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1
Q
A

free-radical halogenation

Rarely an effective method for the synthesis of alkyl halides. It usually produces mixtures of products because there are different kinds of hydrogen atoms that can be abstracted. Also, more than one halogen atom may react, giving multiple substitutions. The chlorination of propane can give a mixture of products.

2
Q
A

free-radical halogenation

Free-radical bromination is highly selective in the synthesis of alkyl halides, and it gives good yields of products that have one type of hydrogen atom that is more reactive than the others. Isobutane has only one tertiary hydrogen atom, and this atom is preferentially abstracted to give a tertiary free radical.

3
Q
A

free-radical halogenation

All the hydrogen atoms in cyclohexane are equivalent, so a medium yield of chlorocyclohexane results. Formation of dichlorides and trichlorides is possible, but these side reactions are controlled by using only a small amount of chlorine and an excess of cyclohexane.

4
Q
A

allylic bromination

Either end of the resonance-stabilized allylic radical can react with bromine. In one of the products, the bromine atom appears in the same position where the hydrogen atom was abstracted. The other product results from reaction at the carbon atom that bears the radical in the second resonance form of the allylic radical. This second compound is said to be the product of an allylic shift.

5
Q

allylic position

A

a carbon atom next to a carbon–carbon double bond

6
Q

initiation step of allylic bromination

A

Initiation Step

Bromine absorbs light, causing formation of radicals.

7
Q

first propagation step of allylic bromination

A

First Propagation Step

A bromine radical abstracts an allylic hydrogen.

8
Q

second propagation step of allylic bromination

A

Second Propagation Step

Either radical carbon can react with bromine.

9
Q
A

For efficient allylic bromination, a large concentration of bromine must be avoided because bromine can also add to the double bond. N-Bromosuccinimide (NBS) is often used as the bromine source in free-radical brominations because it combines with the HBr side product to regenerate a constant low concentration of bromine. No additional bromine is needed because most samples of NBS contain traces of Br2 to initiate the reaction.

10
Q
A

allylic bromination

11
Q
A

allylic bromination

12
Q
A

free-radical halogenation (synthetically useful only in certain cases)

13
Q
A

free-radical halogenation (synthetically useful only in certain cases)

14
Q
A

allylic bromination

15
Q
A

nucleophilic substitution

a nucleophile (Nuc-) replaces a leaving group (X- ) from a carbon atom, using its lone pair of electrons to form a new bond to the carbon atom.

16
Q
A

dehydrohalogenation elimination

Both the halide ion and another substituent are lost. A new π bond is formed. The reagent (B-) reacts as a base, abstracting a proton from the alkyl halide. Most nucleophiles are also basic and can engage in either substitution or elimination, depending on the alkyl halide and the reaction conditions.

17
Q
A

nucleophilic substitution

a nucleophile (-OCH3) replaces a leaving group (Br-) from a carbon atom, using its lone pair of electrons to form a new bond to the carbon atom.

18
Q
A

elimination

when OH is protonated, H2O is the leaving group

19
Q
A

elimination

both Br atoms are lost, iodide ion is a nucleophile that reacts at Br.

20
Q
A

SN2 (second-order nucleophilic substitution)

Hydroxide ion is a strong nucleophile (donor of an electron pair) because the oxygen atom has unshared pairs of electrons and a negative charge. Iodomethane is called the substrate, meaning the compound that is attacked by the reagent. The carbon atom of iodomethane is electrophilic because it is bonded to an electronegative iodine atom. Electron density is drawn away from carbon by the halogen atom, giving the carbon atom a partial positive charge. The negative charge of hydroxide ion is attracted to this partial positive charge.

21
Q
A

SN2 (second-order nucleophilic substitution)

Hydroxide ion attacks the back side of the electrophilic carbon atom, donating a pair of electrons to form a new bond. This one-step mechanism is supported by kinetic information. The rate is found to double when the concentration of either reactant is doubled. The reaction is therefore first order in each of the reactants and second order overall. The rate equation has the following form:

rate = kr[CH3I][-OH]

22
Q

explain the reaction-energy diagram for the SN2 reaction of methyl iodide with hydroxide

A

The reaction-energy diagram for the SN2 reaction of methyl iodide with hydroxide shows only one energy maximum: the transition state. There are no intermediates.

The electrostatic potential maps of the reactants, transition state, and products show that the negatively charged nucleophile (red) attacks the electrophilic (blue) region of the substrate. In the transition state, the negative charge (red) is delocalized over the nucleophile and the leaving group. The negative charge leaves with the leaving group.

23
Q
A

SN2 reaction

Takes place in a single (concerted) step. A strong nucleophile attacks the electrophilic carbon, forcing the group to leave. The order of reactivity for substrates is CH3X > 1° > 2°. (3° alkyl halides cannot react by this mechanism.)

24
Q

which is a stronger nucleophile, a base or its congugate acid?

A

base

25
Q

which is a stronger nucleophile

R-O-

R-OH

A

R-O-

26
Q
A

SN2 halogen exchange reaction

Alkyl fluorides are difficult to synthesize directly, and they are often made by treating alkyl chlorides or bromides with KF under conditions that use a crown ether to dissolve the fluoride salt in an aprotic solvent, which enhances the normally weak nucleophilicity of the fluoride ion.

27
Q
A

SN2 halogen exchange reaction

Iodide is a good nucleophile, and many alkyl chlorides react with sodium iodide to give alkyl iodides.

28
Q
A

SN2 halogen exchange reaction

Alkyl fluorides are difficult to synthesize directly, and they are often made by treating alkyl chlorides or bromides with KF under conditions that use a crown ether to dissolve the fluoride salt in an aprotic solvent, which enhances the normally weak nucleophilicity of the fluoride ion.

29
Q
A

SN2 halogen exchange reaction

Iodide is a good nucleophile, and many alkyl chlorides react with sodium iodide to give alkyl iodides.

30
Q

describe the periodic trend in nucleophilicity

A

Nucleophilicity increases down the periodic table, following the increase in size and polarizability, and the decrease in electronegativity.

  • stronger* **I- ** > **Br- ** > Cl- > F- weaker
  • stronger* -SeH > -SH > -OH weaker
  • stronger* (CH3CH2)3P > ** (CH3CH2)3N** weaker
31
Q

rate the nucleophilicity

carboxylate ion

alkoxide ion

alcohol

A

Alkoxide ion is the strongest nucelophile (similar to hydroxide ion), a species with a negative charge is a stronger nucleophile than a similar neutral species. In particular, a base is a stronger nucleophile than its conjugate acid.

Carboxylate ion is a moderate nucleophile, as carboxylic acids easily dissociate into a carboxylate anion and a positively charged hydrogen ion (proton), much more readily than alcohols do (into an alkoxide ion and a proton), because the carboxylate ion is stabilized by resonance. The negative charge that is left after deprotonation of the carboxyl group is delocalized between the two electronegativeoxygen atoms in a resonance structure.

Alcohol (similar to water) is the weakest nucleophile of the group, as the conjugate acid of alkoxide.

32
Q

contrast basicity and nucleophilicity

A

Basicity is defined by the equilibrium constant for abstracting a proton. Nucleophilicity is defined by the rate of attack on an electrophilic carbon atom. In both cases, the nucleophile (or base) forms a new bond. If the new bond is to a pro- ton, it has reacted as a base; if the new bond is to carbon, it has reacted as a nucleophile.

33
Q

describe the periodic trend in polarisability as it relates to the SN2 reaction

A

Trends in size and polarisability reflect an atom’s ability to engage in partial bonding as it begins to attack an electrophilic carbon atom. As we go down a column in the periodic table, the atoms become larger, with more electrons at a greater distance from the nucleus. The electrons are more loosely held, and the atom is more polarisable: Its electrons can move more freely toward a positive charge, resulting in stronger bonding in the transition state. The increased mobility of its electrons enhances the atom’s ability to begin to form a bond at a relatively long distance.

34
Q

contrast flouride ion with iodide ion in the SN2 reaction

A

Fluoride has tightly bound electrons that cannot begin to form a C—F bond until the atoms are close together. Iodide has more loosely bound outer electrons that begin bonding earlier in the reaction.

35
Q

steric hindrance

A

when bulky groups interfere with a reaction by virtue of their size

36
Q

protic solvent

A

A protic solvent is one that has acidic protons, usually in the form of O - H or N - H groups.

37
Q

describe the effect of anion size on solvation

A

Small anions are solvated more strongly than large anions in a protic solvent because the solvent approaches a small anion more closely and forms stronger hydrogen bonds. When an anion reacts as a nucleophile, energy is required to “strip off” some of the solvent molecules, breaking some of the hydrogen bonds that stabilized the solvated anion. More energy is required to strip off solvent from a small, strongly solvated ion such as fluoride than from a large, diffuse, less strongly solvated ion like iodide.

38
Q

aprotic solvent

A

solvents without O - H or N - H groups, most polar, ionic reagents are insoluble in simple aprotic solvents such as alkanes

39
Q

describle solvent effects on nucleophilicity of anions

A

In contrast with protic solvents, aprotic solvents (solvents without O - H or N - H groups) enhance the nucleophilicity of anions. An anion is more reactive in an aprotic solvent because it is not so strongly solvated. There are no hydrogen bonds to be broken when solvent must make way for the nucleophile to approach an electrophilic carbon atom.

40
Q

polar aprotic solvent

A

Polar aprotic solvents have strong dipole moments to enhance solubility, yet they have no O - H or N - H groups to form hydrogen bonds with anions. Examples of useful polar aprotic solvents are acetonitrile, dimethylformamide, and acetone.

41
Q
A

Fluoride ion, normally a poor nucleophile in hydroxylic (protic) solvents, can be a good nucleophile in an aprotic solvent. Although KF is not very soluble in acetonitrile, 18-crown-6 solvates the potassium ions, and the poorly solvated (and therefore nucleophilic) fluoride ion follows.

42
Q

examples of ions that are strong bases and poor leaving groups in the SN2 reaction

A
43
Q

list any stereochemical effects of the SN2 reaction

A

Walden inversion

an asymmetric carbon atom will undergo inversion of configuration

44
Q

solvolysis

A

A SN1 reaction that takes place with the solvent acting as the nucleophile, its rate does not depend on the concentration of the nucleophile (usually methanol). The rate depends only on the concentration of the substrate.

45
Q

What is the first step in the SN1 mechanism

A

formation of carbocation (rate limiting)

46
Q

What is the second step in the SN1 reaction

A

nucleophilic attack on the carbocation

47
Q

What is the third step in the SN1 reaction

A

loss of proton to solvent (only if the nucleophile is an uncharged molecule like water or an alcohol, the positively charged product must lose a proton to give the final uncharged product)

48
Q

which compounds do not undergo SN1 or SN2 reactions?

A

Vinyl halides and aryl halides generally do not undergo SN1 or SN2 reactions. An SN1 reaction would require ionization to form a vinyl or aryl cation, either of which is less stable than most alkyl carbocations. An SN2 reaction would require back-side attack by the nucleophile, which is made impossible by the repulsion of the electrons in the double bond or aromatic ring.

49
Q

what is the effect of a solvent’s dielectric constant on ionisation rate of tert-butyl chloride and what does that mean in terms of substitution reactions?

A

Ionization of an alkyl halide requires formation and separation of positive and negative charges, similar to what happens when sodium chloride dissolves in water. Therefore, SN1 reactions require highly polar solvents that strongly solvate ions. Note that ionisation occurs much faster in highly polar solvents such as water and alcohols. Although most alkyl halides are not soluble in water, they often dissolve in highly polar mixtures of acetone and alcohols with water.

50
Q

list any stereochemical effects of the SN1 reaction

A

The SN1 reaction is not stereospecific. In the SN1 mechanism, the carbocation intermediate is sp2 hybridized and planar. A nucleophile can attack the carbocation from either face, because the carbocation is planar and achiral; and attack from both faces gives both enantiomers of the product. Such a process, giving both enantiomers of the product (whether or not the two enantiomers are produced in equal amounts), is called racemization. The product is either racemic or at least less optically pure than the starting material.

51
Q

methyl shift

A

Primary halides and methyl halides rarely ionize to carbocations in solution. If a primary halide ionizes, it will likely ionize with rearrangement. The methyl shift occurs while bromide ion is leaving, so that only the more stable tertiary carbocation is formed.

52
Q

SN1 vs SN2: effect of the nucleophile

A

SN1: Nucleophile strength is unimportant (usually weak).

SN2: Strong nucleophiles are required.

The nucleophile takes part in the slow step (the only step) of the SN2 reaction but not in the slow step of the SN1. Therefore, a strong nucleophile promotes the SN2 but not the SN1. Weak nucleophiles fail to promote the SN2 reaction; therefore, reactions with weak nucleophiles often go by the SN1 mechanism if the sub- strate is secondary or tertiary.

53
Q

SN1 vs SN2: effect of the substrate

A

SN1 substrates: 3° > 2° (1° and CH3X are unlikely)

**SN2 substrates: CH3X > 1° > 2° (3° unstable) **

Most methyl halides and primary halides are poor substrates for SN1 substitutions be- cause they cannot easily ionize to high-energy methyl and primary carbocations. They are relatively unhindered, however, so they make good SN2 substrates. Tertiary halides are too hindered to undergo SN2 displacement, but they can ionize to form tertiary carbocations. Tertiary halides undergo substitution exclusively through the SN1 mechanism. Secondary halides can undergo substitution by either mechanism, depending on the conditions.

54
Q

SN1 vs SN2: effect of the solvent

A

SN1: Good ionizing solvent required.

**SN2: May go faster in a less polar solvent **

The slow step of the SN1 reaction involves formation of two ions. Solvation of these ions is crucial to stabilizing them and lowering the activation energy for their formation. Very polar ionizing solvents such as water and alcohols are needed for the SN1. The solvent may be heated to reflux (boiling) to provide the energy needed for ionization.

Less charge separation is generated in the transition state of the SN2 reaction. Strong solvation may weaken the strength of the nucleophile because of the energy needed to strip off the solvent molecules. Thus, the SN2 reaction often goes faster in less polar solvents if the nucleophile will dissolve. Polar aprotic solvents may enhance the strength of weak nucleophiles.

55
Q

SN1 vs SN2: kinetics

A

SN1 rate = kr[R-X]

**SN2 rate = kr[R-X][Nuc:-] **

The rate of the SN1 reaction is proportional to the concentration of the alkyl halide but not the concentration of the nucleophile. It follows a first-order rate equation. The rate of the SN2 reaction is proportional to the concentrations of both the alkyl halide [R - X] and the nucleophile [Nuc:-]. It follows a second-order rate equation.

56
Q

SN1 vs SN2: effect on stereochemistry

A

SN1 stereochemistry: Mixture of retention and inversion; racemization.

SN2 stereochemistry: Complete inversion.

The SN1 reaction involves a flat carbocation intermediate that can be attacked from either face. Therefore, the SN1 usually gives a mixture of inversion and retention of configuration.

The SN2 reaction takes place through a back-side attack, which inverts the stereochemistry of the carbon atom. Complete inversion of configuration is the result.

57
Q

SN1 vs SN2: effect of rearrangements

A

SN1: Rearrangements are common.

SN2: Rearrangements are impossible.

The SN1 reaction involves a carbocation intermediate. This intermediate can rearrange, usually by a hydride shift or an alkyl shift, to give a more stable carbocation. The SN2 reaction takes place in one concerted step with no intermediates. No rearrangement is possible in the SN2 reaction.

58
Q

weak nucleophiles are OK

A
59
Q

strong nucleophile needed

A
60
Q

3° > 2° carbocation

A
61
Q

CH3X > 1° > 2°

A
62
Q

good ionising solvent needed

A
63
Q

wide variety of solvents, like to be less polar

A
64
Q

good leaving group required

A
65
Q

AgNO3 forces ionisation

A
66
Q

first order reaction

A
67
Q

second order reaction

A
68
Q

mixture of inversion and retention

A
69
Q

complete inversion

A
70
Q

common rearrangements of carbocations

A
71
Q

impossible to have rearrangements

A

SN2

72
Q

T/F: I- is a stronger nucleophile than Cl-

A

true, it is less electronegative

73
Q

nucleophile

A

nucleophile is a species that donates a pair of electrons to form a new covalent bond

74
Q
A

SN1: a reluctant first-order substrate can be forced to ionize by adding some silver nitrate (one of the few soluble silver salts) to the reaction. Silver ion reacts with the halogen to form a silver halide (a highly exothermic reaction), generating the cation of the alkyl group.

75
Q

elimination

A

An elimination involves the loss of two atoms or groups from the substrate, usually with formation of a π bond. Elimination reactions frequently accompany and compete with substitutions. By varying the reagents and conditions, we can often modify a reaction to favor substitution or to favor elimination.

76
Q

why is E1 termed unimolecular?

A

The rate-limiting transition state involves a single molecule rather than a collision between two molecules. The slow step of an E1 reaction is the same as in the SN1 reaction: unimolecular ionization to form a carbocation. In a fast second step, a base abstracts a proton from the carbon atom adjacent to the C+. The electrons that once formed the carbon–hydrogen bond now form a π bond between two carbon atoms.

77
Q

first step of E1 mechanism

A

Unimolecular ionization to give a carbocation (rate-limiting)

78
Q

second step of E1 mechanism

A

Deprotonation by a weak base (often the solvent) gives the alkene (fast).

79
Q
A

The E1 reaction almost always competes with the SN1 reaction. Whenever a carbocation is formed, it can undergo either substitution or elimination, and mixtures of products often result.

80
Q

dehydrohalogenation

A

an elimination of hydrogen and a halogen atom

81
Q

E1 product:

A

Ionization of the alkyl halide gives a carbocation intermediate, which loses a proton to give the alkene. Ethanol serves as a base in the elimination.

82
Q

SN1 product:

A

SN1 results from nucleophilic attack on the carbocation. Ethanol serves as a nucleophile in the substitution.

83
Q

Provide the next step in the SN1 mechanism

A

Solvent (a weak nucleophile) attacks the rearranged (in a hydride shift, not shown) carbocation, deprotonation (not shown) gives the rearranged product.

84
Q

Provide the next step in the E1 mechanism

A

The weakly basic solvent removes either adjacent proton on the rearranged (in a hydride shift, not shown) carbocation (fast).

85
Q

What are the four ways a carbocation can react to become more stable?

A
  1. React with its own leaving group to return to the reactant: R+ + :X- -> R - X
  2. React with a nucleophile to form a substitution product (SN1): R+ + Nuc:- -> R - Nuc
  3. Lose a proton to form an elimination product (an alkene) (E1) (shown)
  4. Rearrange to a more stable carbocation, then react further.

The order of stability of carbocations is: resonance-stabilised, 3° > 2° > 1°.

86
Q

Zaitsev’s rule

A

In elimination reactions, the most substituted alkene usually predominates.

87
Q

Zaitsev orientation

A

E1 and E2 elimination reactions that produce the most substituted alkene

88
Q

E2 reaction

A

In the general mechanism of the E2 reaction, a strong base abstracts a proton on a carbon atom adjacent to the one with the leaving group. As the base abstracts a proton, a double bond forms and the leaving group leaves. Like the SN2 reaction, the E2 is a concerted reaction in which bonds break and new bonds form at the same time, in a single step. (shown, methoxide reacts as a base rather than as a nucleophile).

89
Q

describe the reactivity of an alkyl halide substrate in the E2 reaction

A

The order of reactivity of alkyl halides toward E2 dehydrohalogenation is found to be

> >

90
Q

E2 mechanism

A

The concerted E2 reaction takes place in a single step. A strong base abstracts a proton on a carbon next to the leaving group, and the leaving group leaves. The product is an alkene.

91
Q

predict the major product and list the mechanism used

A

E2

92
Q

predict the major product and list the mechanism used

A

E2

93
Q

Name and describe the two conformations required for an E2 dehydrohalogenation

A
94
Q
A

The E2 is a stereospecific reaction, because different stereoisomers of the starting material react to give different stereoisomers of the product. This stereospecificity results from the anti-coplanar transition state that is usually involved in the E2.

95
Q
A

The E2 is a stereospecific reaction, because different stereoisomers of the starting material react to give different stereoisomers of the product. This stereospecificity results from the anti-coplanar transition state that is usually involved in the E2.

96
Q

E1 vs E2: effect of the base

A

E1: Base strength is unimportant (usually weak).

E2: Requires strong bases.

The nature of the base is the single most important factor in determining whether an elimination will go by the E1 or E2 mechanism. If a strong base is present, the rate of the bimolecular reaction will be greater than the rate of ionization, and the E2 reaction will predominate (perhaps accompanied by the SN2).

If no strong base is present, then a good solvent makes a unimolecular ionization likely. Subsequent loss of a proton to a weak base (such as the solvent) leads to elimination. Under these conditions, the E1 reaction usually predominates, usually accompanied by the SN1.

97
Q

E1 vs E2: effect of the solvent

A

E1: Requires a good ionizing solvent.

**E2: Solvent polarity is not so important. **

The slow step of the E1 reaction is the formation of two ions. Like the SN1, the E1 reaction critically depends on polar ionizing solvents such as water and the alcohols.

In the E2 reaction, the transition state spreads out the negative charge of the base over the entire molecule. There is no more need for solvation in the E2 transition state than in the reactants. The E2 is therefore less sensitive to the solvent; in fact, some reagents are stronger bases in less polar solvents.

98
Q

E1 vs E2: effect of the substrate

A

**E1, E2: 3° > 2° > 1° (1° usually will not go E1) **

In the E1 reaction, the rate-limiting step is formation of a carbocation, and the reactivity order reflects the stability of carbocations. In the E2 reaction, the more substituted halides generally form more substituted, more stable alkenes.

99
Q

E1 vs E2: kinetics

A

**E1 rate = kr[RX]
E2 rate = kr[RX][B:-] **

The rate of the E1 reaction is proportional to the concentration of the alkyl halide [RX] but not to the concentration of the base. It follows a first-order rate equation. The rate of the E2 reaction is proportional to the concentrations of both the alkyl halide [RX] and the base [B:-]. It follows a second-order rate equation.

100
Q

E1 vs E2: orientation of elimination

A

E1, E2: Usually Zaitsev orientation.

In most E1 and E2 eliminations with two or more possible products, the product with the most substituted double bond (the most stable product) predominates. This principle is called Zaitsev’s rule, and the most highly substituted product is called the Zaitsev product.

101
Q

E1 vs E2: effect on stereochemistry

A

E1: No particular geometry required for the slow step.

E2: Coplanar arrangement (usually anti) required for the transition state.

The E1 reaction begins with an ionization to give a flat carbocation. No particular geometry is required for ionization.

The E2 reaction takes place through a concerted mechanism that requires a coplanar arrangement of the bonds to the atoms being eliminated. The transition state is usually anti-coplanar, although it may be syn-coplanar in rigid systems.

102
Q

E1 vs E2: rearrangement

A

E1: Rearrangements are common.

**E2: No rearrangements. **

The E1 reaction involves a carbocation intermediate. This interme- diate can rearrange, usually by the shift of a hydride or an alkyl group, to give a more stable carbocation.

The E2 reaction takes place in one step with no intermediates. No rearrangement is possible in the E2 reaction.

103
Q
A

SN1

104
Q
A

SN2

105
Q
A

E1

106
Q
A

E2