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Chem 110 > Block 4: Functional Groups I > Flashcards

Block 4: Functional Groups I Flashcards

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1
Q

Why are alkenes good nucleophiles?

A

The double bond holds four electrons between 2 carbons, giving them a high electron density. Therefore they are attracted to nuclei and can react as a nucleophile

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2
Q

What are the 3 ways of preparing an alkene?

A

Elimination.

  1. Catalytic dehydrogenation
  2. Acid-catalysed dehydration
  3. Base promoted dehydrohalogenation
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3
Q

What reagent is needed for calatlytic dehydrogenation of an alkane to form an alkene?

A

Platinum catalyst and at lease 60 degree C heat

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4
Q

What reagent is needed for acid-catalysed dehydration of an alcohol to form an alkene?

A

Conc. H2SO4

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5
Q

What reagent is needed for base promoted dehydrohalogenation of a haloalkane to form an alkene?

A

A strong base (KOH etc) in alcohol

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6
Q

What is Zaitsev’s rule?

A

The major product is the most substituted alkene- the alkene with the least number of hydrogens directly attached to the carbons of C=C (poor get poorer)

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7
Q

What are the reactions alkenes can undergo?

A

Hydrogenation and reactions using H+ as an electrophile

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8
Q

What is the reagent to hydrogenate an alkene to an alkane?

A

H2 and a Pt Catalyst

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9
Q

In what stereochemistry does the H2 add to the alkene?

A

Syn stereochemistry- same side of the molecule

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10
Q

What are the steps of hydrogenation of an alkene on the Pt catalyst?

A
  1. 2 H atoms are adsorbed onto the catalyst.
  2. The alkene arrives and complexes with the catalyst temporarily and non-covalently.
  3. The H and C next to each other bond, leaving an empty bond for the other H to attach to.
  4. The alkane is produced and the catalyst is regenerated.
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11
Q

How does halogenation occur?

A

When X2 is added, it acts as an electrophile, where one X bonds to the old double bond and breaks its covalent bond. This forms a carbocation and an X anion. Then, the other X atom bonds to the now positive C+ charge. This only occurs anti as the first X takes up all the space on its side of the molecule, preventing the other X to bind syn. (NB with bromine, this is called a bromonium ion intermediate).

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12
Q

What should be remembered about halogenation of an alkene?

A

If other nucleophiles are present they can compete to give a different product.
Discharge of Br2 color is used as a lab test to detect alkenes.

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13
Q

In what reagents does H+ act as an electrophile when adding to alkenes?

A

HCl to form monosubstituted haloalkane.
H+/OHR to form Ether
H+/H2O to form Alcohol.

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14
Q

What is markovnikov’s rule?

A

Addition of an unsymmetrical reagent to an unsymmetrical alkene gives as the major product the compound in which the electropositive part of the reagent (H+) has bonded to the carbon of the C=C directly bonded to the greater number of H atoms (rich get richer)

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15
Q

What is the reaction mechanism for the addition of HZ to an alkene?

A
  1. H+ is added across the C=C bond, forming a carbocation on the carbon it does not bond to.
  2. The nucleophile (Z) attaches to the carbocation. If it is H2O/H+, the whole H2O molecule adds before the extra H+ is lost to form an alcohol
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16
Q

Why does markovnikov’s rule work in terms of carbocations?

A

When there are two possible products from an addition reaction, the C with more Cs attached to it has less Hs. Therefore, a 2nd or 3ry carbocation will be formed, which is more stable than the 1my one it could have formed. Therefore it is more likely to remain in solution, and be the dominant product.

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17
Q

How do you prepare an alkyne?

A

Didehydrohalogenations of dihaloalkanes. The reagent used is alcoholic OH-. Firstly a single HX is removed, resulting in a vinyl halide. Finally the second HX is removed, allowing an alkane to form.

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18
Q

What reactions do alkynes undergo?

A

Hydrogenation, Electrophilic addition of HX and X2, Hydration and formation of alkynide anions (on terminal alkynes).

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19
Q

What products can alkynes form when they undergo hydrogenation?

A

They can form alkanes, E-alkenes or Z-alkenes.

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20
Q

How do akynes form alkanes?

A

With addition of 2H2 and a Pt or Pd catalyst

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21
Q

How do alkynes form E-alkenes?

A

With addition of Li and Liquid NH3 and then H2O

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22
Q

How do alkynes form Z-alkenes?

A

With a lindlar catalyst (Pd/Pb(OAc)2)/H+)

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23
Q

What needs to be remembered about electrophilic addition of HX or X2 to alkynes?

A
  1. Markovnikov’s rule is followed
  2. The reaction can be stopped after 1 mol of reagent (to make an alkene) or let all react (for an alkane)
  3. Anti stereochemistry of addition is observed.
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24
Q

What needs to be remembered about hydration of alkynes?

A
  1. Only one mol of water reacts. Therefore the product always has a double bond.
  2. For all alkynes except ethyne, the product is a ketone (in ethyne the product is an aldehyde).
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25
Q

What is the reaction for the hydration of propyne? (Assume markovnikov’s rule is carried out)

A

CH3CCH +H2O –> CH3C(OH)=CH2

CH3C(OH)=CH2 CH3COCH3

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26
Q

What is tautomerisation?

A

The transformation of an unstable enol into a stable ketone. It forms a tautomeric equilibrium and the species involved are tautomers

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27
Q

How do you form an alkynide anion?

A

This only works for terminal alkynes. By adding a strong base (Na+NH2-) an anion is formed as the terminal H leaves.

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28
Q

Why are alkynide anions useful?

A

On addition of a primary haloalkane, a longer carbon chain can be synthesized. (Haloalkane leaves in substitution reaction- C- end is a greater nucleophile)

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29
Q

What is different about benzene’s electrons?

A

Its double bonds are not fixed, meaning that its pi electrons can freely circulate between Carbons in the ring.

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30
Q

What is a resonance structure/hybrid?

A

The same illustration of benzene, but double bonds in the opposite places.

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31
Q

What is unique about benzene?

A

All bond lengths are equal- 1.39 Angstroms. Normally C-C bonds are longer than C=C bonds.
Benzene does not react like alkenes

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32
Q

What is resonance energy?

A

This is the energy than benzene saves. The energy of cyclohexane-cyclohexadiene increases with each addition of a new double bond, so it would be expected that benzene has a large energy, and low stability. However, benzene’s actual energy state is much lower. The energy difference between expected and actual is called resonance energy.

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33
Q

Why is benzene favoured? What does this mean for benzene’s reactions?

A

It has a low energy state- nature wants to keep this energy down to keep it stable, so substitution is favoured where alkenes would typically undergo addition.

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34
Q

How do you decide whether a structure is cyclic or aromatic?

A

Aromatic structure contain (4N+2) = pi electrons. If N is a whole number, the structure is aromatic.

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35
Q

What are some examples of monosubstituted benzenes?

A

Bromobenzene, nitrobenzene.

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36
Q

What orientations can disubstituted benzenes be in?

A

Ortho (1,2), Meta (1,3), or Para (1,4).

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37
Q

What needs to be remembered when naming more than disubstituted benzenes?

A

Use the lowest possible sum of numbers for the substituents.

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38
Q

What are the steps of benzene reacting to become monosubstituted with an electrophile?

A

First, the electron bonds to the E+.
Secondly, whaeland intermediates are formed where the new positive charge (due to an extra bond on C) and the two remaining double bonds are shuffled around.
Thirdly, the Hydrogen leaves and the double bond re-forms, creating a monosubstituted benzene.

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39
Q

What is the RDS in the monosubstitution of benzene?

A

The addition of the electrophile

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40
Q

Why do whaeland intermediates form?

A

They are generated by the movement of the positive charge and electrons. This shares the charge over all the carbons to make the benzene more stable.

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41
Q

Where can the charge be placed in resonance contributors (whaeland intermediates)?

A

Ortho and Para relative to the Electrophile- never meta.

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42
Q

Why does the benzene choose to undergo substitution?

A

It means that the resonance energy of the original benzene is not lost.

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43
Q

What are some different electrophiles able to be formed?

A

Cl+, NO2+, R-C=O (+), an alkyl group with the removal of X- only works with methy, ethyl and 2-propyl.

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44
Q

How do you generate Cl+?

A

Via halogenation.

Cl2 + FeCl3 Cl+ + FeCl4-

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45
Q

How do you generate NO2+?

A

HNO3 + H2SO4 NO2+ + HSO4- + H2O

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46
Q

What needs to be remembered about NO2+?

A

It is resonance stabilized. The double bond and positive charge on the different oxygens are able to swap around.

47
Q

How do you form R-C=O(+)?

A

Acid halide + AlX3 R-C=O(+) —> Benzene-C(=O)R

48
Q

Which haloalkanes can you use to form an alkylated benzene?

A

Methyl, ethyl or 2-propyl haloalkanes.

49
Q

What are the steps to alkylating a benzene?

A
  1. Removal of X using AlX3. This forms a carbocation (electrophile)
  2. Addition of carbocation to benzene.
50
Q

Why can’t 1-propyl haloalkanes work?

A

When the carbocation is formed, the molecule re-shuffles the distribution of Hs so that the more stable secondary carbocation is formed. Therefore, we end up with the same carbocation as if we’d used the secondary one to start with.

51
Q

What reagents add a ketone group to the benzene?

A

An acyl halide and AlX3

52
Q

What reagents add a halogen to the benzene?

A

X2/FeX3

53
Q

what reagents add an alkyl group to the benzene?

A

CH3X/AlX3

54
Q

What reagents add a nitro group to the benzene?

A

HNO3/H2SO4

55
Q

How does the nitro group continue to react after this?

A

Fe/H+ forms an amine group.

Further addition of HNO2 at 0C forms a nitrile group (Diazonium ion)

56
Q

What needs to be remembered when drawing a diazonium ion?

A

Draw a + on the 4-bonded N

57
Q

When turning a benzene from mono to di-substituted, what is the importance of the existing group (G)?

A

It determines where the incoming electrophile will be placed, and whether the electrophile will react more or less quickly relative to the same electrophile with benzene (activating or deactivating).

58
Q

What determines whether G will be ortho/para or meta directing?

A

The number of plausible resonance contributors which can be formed using either route- whichever has more contributors will form.

59
Q

What are the strongly activating ortho/para directors?

A

-OH, -OR, -NHR, -NR2, NH2

60
Q

What is the key to remembering what ortho/para groups are strongly activating?

A

The strongly activating ones turn ON the molecule (Oxygen and Nitrogen based).

61
Q

What is the weakly activating ortho/para director?

A

Alkyl groups, such as -CH3.

62
Q

What is the deactivating ortho/para director?

A

-X

63
Q

How do you remember the deactivating o/p director?

A

-X stops the reaction

64
Q

Why are the o/p groups o/p directors?

A

They have unshared electron pairs able to be donated into the aromatic ring by resonance (or induction for alkyls).

65
Q

What is the strongly deactivating meta director?

A

-N(+)(=O)-O (NO2)

66
Q

What are the moderately deactivating meta directors?

A

-CHO, -COOH, -C_=N

67
Q

How do you remember what groups are meta directors?

A

Meta groups have Multiple bonds

68
Q

What resonance contributors can o/p groups form when E is put o/p?

A

+ charge goes ortho and para to the electrophile. When it sits underneath G, it can transfer up, forming a double bond as the non bonding pairs are donated from the G’s electronegative element. This makes 4 resonance contributors.

69
Q

Why can’t a fourth resonance contributor form when E is put meta with an o/p group?

A

The positive charge never sits directly under G as it has to stay ortho and para to the E. This means the double bond can’t form, resulting in only 3 resonance contributors.

70
Q

What resonance contributors can form when E is placed meta to a meta G?

A

3- + ortho and para to E.

71
Q

Why can’t a third resonance contributor form when E is placed o/p to a meta G?

A

+ sits beneath the G in one of its contributors. As + is positive and so is the atom directly attached to the benzene ring (d+) the two positive charges repel. Therefore, this structure does not form and there are only 2 resonance contributors.

72
Q

How can alkyl halides be prepared?

A

Addition of HX or X2 to an alkene

Substitution from an alcohol

73
Q

How do alkyl halides form from an alkene?

A

Firstly, the H-X or X-X bond breaks, and the more electropositive ion moves to the least substituted Carbon. Then, the remaining X- is able to join to the created carbocation.

74
Q

When HX (X2) is added to an alkene to form a chiral product, will it be R or S and why?

A

It will be a mixture of R and S, as in the carbocation stage, the molecule is planar. The X- can add either above or below the plane of the molecule, creating two different isomers in equal concentrations.

75
Q

Which direction do racemic optical isomers rotate in?

A

No rotation as the different directions cancel

76
Q

What should be remembered about halogenation of an alkene?

A

Addition can be both syn and anti. Also, remember that markovnikov’s rule applies

77
Q

How does a haloalkane form from an alcohol?

A

Either by substitution by SOCl2 (for primary and secondary alcohols) or HX (for tertiary alcohols).

78
Q

Why can haloalkanes undergo substitution so readily?

A

Halogens are weak bases, due to the fact that their conjugate acids are strong acids. Therefore, they are excellent leaving groups. The ease of substitution increases as the molar mass increases.

79
Q

Why are H-, NH2-, OH- and RO- good nucleophiles for haloalkane substitution?

A

Their conjugate acids are weak acids, whereas they are strong bases. This makes them poor leaving groups.

80
Q

What is the Sn1 unimolecular mechanism of substitution?

A

The rate is proportional only to the concentration of the haloalkane. It occurs when the nucleophile reacts after the RDS. This is favoured when intermediate carbocations formed are relatively stable.

81
Q

What species undergo substitution via Sn1?

A

Tertiary haloalkanes greater than secondary haloalkanes much more than primary haloalkanes- unless benzyl halides.

82
Q

Describe the reaction energy profile for Sn1 substitution

A

The reactant energy increases largely (RDS) before a dip (where the carbocation intermediate is), and the another rise (smaller Ea for the Nucleophile bonding) and then a drop to products’ energy: the two humped camel profile.

83
Q

If we begin with an optically active haloalkane isomer, what will be the optical product of an Sn1 mechanism?

A

A racemic mixture will always form. This is because the carbocation intermediate is planar, and so equal amounts of each isomer are formed

84
Q

Why is Sn1 less favoured in the body?

A

The body doesn’t like to lose chirality. As Sn1 converts optically active compounds to non-active, it loses this.

85
Q

What is the Sn2 bimolecular mechanism?

A

The rate of reaction is proportional to the concentration of both the alkyl halide and the nucleophile. The nucleophile bonds as the halogen leaves- all in one step.

86
Q

What species undergo haloalkane substitution via Sn2?

A

Primary and some secondary haloalkanes.

87
Q

What is the mechanism of Sn2 substitution?

A

The energy of the reactants increases into a single Ea hump, at the tip of which is a double crossed line. This marks the transition state, which is [Nu—-CH3—-Br]. At this point, the mechanism can go back to a haloalkane or continue to react with the nucleophile. There is no carbocation formed and the mechanism is described ad concerted or synchronous.

88
Q

What results if a chiral non-racemic species reacts via Sn2?

A

The chirality will be inverted. This is because the Nucleophile can only add to the opposite end as the leaving group. Therefore if you start with R you will end up with S and vice versa.

89
Q

Why does the body favour Sn2 substitution?

A

Although chirality is reversed, it is still maintained, which the body likes.

90
Q

Why do primary haloalkanes undergo Sn2 while tertiary don’t?

A

As the group of an alcohol increases, the transition state becomes more crowded. This raises the energy of the transition state and therefore Ea. As a result, the more groups attached to the substituting site, the less favoured the reaction is.

91
Q

What other mechanism can haloalkanes react with?

A

Elimination- E1 and E2. This competes with substitution when either are able to occur.

92
Q

What affects whether a haloalkane will undergo substitution or elimination?

A

The category of the alkyl halide, the temperature, solvent, and reagent used.

93
Q

What is E1 mechanism?

A

The rate is proportional to the concentration of the alkyl halide. It occurs via a carbocation intermediate, as the halogen is lost, which then attracts the adjacent H, forming a double bond. It is most favoured in tertiary haloalkanes as their carbocation intermediates are most stable. E1 and Sn1 can compete, leading to mixtures of products.

94
Q

What must be remembered about E1?

A

Saytseff’s rule applied where more than 1 alkene can be formed. Stronger bases, higher temperatures and non-nucleophilic solvents favour elimination over substitution

95
Q

What is E2?

A

The rate is proportional to the concentration of the alkyl halide and the nucleophile (although in this context, the nucleophile is a base).

96
Q

How does E2 work?

A

A base is negative, and attracts the slight positivity of an H atom, causing a bond to begin forming between the two. At the same time, the breaking of the C-H bond causes another C-C bond to begin forming. As this happens, the adjacent C is gaining too many bonds and so begins breaking a bond with X all simultaneously. This is a transition state, before the alkene and X- products are formed.

97
Q

How do you differentiate between a base and a nucleophile?

A

If a species attacks C with an X attached, it is a nucleophile. If it attacks an H on the adjacent C to the one with an X, it is a base, as a base attacks acids (H+)

98
Q

What must be remembered about E2?

A

The nucleophilic attack occurs at the C neighbouring the C bearing X.
It requires a strong base, such as RO- or HO-.
E2 is favoured by tertiary and some secondary haloalkanes.
Saytzeff’s rule applies.

99
Q

When can E2 be observed for primary haloalkanes?

A

If the product will extend pre-existing conjugation, like in benzene rings.

100
Q

Will Elimination or substitution be favoured for primary alkyl halides?

A

Only ever substitution via Sn2 mechanism.

101
Q

What class of reaction will occur when a secondary alkyl halide reacts with a weak nucleophile, with no basicity?

A

Substitution via Sn1 or Sn2

102
Q

What class of reaction will occur when a tertiary alkyl halide reacts with a weak nucleophile, with no basicity?

A

Substitution via Sn1 or Elimination via E1

103
Q

What are some examples of weak, non-basic nucleophiles?

A

H2O, CH3OH, CH3CH2OH

104
Q

What class of reaction will occur when a secondary alkyl halide reacts with a strong nucleophile with low basicity?

A

Substitution via Sn2

105
Q

What class of reaction will occur when a tertiary alkyl halide reacts with a strong nucleophile of low basicity?

A

Substitution via Sn1 or Elimination via E1

106
Q

What are some examples of strong nucleophiles with low basicity?

A

X-, CN-, CH3S-, CH3COO-

107
Q

What class of reaction will occur when a secondary alkyl halide reacts with a strong nucleophile of high basicity?

A

Substitution via Sn2 or Elimination via E2

108
Q

What class of reaction will occur when a tertiary alkyl halide reacts with a strong nucleophile of high basicity?

A

Elimination via E2

109
Q

What are some examples of strong nucleophiles with high basicity?

A

OH-, CH3O-, CH3CH2O-, H3N, (CH3)3N, H2N-

110
Q

What species are not susceptible to nucleophilic substitution?

A

Halogens on sp2 carbons

111
Q

What are grignard reagents?

A

A species which results from the addition of magnesium to an halo-alkane or halo-benzene. The magnesium is electropositive, so the attached carbon is regarded as a carbanion (R-). Therefore, it can act as a nucleophile or base.

112
Q

What reagent is needed to form a grignard reagent?

A

Mg in dry ethyl ether

113
Q

What is the structure of a grignard reagent?

A

The same as the haloalkane, although the Mg inserts itself between the C and the halogen.

Chem 110 (6 decks)

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