6Molecular structure of organic compounds test w solutions

It is a mixture of enantiomers that have the same chemical and physical properties, with the exception of the direction of optical rotation, but different physiological propertiesI. It i

ORGANIC CHEMISTRY TOPICAL:

Molecular Structure of Organic

Compounds

Test 1

Time: 23 Minutes*

Number of Questions: 18

* The timing restrictions for the science topical tests are optional If you are using this test for the sole purpose of content reinforcement, you may want to disregard the time limit

DIRECTIONS: Most of the questions in the following test

are organized into groups, with a descriptive passage preceding each group of questions Study the passage, then select the single best answer to each question in the group Some of the questions are not based on a descriptive passage; you must also 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

Rf

(261)

105

Ha

(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)

GO ON TO THE NEXT PAGE.

Passage I (Questions 1–7)

Until recently, it had been common practice for

pharmaceutical companies to manufacture chiral drugs as

racemates, not as single enantiomers

Usually, only one enantiomer of a given compound

possesses therapeutic value, while the other may have no

beneficial pharmacological properties and may even induce

serious physiological side effects

A typical example of this problem was the use of the

drug Thalidomide in 1961 Administered to reduce nausea

and vomiting during the early stages of pregnancy, the

desired physiological activity lies with the R-isomer The

S-isomer of Thalidomide is a teratogen (an agent that

produces physical defects in developing embryos) As a

result, administration of this drug as the racemate caused

congenital malformations in thousands of infants The

structure of Thalidomide is shown below

N

N C O C O H

Figure 1 Today, advances in the synthesis of specific

enantiomers should severely limit the number of racemic

drugs produced Traditionally, the enantiomer of a chiral

drug could be synthesized via resolution (conversion of the

racemic mixture into diastereomers, separation, and then

reformation of the enantiomers) However, this process is

wasteful because of the numerous steps involved in the

procedure The emergence of a new technique called

enantioselective catalysis may soon be used to avoid this

inefficiency

The catalysts employed in enantioselective catalysis

are usually based on optically active metal complexes and

are highly effective in the production of one enantiomer

over another The success of this technique is attributed to

the production of thousands of chiral products from one

molecule of catalyst A good example is the use of a

nickel-based catalyst to synthesize the S-isomer of the

chiral drug Naproxen (Reaction 1)

MeO

+ HCN L + [Ni]

6-Methoxy-2-vinylnapthalene

CH3 CN H MeO

CH 3

CO 2 H H MeO

Naproxen (L + [Ni] = Metal complex catalyst)

Reaction 1 Another example of enantioselective catalysis is that

of meso breaking For example, azidotrimethylsilane has

been shown to react with cyclohexene oxide in the presence

of the enantioselective catalyst titanium isopropoxide:

O + R 3 SiN 3

chiral catalyst

OSiR 3

N 3

Reaction 2

GO ON TO THE NEXT PAGE.

1 How many chiral centers are present in the

Thalidomide molecule?

A 0

B l

C 2

D 4

2 What is the most likely reaction mechanism between

HCN and 6-methoxy-2-vinylnapthalene?

A Electrophilic addition

B Bimolecular nucleophilic substitution

C Bimolecular elimination

D Free radical addition

3 Reaction 2 proceeds through an SN2 mechanism

Consequently, the attacking nucleophile is:

A the alkyl group.

B the azide ion.

C silicon.

D the chiral catalyst.

4 The double bond of the vinyl group in

6-methoxy-2-vinylnapthalene contains which of the following

orbitals?

I sp orbitals

II sp2 orbitals

III sp3 orbitals

IV p orbitals

A II only

B III only

C I and IV only

D II and IV only

5 Which of the following statements most accurately

describes the properties of Thalidomide?

A It is a mixture of enantiomers that have the same

chemical and physical properties, with the exception of the direction of optical rotation, but different physiological properties

B It is a mixture of diastereomers that have the same

physical and chemical properties, but different physiological properties

C It is a mixture of enantiomers that have different

physiological, chemical, and physical properties

D It is a mixture of diastereomers that have different

physical, chemical, and physiological properties

6 Which of the following is the R-isomer of Naproxen?

A

C CO2 H H

CH3

CH3 H

C CO2 H

MeO

CO2H

CH3 C H

CO2H

CH3 C H MeO

C

7 The term “meso breaking” can be applied to Reaction

2 because:

A the chirality of the catalyst is destroyed.

B chiral centers in cyclohexene oxide are created

when the product is formed

C symmetry is retained when cyclohexene oxide is

converted to the product

D the symmetry of cyclohexene oxide is broken to

form an optically active compound

GO ON TO THE NEXT PAGE.

Passage II (Questions 8–13)

One of the major problems in using solar energy as a

power source is the conversion of the raw material,

sunlight, into a usable form of fuel The transformation of

energy contained in sunlight to chemical energy via

photochemically induced isomerization

(photo-isomerization) is one method that has been suggested in

order to alleviate this problem In this process, a simple

organic molecule is converted to a product that has far more

energy than the starting material This excess energy may

be stored in the product as angle, torsional or non-bonded

strain The reverse reaction, formation of the starting

material, provides a means in which energy can be released

and used in a controlled manner

A good illustration of the photoisomerization process

is the conversion of norbornadiene to quadricyclane

(Reaction 1) Quadricyclane has 62 kcal/mol of excess

strain energy relative to norbornadiene and is therefore

thermally unfavorable; the reverse reaction, quadricyclane to

norbornadiene, is highly favored in the presence of metal

catalysts

Since it is readily synthesized in high yield,

norbornadiene is a convenient starting material in the

laboratory However, further research is required before it

can be mass-produced for large scale adaptation

Norbornadiene

Strain energy =

33 kcal/mol

Quadricylane Strain energy =

95 kcal/mol

hv

Reaction 1 Norbornadiene converts to quadricyclane via a

photochemically allowed [2+2] ring closure, analogous to

the conversion of 1,3-butadiene to cyclobutene (Reaction

2)

hv

Reaction 2

The alkane equivalent of norbornadiene, norbornane,

is shown in Figure 1

4

2 6 3

7

5 1

Norbornane

Figure 1

8 The ring strain that arises in small cycloalkanes, such

as cyclopropane or cyclobutane, is mainly attributed to:

I compression of the bond angles to less than 109.5°

II interaction between eclipsed hydrogens on adjacent carbons

III nonbonding interactions between hydrogens

on non-adjacent carbons

A I only

B II only

C I and II only

D I, II, and III

9 Which of the following will most readily undergo an

intramolecular [2+2] ring closure?

A

B

C

Ph

CH2

CH2 Ph

D

GO ON TO THE NEXT PAGE.

1 0 Catalytic hydrogenation of norbornadiene releases

approximately 57 kcal of heat per mole of

norbornadiene The product, norbornane, is:

A more stable, due to the decrease in angle strain

accompanying saturation

B more stable, due to the increased number of

hydrogens in the product

C less stable, due to the change in hybridization of

the double bonded carbons

D less stable, due to steric effects and an increase in

angle strain

1 1 For substituted norbornanes, it has been found that

those substituted in the 7 position undergo SN2

reactions with the appropriate nucleophile while those

substituted in the 1 position do not This occurs

because:

A it is difficult to form a carbocation intermediate at

carbon 1 in the initial step of the reaction

B it is easier to form a carbocation at the 7 position,

making substitution kinetically favorable

C back-side attack at the 1 position by the

nucleophile is difficult because it is sterically

hindered

D the carbon at position 7 is more electrophilic than

the carbon at position 1

1 2 Which of the following is true of norbornane?

A Norbornane is less stable than cyclopentane but

more stable than cyclohexane

B Norbornane is more stable than both cyclopentane

and cyclohexane

C Norbornane is less stable than chair cyclohexane

but more stable than boat cyclohexane

D Norbornane is less stable than both cyclopentane

and cyclohexane

1 3 A Diels-Alder process ([4+2] cycloaddition) takes place

according to the following scheme:

A Diels-Alder reaction between which of the following would result in the formation of norbornadiene?

A Propene and 1,3-butadiene

B Acetylene and 1,3-cyclopentadiene

C Ethylene and propene

D Ethylene and 1,3-cyclopentadiene

GO ON TO THE NEXT PAGE.

Questions 14 through 18 are NOT

based on a descriptive passage

1 4 Which of the following compounds will exhibit the

greatest dipole moment?

A (Z)-1,2-Dichloro-1,2-diphenylethene

B (E)-1,2-Dichloro-1,2-diphenylethene

C 1,2-Dichloro-1,2-diphenylethane

D 1,2-Difluoroethane

1 5 How many structural isomers of C3H6Br2 are capable

of exhibiting optical activity?

A 1

B 2

C 3

D 4

1 6 C=C, C=O, N=N, and C=N bonds are quite common

in organic compounds However, C=S, C=P, C=Si,

and other similar bonds are not often found The most

probable explanation for this observation is that:

A carbon does not combine with elements found

below the second row of the periodic table

B sulfur, phosphorus, and silicon do not form pi

bonds due to the lack of occupied p orbitals in

their ground state electron configurations

C sulfur, phosphorus, and silicon are incapable of

orbital hybridization

D the comparative sizes of the 2p and 3p atomic

orbitals make effective overlap between them less

likely than between two 2p orbitals.

1 7 Which of the compounds listed below is linear?

A Carbon tetrachloride

B Propyne

C Acetylene

D 1,3-Hexadiene

1 8 What is the order of increasing carbon-carbon bond

length in the molecules listed below?

I Acetylene

II Benzene III Ethylene

A I, II, III

B I, III, II

C II, III, I

D II, I, III

ANSWER KEY:

Passage I

The first three paragraphs talk about the problems associated with manufacturing chiral drugs as their racemates It then goes on to give Thalidomide as a classic example of this problem

N

N C O

C O H

*

A racemic mixture is one that contains equal amounts of two enantiomers, chiral molecules that have opposite configurations from each other at every stereocenter and so are non-superimposable mirror images In organic compounds, a stereocenter or a chiral center is usually defined as a carbon that is attached to four different groups Thalidomide definitely has at

least one chiral center since the text talks about its administration as a racemate and the R– and S– isomers Therefore, choice A

can be eliminated A chiral center is an atom attached to four different functional groups Starting at the benzene ring, you can see that there are no chiral centers, as all of the carbons are attached to only three substituents The same thing applies to the carbonyl groups Moving on to the carbon in the nitrogen-containing ring, you can see that this is a chiral center It is attached

to a hydrogen (not shown explicitly), a nitrogen, a CH2 group, and a carbonyl group Moving clockwise around the ring, the next group is the carbonyl functionality Again, this center is achiral since it is only attached to three substituents; the same thing applies to the carbonyl group opposite Although the nitrogen in the ring is attached to four different groups (the lone pair

of electrons not shown is considered a group), it undergoes rapid inversion of configuration that causes it to not be a chiral center Moving along to the last two carbons in the ring, you can see that neither of these are chiral since they are each attached

to two hydrogens atoms So, there is only one chiral center in the Thalidomide molecule: the carbon in the nitrogen containing ring which we discussed earlier

In Reaction 1, you should be able to see that the first step involves the electrophilic addition of hydrogen cyanide to the double bond in the vinyl group Let’s look at the mechanism of this reaction in more detail Since the double bond is electron rich, it is likely to be attacked by positively polarized molecules or electrophiles The hydrogen in HCN acts as an electrophile and, therefore, adds to the least substituted carbon to form the most stable carbocation intermediate This intermediate is formed as electrons are given up from the double bond to form a new bond to the hydrogen, leaving behind a positive charge The formation of a secondary carbocation is favored over a primary carbocation, and so the intermediate –CH+CH3 is formed The cyanide ion (CN–) then adds to the carbon bearing the positive charge to form the nitrile product shown in Reaction 1 Obviously from the mechanism I have just described, there is addition across the double bond This addition is initially electrophilic and so choice B, which describes the reaction as being nucleophilic substitution, is incorrect Elimination, as stated in choice C, is also incorrect Elimination usually involves the removal of fragments to form a multiple bond; the process that occurs in the first step of Reaction 1 is just the opposite of this This leaves choices A and D Markovnikov’s rule states that in the addition of a HX to an alkene, the hydrogen will add to the least substituted carbon in order to form the most stable carbocation intermediate This rule is followed in Reaction 1, since the hydrogen adds to the –CH2

in the vinyl group, not the –CH group so choice A is the correct response Anti–Markovnikov addition can be obeyed when peroxides are added to the reaction mixture This reaction occurs by a radical mechanism and the hydrogen adds to the most substituted carbon This is not the case in this reaction, so choice D is incorrect

Epoxides, such as the one shown in Reaction 2, are highly strained and, because of the electron withdrawing nature of the oxygen, a nucleophile can attack one of the carbons attached to it resulting in a ring opening reaction Simultaneously, the nucleophile attacks one of the epoxide carbons, and since carbon cannot form 5 bonds, the bond between it and the oxygen is broken The electrons from this bond are taken on by the oxygen forming a negatively charged anion

In Reaction 2, the R SiN molecule forms RSi+ and N The latter is called an azide ion and as it is negatively

N3

N3

O– + SiR3+

N3 OSiR3

The chiral catalyst is involved in the reaction, but not as a nucleophile A catalyst serves to speed up the reaction and

at the end of the reaction, it remains unchanged Therefore, it would not behave as a nucleophile and become incorporated into the product; choice D is wrong

Choices A and C are wrong since these molecules constitute the electrophilic portion of the R3SiN3 molecule Since they are positively polarized, there is no way they could be nucleophilic

Carbon–carbon double bonds are made up of both sp2 hybridized orbitals and p orbitals The electron configuration of carbon is 1s22s22p2 The 2s orbital and two 2p orbitals hybridize to form three sp2 hybrid orbitals This leaves one free p orbital which can overlap with an adjacent orbital to form a pi bond sp2 hybridized orbitals have a geometry of 120°, and these constitute the carbon-carbon and carbon-hydrogen sigma bonds in the molecule Therefore, the double bond in the vinyl group is

formed by the overlap of two sp2 orbitals and two p orbitals.

Thalidomide was administered as a racemate which is an equal mixture of two enantiomers You should know that enantiomers have identical chemical and physical properties with one exception: they rotate plane polarized light in opposite directions However, they do behave differently in chiral environments, and so they exhibit different behavior in the human body One enantiomer reduces nausea and vomiting while the other is a teratogen which is an agent that causes physical defects

in the developing embryo Therefore, Thalidomide is a mixture of enantiomers that have the same chemical and physical properties but completely different physiological effects; choice A is the correct answer

Choices B and D are incorrect because the isomers in Thalidomide are enantiomers, not diastereomers If they were diastereomers, then they would possess different chemical and physical properties and would form a mixture of two different compounds, not a racemic mixture as described in the passage Choice C is incorrect because enantiomers have the same chemical and all of the same physical properties except in the direction that plane polarized light is rotated

First, let’s look at Naproxen drawn in Reaction 1, which you are told is the S-isomer Priorities have to be assigned to

each substituent directly attached to the stereocenter This is done by atomic number, so the lowest priority goes to the hydrogen The other three atoms attached to the stereocenter are carbons, so the atomic weights of the groups attached to these carbons now have to be considered The substituent that will have the next to last priority is the methyl group, since hydrogens are attached to the methyl carbon In the remaining two substituents, the carbons are attached to two other carbons (in the case

of the substituted naphthalene group) and two oxygens (in the case of the carboxyl group) Another rule you need to know, is that in double bonds, the atoms have to be duplicated; a carbonyl group would be classed as a carbon bonded to two oxygen atoms When this is done, the carboxyl group turns out to be of highest priority So to summarize, the order of increasing priority is hydrogen, methyl, naphthalene and then carboxyl In order to assign a configuration, the lowest priority substituent has to be rotated to the back and then arrows are drawn from the highest priority substituent (numbered 1) to the lowest priority substituent (numbered 3) If this is done for Naproxen in Reaction 1, you should see that the arrows are in an anti–clockwise

direction so the molecule is an S-isomer.

In order to qualify as an R–isomer, the answer choice must have an opposite configuration at all chiral centers, making

it a non-superimposable mirror image of the S–isomer Choice C is correct because it has the opposite configuration of the

S-isomer The hydrogen group is already oriented toward the back, and so in drawing arrows from the carboxyl group through to

the methyl group, you can see that the direction is clockwise, or R.

Choices A and D are wrong because they show the chiral carbon and the methoxy group respectively sticking out from the naphthalene ring These choices are wrong because both groups would be in the plane of the ring In addition, the

orientation around the chiral carbon is incorrect in both responses Choice B is incorrect because this is the S-isomer of

Naproxen; the configuration around the chiral center is exactly the same as that shown in Reaction 1

A meso compound is a molecule that contains chiral centers, so you would expect it to be optically active However,