11 reactions of alkyl halides nucleophilic substitutions and eliminations l.
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11. Reactions of Alkyl Halides: Nucleophilic Substitutions and Eliminations. Alkyl Halides React with Nucleophiles and Bases. Alkyl halides are polarized at the carbon-halide bond, making the carbon electrophilic

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alkyl halides react with nucleophiles and bases
Alkyl Halides React with Nucleophiles and Bases
  • Alkyl halides are polarized at the carbon-halide bond, making the carbon electrophilic
  • Nucleophiles will replace the halide in C-X bonds of many alkyl halides(reaction as Lewis base)
  • Nucleophiles that are Brønsted bases produce elimination

d+

Acts as

Nucleophile

d+

Acts as

Base

why this chapter
Why this Chapter?
  • Nucleophilic substitution, base induced elimination are among most widely occurring and versatile reaction types in organic chemistry
  • Reactions will be examined closely to see:
  • How they occur
  • What their characteristics are
  • How they can be used
11 1 the discovery of nucleophilic substitution reactions walden inversion
11.1 The Discovery of Nucleophilic Substitution Reactions:Walden Inversion
  • In 1896, Walden showed that (-)-malic acid could be converted to (+)-malic acid by a series of chemical steps with achiral reagents
  • This established that optical rotation was directly related to chirality and that it changes with chemical alteration
    • Reaction of (-)-malic acid with PCl5 gives (+)-chlorosuccinic acid
    • Further reaction with wet silver oxide gives (+)-malic acid
    • The reaction series starting with (+) malic acid gives (-) acid
reactions of the walden inversion
Reactions of the Walden Inversion
  • The reactions alter the array at the chirality center
  • The reactions involve substitution at that center
  • Therefore, nucleophilic substitution can invert the configuration at a chirality center
  • The presence of carboxyl groups in malic acid led to some dispute as to the nature of the reactions in Walden’s cycle
another example
AnotherExample:
  • (+)-1-phenyl-2-propanol
  • (-)-1-phenyl-2-propanol
  • Used Tosylate as an excellent leaving group
kinetics of nucleophilic substitution
Kinetics of Nucleophilic Substitution
  • Rate (V) = change in concentration with time
    • Depends on concentration(s), temperature, inherent nature of reaction (barrier on energy surface)
  • Arate lawdescribes relationship between the concentration of reactants and conversion to products
    • The rate law is a result of the mechanism
  • A rate constant (k) is the proportionality factor between concentration and rate
  • Kinetics = The study of rates of reactions
  • Rates ↓ as concentrations ↓ but k stays same
  • Rate units: [concentration]/time such as L/(mol x s)
  • The orderof a reaction is sum of the exponents of the concentrations in the rate law
11 2 the s n 2 reaction
11.2 The SN2 Reaction
  • Reaction is with inversion at reacting center
  • Follows second order reaction kinetics
  • Ingold nomenclature to describe characteristic step:
    • S=substitution; N (subscript) = nucleophilic; 2 = both nucleophile and substrate in characteristic step (bimolecular)

Rate is dependant on both Nucleophile & Substrate

Rate = k [CH3-Br] [HO-]

s n 2 process
SN2 Process

The reaction involves a transition state in which both reactants are together

s n 2 transition state
SN2 Transition State
  • The transition state of an SN2 reaction has a planar arrangement of the carbon atom and the remaining three groups
reactant and transition state energy levels affect rate

11.3 Characteristics of the SN2Rxn

Reactant and Transition State Energy Levels Affect Rate

Higher reactant energy level (red curve) = faster reaction (smaller G‡).

Higher transition state energy level (red curve) = slower reaction (larger G‡).

the substrate steric effects on s n 2 reactions
The Substrate: Steric Effects on SN2 Reactions
  • SN2 Sensitive to steric effects

The carbon atom in (a) bromomethane is readily accessible

resulting in a fast SN2 reaction. The carbon atoms in (b) bromoethane (primary), (c) 2-bromopropane (secondary), and (d) 2-bromo-2-methylpropane (tertiary) are successively more hindered, resulting in successively slower SN2 reactions.

the substrate steric effects order of reactivity in s n 2
The Substrate: Steric Effects : Order of Reactivity in SN2
  • SN2 Sensitive to steric effects
  • No reaction at C=C (vinyl or Aryl halides)
the nucleophile in s n 2
The Nucleophile:in SN2
  • Neutral or negatively charged Lewis base
  • Reaction increases coordination at nucleophile
    • Neutral nucleophile acquires positive charge
    • Anionic nucleophile becomes neutral

d+

d+

the nucleophile relative reactivity in s n 2
The Nucleophile: Relative Reactivity in SN2
  • Depends on reaction and conditions
  • More basic nucleophiles react faster
  • Better nucleophiles are lower in a column of the periodic table
  • Anions(-) are usually more reactive than neutrals

Nucleophiles

the leaving group in s n 2
The Leaving Group: in SN2
  • Stable anions that are weak bases are usually excellent leaving groups and can delocalize charge
  • very basic or very small grps are poor leaving groups
  • .

Alkyl fluorides, alcohols, ethers, and amines do not typically undergo SN2 reactions.

the leaving group in s n 218
The Leaving Group: in SN2
  • -OH needs to be turned into a good leaving group.
  • So can convert to a
  • Cl
  • Br
  • Tos
  • Which are excellent leaving groups.

(p-toluenesufonylchloride

p-TosCl)

the leaving group in s n 219
The Leaving Group: in SN2
  • O of the epoxide can be turned into a good leaving group so epoxide can be opened with a weak nucleophile.
  • Addition of an acid (H+) will make epoxide C’s more electrophilic.
  • Cl- attacks less hindered site.
    • (If choice of epoxide C’s is 1o or 2o then major product is from attack at less hindered 1o
    • If choice of epoxide C’s is 1ovs 3o then major product is from attack at more + 3o)

d+

d+

the solvent in s n 2
The Solvent: in SN2
  • Solvents that can donate hydrogen bonds (protic) (-OH or –NH) slow SN2 reactions by associating with reactants
  • Energy is required to break interactions between reactant and solvent

Caged nucleophiles can’t attack so well

the solvent in s n 221
The Solvent: in SN2

Poor for SN2

Good for SN2

Protic solvents (with -OH or –NH) slow SN2 reactions by complexing with reactants

Polar aprotic solvents (no NH, OH, SH) form weaker interactions with substrate and permit fast SN2 reactions

11 4 the s n 1 reaction
11.4 The SN1 Reaction
  • Tertiary alkyl halides react rapidly in protic solvents by a mechanism that involves departure of the leaving group prior to addition of the nucleophile
slide23

SN1 Reaction

The reaction involves a planar carbocation intermediate

s n 1 energy diagram
SN1 Energy Diagram
  • Called an SN1 reaction since rate is dependant only on substrate
  • SN1 occurs in two distinct steps while SN2 occurs with both events in same step
  • The slowest step (Rate-determining step) is formation of the carbocation intermediate

Rate = k [RX]

stereochemistry of s n 1 reaction
Stereochemistry of SN1 Reaction
  • The planar intermediate leads to loss of chirality since a free carbocation is achiral

Nucleophile can attack either face of planar carbocation

Product is racemic or has some inversion

stereochemistry of s n 1 reaction26
Stereochemistry of SN1 Reaction
  • Carbocation is biased to react on side opposite leaving group
  • Suggests reaction occurs with carbocation loosely associated with leaving group (in an ion pair) during nucleophilic addition
stereochemistry of s n 1 reaction effects of ion pair formation
Stereochemistry of SN1 Reaction: Effects of Ion Pair Formation
  • If leaving group remains associated, then product has more inversion than retention
  • Product is only partially racemic with more inversion than retention
  • Associated carbocation and leaving group is an ion pair
learning check
Learning Check:
  • The optically pure tosylate shown was heated in acetic acid to yield a product mixture. If complete inversion had occurred the optically pure acetate product would have [a]D=+53.6o However the product mix has [a]D =+5.3o. What percentage racemization and what percentage inversion has occurred?
solution
Solution:
  • The optically pure tosylate shown was heated in acetic acid to yield a product mixture. If complete inversion had occurred the optically pure acetate product would have [a]D=+53.6o However the product mix has [a]D =+5.3o. What percentage racemization and what percentage inversion has occurred?

+5.3 x 100 = 9.9 % inverted

+53.6

So: 90.1 % racemic

11 5 characteristics of the s n 1 rxn
11.5 Characteristics of the SN1 Rxn

Substrate: in SN1

  • Ability to form stable carbocation intermediate best
  • Tertiary alkyl halide is most reactive by this mechanism
    • Remember Hammond postulate,”Any factor that stabilizes a high-energy intermediate stabilizes transition state leading to that intermediate”
substrate in s n 1 allylic and benzylic halides
Substrate: in SN1Allylic and Benzylic Halides

Allylic and benzylic intermediates stabilized by delocalization of charge

learning check33
Learning Check:
  • Rank the following substances in order of their expected SN1 reactivity:
solution34
Solution:
  • Rank the following substances in order of their expected SN1 reactivity:

1

4

2

3

leaving group in s n 1
Leaving Group: in SN1
  • Critically dependent on leaving group
    • the larger halides ions are better leaving groups
    • H2O, formed when OH of an alcohol is protonated in acid
    • p-Toluensulfonate (TosO-) is excellent leaving group

A leaving group won’t leave unless it’s stable on its own.

(Weak conjugate bases of strong acids make great leaving groups).

leaving group in s n 136
Leaving Group: in SN1

H2O leaving group formed when OH of alcohol is protonated in acid

rds

Nucleophile attacks after rds

nucleophiles in s n 1
Nucleophiles: in SN1
  • SN1 Reaction rate is not normally affected by nature or concentration of nucleophilesince nucleophilic addition occurs after formation of carbocation.

Once the carbocation is formed the rest is quick and easy regardless of nature of nucleophile.

the solvent in s n 1
The Solvent: in SN1
  • Solvents that stabilize the carbocation intermediate and also transition state and speeds rate
  • SN1 reactions go faster with polar protic solvents that cage the carbocation intermediate.
the solvent in s n 1 polar solvents promote ionization
The Solvent: in SN1Polar Solvents Promote Ionization
  • Polar, protic and unreactive Lewis base solvents facilitate R+ formation
learning check40
Learning Check:
  • Predict whether the following reactions is more likely to be SN1 or SN2.
solution41
Solution:
  • Predict whether the following reactions is more likely to be SN1 or SN2.

2o benzylic; forms stable carbocations or can be attacked from behind

Good leaving group

SN1

Good nucleophile

Polar protic solvent

1o; easily attacked from behind; wouldn’t give stable carbocation

Good leaving group

SN2

Good nucleophile

Polar aprotic solvent

learning check42
Learning Check:
  • Predict whether the following reactions is more likely to be SN1 or SN2.
solution43
Solution:
  • Predict whether the following reactions is more likely to be SN1 or SN2.

SN2

SN1

1o allylic; forms stable carbocations or can be attacked from behind

2o allylic; forms stable carbocations or can be attacked from behind

Good leaving group

Good leaving group after protonation with H+

Strong nucleophile

Weak nucleophile

Polar aprotic solvent

Would not stabilize a carbocation intermediate

Polar protic solvent

Stabilizes carbocation intermediate

11 6 biological substitution rxns
11.6 Biological Substitution Rxns
  • SN1 and SN2 reactions are common in biochemistry
  • Unlike in the laboratory, substrate in biological substitutions is often organodiphosphate rather than an alkyl halide
biological substitution examples
Biological Substitution: Examples

Biosynthesis of Geraniol in Roses

SN1

SN1

E1

biological substitution examples46
Biological Substitution: Examples

Biosynthesis of Adrenaline

SN2

11 7 elimination reactions of alkyl halides
11.7 Elimination Reactions: of Alkyl Halides
  • Opposite of addition
  • Generates an alkene
  • Can compete with substitution and decrease yield, especially for SN1 processes
elimination reactions e 1
Elimination Reactions: E1
  • Competes with SN1
  • Favored over SN1 when have poor Nu- that can still be a base
elimination reactions e 1 example
Elimination Reactions: E1 Example
  • Competes with SN1
  • Favored over SN1 when have poor Nu- that can still be a base

SN1

E1

elimination rxns zaitsev s rule
Elimination Rxns: Zaitsev’s Rule
  • In the elimination of HX from an alkyl halide, the more highly substituted alkene product predominates
elimination reactions e 2
Elimination Reactions: E2
  • Competes with SN2
  • Favored over SN2 when have strong base and sterichinderance
11 8 the e2 reaction mechanism
11.8 The E2 Reaction: Mechanism

Base grabs H that is anti-periplanar to leaving group

Transition state combines

leaving of X and transfer of H

Product alkene forms stereospecifically

the e2 rxn deuterium isotope effect
The E2 Rxn: Deuterium Isotope Effect
  • In RDS Breaking of C-H bond is slower than breaking of C-D bond.
elimination rxns e2 stereochemistry
Elimination Rxns: E2 Stereochemistry

Overlap of the developing  orbital in the transition state requires anti-periplanar geometry so electrons can give back-side SN2 type attack.

elimination rxns predicting e 2 products
Elimination Rxns: Predicting E2 Products
  • E2 is stereospecific
  • Meso-1,2-dibromo-1,2-diphenylethane with base gives cis 1,2-diphenyl

cis 1,2-diphenyl product

elimination rxns predicting e 2 products56
Elimination Rxns: Predicting E2 Products
  • E2 is stereospecific
  • RR or SS 1,2-dibromo-1,2-diphenylethane gives trans 1,2-diphenyl

S

S

Trans 1,2 diphenyl product

11 9 the e2 rxn cyclohexene formation
11.9 The E2 Rxn: Cyclohexene Formation
  • Abstracted proton and leaving group should align trans-diaxial to be anti periplanarin approaching transition state
  • Equatorial groups are not in proper alignment
the e2 rxn cyclohexene formation example
The E2 Rxn: Cyclohexene Formation Example

Fast

200 x’s slower since

Ring must flip to less stable ring conformation in order to get anti-periplanar E2

comparing e1 and e2
Comparing E1 and E2

Strong base is needed for E2 but not for E1

E2 is stereospecifc, E1 is not

E1 gives Zaitsev orientation

e1cb reaction
E1cB Reaction
  • Takes place through a carbanion intermediate
11 11 biological elimination rxns
11.11 Biological Elimination Rxns
  • All three elimination reactions occur in biological pathways
  • E1cB very common
  • Typical example occurs during biosynthesis of fats when 3-hydroxybutyryl thioester is dehydrated to corresponding thioester

E1cB

summary of reactivity s n 1 s n 1 e1 e1cb e2
Summary of Reactivity: SN1, SN1, E1,E1cB, E2

Strongly basic nucleophiles give more elimination as steric bulk increases.

Primary halides with strongly basic nucleophiles give mostly SN2 products.

Branched halides with strongly basic nucleophiles give about 50/50 SN2 and E2 products.

Tertiary halides with strongly basic nucleophiles give exclusive E2 products. With neutral or weakly basic nucleophiles SN1 and E1 pathways compete.

important concepts
Important Concepts
  • Unimolecular Substitution in Polar Media -
    • Secondary haloalkanes: slow
    • Tertiary haloalkanes: fast
    • When the solvent is the nucleophile, the process is called solvolysis.
  • Rate Determining Step in Unimolecular Substitution -
    • Dissociation of the C-X bond to form a carbocation intermediate.
    • Added strong nucleophile changes the product but not the reaction rate.
  • Carbocation Stabilization by Hyperconjugation: Tertiary > Secondary. Primary and methyl unstable.
important concepts67
Important Concepts
  • Racemization - Often occurs upon unimolecular substitution at a chiral carbon.
  • Unimolecular Elimination – Alkene formation accompanies substitution in secondary and tertiary system.
  • Bimolecular Elimination - May result from high concentrations of strong base. The elimination involves the anti conformational arrangement of the leaving group and the extracted hydrogen.
  • Substitution Favored - by unhinderedsubstrates and small, less basic nucleophiles
  • Elimination Favored -by hindered substrates and bulky, more basic nucleophiles.
slide68

2.

1.

3.

5.

4.

Which of the following is the product of the SN2 reaction between the hydroxide ion (HO–) and (R)-CH3–CHDI? D = 2H (deuterium)
  • 1
  • 2
  • 3
  • 4
  • 5
slide69

Consider the reaction of (S)-(–)-1-iodo-2-methylbutane to produce (+)-2-methyl-1-butanol. What is the absolute configuration of the product?

  • R
  • S
  • R and S (racemic mixture)
  • R and S (unequal amounts)
  • There is no chiral center in the product
which of the following s n 2 reactions is expected to have the highest rate
Which of the following SN2 reactions is expected to have the highest rate?
  • methyl bromide with water in DMSO
  • ethyl bromide with chloride in methanol
  • ethyl bromide with hydroxide in HMPA
  • ethyl chloride with ammonia in acetonitrile
  • methyl bromide with hydrosulfide (HS–) in HMPA
slide71
When A and B react in t-BuOH, the above rate expression is observed. What is the most likely mechanism of this reaction?
  • E2
  • SN2
  • E1
  • SN1
  • It cannot be determined.
slide72
What is the main reason that polar aprotic solvents are favored over polar protic solvents for SN2 reactions?
  • Polar aprotic solvents dissolve nucleophiles more readily than polar protic solvents.
  • Polar aprotic solvents destabilize anions.
  • Polar aprotic solvents stabilize cations, including carbocations.
  • Polar aprotic solvents form stabilizing hydrogen bonds.
  • Polar aprotic solvents prevent rearrangements from occurring.
what statement about the s n 2 reaction of methyl bromide with hydroxide is incorrect
What statement about the SN2 reaction of methyl bromide with hydroxide is incorrect?
  • The reaction kinetics is first-order in hydroxide.
  • In the transition state the carbon is sp2 hybridized.
  • Absolute configuration is inverted from R to S.
  • The reaction is faster in HMPA than in water.
  • The reaction can be catalyzed by I–.
select the substrate which would react fastest in the substitution reaction

1.

2.

3.

5.

4.

Select the substrate which would react fastest in the substitution reaction.
  • 1
  • 2
  • 3
  • 4
  • 5
which electrophile will react the fastest by the s n 2 mechanism with cyanide nc in dmf
Which electrophile will react the fastest by the SN2 mechanism with cyanide (NC–) in DMF?
  • phenyl iodide (Ph–I)
  • vinyl tosylate (H2C=CH–OTos)
  • ethyl bromide
  • cyclohexyl bromide
  • benzyl tosylate (Ph-CH2-OTos)
which of the following reagents is the best nucleophile for an s n 2 reaction
Which of the following reagents is the best nucleophile for an SN2 reaction?
  • methanol
  • methoxide
  • acetate
  • hydroxide
  • water
which set of reaction conditions represents the best way to carry out the following transformation
Which set of reaction conditions represents the best way to carry out the following transformation?
  • AcOH
  • NaOAc in AcOH
  • NaOAc in H2O
  • NaOAc in DMSO
  • AcOH in HMPA
slide79

In a reaction between an alkyl halide and methoxide, doubling the alkyl halide concentration doubles the rate of the reaction. Which of the following is a reasonable conclusion?

  • The reaction is an SN2 process.
  • The reaction proceeds by the SN1 mechanism.
  • The observation indicates an E2 process.
  • This is a reaction with E1 mechanism.
  • None of these
which compound would serve as the best starting material for the transformation shown below

1.

2.

3.

4.

5.

Which compound would serve as the best starting material for the transformation shown below?
  • 1
  • 2
  • 3
  • 4
  • 5
slide83
Select the reagent and solvent combination which would result in the fastest rate of substitution (R = CH3 in all cases).
  • ROH, HMPA
  • RS–, H2O
  • RO–, H2O
  • RS–, DMSO
  • RSH, H2O
which of the following is the best nucleophile
Which of the following is the best nucleophile?
  • H2O
  • (CH3)3N
  • (CH3)2P–
  • (CH3)2O
  • CH3O–
slide87
Which of the two stereoisomers of 4-t-butylcyclohexyl iodide (127I) will undergo SN2 substitution with 128I– faster, and why?
  • A will react faster because it is more stable
  • B will react faster because it gives a more stable product.
  • A will react faster because nucleophile’s approach from the bottom face of the molecule is easier for steric reasons.
  • A and B will react with the same rate because the transition states for both reactions are the same.
  • B will react faster because it is less stable than A, and the transition state for both reactions is of the same energy.