Nucleophilic Substitution & Elimination Chemistry 1 SN2 / E2 Topics 1. Strong, nonbulky nucleophiles are best in SN2 reactions (strong decides “2” or “1”)

a. Less bulky is better for SN2 reactions (versus E2)

KOC

H

H

H

OC

H

CH3

HOC

CH3

CH3

HOC

CH3

CH3

CH3

Sterically large potassium t-butoxide mainly produces E2 products, even at 1o RX centers

OH

b. A less basic electron pair donor is better for SN2 reactions (versus E2). A conjugate acid with a lower pKa is “less basic”.

CN H

CCR H

CN Cyanide is a good SN2 nucleophile at methyl, primary and secondaryRX centers, but tertiary RX centers give only E2 reaction. pKa of conjugate acid ( ) = 9.

A terminal acetylide is a good SN2 nucleophile at methyl and primary, but secondary and tertiary RX centers give mainly or only E2. pKa of conjugate acid ( ) = 25 (R = carbon or hydrogen)

CCR

Carboxylates are good SN2 nucleophiles at methyl, primary and secondary RX centers, but tertiary RX centers give only E2

OR

CRO

O

R = carbon or hydrogen

Alkoxides are good SN2 nucleophile at methyl and primary, but secondary and tertiary RX centers give mainly or only E2

CRO

O H

OR H

pKa's of conjugate acid = 5

pKa's of conjugate acid = 16-19 Examples of oxygen nucleophiles

pKa (conjugate acid)

OH H

OR H

CR

O

O H

OH

OR

CR

O

O

(conjugate base)

hydroxide

alkoxides

carboxylates

water

alcohols

carboxylic acids

weak nulcleophile / bases strong nulcleophile / bases

(16)

(16-19)

(5)

2. The role of RX in SN2 reactions. a. The Cα position – Anything that hinders approach of nucleophile slows the SN2 reaction.

Cα

H

HH

X

methyl RX (≈ 30)

primary RX (≈ 1)

secondary RX (≈ 1/40)

tertiary RX (≈ 0)

Cα

R

HH

X Cα

R

HR

X Cα

R

RR

X

Complete substitution at Cα stops all SN2.

Cα X

?

??

> > >>

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Nucleophilic Substitution & Elimination Chemistry 2

b. The Cβ position– Anything that hinders approach of nucleophile slows the SN2 reaction.

Cα

Cβ

HH

X

unsubstituted Cβ (≈ 1)

H

HH

Cα

Cβ

HH

X

monosubstituted Cβ (≈ 0.4)

H

H

Cα

Cβ

HH

X

disubstituted Cβ (≈ 0.03)

H

Cα

Cβ

HH

X

trisubstituted Cβ backside of Cα is blocked (≈ 0)

C

H

HHR

RR

R

R

Complete substitution at Cβ stops all SN2.

> > >>

c. An adjacent pi bond (allyl and benzyl) attached to C α greatly accelerates SN2 reactions.

C C

C

X

represents allyl or benzyl pi system

The pi bond is parallel to 2p orbital in transition state having excess electron density. This allows the excess charge to be delocalized (resonance), which lowers the activation energy and makes the reaction go faster.

C C

C

X

Nu Nu

transition state

allyl RX SN2 ≈ x 40

benzyl RX SN2 ≈ x 100

3. Inversion of configuration in SN2 reactions always occurs, but we only “see it” at chiral centers or in rings with “cis/trans” substitution. a. R/S

CNu

2

34

C X

2

34

Nu X

"S" configuration "R" configuration

(R S) or (S R)

b. Rings (cis/trans)

R H

H X

Nu:

transition state

H

Nu

X

R Nu

H Htrans ring cis ring

(cis ring trans ring) or (trans ring cis ring), May or may not have R/S configurations.

R

H

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Nucleophilic Substitution & Elimination Chemistry 3 4. The role of solvent in SN2 reactions

a. Polar, aprotic solvents are best for SN2 reactions. They tend to strongly solvate cations and keep them away from the electron pair donor (nucleophile). They only weakly interact with the anion nucleophiles. Some more common possibilities are listed below.

dimethylsulfoxide DMSO dimethylformamide

DMF

acetonitrile AN hexamethylphosphoramide

HMPA

SH3C CH3

O

CH3C CH3

OC

H

O

NCH3

CH3

CH3C NP

O

NN

NH3C

H3CH3C CH3

CH3

CH3

acetone

S

O

C CH

HH H

H

H

δ+

δ-

strong solvent dipole = polar solventno polarized hydrogen atoms = aprotic

S

O

C CH

HH H

H

H

Very strong interaction with cations.

Very weak interaction with anions.

S

O

C CH

HH H

H

H

Cations tend to be tightly bound to the solvent and anions tend to be unbound and relatively free to react in polar aprotic solvents. Nucleophilic lone pairs are available to react (donate at Cα center).

b. Polar, protic solvents are best for E2 reactions. See point 1 of E2 reactions (next).

5. Common leaving groups (for our course) are stable as anions (which start as neutral) or they are neutral molecules (which start as cations).

chlorideCl

bromideBr

iodide

I

OHH

water

These can be leaving groups in basic, neutral or acidic conditions

Water is a leaving group from a protonated alcohol in acid conditions. In a similar way an alcohol is a leavinggroup from a protonated ether in acid conditions.

ORH

alcoholSOO

Otosylate

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Nucleophilic Substitution & Elimination Chemistry 4 1. Strong, bulky bases are best for E2 reactions (often these are polar, protic solvents and their conjugate

bases). The conjugate bases are highly basic and have a strong attraction for hydrogen (the target in an E2 reaction). This tendency can be emphasized by introducing steric bulk into the base (electron pair donor) such as the case with potassium t-butoxide. Also used as bases are hindered tertiary amines.

O HHOC

H

H

H

H OC

H

CH3

H

H OC

CH3

CH3

H

H OC

CH3

CH3

CH3

H

solvent = water methanol ethanol isopropyl alcohol t-butyl alcohol

conjugate bases = K

OC

H

H

HOC

H

CH3

HOC

CH3

CH3

H

OC

CH3

CH3

CH3

Potassium t-butoxide mainly produces E2 products, even at 1o RX centers.

OH

Conjugate bases are pretty basic as indicated by moderately high pKa of conjugate acids (16-19 range).

2. The role of RX in E2 reactions. a. The Cα position – more substitution here slows SN2 and makes E2 more competitive. However, you always need a Cβ-H (almost always anti).

CH3 (methyl)

No E2 is possible at methyl.

Cβ

H

Cα

X

= fill in the blank

Anti Cβ-H and Cα-X is the usual mode for E2 reactions.

1o < 2o << 3o (only E2 at 3o RX with strong base/nucleophile)

b. The Cβ position – more substitution here slows SN2 and makes E2 more competitive. However, you always need a Cβ-H (almost always anti).

E2 reaction correlates with number of groups at Cβ = 0 < 1 < 2 . (ethyl)

Cβ

H

HH

Cα

X

HH

The Cβ carbon has no additional groups beyond Cα.

Cβ

H

H

Cα

X

HH

The Cβ carbon has one additional group beyond Cα.

Cβ

H

Cα

X

HH

The Cβ carbon has two additional groups beyond Cα.

Cβ

C

Cα

X

HH

The Cβ carbon has three additional groups beyond Cα. Cβ is fully substituted.

H

H

H

Backside approach by nucleophile is blocked.

If there are 3 "R": groups on Cβ, there is no E2 because there is no Cβ-H.

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Nucleophilic Substitution & Elimination Chemistry 5

c. An adjacent pi bond is helpful to E2 reactions because the new pi bond can conjugate with the existing pi bond, BUT generally SN2 is helped more (allyl RX and benzyl RX). Tertiary allylic and benzylic still react by E2 (no SN2).

X Xallyl = benzyl =

3. For us, E2 reactions always attack an anti C β-H (relative to the C α carbon) in competition with SN2. Each Cβ-H must be considered and many outcomes may be possible. A simple beta-CH2 may lead to both E and Z stereoisomers. The major E2 product is decided on the basis of alkene substitution patterns (tetra > tri > di(trans) > di(cis) = di(gem) > monosubstituted alkenes). a. Open chains must be rotated to the desired conformation for reaction. Look especially at beta CH2's for the possibility of E/Z stereoisomers. Cβ-H and Cα-X are always anti for us in E2 reactions.

Cα

Cβ

HH X

H

B

B

Cα

Cβ

Cβ

H X

H

H

B

primary RXsecondary RX tertiary RX

Cα

Cβ

Cβ

C X

H

H

H

b. Two chair conformations rapidly equilibrate in cyclohexane rings. The following is only suggestive, as there are many possibilities for substitution patterns.

interconvert infast equilibrium

X

X

1. In cyclohexanes, the leaving group must be axial to have an anti Cβ-H .2. If a nonhydrogen group is at one of the Cβ positions, then no E2 can occur at that position.3. An external Cβ-H is also possible (shown in the example above).

These are only suggestive of the many possible E2 variations in a cyclohexane.

H

H H

H H

SN2 is difficult when X is equatorial, even if not at a tertiary center.

= anti Cβ-H to Cα-X SN2 is OK when X is equatorial, but not at tertiary center.

Possible E2 approach. Possible E2 approach.

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Nucleophilic Substitution & Elimination Chemistry 6 4. The bases role in E2 reactions - strong bases are necessary (most often bulky alkoxides or bulky

tertiary amines are used).

CH3C

CH3

CH3

O K

Potassium t-butoxide mainly produces E2 products, even at 1o RX centers.

N

N

N

N

DBN = 1,5-diazobicyclononane DBU = 1,5-diazobicycloundecane

Nitrogen bases used to favor E2 reactions.

5. The role of solvent in E2 reactions (same as #1 in E2 reactions). a. Polar, protic solvents are best for E2 reactions. Their conjugate bases are highly basic and have a strong attraction for hydrogen (the target in an E2 reaction). This tendency can be emphasized by introducing steric bulk into the base (electron pair donor), such as is the case with potassium t-butoxide.

O HHOC

H

H

H

H OC

H

CH3

H

H OC

CH3

CH3

H

H OC

CH3

CH3

CH3

H

solvent = water methanol ethanol isopropyl alcohol t-butyl alcohol

conjugate bases = HO

OC

H

H

H

OC

H

CH3

HOC

CH3

CH3

HOC

CH3

CH3

CH3

potassium t-butoxide mainlyproduces E2 products, even at 1o RX centers

6. Common leaving groups (for our course) are stable as anions (which start as neutral) or they are neutral

molecules (which start as cations). (Same as point 5 in SN2 items above.)

chlorideCl

bromideBr

iodide

I

OHH

water

These can be leaving groups in basic, neutral or acidic conditions

Water is a leaving group from a protonated alcohol in acid conditions. In a similar way an alcohol is a leavinggroup from a protonated ether in acid conditions.

ORH

alcoholSOO

Otosylate

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Nucleophilic Substitution & Elimination Chemistry 7 Major expected product(s) in SN2 / E2 competition.

CN

CCR

OR

nucleophile methyl primary secondary tertiary

"bold" = different "typical" outcomes

KOC

CH3

CH3

CH3

compare these

compare these

CRO

O

OH(only)

(only)

(only)

(only)

(only)

(only)

(only)

(only)

(only)

(only)

SN2 > E2SN2 E2

SN2

SN2

SN2

SN2

SN2

SN2 > E2

SN2 > E2

SN2 > E2

E2 > SN2

SN2 > E2

SN2 > E2

E2 > SN2

E2 > SN2

E2 > SN2

E2 > SN2

SN2 > E2

(only)

(only)E2

E2

E2

E2

E2

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Nucleophilic Substitution & Elimination Chemistry 8 SN1 and E1 Factors 1. We write nucleophiles (H-Nu:) and bases (H-B:) differently. A proton is included and they are

written as neutral molecules. Those most often used by us are solvent molecules of polar, protic solvents (water, alcohols and liquid carboxylic acids).

Weak nucleophile bases used in our course: OH H OR H C

R

O

OH

water alcohols carboxylic acids

2. Role of RX compound – RX needs to stabilize a carbocation carbon. a. “R” groups on the Cα carbon are inductively donating and help stabilize positive charge. Tertiary is better than secondary and we will not propose primary or methyl carbocations.

CH

H

Hvery poor

CR

H

Hpoor

CR

R

H

CR

R

Rrelatively goodOK

= donating inductive effect

b. Hyperconjugation stabilizes positive charge. Tertiary is better than secondary and we will not propose primary or methyl carbocations.

C C

H

C C

H

sigma resonance?

carbocation Carbocation stabilized by the electrons in an parallel adjacent sigma bond.

c. Resonance is very good at stabilizing charge (positive, negative, free radicals or neutral conjugate pi systems). i. Pi bonds allow delocalization of electrons via parallel overlap of adjacent 2p orbitals. Most often these pi electrons are from an alkene or aromatic pi system.

carbocation next to a pi bond (alkene or aromatic)

C C

C

C C

C

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Nucleophilic Substitution & Elimination Chemistry 9

ii. Lone pairs adjacent to empty orbitals can share their electron density to help stabilize positive charge (they also delocalize into neutral pi bonds, like enols and amides). Most often nitrogen or oxygen is the donor atom. (Under very different conditions, carbon can do so when present as a carbanion.)

carbocation next to a lone pair (usually nitrogen or oxygen)

X C X C

X = "N" or "O" X = "N" or "O" iii. Really poor carbocations – We usually won’t propose these, though there are rare occasions where we do invoke some of them.

C CR

H

Phenyl carbocation has an empty sp2 orbital.

R

Vinyl carbocation has an empty 2p orbital, but on a sp hybridized carbon atom.

C CR

Terminal alkyne carbocation, has an empty sp orbital.

3. Attack of nucleophile/base a. Attack at Cα carbon (top and bottom)

Nu HH Cα

R

R

If R = R, then we cannot detect top/bottom attack.

top

bottom

H Cα

R2

R1

If R1 ≠ R2, then we can detect top/bottom attack because opposite configurations will form at the Cα carbon, which are enantiomers. We assume racemization, but is not always true.

H Cα

R2

R1

If R1 ≠ R2, and R1 and/or R2 have chiral centers, then we can detect top/bottom attack because opposite configurations will form at the Cα carbon and will form diastereomers in combination with the other chiral centers.

* = has chiral center

*

CR

R

H

There are no chiral centers,but top/bottom attack is seen in cis/trans products which are diastereomers.

b. Attack at Cβ -H: anything goes (can rotate Cβ-H up or down relative to the carbocation 2p orbital).

Cβ

H

R1R2

Cα

R3

R4

Cβ Cα

R3

R4Rotate Cβ-H 180o.

R1

R2

H

E and Z possible stereochemistry.CαCβ

R1

R2

R3

R4

CαCβ

R2

R1

R3

R4

BH

BH

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Nucleophilic Substitution & Elimination Chemistry 10 4. Role of solvent – Polar, protic solvent stabilizes charges of both types in first step of the SN1 and E1

reaction possibilities (ionization of the C-X bond).

OR

H

OR

H

O

R

H

O

R

H

Oxygen end of solvent molecules stabilizes the cations.

O

R

H O

R

H

OR

H

OR

H

Hydrogen end of solvent molecules stabilizes the anions.

5. Leaving groups – Same as point 5 in SN2 reactions.

6. Special cases of SN1 and E1 reactions. a. Assume mainly SN product over E product when alcohols (ROH) are mixed with HX acids (H-Cl, H-Br and H-I). SN2 reactions are assumed at methyl and primary alcohols and SN1 reactions are assumed at secondary, tertiary, allylic and benzylic alcohols.

OR H XH

(X = Cl, Br, I) XR

R = methyl = SN2R = primary = SN2 R = secondary = SN1R = tertiary = SN1R = allylic = SN1R = benzylic = SN1

b. Assume mainly E1 reactions when alcohols are mixed with concentrated sulfuric acid (H2SO4) when heated (∆).

OR H OH

R = methyl = not possibleR = primary = difficult, E1R = secondary = moderate, E1R = tertiary = relatively easy, E1

S

O

O

OH∆ = heat

Alkenes, with possible rearrangement, more substituted alkenes tend to be the major products formed. This is the only reaction where we will propose E1 as the major product.

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Nucleophilic Substitution & Elimination Chemistry 11 Essential clues to make educated guesses about what is occurring.

XR HNu

B H

Nu

NuH H

Nu R

BH H

B

XR

Nu R X

XR

XR

X

B H

X

X

R = carbon portion

methyl = CH3-Xprimary = RCH2-Xsecondary = R2CH-Xtertiary = R3C-Xallylic = CH2=CHCH2-Xbenzylic = C6H5CH2-X

Nu B=strong electron pair donors = SN2 / E2 anions, neutral nitrogen, neutral sulfur

HNu B H=

weak electron pair donors = SN1 / E1neutral solventH2O, ROH, RCO2H

Nu R

substitution products

CaCb

elimination products

X

leaving group

Ca Cb

Ca Cb

Reagents ProductsReaction Conditions

SN1

E1

SN2

E2

Cl

Br

I

SO

O

O

H2O Typical R-X Structures

XH3Cmethyl X(unique)

CH2R Xprimary X(general)

CHRR

Xsecondary X (general)

CRR

RX

tertiary X(general)

CHH2C CH2

X

allylic X(and variations)

CH2

X

benzylic X(and variations)

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Nucleophilic Substitution & Elimination Chemistry 12 Templates for predicting SN2 and E2 reactions

CβCα

X

Cβ

Cβ

HH

HNu

B

Cα

Cβ

HH X

H

Cα

Cβ

Cβ

H X

H

H

interconvert infast equilibrium

C

C

12

3

45

6

possiblereactions

possiblereactions

Cα

CβHH

X

H

Nu

B

Cα

CβCβH

X

HH

Nu

B

two different perspectives

two different perspectives

An axial "X" is necessary for a succesful E2 reaction and also works better for SN2. X can be in many possible positions (there are two conformations for each possibility).

Nu

B

Nu

B

one perspective

primary RX

secondary RX

tertiary RX

cyclohexane RX Cα is carbon with leaving group and Cβ carbons are attached to Cα.

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Nucleophilic Substitution & Elimination Chemistry 13

Problem: Write a 3D structure of (3S,4S)-3-iodo-4-methoxyhexane. (Draw a 2D structure first.)

12

345

6

Cα IC4

C2template for 3D structure

rotated to altenative perspectives C2 Cα

I

C4or

1. What is/are the expected product(s) of this compound with sodium cyanide in DMSO (dimethylsulfoxide

is the solvent). Show all mechanistic details clearly for how each of the possible products is formed (3D structures, curved arrows, lone pairs and formal charges). Indicate major and minor products.

Cα

Cβ

Cβ

I

H

H

Cα

Cβ

Cβ

I

H

H

Cα

Cβ

Cβ

I

H

H

Cα

Cβ

Cβ

I

H

H

CN

CN

CN

CN

1. 3D structure

I

O

H

HCH3

12

345

6(3S,4S)-3-iodo-4-methoxyhexane

2. mechanisms and products

Nu

B

Cα

Cβ

Cβ

H I

HOCH3

Ha

CH3

Hb

CH2CH3

Cα

Cβ

Cβ

H I

HOCH3

Ha

CH3

Hb

CH2CH3

Cα

Cβ

Cβ

H I

HOCH3

Hb

Ha

CH3

CH2CH3

Cα

Cβ

Cβ

H I

HOCH3

Ha

CH3

Hb

CH2CH3

B

C

C

CH3CH2

H CH2CH3

C

C

H

CH3

C

H

H CH2CH3

OCH3

OCH3

B C

C

H

H

C

CH3

H CH2CH3

OCH3

(2Z,4S)-3-methoxy-2-hexene

(2E,4S)-3-methoxy-2-hexene

CNu

Cβ

HOCH3

CH2CH3

HCβHa

CH3

Hb

(3S,4S) (3R,4S)

(3Z)-3-methoxy-3-hexene

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Nucleophilic Substitution & Elimination Chemistry 14

Problem a. Write a 3D structure for the given name: (2R,3S,4R) 2-deuterio-3-bromo-4-methylhexane. Draw 2D first.

CH3C C

CCH2

CH3

(2R,3S,4R) 2-deuterio-3-bromo-4-methylhexane

How would the problem change if the bromine, deuterio and/or methylwere moved to another position?

12

34

56

C CC

2 3 4

All variations are shown below, which are enantiomers, diastereomers?

(2S,3R,4R)

(2R,3R,4S)

(2S,3S,4S)

(2S,3R,4S)

(2R,3R,4R)

(2S,3S,4R)

(2R,3S,4S)

(2R,3S,4R)

b. What are the expected products if hydroxide is the electron donor? How would the expected products differ if hydroxide were changed to ethoxide (?), t-butoxide (?), water (?) or ethanol(?). Write a separate mechanism showing the formation of each possible product.

Cα

Cβ

Cβ

X

H

H Cα

CβCβ

X

HH

Two possible perspectives.

3

3

Nu

B

NuH

BH

or

Nu

B

NuH

BH

or

Possible helpful suggestions.1. Draw conformations staggered.2. Draw anti positions parallel to Cα-X.

1. Draw Cα = C3 (in this problem). Fill it in with a proper configuration (R or S).

2. Add in groups on Cβ carbons, with anti Cβ-H and the “other” groups in any manner. If you are lucky, you

will have the correct configuration. If you are wrong, then switch the two “other” groups.

3. If there is a strong nucleophile/base, then write out all of the SN2/E2 possibilities. The only SN2 product will form by inversion of configuration. Any E2 products require an anti Cβ-H and C α-X conformation. You should draw every possible conformation that allows this, and examine the predicted alkene that forms. This will determine the configuration (E/Z) of any alkene products. More substituted alkenes are generally more preferred products.

4. If there is a weak nucleophile/base (usually H2O, ROH, RCO2H), then write out all of the SN1/E1 possibilities. The first step for both mechanisms is loss of the leaving group which forms a carbocation. If we ignore rearrangement possibilities for now, then there are two choices that can occur (SN1 and E1). For SN1 products add the :Nu-H from the top and bottom. If Cα is a chiral center, there will be two different products. It is possible that there are two different products in a ring with cis/trans possibilities and no chiral centers. You will also have to take off the extra proton via an acid/base reaction to get a neutral product. For E1 products you will have to remove any Cβ-H (no anti requirement). Make a double bond between all different Cβ-Cα's. Switch the two groups on either of the carbons of each double bond to see if different stereoisomers are formed. The possible outcomes are that the switch produces no change, or E/Z diastereomers are formed. All possible outcomes are predicted results in E1 reactions.

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Nucleophilic Substitution & Elimination Chemistry 15 SN / E Worksheet Nu

B

NuH

BH

HCα

X

HH

CβCα

X

HH

H

CβCα

X

HCb

H H

CβCα

X

Cβ

Cβ

HH

H

HCα

X

HH

CβCα

X

HH

H

CβCα

X

HCβ

H H

CβCα

X

Cβ

Cβ

HH

H

Possibilities: SN2 E2 SN1 E1

Possibilities: SN2 E2 SN1 E1

Possibilities: SN2 E2 SN1 E1

Possibilities: SN2 E2 SN1 E1

Possibilities: SN2 E2 SN1 E1

Possibilities: SN2 E2 SN1 E1

Possibilities: SN2 E2 SN1 E1

Possibilities: SN2 E2 SN1 E1

Nu

B

Nu

B

Nu

B

NuH

BH

NuH

BH

NuH

BH

methyl

primary

secondary

tertiary

tertiary

secondary

primary

methyl

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Nucleophilic Substitution & Elimination Chemistry 16 Partial Key a. Write a 3D structure for the given name: (2R,3S,4R) 2-deuterio-3-bromo-4-methylhexane. Draw 2D first.

CH3C C

CCH2

CH3

(2R,3S,4R) 2-deuterio-3-bromo-4-methylhexane

How would the problem change if the bromine, deuterio and/or methylwere moved to another position?

12

34

56

C CC

2 3 4

All variations are shown below, which are enantiomers, diastereomers?

(2S,3R,4R)

(2R,3R,4S)

(2S,3S,4S)

(2S,3R,4S)

(2R,3R,4R)

(2S,3S,4R)

(2R,3S,4S)

(2R,3S,4R)

b. What are the expected products if hydroxide is the electron donor? How would the expected products change if hydroxide were changed to ethoxide (?), t-butoxide (?), water (?) or ethanol(?). Write a separate mechanism showing the formation of each possible product.

Cα

CβCβH

X

CH3H3CD CH2CH3

HH

3

Nu

B

NuH

BH

Cα

CβCβH

X

CH3H3CD CH2CH3

HH

3

Nu

B= hydroxide, ethoxide and many other possibilities.

= water, ethanol and many other possibilities.NuH

BH

The Cα configuration makes no difference in SN1/E1 reactions because it is destroyed in the first step of the reaction.

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Nucleophilic Substitution & Elimination Chemistry 17

CβCα

HH3C

DCβ

CH3H2CCH3Br

HH

(2R,3S,4R) 2-deuterio-3-bromo-4-methylhexaneC

H3C CC

D H

H

CH3

Br

H

CH2CH3

12

34

56

CβCα

HH

H3CCβ

CH3H2CCH3Br

D H

SN2

E2

E2

E2

Nu

B

Nu

B

Nu

B

Nu

B

CβCα

HH

H3CCβ

CH3H2CCH3Br

D H

CβCα

HH

H3CCβ

CH3H2CCH3Br

D H

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Nucleophilic Substitution & Elimination Chemistry 18

CβCα

HH

H3CCβ

CH3H2CCH3Br

D H

(2R,3S,4R) 2-deuterio-3-bromo-4-methylhexaneC

H3C CC

D H

H

CH3

Br

H

CH2CH3

12

34

56

SN1

E1

E1

E1

NuH

BH

more possibilities

NuH

BH

more possibilities

more possibilities

NuH

BH

NuH

BH

more possibilities

CβCα

HH

H3CCβ

CH3H2CCH3Br

D H

CβCα

HH

H3CCβ

CH3H2CCH3Br

D H

CβCα

HH3C

DCβ

CH3H2CCH3Br

HH

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Nucleophilic Substitution & Elimination Chemistry 19

Problem – Predict the possible products (arrow pushing required, lone pairs, formal charges) for the following reactions.

BrH

H

BrCH3

H

H

H

H

HO

CH3O

CN

CCH

BrCH3

H

H

H

H

BrCH3

H

CH3

H

H

consider example at top

BrCH3

H

H

H

H

CH3C

O

O

"careful"

a

b

c

d

e

f

Br

H

BrH

Br

H

H

Br

HH

H3N

C

CH3

H3C

CH3

O

NN N

Br

O

O

HH

Br

H

H

H

H

O

O

SHBr

H

H

(low pKa)

(low pKa)

(high pKasterically bulky)

(low pKa)

F

Br

H

H

C

CH3

H3C

CH3

O

(high pKasterically bulky)

g

h

i

j

k

l

m

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Nucleophilic Substitution & Elimination Chemistry 20 Possible Key – predict products

BrH

BrHO

CCH H

Br

OH

OH

transcistrans cis

Br

Br

HO

Br

H

trans

SN2 = minor

E2 = major

the only anti Cβ-H

Br

Br

OH

OH

transcistrans cis

Br

Br

H

trans

the only anti Cβ-H

CO

OCH3

CO

OCH3

CCHE2

SN2 SN2 = major E2 = minor

In our course SN2 > E2 at primary RX(except t-butoxide).

E2 is in competition with SN2. Because hydroxide is strongly basic we expect E2 > SN2. Similar to "c", next, except for basicity of "O ".

CH3OOCH3

H

H

R

S S

S BrBr

H

H

H

H

Br

HH

H

CH3

H

C-C rotation

C-C rotationH

Br

HH

CH3

H

CH3

Br

HH

HH

E2a E2b E2c

C-C rotation

H

CH3

H

H

H

H

CH3

H

CH3

CH3

H

H

H

"Z" configuration (least = cis)

"E" configuration (middle = trans)

"E" configuration(most = trisubstituted)

CH3OCH3O CH3O

a

b

c

d

SN2 = major

E2 = minor

E2 is in competition with SN2. Because carboxylate is less basic we expect SN2 > E2. Similar to "b", above, except for basicity of "O ".

SN2 = minor

E2 = major

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Nucleophilic Substitution & Elimination Chemistry 21

BrCH3

CH3

H

BrCH3

HH

CH3

H

CH3 H

CH3H

Br

C N

This is a sneaky one. A Cβ carbon is fully substituted so SN2 reaction is greatly inhibited. There is an anti Cβ-H possibility so E2 can occur.

C NH

e

Only the MAJOR product is shown on the remaining problems. It is assumed that you can generate all possible products, including the major and minor possibilities.

BrNN N

(low pKa)

(low pKa)

(low pKa)

Br

H

H

H

H

O

O

SHBr

H

mainly SN2

Br

H

BrO

H

H

H

H O

mainly SN2

H

N N N

N3

transcis

Br

mainly SN2

H

S H

Br

H

H

HH

C

CH3

H3C

CH3

O

(RO has a high pKa ,is sterically bulky and should be only E2.Br cannot significantly rotate to an axial position since the very large t-butyl group locks the ring into the confromation having t-but yl equatorial.)

Br

HH

H

C

CH3

H3C

CH3

O

Only E2 reaction, t-butoxide is too basic and too big and bulky for SN2 reactions

H

HH

H H

HH

H

enantiomers(back axial CH too)

Br

BrCH3

H

H

H

H HH

H3NOnly E2 reaction at tertiary RX center when .an anti Cb-H is present.

Br

H

H

H

H

HH

H3C

H

H

HH

HH3CH

H

H CH2

H3C Br

(High pKa , sterically bulky, only E2 when an anti Cβ-H is present.)

No reaction

f

g

h

i

j

k

cistrans

Acetate is a good nucleophile and not too basic.

Azide is a good nucleophile and not too basic.

Sulfur is a good nucleophile and not too basic.

H3N

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Nucleophilic Substitution & Elimination Chemistry 22

F

SH

BrH

H

H

H

H HH

H3CBr

H

H

H

H

HH

H

CH3

H

HH

CH3CH3

H

H

H CH2

H

CH3H

CH3CH3

H

H

ab c

a b c

Br

Br

O

O

HH

H

HO

O

Br

BrH

H

H

H

H HH

H3CBr

HH

H

H

H

HHCH3

Bonus problem: Sulfur is a very good nucleophile and not very basic, but at tertiary RX we only see E2.

H

HH

H

CH3

H

H

CH2CH3

ab c

H

H

HHCH3 a b c

Only E2 reaction at tertiary RX center and only with an anti Cβ-H.

l

m

SH

F

Intramolecular SN2 reaction. Br must be in an axial position for backside attack.

Only E2 reaction at tertiary RX center and only with an anti Cβ-H.

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