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Choice A, aldehyde formation , is wrong because the open chain form has a carbonyl group and therefore an aldehyde is already formedA. Choice B--hemiketals--are formed as a result of the

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Discretes Test

Time: 30 Minutes

Number of Questions: 30

This test consists of 30 discrete

questions—questions that are NOT based on a descriptive passage These discretes comprise

15 of the 77 questions on the Physical Sciences and Biological Sciences sections of the MCAT

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ORGANIC CHEMISTRY DISCRETES TEST DIRECTIONS: The following questions are not

based on a descriptive passage; you must select the best answer to these questions If you are unsure of the best answer, eliminate the choices that you know are incorrect, then select an answer from the choices that remain Indicate your selection by blackening the corresponding circle on your answer sheet A periodic table is provided below for your use with the questions

PERIODIC TABLE OF THE ELEMENTS

1

H

1.0

2

He

4.0 3

Li

6.9

4

Be

9.0

5

B

10.8

6

C

12.0

7

N

14.0

8

O

16.0

9

F

19.0

10

Ne

20.2 11

Na

23.0

12

Mg

24.3

13

Al

27.0

14

Si

28.1

15

P

31.0

16

S

32.1

17

Cl

35.5

18

Ar

39.9 19

K

39.1

20

Ca

40.1

21

Sc

45.0

22

Ti

47.9

23

V

50.9

24

Cr

52.0

25

Mn

54.9

26

Fe

55.8

27

Co

58.9

28

Ni

58.7

29

Cu

63.5

30

Zn

65.4

31

Ga

69.7

32

Ge

72.6

33

As

74.9

34

Se

79.0

35

Br

79.9

36

Kr

83.8 37

Rb

85.5

38

Sr

87.6

39

Y

88.9

40

Zr

91.2

41

Nb

92.9

42

Mo

95.9

43

Tc

(98)

44

Ru

101.1

45

Rh

102.9

46

Pd

106.4

47

Ag

107.9

48

Cd

112.4

49

In

114.8

50

Sn

118.7

51

Sb

121.8

52

Te

127.6

53

I

126.9

54

Xe

131.3 55

Cs

132.9

56

Ba

137.3

57

La *

138.9

72

Hf

178.5

73

Ta

180.9

74

W

183.9

75

Re

186.2

76

Os

190.2

77

Ir

192.2

78

Pt

195.1

79

Au

197.0

80

Hg

200.6

81

Tl

204.4

82

Pb

207.2

83

Bi

209.0

84

Po

(209)

85

At

(210)

86

Rn

(222) 87

Fr

(223)

88

Ra

226.0

89

Ac †

227.0

104

Unq

(261)

105

Unp

(262)

106

Unh

(263)

107

Uns

(262)

108

Uno

(265)

109

Une

(267)

*

58

Ce

140.1

59

Pr

140.9

60

Nd

144.2

61

Pm

(145)

62

Sm

150.4

63

Eu

152.0

64

Gd

157.3

65

Tb

158.9

66

Dy

162.5

67

Ho

164.9

68

Er

167.3

69

Tm

168.9

70

Yb

173.0

71

Lu

175.0

90

Th

232.0

91

Pa

(231)

92

U

238.0

93

Np

(237)

94

Pu

(244)

95

Am

(243)

96

Cm

(247)

97

Bk

(247)

98

Cf

(251)

99

Es

(252)

100

Fm

(257)

101

Md

(258)

102

No

(259)

103

Lr

(260)

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GO ON TO THE NEXT PAGE.

1 On the graph below, which is the titration curve

for an amino acid, what does point A represent?

H

nits of base

A The isoelectric point

B pH = pKb2

C pH = 14 – pKb2

D pH = pKb2 – 14

2 The diagram below shows a step in which of the

following processes?

OH

O

CH2OH

H

O

CH2OH

OH H

O

H OH

CH2OH

A Aldehyde formation

B Hemiketal formation

C Mutarotation

D Anomerization

3 Which of the following products might be

formed if benzoyl chloride was treated with

excess CH3MgBr?

A C6H5COOH

B C6H5COOCH3

C C6H5CHO

D C6H5C(CH3)2OH and C6H5COCH3

4 Which of the following reactions would produce

an ester?

A 2CH3OH H2SO4

B CH3CO2H + SOCl2

C CH3CO2H + C2H5OHH2SO4

D C6H5OH + CH3CH2Br NaOH

H2O

5 Which of the following reactions will NOT yield

C6H5CH2COCH3?

A C6H5CH2C CH + HgSO4 / H2O

B C6H5CH2CH2CH2OH + cold, dil KClO3

C C6H5CH2(CH3)C CH2 + 1 O3

2 Zn / H2O

D C6H5CH2(CH3)CHOH + cold, dil Na 2Cr2O7

6 Which of the following reactions will NOT

result in the formation of a carboxylic acid?

A Carbonation of Grignard reagents

B Hydrolysis of nitrites

C Reduction of aldehydes

D Oxidation of primary alcohols

7 The following reaction is an example of:

H2C

A esterification.

B tautomerism.

C elimination.

D dehydration.

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GO ON TO THE NEXT PAGE.

8 Which of the following compounds is the most

basic?

9 Which of the following can be synthesized from

an arenediazonium salt?

I C6H5Br

II C6H5CN

III C6H5OH

A I only

B II and III

C I and III

D I, II, and III

1 0 In the following reaction, what is the final

product?

R2C=O + H2NNHC6H5→ R2COHNHNHC6H5

→ R2C=NNHC6H5

A An oxime

B A phenaylhydrazone

C A semicarbazone

D An aromatic nitrile

1 1 Which of the following processes describes the

reaction below?

A Alkylation

B Hoffman elimination

C Salt formation

D Conversion into amides

1 2 The reaction below will primarily yield which of

the following products?

A CH3

NO2

B CH3

NO2

C CH3

NO2

C6H5CH3 HNO3

H2SO4

D A and B

1 3 Which of the following describes the reaction

below?

H5C2 C C CH3H2, Pd

CH3

H5C2

A Catalytic hydration

B Substitution

C Stereospecific reduction

D Racemization

1 4 Which of the following reactions is NOT an

electrophilic aromatic substitution reaction?

A C6H6 + CH3CH2COCl/AlC13→

B C6H6 + H2, Rh/C →

C C6H6 + C6H5CH2Cl/AlCl3→

D C6H6 + Br2/FeBr3→

1 5 Proteins can be characterized by the fact that

they:

A are composed of a single peptide chain.

B have a primary structure formed by covalent

linkages

C retain their conformation above 40°C.

D always have quaternary structures.

R2CHCR2– N

+

R3O –

H →heat R2C=CR2+R3N:+H2O

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GO ON TO THE NEXT PAGE.

1 6 Halogenation of alkanes proceeds partly by the

mechanism depicted below

I X2→hv 2X•

II X • + RH → HX + R •

III R • + X2 → RX + X •

Which of these steps can be described as chain

propagation?

A I, II, and III

B III only

C II and III

D I and II

1 7 Which of the following compounds is least

susceptible to electrophilic aromatic

substitution?

A C6H5CH3

B p-Br– C6H4– NH3+

C p-O2N–C6H4–NH2

D p-H3CO–C6H4–OCH3

1 8 How many different stereoisomers can the

following compound have?

COOH

H

H HO

A 2

B 3

C 4

D 8

1 9 In thin-layer chromatography, which of the

following best describes the behavior of the

solvent?

A It moves downward because of gravity.

B It moves upward because of capillary action.

C It moves downward because of capillary

action

D It does not move.

2 0 Which of the following statements is correct?

I CH2OH

H HO

COOH

OH

H3C

OH

H3C

II. CH2OH

H HO

H HO

H HO

A Compounds I and II are diastereomers

B Compounds I, II, and IV are meso structures.

C Compounds II and IV are optically inactive.

D Compounds III and IV are optically active.

2 1 Which two of the following compounds are

geometric isomers?

A I and III

B I and IV

C II and III

D II and IV

2 2 Which of the following is a result of the reaction

below?

Cl + Br– H

H

C2H5

CH3

A Inversion of configuration

B Retention of optical activity

C Mutarotation

D Loss of optical activity

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GO ON TO THE NEXT PAGE.

2 3 When placed in an electric field, which of the

following compounds will migrate toward the

cathode at pH 7.0?

A H2NCH2COOH

B H2NCH2CHNH2COOH

C HOOCCH2CHNH2COOH

D HOOCCHNH2CH2–S–S–CH2CHNH2COOH

2 4 At atmospheric pressure, a certain organic liquid

has a boiling point of 185°C, and decomposes at

180°C, while its isomer, also a liquid, boils at

215°C and decomposes at 190°C These isomers

can be separated by:

A vacuum distillation.

B simple distillation.

C sublimation.

D fractional distillation.

2 5 Infrared spectroscopy provides a chemist with

information about:

A functional groups.

B conjugated bonds.

C molecular weights.

D distribution of protons.

2 6 A compound with the molecular formula C8Hl8,

produces a single NMR signal What is its

structural formula?

A CH3(CH2)6CH3

B (CH3)2CHCH2(CH2)3CH3

C (CH3)2CHCH2CH(CH3)CH2CH3

D (CH3)3CC(CH3)3

2 7 Which of the following compounds will give the

greatest number of proton NMR peaks?

A

B

C

D

CH3CH3

(CH3)2C=CH2

CHCH3

CH3CHBrCH2CH3

2 8 A clear liquid is subjected to infrared

spectroscopy and produces a spectrum with prominent, sharp peaks at approximately 2950

cm–1 and 1700 cm–1, as well as a number of smaller peaks between 1460 cm–1 and 900 cm–1 This substance is most likely:

A a ketone.

B an aldehyde.

C an alcohol.

D an alkane.

2 9 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

I V CHO

CHO

H HO

A I and III

B I and IV

C II and III

D II and IV

3 0 What is the most likely mechanism for the

reaction between 1-chloropropane and sodium cyanide?

A S N2

B E2

C SN1

D El

END OF TEST

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THE ANSWER KEY IS ON THE NEXT PAGE

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ANSWER KEY:

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ORGANIC CHEMISTRY DISCRETES TEST EXPLANATIONS

1 The correct answer for question 1 is choice C This problem deals with the titration of amino acids The titration given begins with a strongly acidic solution As we add base, the pH rises at first But then as the pH

approaches the pKa1, protons begin to dissociate from the carboxyl group The value of the pH then levels off

because the carboxyl group buffers the solution when the pH is close to its pKa1 Therefore, point A on the graph

represents the value of pH equal to the pKa1 That is, point A shows deprotonation of the amino acid carboxyl

group But the reverse reaction—protonation of the amino acid, is represented by pKb2 And by the definition of

pK, the pH at point A is equal to 14 minus the pKb2 Therefore, choice C is correct Choice A the isoelectric point is in fact point B When we continue to add base after point A, the pH gradually rises until all the molecules

of the amino acid are in the neutral form

2 The correct answer for question 2 is again choice C The question calls upon your understanding of the phenomenon of mutarotation Cyclic hemiacetal forms of monosaccharides with different configurations around the first carbon that is, anomers are easily broken in aqueous solutions In such reactions either the alpha or beta anomer becomes an open chain In an aqueous solution, especially if it is slightly acidic, this open chain is easily recyclized, forming a mixture containing both anomers in their equilibrium concentrations Thus the initial opening and the subsequent closing of the chain results in a mixture of anomers This phenonenon is called mutarotation and

so choice C is correct Choice A, aldehyde formation , is wrong because the open chain form has a carbonyl group and therefore an aldehyde is already formed Choice B hemiketals are formed as a result of the nucleophilic addition

of a hydroxyl donated by an alcohol to a ketone carbonyl group Since in this question you have an aldehyde carbonyl group, this choice is also wrong Although this reaction is an example of anomerization, choice D is wrong since mutarotation is a more accurate description of this process Again then, choice C is the correct answer

3 Now for question 3 The correct answer here is choice D The reagent here benzoyl chloride is an acyl halide which will form a ketone on reaction with an equimolar quantity of a Grignard reagent like this one in the question With excess of the Grignard reagent however, acyl halides react further to produce primarily tertiary alcohols along with a small amount of ketone Therefore, the product of this reaction will be a mixture of a tertiary alcohol and acetophenone This corresponds to choice D and so this is the correct answer As for the other choices, Choice A benzoic acid would be formed by hydrolysis of benzoyl chloride Choice B methyl benzoate would be formed by reaction of benzoyl chloride with methanol Finally, choice C benzaldehyde would be formed by reduction of benzoyl chloride Again then, the correct answer is choice D

4 The correct answer here is choice C Esterification is a process in which a carboxylic acid reacts with an alcohol in the presence of an acidic or basic catalyst to form an ester plus water Therefore choice B formation of an acyl halide can be rejected at once Choice A illustrates the formation of dimethyl ether from two molecules of methanol, therefore this choice is also wrong Choice D is wrong as this is an example of the Williamson Ether synthesis which involves reaction of a phenoxide ion with an alkyl halide to produce an ether So the only choice that results in ester formation is choice C

5 Now for question 5 The correct choice is B This question asks you to find the correct answer choice that will NOT lead to the formation of a particular compound, which happens to be benzyl methyl ketone Choice A shows an alkyne being treated with an acidic mercury sulfate solution This leads to Markownikov addition of water—that is, hydrogen adds to the less substituted triple bonded carbon and hydroxyl adds to the other triple bonded carbon This results in the formation of an enol which will then spontaneously rearrange to form the ketone we are looking for So, choice A DOES yield benzyl methyl ketone and therefore it's NOT the answer we're looking for

In choice B, a primary alcohol is oxidized with cold, dilute potassium chlorate Although ketones can be synthesized

by oxidizing secondary alcohols, oxidation of primary alcohols results in the formation of aldehydes NOT ketones Therefore, B is the correct answer Choice C ozonolysis of an alkene will yield our ketone plus formaldehyde and choice D oxidation of a secondary alcohol also gives our desired ketone, so both of these choices are wrong Again, B is the correct answer choice

6 Choice C is correct This question deals with common methods of preparing carboxylic acids Let's go through the choices given The carbonation of a Grignard reagent leads to the formation of the magnesium salt of a carboxylic acid When treated with mineral acid, the magnesium salt is then converted to a carboxylic acid

Therefore choice A CAN be used to prepare carboxylic acids, so this is wrong The acidic or basic hydrolysis of nitriles also yields carboxylic acids, so choice B is also wrong The oxidation of primary alcohols by various strong

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oxidizing agents such as potassium permanganate yields carboxylic acids, so choice D can be rejected Choice C is correct because carboxylic acids are formed by the oxidation of aldehydes, not their reduction

7 For question 7, the correct choice is B This diagram represents the tautomeric transformations of a ketone Ketones exist in two spontaneously-interconvertible forms the keto form and the enol form These differ in the placement of a hydrogen atom and a double bond The keto structure is more stable because a carbon-oxygen double bond is more stable thermodynamically than a carbon-carbon double bond with a hydroxyl group attached and the chemical properties of ketones are determined mostly by the keto form

None of the other choices describes this process Choice A esterification is a reaction between a

carboxylic acid and an alcohol resulting in the formation of an ester Choice C elimination is a reaction

characteristic of alkyl halides and leads to the formation of a stable double bond Choice D dehydration is the loss

of a hydroxyl group and a proton to yield water and usually a stable double or single bond is left behind So again, the correct choice is B

8 Now for question 8 The correct answer is choice A This question deals with the effect of substituents on the relative basicities of aromatic amines The basicity of aromatic amines or anilines manifests itself in

protonation of the amino-group nitrogen atom to make positively charged anilinium ions In general, electron withdrawing substituents decrease the basicity of anilines because the withdrawal of electron density increases the positive charge on the anilinium ion, making it less stable On the other hand, electron-donating substituents decrease the positive charge of the anilinium ion and stabilize it Now look at the choices B, C, and D have electron withdrawing substituents, while A has an electron donating methoxy substituent Since electron donating substituents increase basicity, choice A is the most basic among the compounds shown and therefore this choice is correct

9 Let's now look at question 9 The correct choice is D Arenediazonium salts, which are synthesized from primary aromatic amines, are compounds with an N2+ group attached to the aromatic ring They are useful for synthesizing a wide variety of compounds due to the fact that they can easily be made to undergo replacement reactions in which molecular nitrogen is released and a nucleophilic substituent attaches to the aromatic ring in its place For example, in the presence of cuprous halides, diazonium salts release molecular nitrogen and form

halogen-substituted arenes often called the Sandmeyer reaction Similarly, in the presence of cuprous cyanide, nitrogen is released and replaced by the cyanide ion, thus forming aromatic nitriles In the same way, in cold aqueous solutions, the hydroxyl group replaces the nitrogen to form phenol Therefore, compounds I, II, and III can all be obtained by the replacement of nitrogen in diazonium salts, and so the correct choice is D As all three choices are correct, you can discard A, B, and C which state that only one or two of the compounds will be formed

So again, D is the right answer

10 For question 10, the correct answer is choice B This question requires you to know the most important reactions between ammonia derivatives and aldehydes In these reactions, the nucleophilic nitrogen of an ammonia derivative attacks the electrophilic carbonyl carbon, while hydrogen released from this nitrogen attacks the

nucleophilic carbonyl oxygen This reaction results in the formation of a single carbon-nitrogen bond, as well as a hydroxyl group on the formerly carbonyl carbon Since this system is quite unstable, the addition is always

followed by dehydration that is, a water molecule is formed from the carbons hydroxyl and the neighboring

nitrogen's hydrogen The release of the water molecule results in the formation of a carbon-nitrogen double bond In this case, the final product of the reaction between the carbonyl compound and phenylhydrazine is phenylhydrazone and therefore choice B is correct Choices A and C are synthesized in reactions with other ammonia derivatives Choice A is wrong because oximes are synthesized by reactions between carbonyl compounds and hydroxylamine Choice C is wrong because semicarbazones are synthesized by reactions between carbonyl compounds and

semicarbazides Finally, aromatic nitriles are obtained by reacions between diazonium salts and cuprous cyanide, and

so choice D is also wrong Again, the correct choice is B

11 Now look at question 11 The right answer here is choice B The reaction given is an example of Hofman elimination When a basic solution of a quaternary ammonium hydroxide is heated, it undergoes decomposition forming an alkene, a tertiary amine and water This reaction proceeds by an E2 mechanism The base abstracts a hydrogen ion from carbon and the subsequent release of the molecule of tertiary amine results in the formation of the double bond So, choice B is correct Let's now look at the wrong answers Choice A alkylation is the insertion

of an alkyl group into a molecule Choice C formation of ammonium salts is possible only in acidic solutions because it requires protonation of of nitrogen Moreover, in this case nitrogen doesn't have unshared electron pairs, therefore protonation and subsequent formation of a salt may occur only upon the breakdown of a carbon-nitrogen

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