Nucleophilic Substitution & Elimination Chemistry Beauchamp 1 This current chapter continues the theme of the previous acid/base chapter. As much as possible, try to focus on the similarities: electron pair donation and acceptance. The approach taken in this chapter is somewhat simplistic. There is a slight sacrifice in accuracy, but it reduces much of the complicated subtlety and ambiguity that makes this topic so difficult for students. Our approach still leads to mostly correct predictions for the outcome of the reactions under study. Even in trying to keep it simple, there are a lot of details to learn and keep track of.

All of the reactions to be presented in this chapter involve two electron transfers from an electron pair donor to an electron pair acceptor. The electron pair donors look very similar to the bases presented previously. In fact, electron pair donors will still attack protons (where we still call them bases), but there will be a competitor site to attack: carbon atoms. When attack occurs at carbon (or any nonhydrogen atom, in general), the electron pair donors are called nucleophiles.

There are four reactions to be studied in this chapter: two types of substitution reactions (SN2 and SN1) and two types of elimination reactions (E2 and E1). The main competition in these four reactions, as we view them, is “SN2 versus E2” and “SN1 versus E1”. “S” stands for substitution, “E” stands for elimination and “N” stands for nucleophilic. The “1” and “2” are from the kinetics of the reactions, and represent the order of each reaction. A “2” means that there are two participants in the slow step of a bimolecular reaction (an RX compound and an electron pair donor). A “1” means that the slow step of a reaction depends on only one reactant (just the RX compound). The reaction conditions for the “2” reactions will be presented as sharply different from the reaction conditions for the “1” reactions, even though this is not completely accurate. Your choices will be more straight forward when presented this way, and you can fine tune the details if you continue your studies beyond this first year course.

The “2” reaction conditions will have strong electron pair donors present. We will view all anions as strong electron pair donors. This allows a clear demarcation from weak electron pair donors, which we will view as neutral molecules. There are two neutral exceptions that we will have to learn. Neutral nitrogen compounds (ammonia, primary, secondary and tertiary amines) and neutral sulfur compounds (thiols and sulfides) are considered strong electron pair donors in this chapter. Polar aprotic solvents will enhance the reactivity of strong nucleophiles in SN2 reactions. Typical “1” conditions generally involve a neutral polar protic solvent, such as H2O, ROH and RCO2H (or mixtures of these).

Common to all four reactions (SN2/E2 and SN1/E1) is a carbon atom with a leaving group, which we will identify as the “R-X” compound. We will present just five types of leaving groups, “X”, in this chapter. They are chloride, bromide, iodide, tosylate and water. There are other possible leaving groups that will appear during the rest of the book, but by the time you see them, they will just be a part of your general approach to mechanisms. Our reason for calling the leaving group “X” is to think of them all just alike, and only learn one strategy for the entire group. There are actually small differences among the leaving groups, but at this point of your organic career it is more important for you to see the similarities. The feature that makes something a good leaving group, is that it is stable on its own. We know that chloride, bromide, iodide, sulfonates and water are stable conjugate bases from discussions in the previous acid/base chapter. All of them have very strong conjugate acids (pKa’s < 0). The carbon portion, “R”, will favor or prevent certain of the reaction choices. The carbon atom attached to the leaving group (called the C-alpha carbon, Cα) will be classified by its substitution pattern as methyl, primary, secondary, tertiary, allylic or benzylic. It will also be necessary to examine the features of any carbon atoms attached to Cα. These carbon atoms are called C-beta carbons (Cβ,) and there can be anywhere from zero to three of them.

These are most of the details you will need to sort through in order to make reasonable predictions about how these reactions work. Many of the reactions you will study later in this book are variations of the mechanisms you learn hear. Spend the time to learn this chapter well and it will help make subsequent chapters easier to learn.

Nucleophilic Substitution & Elimination Chemistry Beauchamp 2 Four new mechanisms to learn: SN2 vs E2 and SN1 vs E1 S = substitution = a leaving group (X) is lost from a carbon atom (R) and replaced by nucleophile (Nu:)

N = nucleophilic = nucleophiles {Nu:) donate two electrons in a manner similar to bases (B:)

E = elimination = two vicinal groups (adjacent) disappear from the skeleton and are replaced by a pi bond

1 = unimolecular kinetics = only one concentration term appears in the rate law expression, Rate = k[RX]

2 = bimolecular kinetics = two concentration terms appear in the rate law expression, Rate = k[RX] [Nu: or B:]

SN2 competes with E2

SN1 competes with E1

These electrons always leave with X.SN2

E2

SN1

E1

XRBNu BHNuHCompeting Reactions

Competing Reactions

Carbon Group

LeavingGroup

Nu: / B: = is an electron pair donor to carbon (= nucleophile) or to hydrogen (= base). It can be strong (SN2/E2) or weak (SN1/E1).

(strong) (weak)

R = methyl, primary, secondary, tertiary, allylic, benzylic

X = -Cl, -Br, -I, -OSO2R (possible leaving groups in neutral, basic or acidic solutions)

X = -OH2 (only possible in acidic solutions) Problem 1 - Draw all of the "R" patterns listed above connected to “X”. SN2 and E2 reactions are one step reactions. The key bonds are broken and formed simultaneously, without any intermediate structures. These are referred to as concerted reactions. The SN2 and E2 mechanisms compete with one another in consuming the R-X compound. Approach of the nucleophile/base is always from the backside in SN2 reactions and mainly from the backside in E2 reactions (always from the backside in this chapter).

Nucleophilic Substitution & Elimination Chemistry Beauchamp 3 SN2 and E2 Competition – One Step Reactions

CβH

Cα X

B

Nu

E2 approach (usually from the backside in an "anti" conformation,"syn" is possible, but not common)

SN2 approach (alwaysfrom the backside, resulting in inversion of configuration) B Nu≈

strong base ≈ strong nucleophile

One step, concerted reactions, from the backside.

CαNu

Cβ

H

SN2 product (a nucleophile substitutes for a leaving group)

E2 product (a pi bond forms)

Cβ

Cα

X

Terms

S = substitution E = elimination Nu: = strong nucleophile B: = strong base X = leaving group

2 = bimolecular kinetics (second order, rate in slow step depends on RX and Nu: /B: )

R-X = R-Cl, R-Br, R-I, R-OTs, ROH2+

stableleaving group

SN1 and E1 reactions are multistep reactions and also compete with one another. Both of these reactions begin with the same rate-limiting step of carbocation formation from an R-X compound. Carbocations (R+) are very reactive electron deficient carbon intermediates that typically follow one of three possible pathways leading to two ultimate outcomes: 1. add a nucleophile or 2. lose a beta hydrogen atom. The additional competing pathway for carbocation intermediates is rearrangement, in which atoms in a carbocation change positions to form a similar or more stable carbocation. Once formed, a new carbocation is analyzed in a similar manner to the previous one it came from. It may possibly rearrange again, but ultimately, the final step will be to either add a nucleophile at the carbocation carbon (SN1) or to react with a base at a beta hydrogen atom (E1).

Nucleophilic Substitution & Elimination Chemistry Beauchamp 4 SN1 and E1 Competition – Multistep Reactions

E1 approach comes from parallel Cβ-H with either lobe of 2p orbital.

SN1 approach is from either the top face or the bottom face.

Bweak base

SN1 and E1 reactions are multistep reactions.

E1 products(pi bond forms)

X stableleaving group

Terms

S = substitution E = elimination H-Nu: = weak nucleophile H-B: = weak base X = leaving group

1 = unimolecular kinetics (first order reaction, the rate in the slow step depends only on RX)

R-X = R-Cl, R-Br, R-I, R-OTs, ROH2+

SN1 and E1 reactions begin exactly the same way. The leaving group, X, leaves on its own, forming a carbocation.

CαCβ

X

H

The R-X bond ionizes with help from the polar, protic solvent, which is alsousually the weak nucleophile/base.

CαCβ

H

X

solvated carbocation

addition of a weak nucleophile (SN1 type reaction)

Loss of a beta hydrogen atom, (Cβ-H in most E1 reactions that we study).

rearrangement to a new carbocation of similar or greater stability

add Nu: (SN1)

lose beta H (E1)

rearrange

H

Nu H

weak nucleophile

CαCβ

H

Cα

Nu

Cβ

H

CαCβ

H

Nu H

H

Nu H

Nu H

Cβ Cα

CαCβ

H

Nu

Cα

Nu

Cβ

H

solvated carbocation

Bweak base

H Nu H weak nucleophile

=

SN1

E1

Often a final proton transfer is necessary.

SN1 product (a nucleophile substitutes for a leaving group)

BHNu H = usually a polar, protic solvent (or mixture) of H2O, ROH or RCO2H=

NuH

BH

R

The above pairs of reactions (SN2/E2 and SN1/E1) look very similar overall, but there are some key differences. The nucleophile/base is a strong electron pair donor in SN2/E2 reactions (that’s why they participate in the slow step of the reaction) and a weak electron pair donor in SN1/E1 reactions (that’s why they don’t participate in the slow step of the reaction). This leads to differences in reaction mechanisms, which show up in the kinetics of the rate law expression (bimolecular = 2 and unimolecular = 1). It is recommended that you look at the reaction conditions first to decide what mechanisms are possible. You cut you choices in half when you decide that the electron pair donor is strong (SN2/E2) or weak (SN1/E1). Problem 2 - Draw each of the “X” groups, as they would appear as products. Indicate why they are good leaving groups.

Nucleophilic Substitution & Elimination Chemistry Beauchamp 5 Problem 3 - Fill in an example of each of the following patterns below.

-Cl -Br -I -OTs -OH2

Me = methyl = CH3-

1o = primary = RCH2-

2o = secondary = R2CH-

3o = tertiary = R3C-

allyl = CH2=CHCH2-

benzyl = C6H5CH2-

(tosylate) (protonated alcohol)

Important details to be determined in deciding the correct mechanisms of a reaction. 1. Is the nucleophile/base considered to be strong or weak? We simplistically view strong electron pair

donation as coming from anions of all types and neutral nitrogen and sulfur atoms. Weak electron pair donors will typically be neutral solvent molecules, usually water (H2O), alcohols (ROH), mixtures of the two, or simple, liquid carboxylic acids (RCO2H).

2. What is the substitution pattern of the R-X substrate at the Cα carbon attached to the leaving group, X? Is it a methyl, primary, secondary, tertiary, allylic, or benzylic carbon? What about any Cβ carbon atoms? How many additional carbon atoms are attached at a Cβ position (none, one, two or three)? Answers to these questions will determine SN2, E2, SN1 and E1 reactivities and alkene substitution patterns and relative stabilities in E2 and E1 reactions.

XH3CH2CR X

HCR X

R

CR X

R

R

CH

H2CH2C X

H2C X

methyl (Me) primary (1o) secondary (2o) tertiary (3o) allylic benzylic

Cβ Cα X

C-beta carbon (0-3 of these)

C-alpha carbon (only 1 of these)

Nucleophilic Substitution & Elimination Chemistry Beauchamp 6 3. Are the necessary “ anti ” Cβ-H/Cα-X bond orientations possible to allow E2 reactions to occur? Also

possible, but much less common, are “ syn ” E2 reactions, which are not emphasized in this book.

In E2 reactions the base approach occurs most commonly from an anti direction at Cβ-H because the conformation is staggered. The electrons from the Cβ-H bond force off the leaving group, X, as the base, B:, forces them away from the Cβ hydrogen atom.

Bstrong base

Different E2 products are possible.

CαCβ

X

H

In E2 reactions the base approach occurs rarely from a syn direction at Cβ-H because the conformation is eclipsed.

Two possible conformations for an E2 reaction to occur.

The Cβ-H bond and the Cα-X bonds need to be parallel so the pi bond of the product can form as the reaction proceeds. Only two conformations allow this, anti and syn. Anti is highly preferred (> 99%) due to its staggered conformation. We will emphasize anti eliminations in E2 reactions in this book.

anti Cβ-H/Cα-X bonds

concerted reaction (anti elimination is usual and common)

syn Cβ-H/Cα-X bonds

syn elimination is rare

strong base

B

Cβ Cα

CαCβ

XH

SN2 / E2 mechanisms The rate laws for both SN2 and E2 reactions are bimolecular (that’s what the “2” means). In fact they look very similar, except that the values of the rate constants; kSN2 and kE2, are different. That means the activation energies (Ea) are different [k = exp –(Ea/RT)]. The reaction with the larger rate constant (lower Ea) will occur faster and will form more product. That’s because the reactions are usually occurring under kinetic control (what forms fastest in a single step), instead of thermodynamic control (what is most stable in repeated steps).

kSN2 ≠ kE2 and (Ea)SN2 ≠ (Ea)E2

SN2 rate law Rate = kSN2 [RX][Nu: ]

E2 rate law Rate = kE2 [RX][B: ]

B: is the same as Nu: , only the position of attack is different.

...except by coincidence Simplistic pictures of an SN2 mechanism and an E2 mechanism are shown below. Initially, both the strong nucleophile/base and the R-X electrophile must shed portions of their solvation shells to encounter one another. In our approach, solvation factors are usually ignored (though we will briefly discuss them later in this chapter). The effects on the reactions from many solvent molecules are very complicated and since the solvent does not itself undergo change in the overall reaction, it is often easier to leave it out at first. However, it is often one of the most important details of a reaction mechanism.

Nucleophilic Substitution & Elimination Chemistry Beauchamp 7 Simplistic Picture of a General SN2 mechanism (with the solvent)

The reactants must encounter one another with the proper orientation (backside attack by nucleophile in SN2 reactions).

The SN2 transition state is highly conjested as the sp2 carbon inverts in configuration. The carbon atom has high PE because it has 10 electrons around the carbon, Cα. It is very sensitve to steric effects.

The products diffuse away from one another after the nucleophile is added and the leaving group is gone. The Cα carbon has inverted its configuration at the reaction site. The reaction succeeds because the products are more stable than the reactants and there is sufficient energyto crossthe transition state energy barrier (Ea)

"Transition State"

BrC

H

H H

HOS

SS

S

S

S

SS

S SS

S

S

S

S

S

SS

S

S

S

S

C

H

HH

BrOH

SS

S

S

S S

S S

SSS

SS

S

S

S

SS

SS

S

C

H

HH

HO

SS

SS S

S S

S

S

SS

S

SS

Br

S

S

S

S

S

S = solvent molecule

Same picture without the solvent

"Transition State"

C

H

HH

BrOH BrC

H

H H

HO C

H

HH

HO Br

Potential Energy vs. Progress of Reaction Diagram

sp2 carbon transition state as carboninverts configuration forms a high PE carbon with 10 electrons at carbon. This is a concerted, one-step reaction.

"Transition State"

C

H

HH

BrOH

BrC

H

H H

HO

C

H

HH

HO Br

reactants

products

Ea - this energy difference determines how fast the reaction occurs: "kinetics".

Ea

∆G∆G - this energy difference determines the extent of the reaction; the ratio of products versus reactants at equilibrium (when kinetics allows the reaction to proceed. Thermodynamics is determined mostly by the strengths of the bonds and solvation energies of the reactant and product speces: "thermodynamics".

POR = progress of reaction

PE Rate = kSN2 [RX][Nu] = bimolecular reaction

kSN2 = 10

Ea = -2.3RT log(kSN2)

-Ea

2.3RT

∆Go = -2.3RT log(Keq)

Keq = 10 -∆Go

2.3RT

Nucleophilic Substitution & Elimination Chemistry Beauchamp 8 Important Details of SN2 Reactions 1. SN2 reactions occur in one step. They are concerted reactions that go directly from reactants to

products.

2. Both reactants (RX and Nu:-/B:-) are involved in the slow (…and only) step. The transition state, (T.S.) is the high potential energy point of the reaction and determines the activation energy (Ea). This is called the rate determining step (R.D.S.).

3. Bond formation and bond breakage are occurring at the same time, which keeps the Ea lower.

4. The high potential energy T.S. carbon is approximately sp2, trigonal planar, where carbon is surrounded by ten electrons in violation of the octet rule.

5. The reacting carbon (Cα) inverts its configuration. Generally, if this is a stereogenic center, R S or S R, or if in a disubstituted ring, cis trans or trans cis. Inversion is also assumed to occur at achiral RX centers, even if it is not observable (such as a methyl or primary RX center).

6. Generally, strong base/nucleophiles are used in SN2 reactions. We simplistically indicate this with negatively charged nucleophile to represent anions of all types, (and neutral nitrogen and sulfur atoms).

7. The HOMO/LUMO overlap is important in molecular orbital theory (but not discussed here). Relative rates of SN2 reactions with different substitution patterns at Cα of the R-X structure Substitution at Cα and Cβ (relative to the carbon with X) slows the rate of SN2 reactions because the substituent groups block the backside approach of the nucleophile to the Cα-X carbon (…and increase the activation energy by increasing the congestion in the pentavalent transition state). Both Cα and Cβ substitution patterns show their most dramatic decreases in rate of reaction when all possible carbon positions are switched from hydrogen to carbon (or other large group). As long as there is at least one hydrogen atom at the Cα and Cβ positions, there remains a possible path of approach by the nucleophile to the backside of the carbon.

Cα

H

H

XH Cα

H

H

XCH3 Cα

CH3

H

XCH3 Cα

CH3

CH3

XCH3

k 0t-butyl (tertiary)

k 0.025isopropyl (secondary)

k 30methyl (unique)

k 1 ethyl (primary)Reference compound

Relative Rates of SN2 Reactions - Steric hindrance at the Cα carbon slows down the rate of SN2 reaction.

140 (very low)

Nucleophilic Substitution & Elimination Chemistry Beauchamp 9

Primary substitution allows two easy pathsof approach by the nucleophile. It is the least sterically hindered "general" substitution pattern for SN2 reactions.

Cα XR

H

H

Nu

methyl RX primary RX secondary RX

Secondary substitution allows one easy path of approach by the nucleophile. It reacts slowest of the possible SN2 substitution patterns.

Cα X

R

R

H

Nu

Tertiary substitution has no easy path of approach by the nucleophile from the backside. We do not propose any SN2 reaction at tertiary RX centers.

Cα X

R

R

R

Nu =

When the Cα carbon is completely substituted the nucleophile cannot get close enough to make a bond with the Cα carbon. Even Cβ-H sigma bonds block the nucleophile's approach

Cα

Cβ

Cβ

X

H

H

H

H

H H

H

HH

Nu

tertiary RX

Methyl has three easy paths of approach by the nucleophile. It is the least sterically hindered carbon in SN2 reactions, but it is unique.

Cα X

H

H

H

Nu

Cβ

Relative rates of SN2 reactions with different substitution patterns at Cβ of the R-X structure

Steric hindrance at the Cβ carbon also slows down the rate in SN2 reactions.

CH2Cβ

H

H

H

X CH2Cβ

H

CH3

H

X CH2Cβ

CH3

CH3

H

X CH2Cβ

CH3

CH3

CH3

X

k 1 ethyl

k 0.4 propyl

k 0.03 2-methylpropyl

k 0.00001 2,2-dimethylpropyl (neopentyl)

All of these structures are primary R-X compounds at Cα, but substituted differently at Cβ.

Reference compound

Cα

Cβ

HH X

CH3CH3

H

Nu

A completely substituted Cβ carbon atom also blocks the Nu: approach to the backside of the C-X bond. A large group is always in the way at backside of the C-X bond.

If even one bond at Cβ has a hydrogen then approach by Nu: to backside of the C-X bond is possible and an SN2 reaction is possible.

Cα

Cβ

HH X

CCH3

CH3HH

HNu

Nucleophilic Substitution & Elimination Chemistry Beauchamp 10 Especially reactive SN2 patterns

The excess electron density in the sp2 transition state of an SN2 reaction (ten electrons at carbon) can be delocalized into an adjacent pi system, when present. This helps lower the activation energy, Ea. This is an especially powerful effect when an oxygen can take on some of the excess negative charge. Especially reactive SN2 patterns include allyl RX, benzyl RX and X at a Cα position of a carbonyl group.

CH2CH3 X CH2CΗ XH2C CH2 X CH2 XCR

O

All of these structures are primary R-X compounds, but the pi patterns are much more reactive than a simple ethyl RX.

k 40 allyl RX

k 1 ethyl k 100,000

α substituted carbonyl k 120 benzyl RXReference compound

The adjacent pi bond can take on some of the excess negative charge density in the transition state, because of its parallel orientation with the transient 2p orbital at Cα in the transition state. This will lower the activation energy, Ea, and increase the rate of the reaction.

C

CO

Nu Xtransition state

Delocalization of excess negative charge in an SN2 transition state,lowers Ea and makes the rate constant larger (faster rate).

See data above.

CR

O

CH2X

CH3 CH2X

Relative Rates = 100,000

1

Inversion of Configuration

Inversion of configuration at the Cα position (R S or S R) in SN2 reactions is so important, it gets its own name, “Walden Inversion”.

Nu Cα X

2

3

4

XCαNu

4

3

2

S R

If Cα is a chiral center and X is #1 in priority (the other groups are #2, #3 and #4 above), then S absolute configuration typically inverts to R (and vice versa). This observation assumes there is no relative change in the order of priorities of the groups at the chiral center (X = #1 priority and Nu = #1 priority).

Nucleophilic Substitution & Elimination Chemistry Beauchamp 11 SN2 Inversion of Configuration in Rings A cyclic RX compound with no other substituents does not allow us to detect inversion of configuration (versus retention of configuration or racemization of configuration).

We cannot tell the top from the bottom without some other marker in the ring.

same product

Nu

C

X

H

Nu

C

X

H

inversion of configuration

retention of configuration

??

However, if the RX ring were substituted with another group that did not interfere with the SN2 reaction, it would allow the observation of backside attack by the nucleophile. A very simple methyl substituent would allow us to see which side the nucleophilic attack occurs from. We can observe if the opposite side or the same side…or both sides are attacked by comparing the initial relative position of the leaving group to the final relative position of the nucleophile.

H3C

H

C

H

Nu

cis = initial stereochemistry (of methyl and X)

trans = final stereochemistry (of methyl and Nu) from backside attack by nucleophile

...leads to...

Nu

H3C

H

C

X

H X

Problem 4 - How can you tell whether the SN2 reaction occurs with front side attack, backside attack or front and backside attack? Use the two molecules to explain you answer. Follow the curved arrow formalism to show electron movement.

HH3C

H Br

C BrH

H3CCH3CH2

ICH3C

O

O

a b

Nucleophilic Substitution & Elimination Chemistry Beauchamp 12 Ring Size Effects on the Rate of SN2 Reactions Each type of ring (C3, C4, C5, C6,…and substituted variations) will have its own peculiar effect on how the SN2 reaction behaves. As previously shown, if there is some conformation that blocks approach of the nucleophile from the backside of the C-X bond, this will slow the SN2 reaction. If the sp2 hybridization of the reaction carbon in the transition state (TS) is restricted, this will also slow down the rate of SN2 reaction. Since rings only have a limited number of possible conformations, each ring system will have to be examined individually to understand experimental data on its relative rate of SN2 reaction. Bond angle changes in small rings can be very inhibiting to the SN2 reaction, because of the angle strain [sp3 (109o) sp2 (120o) = T.S.]. Some typical SN2 rate data are shown below.

krelative (SN2)

Br Br Br Br Br Br Br

1.0 0.00001 0.008 1.6 0.01 0.97 0.22

Problem 5 - Why might C3 and C4 rings react so slowly in SN2 reactions? (Hint-think about bond angles in the transition state versus bond angles in the starting ring structure.) Problem 6 - Why might C6 rings react slower in SN2 reactions? What are the possible conformations from which a reaction is expected? Trace the path of approach for backside attack across the cyclohexane ring to see what positions block this approach. Which conformation would have the leaving group in a more reactive position (axial or equatorial)? Is this part of the difficulty (which conformation is preferred)?

These two chair conformations interconvert in a very fast equilibrium.

Problem 7 - In each of the following pairs of nucleophiles one is a much better nucleophile than its closely related partner. Propose a possible explanation.

N N O O CC

C

C

O

HH

HH

HH

HH

H

b. c.a. relative rates 250/1

OCH

HH

methoxidet-butoxide

Nucleophilic Substitution & Elimination Chemistry Beauchamp 13 Problem 8 - The following two reaction sequences have not been specifically studied, but are analogous to reactions we are studying. Part of the fun of organic chemistry is to discover the mechanistic logic in a newly seen reaction. See if you can explain the observations below. a. The first reaction occurs in two steps, forming initially two ionic species which react further with one another. Propose an arrow pushing mechanism and list the related analogy mechanism. Look at the new bonds formed in the products and the old bonds broken in the reactants to decide what atoms joined together and split apart. You also have to decide which atoms donated and accepted the electron pairs.

CH2R' S CH2 R'

What is themechanism?

PR1

R1 R1

SP

R1

R1 R1

Sresonance

PR1

R1 R1

CH2R' S S CH2 R'What is themechanism?

Add in any formal charge.

Hint: The first mechanism type is the same as the second mechanism type and the same as the topic we are currently discussing.

step 1

step 2

b. A single dipolar species (two charges) is formed in the first step of the following reaction. What is the leaving group in this first step? Normally such a leaving group would be poor, but in this reaction it is OK. Why? (How small is the ring?) This intermediate undergoes a rotation about the center bond and then reacts in a second step. The two ionic sites bond together and make a four-membered ring, which makes the phosphorous exceed the octet rule. This is acceptable because of phosphorous’ available 3d orbitals. The ring ruptures, forming two pi bonds (P=O and C=C). Propose an arrow pushing mechanism, which rationalizes the overall transformation. Be specific in your third step to indicate whether the elimination occurs in a syn or anti manner.

add in any formal charge

PR1

R1 R1

OP

R1

R1 R1

Oresonance

PR1

R1 R1"SN2 likemechanism"

C C

O

H CH3

HH3C rotate center bond by 180o

CCHH

H3C CH3 internaleliminationmechanism

two ionic charges in a single intermediate structure, (P+ and O-)

still two ionic charges, P+ and O-

O- attacks P+ to form a four atom ring

step 1

step 2

step 3

Nucleophilic Substitution & Elimination Chemistry Beauchamp 14 Simplistic Picture of a General E2 mechanism – with the solvent

The reactants must encounter one another with the proper orientation (anti conformation of Cβ-H and Cα-X) for the E2 reaction to occur. There may be more than one type of anti Cβ-H that can react leading to different E2 products. The high potential energy of the strong base drives the reaction to completion.

The transition state developing the pi bondrequires parallel overlap for developing 2p orbitals. Anti and syn allow this, but anti is much preferred due to its staggered conformation.

The products diffuse away from one another after reacting. The carbon hybridizations have changed from sp3 to sp2. Less common, but possible is a change from sp2 to sp. The most stable alkene (most substituted) has the lower Ea and froms at a faster rate (usually).

S

S S S

S

S

SS

S

S

S

SSS

S

S

C

CH3C

H H

CH3

Br

HOH

"Transition State"SS

S

S

S

S

OH

S

S

S

S

S

S

S S

SS

S

S

S

S S

S

S

S

SSS

CCH3CH3 Br

C HHH

S

S

S

S

S

S

S

S

SS

S

S

Br

S

S

S

S

S

C

CH3C

H H

CH3

O HH

S

S

S

Same picture without the solvent

The transition State - requires parallel overlapof the two 2p orbitals forming the pi bond. This is easiest when Cβ-H is anti to Cα-X.

OH

C

CH3

H3C

Br

C HHH

C

CH3C

H H

CH3

Br

HOH Br

C

CH3C

H H

CH3

OHH

When present E or Z configurations are fixed by the anti Cβ-H/Cα-X conformation(s).

Important Details 1. E2 reactions occur in one step. They are concerted reactions that go directly from reactants to products.

2. Both reactants (RX and Nu:-/B:-) are involved in the slow (…and only) step. The transition state, (T.S.)

is the high potential energy (PE) point of the reaction and determines the activation energy (Ea). This is called the rate determining step (R.D.S.).

3. Bond formation (O-H and C=C) and bond breakage (Cβ-H and Cα-Br) are occurring at the same time, which keeps the activation energy (Ea) lower. This requires parallel overlap of these two bonds.

4. Two sp3 carbon atoms are both becoming the sp2 carbon atoms of the pi bond.

5. Stereochemistry (E/Z) at C=C is fixed by anti orientation of Cβ-H and Cα-Br bonds

6. Generally, strong base/nucleophiles are used. (This is simplistically shown as negative charge, although sterically hindered neutral nitrogen is also a good base. Neutral sulfur is not a good base.) To emphasize E2 > SN2, bulky sterically hindered, strong bases are used (such as potassium t-butoxide).

Nucleophilic Substitution & Elimination Chemistry Beauchamp 15

The competition in SN2 versus E2 reactions is between electron donation to a carbon atom (SN2) or electron donation to hydrogen atom (E2). We will see competition between these two sites many times in organic chemistry. Problem 9 - a. What dihedral angles are possible for Cβ-H versus Cα-X in typical conformations. Use Newman projections to illustrate these.

θ = dihedral angle 0o 60o 120o 180o 240o 300o

θ

b. Considering the fact that the new pi bond has to form simultaneously with (Base-H) bond formation and Cβ-H and Cα-X bond breakage, which of the above conformations look best to allow all of this to occur at once? If more than one conformation permits the two 2p orbitals forming the pi bond to overlap, is one conformation better than another? Why? Problem 10 - Draw a Newman projection of the two possible conformations leading to E2 reaction. Show how the orientation of the substituents about the newly formed alkene compares. Are they the same? Assume priorities are R2 > R1 and R4 > R3. Specify the absolute configuration of each alkene as E or Z.

reactant conformation 1

reactant conformation 2

alkene stereochemistry?

alkene stereochemistry?

anti E2reaction

syn E2reaction

C

R2

R1

CR4

R3

X

H

rotate about center bond

Newmanprojections

Nucleophilic Substitution & Elimination Chemistry Beauchamp 16 Keeping Track of the Cβ Hydrogens At least one Cβ position, with a hydrogen atom, is necessary for an E2 reaction to occur. In more complicated systems there may be several different types of hydrogen atoms on different Cβ positions. After all, in E2 reactions there can be anywhere from one to three Cβ carbons, and each Cβ carbon can have up to three hydrogen atoms.

Cα

H

X

HH

zero Cβ position(unique methyl, only SN2)

Cα

H

Cβ

X

H

one Cβ position(1o RX, usually SN2 > E2)

two Cβ positions(2o RX, SN2/E2 both competitive)

Cα

Cβ

Cβ

X

H

three Cβ positions(3o RX, only E2)

Cα

Cβ

Cβ

X

Cβ

With proper rotations, each Cβ-H may potentially be able to assume an anti conformation necessary for an E2 reaction to occur. There are a lot of details to keep track of and you must be systematic in your approach to consider all possibilities. Let’s consider a moderately complicated example.

Cα

CH2

CH3

CH3

X

CH

H3C

CH2H3C

Redraw structure to explicitely show all of the Cβ hydrogen atoms. Notice the base/nucleophile is strong, so SN2 or E2 reaction are possible. The RX pattern is tertiary, so no SN2 is possible. There are three Cβ carbon atoms and all of them have hydrogen atoms on them. That leaves E2 reaction as the only possible choice.

OR OHR

Na

Cα

Cβ2

CH3

Cβ3

X

Cβ1

H3C

CH2H3C

HHH

H

H

H

There are two chiral centers, Cα and Cβ1, that make this structure more complicated than we are treating it. Possible stereoisomers are RR, RS, SR and SS.

Problem 11 - How many total hydrogen atoms are on Cβ carbons? How many different types of hydrogen atoms are on Cβ carbons? How many different products are possible? Hint - Be careful of the simple CH2 above. The two hydrogen atoms appear equivalent, but E/Z (cis/trans) possibilities are often present. (See below for relative expected amounts.)

CαCβ

H

X

BThis sketch will work for every possibility above, IF you fill in the blank positions correctly.

Nucleophilic Substitution & Elimination Chemistry Beauchamp 17 Stability of Pi Bonds In elimination reactions, more substituted alkenes are normally formed in greater amounts. Greater substitution of carbon groups in place of hydrogen atoms at alkene carbons (and alkyne carbon atoms too) translates into greater stability (lower potential energy). Alkene substitution patterns are shown below. There are three types of disubstituted alkenes and their relative stabilities are usually as follows: geminal ≈ cis < trans.

C

H

H

C

H

H

C

R

H

C

H

H

C

R

H

C

R

H

C

R

R

C

H

H

C

R

H

C

H

R

C

R

R

C

R

H

C

R

R

C

R

R

< < < < <≈

The more substituted alkenes are usually more stable than less substitued alkenes. Substitution here, means an R group for a hydrogen atom at one of the four bonding positions of the alkene.

1 2 3 4 5 67

Relative stabilities of substitued alkenes.

1 = unsubstituted alkene (ethene is unique)2 = monosubstituted alkene3 = cis disubstituted alkene4 = geminal disubstituted alkene5 = trans disubstituted alkene6 = trisubstituted alkene7 = tetrasubstituted alkene

Saytzeff's rule: More stable alkenes tend to form faster (because of lower Ea) in dehydrohalogenation reactions. They tend to be the major alkene product,though typically a little of every alkene product possible is obtained.

Possible explanations for greater stability with greater substitution of the pi bond 1. A fairly simple-minded explanation for the relative alkene stabilities is provided by considering the

greater electronegativity of an sp2 orbital over an sp3 orbital. An alkyl group (R ) inductively donates electron density better than a simple hydrogen. From the point of view of a more electronegative sp2 alkene carbon, it is better to be connected to an electron donating alkyl carbon group than a simple hydrogen atom. The more R groups at the four sp2 positions of a double bond, the better. However, be aware that other features such as steric effects or resonance effects can reverse expected orders of stability.

C C C C

HR

...inductively donating "R" substituentis more stable than unsubstituted "H"...

Nucleophilic Substitution & Elimination Chemistry Beauchamp 18 2. Hyperconjugation can be viewed as a form of “sigma” resonance from a C-H sigma bonding orbital

into the adjacent pi-bond. (HOMO/LUMO orbitals can be used to explain this donation too.)

CCC

H

C

HC

H

CH

H

C

HC

H

CH

H hyperconjugation(sigma resonance?)

H

HCC

C

H

H

H

2D 3D

Problem 12 – Order the stabilities of the alkynes below (1 = most stable).

H C C H R C C H R C C R

"R" = a simple alkyl group Problem 13 - Match the heats of formation, ∆Hf

o’s, with the appropriate isomeric alkenes. Which alkene is most stable in each group?

a b∆Hof's = -0.2 , -1.9 , -3.0 ∆Ho

f's = -8.6 , -10.1

Problem 14 - Calculate the following heats of hydrogenation, ∆Ho

rxn’s, (hydrogenation = addition of H2 to a pi bond). Do these values make sense, given the above patterns of alkene stabilities? Explain your answers.

H2catalyst

H2catalyst

H2catalyst H2

catalyst

∆Hof = -8.6 ∆Ho

f = -36.9∆Ho

f = -35.1

∆Hof = -35.1

∆Hof = -5.3

∆Hof = -6.9∆Ho

f = -35.1∆Hof = -7.9

a b

c d

Problem 15 -Provide an explanation for any unexpected deviations from our general rule for alkene stabilities above.

-19.9∆Hof = -22.7∆Ho

f = ∆Hof = -18.7

Nucleophilic Substitution & Elimination Chemistry Beauchamp 19 Problem 16 - Reconsider the elimination products expected in the problem on page 16 and identify the ones that you now expect to be the major and minor products.

OR OHR

NaCα

Cβ2

CH3

Cβ3

X

Cβ1

H3C

CH2H3C

HHH

H

H

H

Problem 17 - What are all of the possible products resulting from E2 elimination of 2-bromobutane? Assume B:-- is the base/nucleophile. Consider the total number of beta C-H’s and the different types of beta C-H. Which products are expected to be more stable? What is the most likely competing reaction? Possible sawhorse and Newman projections are provided below. Fill in the appropriate atoms for each possibility, according to our understanding of the E2 mechanism.

H

HCC

H

HCC

H

HCC

Problem 18 - (2R,3S)-2-bromo-3-deuteriobutane when reacted with potassium ethoxide produces cis-2-butene having deuterium and trans-2-butene not having deuterium. The diastereomer (2R,3R)-2-bromo-3-deuteriobutane under the same conditions produces cis-2-butene not having deuterium and trans-2-butene having deuterium present. Explain these observations by drawing the correct 3D structures, rotating to the proper conformation for elimination and showing an arrow pushing mechanism leading to the observed products. (Protium = H and deuterium = D; H and D are isotopes.)

(2R,3S)trans-2-butene(no D present)

cis-2-butene (D present)

cis-2-butene(no D present)

trans-2-butene (D present)

and and

(2R,3R)

O O

K K

Nucleophilic Substitution & Elimination Chemistry Beauchamp 20 Substituted Cyclohexanes and E2 reactions Studies of substituted cyclohexane structures have reinforced the idea that an anti Cβ-H/Cα-X conformation is necessary to allow an E2 reaction to succeed. Chair conformations of cyclohexane structures have well defined positions for all of their atoms. When a leaving group is in an axial position, the adjacent axial positions are anti as required for E2 elimination. A hydrogen atom at either of these positions will allow for easy E2 elimination. The other cyclohexane conformation will put the leaving group into the equatorial position. In this conformation the anti positions are the carbons of the cyclohexane ring. No elimination is possible from this conformation. The combined effect of all substituents on the ring, as well as with each other, that will determine the preferred position for X at equilibrium (axial or equatorial). One cyclohexane conformation is potentially reactive (X when axial) and one conformation is nonreactive (X when equatorial).

represents anti positions

XH

H

H

If is a hydrogen atom, then E2 is possible because is in an anti position from X. SN2 also works better in this conformation as long as Cα is not tertiary and Cβ positions are not quaternary.

E2 is not possible with in an equatorial position because the ring carbons C3 and C5 are anti to X (bolded bonds). SN2 is inhibited because of the axial positions at C3 and C5.

12

3 45

6

H H

H H

XH

HH

123

4 5 6

H

H

H

H

Problem 19 - What are the cyclohexane conformational relationships in adjacent axial/axial positions, axial/equatorial positions and equatorial/equatorial positions (syn, gauche or anti)? Which combination(s) allow for easy E2 reaction to occur? Draw these representations clearly as bond-line formulas and as Newman projections. Problem 20 - The structures of neomenthyl chloride and menthyl chloride are given below. Neomenthyl chloride reacts very quickly and produces two cyclohexane E2 products in the indicated amounts, while menthyl chloride reacts very slowly and only produces one cyclohexane E2 product as indicated. Explain these observations. What would be the observed product if any SN2 product were observed? Redraw the structures in a more realistic 3D representation (two chair conformations for each isomer) and point out the problems leading to these observations.

H

CHCH3

CH3

Cl

H

H

H

CH3

H

menthyl chloride

H

CHCH3

CH3

H

Cl

H

H

CH3

H

neomenthyl chloride 2-menthene 25%

H

CHCH3

CH3

CH3

H

3-menthene 75%

H

H

CH3

H

CHH3C

H3Cfast

OROR H

2-menthene 100%

H

CHCH3

CH3

CH3

H

veryslow

OROR H

Nucleophilic Substitution & Elimination Chemistry Beauchamp 21 Problem 21 - A stronger base (as measured by a higher pKa of its conjugate acid) tends to produce more relative amounts of E2 compared to SN2, relative to a second (weaker) base/nucleophile. Greater substitution at Cα and Cβ also increases the proportion of E2 product, because the greater steric hindrance slows down the competing SN2 reaction. Use this information to make predictions about which set of conditions in each part would produce relatively more elimination product. Briefly, explain your reasoning. Write out all expected elimination products. Are there any examples below where one reaction (SN2 or E2) would completely dominate? a.

I

Cl

...versus...I

Cl

b.

Cl

...versus...

ClO

O OpKa = 5 pKa = 16

c.

Cl

...versus...

Cl

d.

N C...versus...

N C

ClCl

e.

OR

Br

...versus...

OR

Br

f.

Cl

...versus...N C

Cl

pKa = 25 pKa = 9 Problem 22 – One of the following reactions produces over 90% SN2 product and one of them produces about 85% E2 product. Match these results with the correct reaction and explain why.

ClO

pKa = 16

ClO

pKa = 19...versus...

Nucleophilic Substitution & Elimination Chemistry Beauchamp 22 Problem 23 - Propose an explanation for the following table of data. Write out the expected products and state by which mechanism they formed. Nu: -- /B: -- = CH3CO2 -- (a weak base, but good nucleophile).

H2CH3C Br

HCH3C Br

CH3

HCCH Br

CH3

CH3C Br

CH3

CH3

H3C

H3C

percent substitution

percent elimination

100 %

100 %

11 %

0 %

0 %

0 %

89 %

100 %

O

O

O

O

O

O

O

O

Problem 24 - Explain the following observations. Models would probably help.

O

H Br

H

OO

H

H

O

CH3OCH3OH

O

H CH3Br

H

OO

H

O

CH3CH3O

CH3OH

a bCH3

Problem 25 - Provide a step-by-step mechanism (curved arrows, formal charge, electron pairs, etc.) for each of the following observations and explain the observed differences in product formation. Hint - What cyclohexane conformation is present in the bridged ether product below? How could it get that way?

a. trans-4-chlorocyclohexanol + HO OHO

OHb. cis-4-chlorocyclohexanol + HO HO OH