benzene

C 1,3-cyclohexadiene

D 1,4-cyclohexadiene

E cyclohexene

6 Which of the compounds below could be the reagents used to carry out the conversion shown?

O

OH HO

OH

A X = ethanol; Y = H2/Pt

B X = H+/H2O; Y = LiAlH4/ether

C X = ethanol; Y = LiAlH4/ether

D X = H+/H2O; Y = NaBH4/H2O

E X = ethanol; Y = NaBH4/H2O

7 Which of the following would yield more than one major organic product upon reaction with HCl in the absence

of peroxides?

A 1,2-dichloroethane

B 1,1-dichloropropane

C 3-hexene

D 2-hexene

E 2-methyl-1-hexene

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OR G A N I C CH E M I S T R Y SU B J E C T TE S T 5

8 The α-helix is an example of a protein's _

structure

A lewis

B resonance

C primary

D secondary

E tertiary

9 What reaction type is responsible for the biochemical

formation of peptide bonds?

A rearrangement

B addition

C resolution

D elimination

E condensation

10 What is/are the product(s) if 1-ethyl-2-methylcyclohexene is hydrogenated over platinum?

A

H H

CH2 CH3

CH3

B

H CH2 CH3

H

CH3

C

H CH3

H

CH3 CH2

D

H CH2 CH3

H

CH3

H CH3

H

CH3 CH2

and

E

H H

CH2 CH3

CH3

H H

CH3

CH3 CH2

and

2 K A P L A N

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11 The compound below is heated in a deuteroacid solution

(D+/D2O) and allowed to reach equilibrium Which of

the labeled hydrogens will be exchanged for deuterium?

O

H

1 1

2

3

3 4 5

A 2 and 3

B 1, 2 and 3

C 1, 3 and 5

D 1, 4 and 5

E 2 and 4

12 What are the major products of the reaction sequence

shown below?

CH3 CH2 C

O OCH 3

LiAlH 4 THF

HCl

H 2 O

A CH3CH2COOH and CH3OH

B CH3CH2COOH and HCOOH

C CH3CH2CH2OH and CH3OH

D CH3CH2CHO and CH3OH

E CH3CH2CH2OH and CH2O

13 Which of the following would yield one major organic

product upon reaction with HBr in the presence of

peroxides?

A 1,2-dichloroethane

B 1,1-dichloropropane

C 2-hexene

D 3-hexene

E 2-methyl-3-hexene

14 Which of the following statements is true of phenol, pictured below?

OH

A It exists predominantly in the enol form because its keto tautomer is nonaromatic

B It exists predominantly in the enol form because its keto tautomer is antiaromatic

C It exists predominantly in the keto form because its enol tautomer is nonaromatic

D It exists predominantly in the keto form because its enol tautomer is antiaromatic

E It exists predominantly in the keto form because its keto tautomer is aromatic

15 Which of the compounds below could be the reagents used to carry out the conversion shown?

H 3 C C

O Et

O

X Y

O

O Et

O

A X = H+/H2O , Y = NaOEt/EtOH

B X = KMnO4/OH-, Y = NaOEt/EtOH

C X = NaOEt/EtOH, Y = NH3/H2O

D X = NH3/H2(10 atm), Y = Zn/H+

E X = NaOEt/EtOH, Y = H+/heat

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16 Which of the following molecules would produce the

NMR spectra shown below?

CH3 C Br

Br CBr 3

CH3 C H

Br

CH3

CH3 C Br

Br

CH3

CH3 C H

H

CH3

CH3 C H

Br CBr 3

A

B

C

D

E

17 How many chiral carbons are there in morphine?

HO

O

H

H HO

N

H

CH3

Morphine

A 3

B 4

C 5

D 6

E 7

18 The two products formed by the following racemization reaction could best be described as:

C

O

CH3

C 2 H 5

H

H 3 C

C

O

CH3

C 2 H 5

H 3 C

H

C

O

CH3

C 2 H 5

H

H 3 C

EtO - , EtOH

R

S

R

A enantiomers

B diastereomers

C conjugate acid and base

D conformational isomers

E same compound

4 K A P L A N

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19 Which of the following reactions will lead to the

formation of an aldehyde?

A

Cl

O

H 2 0

B

OH

H

PCC

C

OH KMnO4

D

CH3 CH2 Cl 1 CN -

2 H+ /H2 O

E

KMnO4 / H + heat

20 Which of the following species will form the most stable

radical?

A

B

C

D

E

21 The diagram below shows a step in which of the following processes?

OH

O

CH 2 OH

H

O

CH2 OH

OH

H

O

H OH

CH2 OH

A Aldehyde formation

B Hemiketal formation

C Mutarotation

D Anomerization

E Tautomerization

22 Which of the following reactions would produce an ester?

A 2 CH3OH + H2SO4 ∅

B CH3COOH + SOCl2 ∅

C CH3COOH + C2H5OH + H2SO4 ∅

D C6H5OH + CH3CH2Br ∅

E CH3CH2Br + CH3CH2O-Na+ ∅

23 Which of the following describes the reaction below?

H 5 C 2 C C CH3 H 2 , Pd

CH3

H 5 C 2

A Catalytic hydration

B Substitution

C Stereospecific reduction

D Racemization

E Grignard reduction

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24 Which of the following is a result of the reaction below?

Cl + Br -

H

C 2 H 5

CH3

A Inversion of configuration

B Retention of optical activity

C Mutarotation

D Double bond formation

E Loss of optical activity

25 Infrared spectroscopy provides a chemist with

information about:

A functional groups

B conjugated bonds

C molecular weights

D distribution of protons

E ionization patterns

26 Which of the following compounds form a racemic

mixture?

I. CHO

CHO

H HO

III. CHO

CHO

OH

H

OH

H

II CHO

CHO

OH

H

H HO

IV CHO

CHO

H HO

H OH

A I and III

B I and IV

C II and III

D II and IV

E I and II

27 Which of the following compounds is the most

susceptible to nucleophilic attack by OH–?

A Propanoic acid

B Ethanal

C Benzyl chloride

D Propionyl chloride

E Propanal

28 Reaction of benzoic acid with thionyl chloride followed

by treatment with ammonia will yield which of the following compounds?

A Benzamide

B p-Aminobenzoic acid

C p-Chlorobenzamide

D 3-Chloro-4-aminobenzaldehyde

E m-Aminobenzoic acid

29 Which of the following bonds is a peptide linkage? A

B

C

D

E

C

O

N

H

C

O

H

C O

O

C N

H

C

6 K A P L A N

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30 Which of the following pairs of side chain –R groups would tend to associate with each other?

A

B

C

D

E

and (CH 2 ) 2 COOH

CH3

CH2 CH3 and

CH3

N

H H

and CH3 CH2 NH2

STOP! END OF TEST.

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THE ANSWER KEY AND EXPLANATIONS BEGIN ON THE FOLLOWING PAGE

8

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ORGANIC CHEMISTRY SUBJECT TEST 5

ANSWER KEY

1 C

2 A

3 B

4 B

5 B

6 C

7 D

8 D

9 E

10 E

11 D

12 C

13 D

14 A

15 E

16 E

17 C

18 A

19 B

20 A

21 C

22 C

23 C

24 E

25 A

26 D

27 D

28 A

29 D

30 C

K A P L A N _ 9

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EXPLANATIONS

1 C

This is an example of a reaction utilizing a Grignard reagent Whenever you see Mg as a reactant, the first thing that should come to mind is a Grignard synthesis In Grignard reactions, magnesium is used to turn an alkyl halide into a good nucleophile Grignard reagents are utilized in many reactions, because the carbon associated with the magnesium has a partial negative charge, and is therefore a good nucleophile In this reaction, the magnesium nucleophilically attacks the ethylmagnesium iodide (very similar to an SN2 reaction) and forms our Grignard reagent (an organomagesium compound) The Grignard reagent is a strong base (it is proton deficient), and undergoes an acid/base reaction with heavy water (the hydrogen isotope deuterium, D, has a neutron and so is heavier than H) Specifically, the ethyl group abstracts a proton from heavy water, and the magnesiumidodide is displaced as a result

Grignard reagent

2 A

This question asks what happens when you react an alkyne with a strong base The alkyne hydrogens are relatively acidic, owing to resonance stabilization of the conjugate base, and so can undergo acid/base reactions NaNH2 (sodium amide) is a strong base, and so will remove the terminal hydrogen The resulting carbanion is a strong nucleophile, and will undergo an SN2 reaction with methylbromide The bromine will be displaced, and the methyl group will add to the terminal carbon, giving us 2-butyne

3 B

Most substances tend to become more soluble at higher temperatures and so statement I is correct Statement II is also correct Recalling the phrase “like dissolves like” compounds that are polar or ionic will dissolve in solvents that are also polar, whereas compounds that are non-polar will dissolve in non-polar solvents Therefore, similar polarities will increase the solubility Statement III molecular weight and statement IV density of the solvent do not determine solubilities, so I and II are the only true statements, and choice B is the correct response

4 B

This question is describing an acid catalyzed dehydration reaction When a 3_ alcohol is treated with strong acid, an

E1 reaction occurs The hydroxyl group gets protonated and then leaves in the form of water, giving us a carbocation The released water then abstracts a proton from the carbocation, reforming the acid (that is why it is a catalyst) and converting the carbocation into an alkene Since the reaction proceeds via E1, the reaction rate depends only on the concentration of the substrate, and so is first order with respect to the alcohol only

5 B

The question stem informs us that the unknown cyclic hydrocarbon does not react with bromine in dichloromethane, carbon tetrachloride, or water This means that the unknown cannot contain any non-conjugated double or triple bonds, otherwise the bromine would have added across the π bonds Since the unknown did react when FeBr3 was added, it must

be benzene The FeBr3 acts as a Lewis acid and catalyzes aromatic electrophilic addition reactions Bromine is not a strong enough electrophile to disrupt the conjugated double bonding in the benzene ring, but the addition of the

iron(III)bromide catalyst converts the bromine into a strong enough electrophile that it can add to the benzene ring

6 C

In this reaction a cyclic anhydride is transformed into an ester, then reduced to an alcohol Cyclic anhydrides are very reactive (they undergo the same reactions as acyl chlorides) because there are two electron withdrawing groups deshielding

10 _ K A P L A N

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the carbon nucleus; the carbonyl carbon geometry is trigonal planar, so there is little steric hindrance to nucleophilic attack;

and there is a good leaving group attached Choices B and D are wrong because if an anhydride is reacted with water it will

become a dioic acid (the anhydride was formed by condensing a dioc acid) When the anhydride is reacted with ethanol,

the oxygen of the alcohol will add to the carbonyl carbon, and the other side of the molecule acts as a leaving group The

product formed will be an ester The carbonyl carbons are then reduced using LiAlH4, a strong reducing agent, to form a

diol Choice E is wrong because NaBH4 can only reduce aldehydes and ketones, it is not a strong enough reducing agent to

effect an acid (or it's derivatives) If platinum were used, as in choice A, the double bond would be reduced to a single

bond The product has the double bond intact, so A is incorrect

O

OH

OCH 2 CH3 HOCH2 CH3

O

OH HO

OH

*

7 D

The addition of acid halides across double bonds has to follow Markovnikov's rule the H adds to the less substituted

carbon (the carbon with the most H's) and the halide adds to the other Choices A and B will not react with HCl Answer

choice C, hexene, is symmetrical, it does not matter to which side of the double bond the H adds, and only

3-chlorohexane can form Answer choice D, 2-hexene, can form more than one product Each carbon that is part of the

double bond is equivalent, so the hydrogen can add to either, followed by the chlorine adding to the other The resulting

products can be either 2-chlorohexane or 3-chlorohexane The only product that can result from choice E is

2-chloro-2-methylhexane

8 D

α-helixes and β-pleated sheets are the two common forms the primary amino acid sequence of a protein can assume

These secondary structures are held together by hydrogen bonds A protein's primary structure is the sequence of amino

acids it contains Tertiary structure is the final 3-dimensional form the functioning protein assumes, and is a result of

hydrophilic and hydrophobic interactions between the amino acid R groups

9 E

Peptide bonds are the covalent links between two amino acids It is formed by a nucleophilic acyl substitution of the

carbonyl carbon of one amino acid by the amine of the other The hydroxyl group of the carboxylic acid leaves in the form

of water, and so the reaction is called a condensation

10 E

Hydrogenation on a platinum surface leads to syn addition of H atoms Choices B, C, and D can thus be eliminated

Choice A may have been tempting, but keep in mind that the hydrogen can add to either side, as long as the two atoms are

syn

K A P L A N 11

H 3 C CHH H 2 CH3 H 3 CH2 C CHH H 3

versus

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