Presentation on theme: "11. Reactions of Alkyl Halides: Nucleophilic Substitutions and Eliminations."— Presentation transcript:
11. Reactions of Alkyl Halides: Nucleophilic Substitutions and Eliminations
2 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 ++++ ++++ Acts as Nucleophile Acts as Base
3 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
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 PCl 5 gives (+)-chlorosuccinic acid Further reaction with wet silver oxide gives (+)-malic acid The reaction series starting with (+) malic acid gives (-) acid
5 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: 6 (+)-1-phenyl-2-propanol (-)-1-phenyl-2-propanol Used Tosylate as an excellent leaving group
7 Kinetics of Nucleophilic Substitution Rate (V) = change in concentration with time Depends on concentration(s), temperature, inherent nature of reaction (barrier on energy surface) A rate law describes 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 order of a reaction is sum of the exponents of the concentrations in the rate law
The S N 2 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 = k [CH 3 -Br] [HO - ] Rate is dependant on both Nucleophile & Substrate
9 S N 2 Process The reaction involves a transition state in which both reactants are together
10 S N 2 Transition State The transition state of an S N 2 reaction has a planar arrangement of the carbon atom and the remaining three groups
11 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 ‡ ) Characteristics of the S N 2 Rxn
12 The Substrate: Steric Effects on S N 2 Reactions The carbon atom in (a) bromomethane is readily accessible resulting in a fast S N 2 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 S N 2 reactions. S N 2 Sensitive to steric effects
13 The Substrate: Steric Effects : Order of Reactivity in S N 2 S N 2 Sensitive to steric effects No reaction at C=C (vinyl or Aryl halides)
14 The Nucleophile: in S N 2 Neutral or negatively charged Lewis base Reaction increases coordination at nucleophile Neutral nucleophile acquires positive charge Anionic nucleophile becomes neutral ++++ ++++
15 The Nucleophile: Relative Reactivity in S N 2 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
16 The Nucleophile: Examples
17 The Leaving Group: in S N 2 Stable anions that are weak bases are usually excellent leaving groups and can delocalize charge Alkyl fluorides, alcohols, ethers, and amines do not typically undergo S N 2 reactions. very basic or very small grps are poor leaving groups.
18 The Leaving Group: in S N 2 -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)
19 The Leaving Group: in S N 2 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 1 o or 2 o then major product is from attack at less hindered 1 o If choice of epoxide C’s is 1 o vs 3 o then major product is from attack at more + 3 o ) ++++ ++++
20 The Solvent: in S N 2 Solvents that can donate hydrogen bonds (protic) (-OH or –NH) slow S N 2 reactions by associating with reactants Energy is required to break interactions between reactant and solvent Caged nucleophiles can’t attack so well
21 The Solvent: in S N 2 Polar aprotic solvents (no NH, OH, SH) form weaker interactions with substrate and permit fast S N 2 reactions Protic solvents (with -OH or –NH) slow S N 2 reactions by complexing with reactants Poor for S N 2Good for S N 2
The S N 1 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
23 S N 1 Reaction The reaction involves a planar carbocation intermediate
24 S N 1 Energy Diagram The slowest step (Rate-determining step) is formation of the carbocation intermediate Rate = k [RX] Called an S N 1 reaction since rate is dependant only on substrate S N 1 occurs in two distinct steps while S N 2 occurs with both events in same step
25 Stereochemistry of S N 1 Reaction The planar intermediate leads to loss of chirality since a free carbocation is achiral Product is racemic or has some inversion Nucleophile can attack either face of planar carbocation
26 Carbocation is biased to react on side opposite leaving group Stereochemistry of S N 1 Reaction Suggests reaction occurs with carbocation loosely associated with leaving group (in an ion pair) during nucleophilic addition
27 Stereochemistry of S N 1 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
28 Stereochemistry of S N 1 Reaction: Example:
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 [ ] D =+53.6 o However the product mix has [ ] D =+5.3 o. What percentage racemization and what percentage inversion has occurred? 29
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 [ ] D =+53.6 o However the product mix has [ ] D =+5.3 o. What percentage racemization and what percentage inversion has occurred? x 100 = 9.9 % inverted So: 90.1 % racemic
Characteristics of the S N 1 Rxn Substrate: in S N 1 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”
32 Substrate: in S N 1 Allylic and Benzylic Halides Allylic and benzylic intermediates stabilized by delocalization of charge
Learning Check: Rank the following substances in order of their expected S N 1 reactivity: 33
Solution: Rank the following substances in order of their expected S N 1 reactivity:
35 Leaving Group: in S N 1 Critically dependent on leaving group the larger halides ions are better leaving groups H 2 O, 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).
36 Leaving Group: in S N 1 H 2 O leaving group formed when OH of alcohol is protonated in acid Nucleophile attacks after rds rds
37 Nucleophiles: in S N 1 S N 1 Reaction rate is not normally affected by nature or concentration of nucleophile since nucleophilic addition occurs after formation of carbocation. Once the carbocation is formed the rest is quick and easy regardless of nature of nucleophile.
38 The Solvent: in S N 1 Solvents that stabilize the carbocation intermediate and also transition state and speeds rate S N 1 reactions go faster with polar protic solvents that cage the carbocation intermediate.
39 The Solvent: in S N 1 Polar Solvents Promote Ionization Polar, protic and unreactive Lewis base solvents facilitate R + formation
Learning Check: 40 Predict whether the following reactions is more likely to be S N 1 or S N 2.
Solution: 41 Predict whether the following reactions is more likely to be S N 1 or S N 2. SN1SN1 SN2SN2 2 o benzylic; forms stable carbocations or can be attacked from behind Polar protic solvent Good leaving group 1 o ; easily attacked from behind; wouldn’t give stable carbocation Polar aprotic solvent Good leaving group Good nucleophile
Learning Check: Predict whether the following reactions is more likely to be S N 1 or S N 2. 42
Solution: Predict whether the following reactions is more likely to be S N 1 or S N SN1SN1 2 o allylic; forms stable carbocations or can be attacked from behind Polar protic solvent Stabilizes carbocation intermediate Good leaving group after protonation with H + SN2SN2 1 o allylic; forms stable carbocations or can be attacked from behind Polar aprotic solvent Would not stabilize a carbocation intermediate Good leaving group Weak nucleophile Strong nucleophile
Biological Substitution Rxns S N 1 and S N 2 reactions are common in biochemistry Unlike in the laboratory, substrate in biological substitutions is often organodiphosphate rather than an alkyl halide
Biological Substitution: Examples 45 Biosynthesis of Geraniol in Roses SN1SN1 SN1SN1 E1
Biological Substitution: Examples 46 Biosynthesis of Adrenaline SN2SN2
Elimination Reactions: of Alkyl Halides Opposite of addition Generates an alkene Can compete with substitution and decrease yield, especially for S N 1 processes
Elimination Reactions: E 1 48 Competes with S N 1 Favored over SN1 when have poor Nu - that can still be a base
Elimination Reactions: E 1 Example 49 Competes with S N 1 Favored over S N 1 when have poor Nu - that can still be a base SN1SN1E1
50 Elimination Rxns: Zaitsev’s Rule In the elimination of HX from an alkyl halide, the more highly substituted alkene product predominates
51 Elimination Reactions: E 2 Competes with S N 2 Favored over S N 2 when have strong base and steric hinderance
The E2 Reaction: Mechanism Product alkene forms stereospecifically Transition state combines leaving of X and transfer of H Base grabs H that is anti-periplanar to leaving group
The E2 Rxn: Deuterium Isotope Effect In RDS Breaking of C-H bond is slower than breaking of C-D bond. 53
54 Elimination Rxns: E2 Stereochemistry Overlap of the developing orbital in the transition state requires anti- periplanar geometry so electrons can give back-side S N 2 type attack.
55 Elimination Rxns: Predicting E 2 Products E2 is stereospecific Meso-1,2-dibromo-1,2-diphenylethane with base gives cis 1,2- diphenyl cis 1,2-diphenyl product
56 Elimination Rxns: Predicting E 2 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
The E2 Rxn: Cyclohexene Formation Abstracted proton and leaving group should align trans-diaxial to be anti periplanar in approaching transition state Equatorial groups are not in proper alignment
58 The E2 Rxn: Cyclohexene Formation Example 200 x’s slower since Ring must flip to less stable ring conformation in order to get anti-periplanar E2 Fast
59 Comparing E1 and E2 Strong base is needed for E2 but not for E1 E1 gives Zaitsev orientation E2 is stereospecifc, E1 is not
The E1cB Reactions
61 E1cB Reaction Takes place through a carbanion intermediate
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: S N 1, S N 1, E1,E1cB, E2 Strongly basic nucleophiles give more elimination as steric bulk increases. 64 Primary halides with strongly basic nucleophiles give mostly S N 2 products. Branched halides with strongly basic nucleophiles give about 50/50 S N 2 and E2 products. Tertiary halides with strongly basic nucleophiles give exclusive E2 products. With neutral or weakly basic nucleophiles S N 1 and E1 pathways compete.
65 Summary of Reactivity: S N 1, S N 1, E1,E1cB, E2
Important Concepts 1.Unimolecular Substitution in Polar Media - Secondary haloalkanes: slow Tertiary haloalkanes: fast When the solvent is the nucleophile, the process is called solvolysis. 2.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. 3.Carbocation Stabilization by Hyperconjugation: Tertiary > Secondary. Primary and methyl unstable.
Important Concepts 4.Racemization - Often occurs upon unimolecular substitution at a chiral carbon. 5.Unimolecular Elimination – Alkene formation accompanies substitution in secondary and tertiary system. 6.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. 7.Substitution Favored - by unhindered substrates and small, less basic nucleophiles 8.Elimination Favored - by hindered substrates and bulky, more basic nucleophiles.
Which of the following is the product of the S N 2 reaction between the hydroxide ion (HO–) and (R)-CH 3 –CHDI? D = 2 H (deuterium)
Consider the reaction of (S)-(–)-1- iodo-2-methylbutane to produce (+)-2-methyl-1-butanol. What is the absolute configuration of the product? 1. R 2. S 3. R and S (racemic mixture) 4. R and S (unequal amounts) 5. There is no chiral center in the product
Which of the following S N 2 reactions is expected to have the highest rate? 1. methyl bromide with water in DMSO 2. ethyl bromide with chloride in methanol 3. ethyl bromide with hydroxide in HMPA 4. ethyl chloride with ammonia in acetonitrile 5. methyl bromide with hydrosulfide (HS – ) in HMPA
When A and B react in t-BuOH, the above rate expression is observed. What is the most likely mechanism of this reaction? 1. E2 2. S N 2 3. E1 4. S N 1 5. It cannot be determined.
What is the main reason that polar aprotic solvents are favored over polar protic solvents for S N 2 reactions? 1. Polar aprotic solvents dissolve nucleophiles more readily than polar protic solvents. 2. Polar aprotic solvents destabilize anions. 3. Polar aprotic solvents stabilize cations, including carbocations. 4. Polar aprotic solvents form stabilizing hydrogen bonds. 5. Polar aprotic solvents prevent rearrangements from occurring.
What statement about the S N 2 reaction of methyl bromide with hydroxide is incorrect? 1. The reaction kinetics is first-order in hydroxide. 2. In the transition state the carbon is sp 2 hybridized. 3. Absolute configuration is inverted from R to S. 4. The reaction is faster in HMPA than in water. 5. The reaction can be catalyzed by I –.
Select the substrate which would react fastest in the substitution reaction
Which electrophile will react the fastest by the S N 2 mechanism with cyanide (NC – ) in DMF? 1. phenyl iodide (Ph–I) 2. vinyl tosylate (H 2 C=CH– OTos) 3. ethyl bromide 4. cyclohexyl bromide 5. benzyl tosylate (Ph- CH 2 -OTos)
Which of the following reagents is the best nucleophile for an S N 2 reaction? 1. methanol 2. methoxide 3. acetate 4. hydroxide 5. water
Which set of reaction conditions represents the best way to carry out the following transformation? 1. AcOH 2. NaOAc in AcOH 3. NaOAc in H 2 O 4. NaOAc in DMSO 5. AcOH in HMPA
Which of the following will give the fastest S N 1 reaction?
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? 1. The reaction is an S N 2 process. 2. The reaction proceeds by the S N 1 mechanism. 3. The observation indicates an E2 process. 4. This is a reaction with E1 mechanism. 5. None of these
Which compound would serve as the best starting material for the transformation shown below?
Including stereoisomers, how many products are possible from the following reaction?
What mechanism is most likely to operate for the following reaction? 1. S N 1 2. S N 2 3. E1 4. E2 5. E1cB
Select the reagent and solvent combination which would result in the fastest rate of substitution (R = CH 3 in all cases). 1. ROH, HMPA 2. RS –, H 2 O 3. RO –, H 2 O 4. RS –, DMSO 5. RSH, H 2 O
Which of the following is the best nucleophile? 1. H 2 O 2. (CH 3 ) 3 N 3. (CH 3 ) 2 P – 4. (CH 3 ) 2 O 5. CH 3 O –
What is the major product of the following reaction?
What is the most likely product of the following reaction?
Which of the two stereoisomers of 4-t-butylcyclohexyl iodide ( 127 I) will undergo S N 2 substitution with 128 I – faster, and why? 1. A will react faster because it is more stable 2. B will react faster because it gives a more stable product. 3. A will react faster because nucleophile’s approach from the bottom face of the molecule is easier for steric reasons. 4. A and B will react with the same rate because the transition states for both reactions are the same. 5. B will react faster because it is less stable than A, and the transition state for both reactions is of the same energy.