2 always leave with x. n1

Nucleophilic Substitution & Elimination Chemistry Beauchamp 1

y:\files\classes\315\315 Handouts\315 Fall 2013\2 315 SN and E & chem catalog.doc

Four 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)

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, sulfur and phosphorous 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 C carbon attached to the leaving group, X? Is it a methyl, primary, secondary, tertiary, allylic, or benzylic carbon? What about any C carbon atoms? How many additional carbon atoms are attached at a C 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.



R-Br examples C1 C7 (infinite possibilities exist)

Br

Br

Br

Br Br Br

Br

BrBr

Br Br

Br Br

Br

Br

Br

Br

Br Br Br

Br

Br

Br

Br

Br

Br

Br

Br

Br

Br

BrBr

H3CBr

1C 2C 3C 4C

5C

6C

6C

7C

6C

Br

Br

Br

Br

BrBr

Br

Br

Br

Br

Br

BrBr

Br

Br

Br

Br Br Br

Br

Br Br

Br

BrBr

BrBr

Br

BrBr

Br

Br

BrBr

BrBr

BrBr

Br

7C

7C

7C

7C

7C Extra patterns to know (allylic and benzylic RX are very fast SN2 patterns) (1o neopentyl, vinyl and phenyl RX patterns are unreactive).Br Br Br Br Br Br

BrBr

allylic allylic allylic

benzylic benzylic

1o neopentyl RBr vinyl X phenyl X

a b ab

c d

ab c d e f

g h i j k l m

no p

a b c d e f g h

ab c d e f

g h i j k lm

no p q

q r s t

u v w x y z aa bb

cc dd ee ff gg hh

kk ll mm

ii jj

a b c a b

* = chiral center

*

*

* *

*

*

**

** *

* *

*

*

*

* * *

*

**

***

*

****

* **

* **

* *

1o neopentyl

2o1o1o

1o

1o 1o3o

2o

2o3o 1o1o2o2o1o

methyl

1o 2o 2o 1o3o 2o

3o 1o 1o neopentyl2o 2o1o 1o

2o neopentyl 1o 1o 3o

2o 2o1o2o 1o 3o

3o2o1o1o2o

1o 1o1o neopentyl 2o

2o

2o 1o1o1o

3o

3o2o

3o

2o 1o neopentyl 1o1o

2o2o

2o 2o

1o 1o

3o2o

1o neopentyl 3o 1o



Strong electron pair donation in our course (SN2 / E2).

Na

O H

Na

S HO R

Na K

ONa

O

O

hydroxide alkoxides potassium t-butoxide carboxylates (make, acetate)

hydrogen sulfide (thiolate)(buy) (make) (make, strong base) (buy)

Nu

B

=

SN2/E2 concerted reactions (one step) SN2 = always backside attack at C (inversion of configuration)

E2 = anti C-H / C-X elimination (forms pi bonds)

Na

S R

alkyl sulfide(thiolate)(make)

Na

C NN

O

Ophthalimidate (an imidate)

C C H

Na CH2

OLi

ketone enolates

LiNa

cyanideacetylides

ester enolates

H2C

O

O

(make)

(make)(make)(buy)

(make)

Na

NN

N

(buy)azide

BH

H

H

H

Na

AlH

H

H

H

Li

sodiumborohydride

lithiumaluminium

hydride

S

S

H

Li

dithiane anion(make)

BD

D

D

D

Na

sodiumborodeuteride

AlD

D

D

D

Li

lithiumaluminiumdeuteride

CH2

Li

n-butyl lithium (very strong baseor nucleophile, useanytime)

Na Cl

Br

INa

Na

good nucleophileswith tosylates and

in strong acid(buy) (buy) (buy) (buy) (buy) (buy)

S

Ph

Ph

CH2

sulfurylids

(MgBr)

Grignard reagents(very strong basesand nucleophiles)

R Li

alkyl lithium (very strong basesand nucleophiles)

R R2Cu

organocuprates(good carbonnucleophiles)

Li

Na HN

lithiumdiisopropylamide(very strong base)

sodium hydride (very strong base)

Na

sodium amide (very strong base)

SCl

O

O= Ts-Cl (tosyl chloride)makes ROH into tosylates

N

pyridine = proton sponge

KH

potassium hydride (very strong base)

R = C or HNR2

= py

BrCu

cuprousbromide

Li

(buy) (buy) (buy)

(make)

(make) (make) (make) (buy)

(buy)

Important additions for us.

SPh Ph

Ph = phenylP

Ph Ph

diphenylsulfide,used with C=O

to make epoxides

triphenylphosphineused with C=O to make alkenes

Ph

CrO3 / pyridine

pyridinium chlorochromatePCC

oxidizing E2 reaction

CrO3 / H2O

Jones reagentoxidizing E2 reaction

(buy)(buy)



These two reactions look similar, but there are important differences.

Br O

majorminorH

O

HO

H

H

Br

Br

HO

H

O

majorminor

H

Br

HO

H

Rate = kSN2[HO ]1[RBr]1 Rate = kE2[HO ]1[RBr]1

Rate = kE1[RBr]1Rate = kSN1[RBr]1

SN2 versus E2 overview (essential features) Example: 1o RX, requires strong nucleophile/base, SN2 > E2, exceptions: potassium t-butoxide or sodium amide.

CH

CH3H

C Br

HH

1-bromopropane

Nu

CH

CH3H

C Br

HH

1-bromopropane

O

C

CNu

HH

H

CH3

H

E2 > SN2

(when t-butoxide) C

C

H

H

H3C

H

alkene

E2

Nu B=

strong = anything with negative charge,and neutral sulfur, phosphorous or nitrogen

SN2 > E2

(unless t-butoxide)

C

H3C

H3C

H3C

(also R2N = E2)

SN2 alwaysbackside attack

E2 alwaysanti C-H C-X

SN2 reactions are the most important reactions – always backside attack at C-X carbon

XH3CH2CR X

HCR X

R

CR X

R

R

CH

H2CH2C X

H2C X

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

(1o, 2o, 3o versions)

C C X

C-beta carbon (0-3 of these)

C-alpha carbon (only 1 of these)

benzylic(1o, 2o, 3o versions)



Relative rates of SN2 reactions - steric hindrance at the C carbon slows down the rate of SN2 reactions.

XC

H

H

Hk 30

methyl (unique)

XC

H

H3C

Hk 1

ethyl (primary)reference compound

XC

CH3

H3C

Hk 0.025

isopropyl (secondary)

XC

CH3

H3C

CH3

k 0t-butyl (tertiary)

(very low)140

C

H

HH

X

methyl 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.

C

H

RH

X

primary RX

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

C

R

RH

X

secondary RX

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

tertiary RX

C

R

RR

X = C

C

X

C

H

HH

H

HH

CH

H

H

Nu 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.

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

All of these are primary R-X structures at C, but substituted differently at C.

H2CC

H

H

Hk 1ethyl

reference compound

H2CC

H

H3C

Hk 0.4propyl

H2CC

CH3

H3C

Hk 0.03

2-methylpropyl

H2CC

CH3

H3C

CH3

k 0.00001 02,2-dimethylpropyl

(1o neopentyl)

X X XX

C

C

X

H

C

CH3CH3

HH

H

H

Nu

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

C

C

X

H

H

CH3CH3

HNu

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



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

GG - 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

PERate = 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

Problem 1 - 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 for how the reaction actually works.

HH3C

H Br

C Br

H

H3CCH3CH2

ICH3C

O

O

a b

Problem 2 - 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.)

krelative (SN2)

Br Br Br Br Br BrBr

1.0 0.00001 0.008 1.6 0.01 0.97 0.22

Problem 3 - 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 chair conformation would have the leaving group in a more reactive position (axial or equatorial)? Is this part of the difficulty (which conformation is preferred)?



H H

H

HH

H

H

H

H

HH

H

H

H

H

H

H

H

H

H

H

H

These two chair conformations interconvert in a very fast equilibrium.

X

X

Problem 4 - 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

H

HH

HH

b. c.a. relative rates 250/1

OC

H

H

H

methoxidet-butoxide

Problem 5 – Write out the expected SN2 product for each possible combination (4x8=32 possibilities).

AlD

D

D

D Li

Nu

X

Nu

Nu

Nu =H

O

H3CO

O

O NC

HS

H3CS

NN

N

?

?

X = Cl , Br , I ...a good leaving group

1 3 4 6 7

X

X

X

?

?

Nu

A

B

C

D

2 58



Problem 6 – Using R-Br compounds from page 2 and reagents from page 3 to propose starting materials to make each of the following compounds. One example is provided. TM-1 is an E2 product (see page 15), all the others are SN2 products.

NN

N

?

Br

problem solution

NaN3

N3

2-azidopropane

TM = target molecule

N

O

O

O

TM-1 TM-2 TM-3

OR

TM-4 TM-5 TM-6 O

SPh Ph

CH3C

O

TM-7 TM-8 TM-9 TM-10 TM-11

P

PhPh

Ph

Br

N

Br

O

SH

TM-12 TM-13 TM-14 TM-15

HO

O

O

SR

D

TM-16 TM-17

S

S

H

Acid/base reactions important to our course (some reagents have to be made, often by acid/base reactions) and subsequent reactions (mostly SN2) Make alkoxides and use as nucleophiles only at Me-X and 1o RCH2-X in SN2 reactions.

alkoxides are good nucleophiles at methyl, primary, allyl and benzyl RX, mostly E2 at secondary and only E2 tertiary RX, they are also used as moderately strong bases

OR

Na

alkoxides

OR H

alcohols Na

H

Br RO

SN2

ethers

Keq =Ka1

Ka2

10-16

10-37= = 10+21

HH

acid/base



Make carboxylates and use as nucleophiles at Me-X, 1o RCH2-X and 2o R2CH-X in SN2 reactions.

O

O

Na

ethanoate(acetate)

O H

Na

O

Oethanoic acid(acetic acid)

HBr

O

Oesters

SN2

OH H

Keq =Ka1

Ka2

10-5

10-16= = 10+11

carboxylates are good nucleophiles at methyl, primary, secondary, allyl and benzyl RX, making esters, only E2 at tertiary RX

acid/base

SN2 with acetate produces esters, then acyl substitution with hydroxide produces the alcohol, if desired. A new type of substitution reaction.

O

O

O H

Na OO

O H OH

OO O

O

OH

Me, 1o, 2o alcoholsSN2 > E2Look at similarities to imide hydrolysis, just below.

NaNa

2

acid/base

Problem 7 – Show the acyl substitution mechanism for each functional group with hydroxide.

Na

O HC

R

O

X

X = possible leaving group

CR

O

X

OH

tetrahedralintermediate

CR

O

OH

X

CR

O

OXH

CR

O

Cl

Functional Groups to use

CR

O

OC

R

O

O

CR

O

NC

O

RR

R

Racid chloride anhydride ester 3o amide

acid/base



Make imidates and use as nucleophile at RX centers to convert to primary amines. 1. Make alkyl imides 2. Hydrolyze in base to make primary amines (acyl substitution) 3. Workup) = Gabriel amine synthesis; duplicates azide amine synthesis (1. SN2 with NaN3 2. SN2 with LiAlH4 at nitrogen 3. workup)

N

O

O

HNa

O HN

O

O

Na SN2rxns

N

O

Oimideimidate

alkyl imide

Keq =Ka1

Ka2

10-8

10-16= = 10+8

Na

O H

N

O

O OH

N

O

O

OH

N

O

O

O

H

acyl substitution #1reaction leads toa primary amine

O H

N

O

O

O

H

O

H

N

O

OH

O

OH

O

O

O

O

N

H

H

primary amine(also made by1. NaN3

2. LiAlH4

3. workup)

throw away

resonance stabilized makesimidate less basic and a betterbehaved nucleophile

1

acid/baserxn

2

3

Look at similarities with ester hydrolysis, just above.

Br

acyl substitution #2

acid/baserxn

acid/base

Alternative azide strategy to make primary amines (SN2 and acid/base reactions)

AlH

H

H

HLi

NN

N

Br

N

N

N

resonanceN

N

N

alkyl azide

N

N

N

N

N

N

H

gas

OH H

H

2. workup

NH

1o amine

H

step 1 - make alkyl azide

step 2 - make primary amine

SN2

SN2

acid/base



Make potassium t-butoxide and use as sterically large, strong base at 1o RCH2-X, 2o R2CH-X and 3o R3C-X in E2 reactions.

OKK

OH

t-butyl alcohol potassium

t-butoxide

H BrH

R E2

R

alkenes

HH

Keq =Ka1

Ka2

10-19

10-37= = 10+18

potassium t-butoxide, sterically bulky base that mostly does E2 reactions with RX compounds (except SN2 with CH3-X).

acid/base

Make thiols using NaSH as the nucleophile at Me-X, 1o RCH2-X and 2o R2CH-X in SN2 reactions.

SH

Na

hydrogen sulf ide

Br HS

thiolSN2

Make thiolates and use as nucleophiles at Me-X, 1o RCH2-X and 2o R2CH-X in SN2 reactions.

acid/base

SR

Na

alkylthiolatesNa

SR H

thiols

O HBr

SN2R

S

sulfides

Keq =Ka1

Ka2

10-9

10-16= = 10+7

Synthesis of lithium dithiane anion (acid/base) 2. SN2 with RX 3. Hydrolyze to make aldehydes or ketones)

A good nucleophile that can be made into aldehydes and ketones

S

S

dithiane anion

LiS

S

dithiane

aldehydesand

ketones(later)

Br

SN2

H2CH

HLi

S

S

H

Keq =Ka1

Ka2

10-50

10-35= = 10+15

acid/base

Make terminal acetylides and use as nucleophiles only at Me-X and 1o RCH2-X in SN2 reactions.

CCR

Na

terminalacetylides

CCR

terminalalkynes

H

NaNR2

H3CBr

SN2

R

CH3alkynes

Keq =Ka1

Ka2

10-25

10-37= = 10+12

terminal acetylides are good nucleophiles at methyl, primary, allyl and benzyl RX, butmostly E2 at secondary and only E2 tertiary RX

acid/base



Zipper reaction – moves triple bond to end of linear chain to form the most stable anionic charge, further chemistry is possible. This reaction is almost identical to tautomers, without any heteroatoms.

Synthesis of lithium diisopropyl amide, LDA, sterically bulky, very strong base used to remove Cα-H proton of carbonyl groups. (acid / base reaction) to make carbonyl enolates (next).

Keq =Ka1

Ka2

10-37

10-50= = 10+13

CH2 Li

N

H

NLi

Think - sterically bulky, very basicthat goes after weakly acidic protons.

LDA = lithiumdiisopropyl amide

given

given

acid/base

React ketone enolate (nucleophile) with R-Br electrophile (Me, 1o and 2o RX compounds)

H2C

O Li

ketone enolates(resonance stabilized)

React ketone enolate with RX compounds (methyl, 1o and 2o RX)

Br

O

1o RX key bondR

S Larger ketone made from smaller ketone.

SN2reactions

-78oC



React ester enolate (nucleophile) with R-Br electrophile (Me, 1o and 2o RX compounds)

H2C

O

O

Li

ester enolates(resonance stabilized)

React ester enolate with RX compounds (methyl, 1o and 2o RX)

Br

O

O

1o RX key bondR

S Larger ester made from smaller ester.

R RSN2reactions

-78oC

diphenylmethylsulfoniumbromide salt

S

Ph

Ph

CH2

sulfur ylidto make epoxides

S

Ph

Ph

CH2make epoxidesfrom aldehydes

and ketones

H2C

Li

Keq =Ka1

Ka2

10-50

10-35= = 10+15

H

S

Ph

Ph

H3C

Br

diphenylsulf ide

SN2acid/base



Clarification of “Hydride” electron pair donors: In our course sodium hydride and potassium hydride are always “basic,” and lithium aluminum hydride and sodium borohydride are always nucleophilic hydride.

Basic hydride

In our course, sodium hydride (NaH) and potassium hydride (KH) are always basic (= electron pair donation by hydride to a proton), never a nucleophile. The conjugate acid of hydride is hydrogen gas (with a pKa = 37, H2 can hardly be considered an acid).

Sodium hydride and potassium hydride (KH)

Na

Aldrich, 2012$39 / 100 grams

60% oil dispersionMW = 24.0 g/mol

H KH


30% oil dispersionMW = 40.1 g/mol

Problem 8 – Write an arrow pushing mechanism for each of the following reactions.

O

H

O

H

H

Na

K

H

pKa

ROH 16-19

H-H 37pKa = 17

pKa = 19

Nucleophilic hydride – Formation of C-H bonds using nucleophilic lithium aluminum hydride and sodium borohydride.

In our course, sodium borohydride (NaBH4) and lithium aluminum hydride (LiAlH4 = LAH) are inorganic salts containing nucleophilic hydride, very unusual nucleophiles. Both reagents supply nucleophilic hydride that can displace X in SN2-like reactions with RX compounds. Sodium borohydride and lithium aluminum hydride are used in many other reactions. They also react with carbonyl compounds (C=O) and epoxides (both to be discussed more later). We will often use the deuterium version of borohydride and aluminum hydride so we can identify where a reaction occurred. In reality, this is not very common because of the expense. But, for purposes of probing your understanding of SN2 reactions, using them shows if you understand what is happening. The deuterium isotope of hydrogen reacts similarly to the proton isotope, but there are experimental methods which allow us to observe where a reaction took place (e.g. NMR).

Sodium borohydride and lithium aluminum hydride (LAH) - 4 equivalents of hydride per anion

B H

H

HH Na Li

Al H

H

HH

B D

D

DD Na

Aldrich, 2012$312 / 100 gramsMW = 37.8 g/mol

Aldrich, 2012$120 / 100 gramsMW = 38 g/mol


MW = 41.8 g/mol


MW = 42 g/mol

Li

Al D

D

DD

The deuteride salts are way more expensive.

The hydride salts are way more common.

All these reagents will undergo SN2 reactions at RX centers. Because there is a greater difference in electronegativity between aluminum and hydrogen than boron and hydrogen, LAH is more reactive than sodium borohydride. This won’t be important for us to know until we study carbonyl reactions.



Problem 9 – Write an arrow pushing mechanism for the following reaction.

Li

Al D

D

DD

C

Cl

CH3

H

+ ?

Na

B H

H

HH

C

Cl

DCH3

+ ?

a. b.

Problem 10 – It is hard to tell where the hydride was introduced since there are usually so many other hydrogens in organic molecules. Where could have X have been in the reactant molecule? There are no obvious clues. Which position(s) for X would likely be more reactive with the hydride reagent? Could we tell where X was if we used LiAlD4?

LAH

Where was "X"?

H2C

CH2

H2C

CH3

E2 Reactions Compete with SN2 Reactions

E2 reactions also occur at the backside relative to the leaving group, but at C-H instead of Cα-X. The C-H proton has to be anti to the Cα-X to allow for parallel overlap of the 2p orbitals that form the new pi bond. This allows elimination to occur in a concerted manner. The syn conformation also has parallel overlap of 2p orbitals, but is present in less than 1% due to an eclipsed conformation (staggered conformations > 99%).

E2 Potential Energy vs. Progress of Reaction Diagram (= concerted, energy picture looks very similar to SN2)

transition state

reactants

products

PEpotential energy

POR = progress of reaction - shows how PE changes as reaction proceeds

higher(lessstable)

lower(morestable)

Ea = -2.3RT log(kE2)

G = -2.3RT log(Keq)

OH

C

C

R1R2

BrC

C

R1 R2

Transition state - requires parallel overlapof the two 2p orbitals forming the pi bond. This is easiest when C-H is anti to C-X.

Br

HO

R3R4

H

R3 R4 H O

H

Br

C

C

R1 R2

R3 R4

If stereochemical priority is R1 > R2 and R3 > R4 then this would be Z configuration, which is fixed by the requirement for anti C-H / C-X elimination.

E2 mechanism depends on steric factors and basicity of the electron pair donor. More steric hindrance and more basic favors E2 over SN2

= negative(exergonic)

Keeping Track of the C Hydrogens in E2 reactions



At least one C 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 C positions. In E2 reactions there can be anywhere from one to three C carbons, and each C carbon can have zero to three hydrogen atoms.

C

H

H H

X

C

H

C H

X

C

C

C H

X

C

C

C C

X

zero C positionsunique methyl,

only SN2 possible

one C position1o RX, usually SN2 > E2(except with t-butoxide)

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

three C positions(3o RX, only E2)

0-3 H at C0-6 H at C 0-9 H at C

With proper rotations, each C-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. Using one of these two perspectives may help your analysis of E2 reactions Let’s consider a moderately complicated example (next problem).

B

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

C

C

H

X

CH

C X?

B

?

horizontal perspectiveC-H / C-Xvertical perspective

C-H / C-X

Problem 11 - How many total hydrogen atoms are on C carbons in the given RX compound? How many different types of hydrogen atoms are on C carbons (a little tricky)? How many different products are possible? Hint - Be careful of the simple CH2. The two hydrogen atoms appear equivalent, but E/Z (cis/trans) possibilities are often present. (See below for relative expected amounts of the E2 products.)

C

CH2

CH3

CH3

X

CH

H3C

CH2H3C

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

OR OHR

Na

C

C

CH3

C

X

C1

H3C

CH2H3C

HHH

H

H

H

There are two chiral centers, C and C1, that make this structure more complicated than we are treating it. Possible stereoisomers are RR, RS, SR and SS.

?

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 56

7

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 (E2 and E1). They tend to be the major alkene product,though typically a little of every alkene product possible is obtained.

"unsubstituted" "mono" "di-cis" "di-geminal" "di-trans" "tri" "tetra"

< < < <

Possible explanations for greater stability with greater substitution of the pi bond

A fairly simple-minded explanation (the one we will use) 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"...

Problem 12 - Reconsider the elimination products expected in problem 1 and identify the ones that you now expect to be the major and minor products. How would an absolute configuration of Cα as R, compare to Cα as S? What about C1 as R versus S?

OR OHR

NaC

C2

CH3

C3

Br

C1

H3C

CH2H3C

HHH

H

H

H

Compare(3S,4R)-3-bromo-3,4-dimethylheptane

(3S,4S)-3-bromo-3,4-dimethylheptane

?

C

C

H

Br

C3

HH3C

H

C1

H3CCH2CH3

one of your possibilities(which one?)

* = chiral center

* *

H

H

H

(3R,4R)-3-bromo-3,4-dimethylheptane

(3R,4S)-3-bromo-3,4-dimethylheptane



Problem 13 -Provide an explanation for any unexpected deviations from our general rule for alkene stabilities above. A more negative potential energy is more stable.

Hof = -18.7 kcal/mole Ho

f = -22.7 kcal/mole Hof = -19.97 kcal/mole

Problem 14 – Order the stabilities of the alkynes below (1 = most stable). Provide a possible explanation. H C C H R C C H R C C R

"R" = a simple alkyl group

Problem 15 - 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

our rules

SN2 > E2

SN2 > E2

SN2 > E2(wrong

prediction)

only E2

Problem 16 – One of the following reactions produces over 90% SN2 product and one of them produces about 85% E2 product in contrast to our general rules (ambiguity is organic chemistry’s middle name). Match these results with the correct reaction and explain why they are different.

ClO

pKa = 16

ClO

pKa = 19...versus...



Problem 17 - 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 C and C 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(RCO2H) = 5 pKa(ROH) = 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(RCCH) = 25 pKa(NCH) = 9



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. Their chemistries are similar, but we can tell them apart.)

(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

H

H

CC

H

H

CC

H

H

CC

Problem 19 - 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.

B

C C

R3

R4

R2

R1

H

X

anti E2reaction

staggered

reactant conformation 1

rotate about center bond

syn E2reaction eclipsed C C

R3

R4

X

reactant conformation 2

alkene stereochemistry?

alkene stereochemistry?

Newmanprojections

B

< 1% > 99%



Alkyne synthesis (via two E2 reactions with RBr2 and NaNR2, two times)

Sample alkynes using double E2 reactions of 1. RBr2 + NaNR2 2. Workup (neutralize mixture)

Possible steps in mechanism (E2 twice then 2. acid/base or 2. RX electrophile)

BrBr

HN

R R

Na

Br

H

H

NR R

H

NR R

Na

Na2. workup

H

H2C

R

X

H2C

R

HO

H

H

2.

a

b

a

b

SN2

2 eqs. Br2h

(314 reaction,see page 58-60

this topic)

Make starting alkynes using two leaving groups and very basic R2N

-- starting from an alkane.

Na sodium amide(very strong base)

Br Br

Br

Brand

Br2 h

free radical substitution

(2 equivalents)(2 equivalents)

NR2

E2 twice

1.

2. workup

Use steric bulk and/or extreme basicity to drive E2 > SN2.

Br2 h


(2 equivalents)Br Br


(2 equivalents)

NR2

E2 twice

1.

2. workup

Br2 h


(2 equivalents)

Br Br


(2 equivalents)

NR2

E2 twice

1.

2. workup



Sample alkynes using SN2 reactions of 1. Terminal alkyne + NaNR2 2. SN2 only at methyl or primary R-X (can do this twice starting with ethyne)

newbond

newbond

newbond

newbond

newbond

newbond

newbond

(1 equivalent)


1.

2. BrH3C

acid / base

SN2 > E2

NR2newbond

(1 equivalent)


1.

2.

acid / base

SN2 > E2Br

NR2

Terminal acetylides are OK nucleophiles at methyl and 1o RX, but E2 > SN2 at 2o RX and only E2 at 3o RX.

newbond

1. RNH Na RNH2

1 equivalent

NaBr

2.

(4C = 2C + 2C)terminal alkyne

SN2

Br

2.

Na

1. RNH Na RNH2

1 equivalent

(5C = 3C + 2C)internal alkyne

SN2

1. RNH Na RNH2

1 equivalent

NaBr

2.


SN2

Br

2.

Na

1. RNH Na RNH2

1 equivalent

(5C = 3C + 2C)

internal alkyne

(zipper reaction)

1. excess NaNRH2. workup (neutralize)

SN2acid/base

like tautomers



1. RNH Na RNH2

1 equivalent

Na

2.

2.

Na


1. RNH Na RNH2

1 equivalent

(6C = 3C + 3C)

internal alkyne

(zipper reaction)

1. excess NaNRH2. workup (neutralize)

Br

Br

like tautomers

SN2

SN2

acid/base

acid/base

terminal acetylides (SN2 only at methyl RX and 1o RX) larger alkynes

BrH3CCCH3C

Na SN2CCH3C CH3

nucleophile = ?electrophile = ?

Na

Br

BrH2CCCH3C

Na SN2CCH3C CH2

H3C CH3nucleophile = ?electrophile = ?

Na

Br

BrH2CCCH3C

Na SN2CCH3C CH2

CH2H2CH3C CH3nucleophile = ?

electrophile = ?

Na

Br



Problem 20 – What are the possible products of the following reactions? What is the major product(s) and what is the minor product(s)? There are 55 possible combinations.

AlD

D

D

D

BD

D

D

D LiNa

Nu

X

X

Nu

Nu

Nu =H

O

RO C

OH3C

H3CCH3

O

O NC C

C

HH

SR

S

NN

N

?

?

?

X = Cl , Br , I ...a good leaving group

1 23 4 5 6

7

X

X

X

?

?

Nu

Nu

A

B

C

D

E

OR OR

8 OR9 O

enolatesS

S10

S

Ph

Ph

CH2

sulfurylids

11

N

O

Oimidate



C

H

HH

XC

D

HT

X

easier harder

Reaction Templates - sideways and vertical perspectives (either one will work)

C

C

HH

X

methyl (Me)

primary (1o)

secondary (2o)

tertiary (3o)

H

H

H

C

C

HX

H

H

H

C

C

HD

X

H

R

D

CHH

H

C

C

HX

H

R1

D

CHD

R2

C

C

X

H

H

H

CHH

H

CH

H

C

C

X

H

R1

D

CHD

R2

CH

D

priority R1 > R2 > R3

Nu:B

iotopes of hydrogenH = protium (proton)D = deuteriumT = tritium

H-Nu: / H-B:

Nu:B

O

O

O

O

=

OH

H-Nu :H-B :

OH

OH

O

=

OH H

C

HH

H

XC

DH

T

X

C

CH

H

X

H

HH

C

CH

D

X

H

RD

C

CH

X

H

HH

C

H

H

H C

C

H

X

H

R1D

C

H

D

R2

C

C

X

H

R1D

C

H

D

R2

C

H

DR3

C

C

X

H

HH

C

H

H

H

C

H

HH

Strong (SN2 / E2)

Weak (SN1 / E1)

SN2/E2 (Nu: / B: ) always backside for SN2 and usually anti C-H/C-X attack for E2

SN1/E1 (H-Nu: / H-B:) - attack from either face of R+ for both reactions (SN1 and E1)

side view vertical view

conjugate acid pKa = 16




strong

weak

Nu:B

H-Nu: / H-B:

strong

weak

Nu:B

H-Nu: / H-B:

strong

Nu:B

H-Nu: / H-B:

strong

Nu:B

H-Nu: / H-B:

strong

weak weak

weak

Nu:B

H-Nu: / H-B:

strong

weak

Nu:B

H-Nu: / H-B:

strong

weak

Nu:B

H-Nu: / H-B:

strong

weak

H R3

Example: 3-bromo-2-methoxyhexane (R,R), (S,S), (R,S), (S,R)

12

34

56 C

C

HBr

Ha

Hb

C2H6

CH

CH3

C

C

H

Br

C

H

H3C

H3CO

Ha

C2H6Hb

template side view

vertical view

(2R,3R)(2S,3R)

OCH3



12

34

Examples - you can use the vertical views or side views presented above. (Never ending supply of problems.)

(R,R)(S,S)(R,S)(S,R)

2-bromo-3-deuteriobutane 2-bromo-3-methylbutane

(R or S)

12

34

5(R,R)(S,S)(R,S)(S,R)

2-bromo-3-deuteriopentane 2-bromo-3-methylpentane


3-bromo-2-methylpentane(R,R)(S,S)(R,S)(S,R)

3-bromo-2-deuteriopentane

(R or S)

12

34

56


2-bromo-3-deuteriohexane 2-bromo-3-methylhexane


3-bromo-2-methylhexane


3-bromo-2-deuteriohexane

(R or S)

12

34

5

6(R,R)(S,S)(R,S)(S,R)

3-bromo-4-deuteriohexane 3-bromo-4-methylhexane


12

34

56

7(R,R)(S,S)(R,S)(S,R)

2-bromo-3-deuterioheptane 2-bromo-3-methylheptane


3-bromo-2-methylheptane


3-bromo-2-deuterioheptane

(R or S)

12

34

5

6

7 (R,R)(S,S)(R,S)(S,R)

3-bromo-4-deuterioheptane 3-bromo-4-methylheptane


4-bromo-3-methylheptane


4-bromo-3-deuterioheptane


Na

O HN

O

O

Na

S H

phthalimidate (an imidate)

C C

Na

O R

NaK

ONa

O

O Na

hydroxide acetylidesalkoxides potassium t-butoxide carboxylates (acetate)

hydrogen sulfide (thiolate)

(buy) (make) (make)(make)

(make)(buy)

Nu

B

=

SN2 / E2Na

NN

N

(buy)azide

BH

H

H

H

Na

AlH

H

H

H

Li

sodiumborohydride

lithiumaluminium

hydride

Na

C N CH2

OLi

ketone enolates

Li

cyanide ester enolates

H2C

O

O

(make)(buy)

S

S

H

Li

(make)

dithiane anion(make)

SN2/E2 concerted reactions (one step)SN2 = always backside attack (inversion of configuration)E2 = anti Cb-H / Ca-X elimination (forms pi bonds)

BD4AlD4

Nu

B=

H

H

HO

H RO

H

O

OH

water alcoholscarboxylic acids

SN1/E1 form carbocation (R+) in first step,R+ has three common choices1. rearrange to similar or more stable R+

2. add nucleophile (top/bottom)3. lose any beta proton (top/bottom) (forms pi bonds)

SN1 / E1



Strong base/nucleophile conditions (negative charge in our course) leads to SN2 and/or E2 reactions. Simple mechanisms.

T

C Br

DH

reaction conditions

T

CNu

DH

Nu

SN2Br

CH

DCH3

C Br

DH

reaction conditions

C

CNu

DH

H

D

CH3

B

SN2

CH

DCH3

C Br

DH

reaction conditions

Nu

E2

Br

Br

C

C

D

H

CH3

D

C

C

CH3

H

H

D

b = anti D

a = Z

b = ECD

CH3H

C Br

DH

reaction conditions

E2

methyl pattern primary pattern (SN2 > E2, except t-butoxide and R2N )

There is no E2 at methyl RX

S R 1S,2R 1R,2R

1S,2R

1S,2R

a = anti H

b

a

B

E2 > SN2 when t-butoxide and R2N

B

CH

DCH3

C Br

C

CH2

H

CH3

CH2CH3

H

CH

DCH3

C Br

C

HH

H

H

reaction conditions

C

CNu

C

H

H

D

CH3

H

H

H

Nu

SN2

CH

DCH3

C Br

C

HH

H

H

reaction conditions

E2

Br

Br

C

C

D

H

CH3

CH3

C

C

CH3

H

H

CH3

C

C

H

H

CHD

H

CH3

a = anti Hb = anti D

c

There is no SN2 at tertiary RX

reaction conditions

E2 Br

C

C

H3C

H3C

CHDCH3

CH2CH3

a = anti Hb = anti D

c

secondary patterntertiary pattern

d

C

C

HC

CHDCH3

H

H

C2CH3

CH3

C

C

D

H3C

CH3

CH

CH2CH3

CH3

C

C

H3C

H3C

H

CH

CH2CH3

CH3

2S,3R

b = E

2R,3R

c = neither

a = Z

a = Z

b = E c = neither d = Z2S,3R

2R,3R,4S

B

(SN2 > E2, except HO , RO , R2N and RCC )



Example Mechanisms shown below.

BrH

D

12

34

56

3R-bromo-2S-deuteriohexane

Na

O H

H secondary RX (2o)

C

C

HBr

H

CH3

D

CHa

Hb

CH2CH3

OHC

C

HHO

C

HCH3

D

Ha CH2CH3

Hb

a

bc

(2S,3S)

C

C

H3C

H

D

CH2CH2CH3

C

C

H

H3CH2C

CHDCH3

Hb

a, b, c

C

C

HBr

D

H

CH3

CHb

CH2CH3

Ha

OH

deC

C

H

H

CH3

CH2CH2CH3

C

C

H

Ha

CHDCH3

CH2CH3

d, e

sigma bond rotations

One SN2 product and four E2 products.

b

a

c

d

e

only one SN2 productfour possible E2 products

(2S,3R)(2E, with "D")

(3E, lost Ha)

(3Z, lost Hb)(2Z, without "D")

1. strong Nu: /B:2. 2o R-X



Important acid/base reactions are shown on pages 15-19. When necessary for a preliminary step of a reaction, you should consult those pages. Some of the strong base/nucleophiles used in our course are listed below along with the usual preferred reactions according to our simplistic rules. Cuprates will be discussed later. Missing are enolates, imidates, magnesium and lithium organometallics and a few others.



cuprates

H3CBr

R

H2C

Br RCH

Br RC

Br

R RR

methyl RBr primary RBr secondary RBr tertiary RBr

RCu

R

Li only SN2 SN2 > E2 SN2 > E2 only E2

R-Brcompounds

strongbase/nucleophiles

too sterically hinderedno C-H

O

CH2

Na

enolates

only SN2 SN2 > E2 SN2 > E2 only E2too sterically hinderedno C-H

S

Ph

Ph

makesulfurylids

S

Ph

Ph

CH2

sulfur ylids

P

Ph

Ph

makephosphorous

ylids

Ph

P

Phphosphorous

ylids

Ph

react with aldehydes

and ketones

N

O

Oimidate


Grignard reagents

R

Li

(MgBr)

Rorganolithium

reagents




and ketones


and ketones


and ketones


and ketones


and ketones


and ketones


and ketones


and ketones


and ketones


and ketones


and ketones


and ketones


and ketones


and ketones


and ketonesCH2

(covered in later topic)




Ph(covered in later topic)



SN1 and E1 Competition – Multistep Reactions Arising From Carbocation Chemistry

SN2/E2 and SN1/E1 reactions look very similar overall, but there are some key differences. In SN2/E2 reactions the nucleophile/base is a strong electron pair donor (negative charge in our course), which is why they participate in the slow step of the reaction and force a concerted, one-step reaction. In SN1/E1 reactions the nucleophile/base is a weak electron pair donor (stable, neutral molecules: H2O, ROH and RCO2H for us) and that’s why they don’t participate in the slow step of the reaction, which is ionization of the Cα-X bond. This leads to differences in reaction mechanisms, which show up in the rate law expression (kinetics is bimolecular = 2 or unimolecular = 1). The rate of “1” reactions only depends on how fast RX forms a carbocation. The unimolecular reactions lead to possible side reactions from carbocation rearrangements. You need to carefully look at the reaction conditions to decide what mechanisms are possible. You cut you choices in half when you decide that the electron pair donor is strong (negatively charged = SN2/E2) or weak (neutral = SN1/E1).

X

X = Cl, Br, I

+H3CO

H

H3CO

weak nucleophil/base

(neutral in our course)

SN1/E1reactions

SN2/E2reactionsstrong

nucleophil/base(negatively charged

in our course)

X

X = Cl, Br, I

O

SN1 product(major)

SN2 product(minor)

E1 product(minor)

E2 product(major)

Rate = kSN2 [RX][CH3O: ] Rate = kE2 [RX][CH3O: ]

Rate = kSN1 [RX] Rate = kE1 [RX]

CH3

OCH3

SN1 and E1 reactions begin the same way, with ionization of the Cα-X bond to form a carbocation. Ion formation is energetically expensive. Protic solvents help by being able to solvate both cations and anions. Carbocations also need some help from inductive and/or resonance effects. If it can form, the initial carbocation intermediate typically follows one of three common paths: 1. addition of a weak nucleophile at carbon and/or 2. loss of a beta hydrogen to form a pi bond and/or 3. rearrangement to a similar or more stable carbocation. Rearrangement merely delays the ultimate reaction products: addition of a nucleophile or loss of a beta hydrogen atom to forma pi bond.

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

CC

X

R2R1

H

R3

R4

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

CC

H

R3

R4R2

R1

X

solvated carbocation

has to be 2o or 3o in our course

(3 reaction choices)

addition of a weak nucleophile (SN1 type reaction)

Loss of a beta hydrogen atom, (Cb-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

Bweak base

H Nu H weak nucleophile

=

NuH

BH

?

next page

R

SN1 and E1 reactions are multistep reactions. (H-Nu: / H-B: = H2O, ROH, RCO2H in our course)



E1 approach comes from parallel C-H with either lobe of 2p orbital,anti or syn is possible

Bweak base

E1 products(pi bond forms)

E or Z is possible.

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+

H

CC

H

R3

R4R2

R1

C

Nu

C

H

R3

R4R1

R2

CC

H

R3

R4 Nu

R1

R2

H

H

Nu H

Nu H

C C

R4

R1

R2

CC

H

R3

R4 Nu

R1

R2

C

Nu

C

H

R3

R4R1

R2

solvated carbocation

SN1

R3

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=

redrawn

NuH H

BH H

SN1 approach is from either the top face or

the bottom face. Racemization is possible

(if observable).

Nu H

weak nucleophile

CC

H

R3

R4R2

R1

solvated carbocation(from previous page)

SN1

E1

Bweak base

E1 products(pi bond forms)

E or Z is possible.

X stableleaving group

H

CCR2

R1

C C

R3

R1

R2

R4BH H

E1

rotate 180o around C-C bond,can lose any beta H, top or bottom

H

R4

R3

:B-H = :Nu-H = H2O or ROH or RCO2H

same C-H shown on both faces(top and bottom)

C-H bottom face

C-H top face

"Z" if R1 > R2 and R3 > R4

"E" if R1 > R2 and R3 > R4

C could be R or S

reactants

E1 products

(PE)

Progress of reaction (POR)

(10)

-Ea (E)

2.3RT

Rate SN1 = kSN1[RBr]1

Rate E1 = kE1[RBr]1 =(10)

-Ea (SN)

2.3RT

(10)

- Ea

2.3RT

Say Ea(SN) = 13 kcal/mol and Ea(E) = 14.6 kcal/mol

=

(10)

- Ea

2.3RT= (10)

-(13 - 14.6)

1.3= (10)

1.6

1.3= = 101.2 =Rate SN1Rate E1

401SN1 products

Ea

E1 rate (slower)

SN1 rate (faster)Ea



R-X Substitution Pattern and rates of SN1 reactions

The order of stability at the electron deficient carbocation carbon is methyl primary secondary tertiary. This is consistent with our understanding of inductive electron donating ability of alkyl groups compared to hydrogen. R groups (alkyl groups) are electron donating (an inductive effect). We observed, previously, that this helps alkene stability and makes it harder to form an anionic conjugate base in acid/base chemistry. A carbocation is extremely electron deficient (the opposite of a carbanion) and is very electronegative. Extra electron donation to a carbocation center proves very helpful. This can occur through an inductive effect or a resonance effect.

Inductive effects are proposed to occur via polarizations of sigma bonds in the organic skeleton, helping (or hurting) a center of reactivity. We can represent these in a carbocation center, simplistically, as shown below. Hyperconjugation is an alternative explanation to rationalize how extra electron density can be donated to the electron deficient carbocation carbon, but we will not use it in our course.

C

X

CHH

H

X

CC

H

HH

H

H

X

Sigma electrons are pulled toward the carbocation carbon. Part of the + is distributed on to the hydrogen atoms, but not typically shown with formal charge.

CHH

H

+ +

+

+

Additional sigma bonds of alkyl substituent(s) allow further polarizations of electrons from more bonds(inductive donating effect), which spreads out + charge through sigma bond polarizations and helpsstabilize the electron deficient carbocation carbon.

CC

H

HH

H

H

+ +

++

+

+

+

Resonance effects also make carbocations more stable. Allylic and benzylic RX compounds are very reactive in SN1/E1 reactions (and SN2 reactions). With resonance, there is actually full “pi electron” donation from an adjacent pi bond, instead of the inductive effect just mentioned above. An adjacent pi bond tends to produce greater stabilization of a carbocation than a single alkyl substituent. Resonance donation from lone pairs of heteroatoms can also be strongly stabilizing for carbocations. We will see such intermediates many times in later chapters.



C C

C H

HH

H

H

C C

C H

HH

H

H

C C

C H

HH

H

H

C C

C H

HH

H

H

resonance from adjacent pi bond

2D

3D

O C

R

RH

resonance from adjacent lone pair

CO

H R

R

CO

H R

R

O C

R

RH

Resonance effects help stabilize carbocations.

2D

3D

C C

C H

HH

X

H H

CO

R

R

A H

strongacid

resonance resonance

resonanceresonance

Gas phase Stabilities as Indicated by Hydride Affinities

Hydride affinity is the energy released when a hydride is added to a carbocation (gas phase reaction). The energy of reaction, H, is very negative because the two reactive species (carbocation and hydride) are very unstable and the product formed is a stable molecule. How much do inductive and resonance effects help a carbocation center? The following gas phase data below show the differences in carbocation stability are enormous. In fact, differences are so large that we will almost never propose methyl or primary carbocation possibilities as reaction pathways in our course. We will consider these two patterns (CH3-X and RCH2-X) as unreactive in SN1 and E1 chemistry, and that should make your life a little bit easier.

Some of these values are esitimated from different sets of data.

We will not proposethese carbocations insolution in this book.

CH3

CH2

CH2

CH2

CH3CH2

methyl & primary

-277

-315

-270

-265

-267

tertiary "C" resonance "X" resonance other / misc.

C

CH3

H3C

CH3

CHH2C CH2

CH2

H

CH C

CH

HC H

-232

-227

-227

-256

-239

-229

-220

-210

CH2H2N

OH CH2 -248

-218

-230

-386

-287

ethynyl carbocation(see problem below)

CH

HC H

phenyl carbocation(see problem below)

-300

vinyl carbocation(see problem below)

secondary

CHH3C

H3C

-249

H -246

Inductive effectsResonance effects

adjacent pi bond adjacent lone pair

CO CH3

Hydride Affinity

H = energy released =

HR

RH

Potential EnergyA larger hydride

affinity means a less stable carbocation.

empty sp orbital

empty 2p orbital

empty sp2

orbital

very unstable



Problem 21 - Explain the differences in stability among the following carbocations (hydride affinities).

CH3CH2

CH2

CH

CH3H3C

CH3

CH3C CH3 H2C

HC

CH2

CH2

HOCH2 H2N

CH2

CO CH3

315

H3C

270reference

249 232 256 239 230248 218

= +45 = -21 = -38 = -14 = -31 = -22 = -52 = -40

All relative energy values in kcal/mole versus a primary carbocation. A positive value is less stable and a negative value is more stable relative to our reference primary carbocation.

Too unstable to propose in our course.

2o R+ 3o R+ 1o allylic R+ 1o benzylic R+

lone pair resonance stabilization of R+1o R+

methyl R+

pi bond resonance stabilization of R+

= 0

A more negative is a more stable carbocation.

Inductive effect stabilization of carbocations (3o > 2o R+)

Problem 22 - Why are vinyl carbocations so difficult to form? (Hint – What is their hybridization?) How does an empty sp2 orbital (phenyl carbocation) or empty sp orbital (ethynyl carbocation) compare to a typical sp2 carbocation carbon with an empty 2p orbital? (An empty 2p orbital is also present on the vinyl carbocation, but the carbon hybridization is sp.)

Problem 23 – The bond energy depends on charge effects in the anions too. Can you explain the differences in bond energies below? (Hint: Where is the charge more delocalized?) We won’t emphasize these differences.

CH3C

CH3

CH3

X

X = Gas Phase B.E.

Cl +157Br +149I +140

The activation energies for ionization in solvents are on the order of 20-30 kcal/mole (SN1 and E1 reactions) It is clear from the difference in the gas phase energies of ionization that the solvent is the most stabilizing factor in ion formation. The magnitudes of these energies are compared in the potential energy diagram below. Because solvent structure is so complex we ignore it, but we do so at our own peril.

Many small solvent/ion interactions make up for a single, large covalent bond (heterolytic cleavage). A typical hydrogen bond is about 5-7 kcal/mole and typical covalent bonds are about 50-100 kcal/mole. In a sense the polar proticsolvent helps to pull the C-X bond apart. The "polarized" protons tug on the "X" end and the lone pairs of the solvent molecules tug on the "C" end. If the carbocation is stable enough, the bond will be broken.

ORH

RO

H

O

R

H

O

R

H

CH3C CH3

CH3

ORH

RO

HO

R

H

OH

R

X

gas phase reactions polar solvent phase reactions

Carbocations are more stable and have smaller energy differences in solution than the gas phase. (But methyl and primary are still too unstable to form in solution and we won't propose them in this book.)

Solvent / ion interactions are the most significant factor (about 130 kcal/mole here).

CH3C

CH3

CH3

Cl

ClC

H3C

H3C

CH3

+160

+20-300

50

100

150

200

PE

CH3C

CH3

CH3

Cl

ClC

H3C

H3C

CH3



Weak Nucleophile/Bases are used in SN1/E1 Reactions

We emphasize the term weak here because if the Nu: were strong (negative charge), the reactions observed would be SN2/E2. Weak, for us, is represented by a neutral solvent molecule with a pair of electrons. For us, this will be a polar solvent molecule such as water (HOH), an alcohol (ROH) or a liquid carboxylic acid (RCO2H). All of these are protic solvents, which are reasonably good at solvating both cations and anions. In every case, there is an extra hydrogen atom on the oxygen atom of a solvent molecule that must be removed in a final acid/base reaction to produce the neutral organic product. This protonated cationic intermediate results from the addition of water (H2O, forms alcohols) or an alcohol (ROH, forms ethers) or a liquid carboxylic acid (RCO2H, forms esters).

XR R HNu NuR H

NuR NuH H

HNu

X X

very reactive carbocation strong Lewis acid = SN1

strong Bronsted acid = E1

weak nucleophile (...or weak base)

(H2O, ROH or RCO2H)

cationic intermediate requires proton transfer to form neutral product

best base in acidic medium

is usually anotehr solvent molecule

substitution product

protonatedsolvent

H-Nu = H2O, forms alcoholsH-Nu = ROH, forms ethersH-Nu = RCO2H, forms esters

Weak electron pair donation in our course.

We will view the attack on an sp2 carbocation as equally accessible from either face (top or bottom). Because carbocations are flat (in our simplistic view), attack is equally accessible from either face (top or bottom). The carbocation carbon, itself, is not chiral since it is sp2 hybridized and only has three atoms attached to it. However, it is prochiral and can become chiral if the addition of a nucleophile brings in a fourth different group. This would lead to enantiomers, if this was the only chiral center. We are assuming that a 50/50 recemic mixture forms (in our course). It is also possible that there may be a stereogenic center somewhere else in the carbocation structure. Top and bottom attack would then lead to the formation of diastereomeric products.



CR1

R2

R3

mirrorplane

achiral carbocation carbon

bottom face attack

top face attack

CR1

R2

R3

achiral carbocation carbon with no other chiral centers in R1, R2 or R3

HNu

top

bottom

CR1

Nu

R2R3

CR1

Nu

R3R2

The new stereogenic center forms a racemic mixture of enantiomers (assuming no other chiral centers exist in the molecule)

(dl) (+/-)

If all three attached groups at a carbocation carbon are different from one another and the attacking nucleophile, then a racemic mixture of enantiomers will form.

bottom face attack

top face attack

CR1

R2

R3

chiral branch in carbocation =

HNu

top

bottom

CR1

Nu

R2R3

CR1

Nu

R3R2

The new stereogenic center forms both R and S absolute configurations. If another chiral center is present that does not change in the reaction,then diastereomers will form, (RR) vs (RS).

If one or more chiral centers were present in the carbocation, the top and bottom attack at the carbocation center would lead to diastereomers formed in unequal amounts.

R/S assumes priorities Nu > R1 > R2 > R3

S

R

Problem 24 – Draw in all of the mechanistic steps in an SN1 reaction of 2R-bromobutane with a. water, b. methanol and c. ethanoic acid. Add in necessary details (3D stereochemistry, curved arrows, lone pairs, formal charge). What are the final products?

Bra.

HO

H

b.

H3CO

H

c.

H3CC

O

O

H

Donate the carbonyl (C=O) electrons to the carbocation.

SN1 products will generally outcompete E1 products, in our course. The only exception for us (presented later) will be when alcohols are mixed with concentrated sulfuric acid at high temperatures to form alkenes (the E1 product).



Keeping Track of the C Hydrogens in E1 Reactions (more possibilities than E2 reactions)

E1 products arise from the same carbocation intermediate formed in SN1 reactions. Just as in E2 reactions, we have to examine each type of C-H. In more complicated R-X molecules there may be several different types of hydrogen atoms on C positions. After all, there can be either two or three C carbons with zero to three hydrogen atoms on each. We will only consider secondary and tertiary RX compounds below, since methyl and primary carbocations do not form (in our course). That still leaves a lot of possibilities.

CC

C

H

X

CC

C

C

X

2o R-X has six C positions 3o R-X has nine C positions

CC

C

H

CC

C

C

XH3CR

H2C

XNo Reactionin our course.

NuH

NuH NuH

methyl primary

HNu: = H2O, ROH, RCO2H (weak, in our course)

E1 Mechanism (unimolecular kinetics) loss of proton from any adjacent Cβ-H position, top or bottom

The high reactivity (low stability) of carbocations forces some very quick choices to try and stabilize the situation. The carbocation needs two electrons to complete its octet (in a hurry!). There are three ways it typically does this. We have studied the two ultimate pathways, SN1 and E1 reactions.

The third possibility, rearrangements, is discussed next. Rearrangements are a temporary solution for an unstable carbocation. Rearrangements transfer the unstable carbocation site to a new position having a similar energy or, better yet, to a site where the positive charge is more stable. If such possibilities exist, this will very likely be one of the observed reaction pathways. However, even with a rearrangement a carbocation will not gain the two needed electrons. The electron deficiency is merely moved to a new position. This process can occur a number of times before a carbocation encounters its ultimate fates, discussed above, SN1 and E1.



SN1 / E1 possibilities –extra complications at Cβ positions, In this problem 2o RX, rearrangements are NOT considered (H2O,ROH,RCO2H)



SN1 / E1 possibilities –extra complications at Cβ positions, 2o RX, rearrangements NOT considered, with deuterium (makes it a little harder)

OR HR

O

OOH H

Examples of weak nucleophile/bases

a b

c

H

water alcohols carboxylic acids

E1 products from left C carbon atom (4 possible alkenes from the lef t C carbon)

H3CC

CC

H2C

CH3

D H

H

D

Br

H

H3CC

CC

H2C

CH3

D H DH

H

Redrawn

(2S,3R,4S) (2S,4S)

(2S,4S)

CCH

C

D

H

H3CH

CH2D

CH3

H2O

CCH

C

H

CH2D

CH3D

H3CH

H2O

same first stepfor SN1/E1

add nuclephile (top/bottom) = SN1

lose -H (top/bottom) = E1

CCH

C

H

H

DH3C

CH2D

CH3

H2O

CCH

C

H

CH2D

CH3H

DH3C

H2O

"D" parallel toempty 2p orbitalon top

"D" parallel toempty 2p orbitalon bottom

"H" parallel toempty 2p orbitalon top

"H" parallel toempty 2p orbitalon bottom

C C

H3C H

H CHD

H2C

CH3

C C

H H

H3C CHD

H2C

CH3

C C

D H

H3C CHD

H2C

CH3

C C

H3C H

D CHD

H2C

CH3

4S,2E-alkenewithout "D"

4S,2Z-alkenewithout "D"

4S,2Z-alkenewith "D"

4S,2E-alkenewith "D"

H2O

(2S,4S)

(2S,4S)

(2S,4S)

(2S,4S)

(4S)

(4S)

(4S)

(4S)

rearrangement to similar or

more stable carbocation (R+)

(start all over)carbocation

choices

rotate bond

rotate bond

rotate bond



E1 products from right C carbon atom (4 possible alkenes from the lef t C carbon)

Redrawn from above (2S,4S)

CCH

C

D

H

H3CH

CH2D

CH3

CCH

CD

H3CH

"D" parallel toempty 2p orbitalon top

"D" parallel toempty 2p orbitalon bottom

"H" parallel toempty 2p orbitalon top

"H" parallel toempty 2p orbitalon bottom

2S,3E-alkenewithout "D"

2S,3Z-alkenewithout "D"

2S,3Z-alkenewith "D"

2S,3E-alkenewith "D"

OH2

H

DH2C

OH2

CCH

C

D

D

H3CH

HCH2

CCH

CD

H3CH

OH2

D

H2C

H

OH2

H3C

H3C

CH3

C

CH2C

HDC

D

H

CH3

H3C

C

C D

CDH

H2C

HH3C

C

CH2C

HDC

H

H

CH3

H3C

CH3

C

C H

CDH

H2C

HH3C

CH3

OH2

CCH

C

D

H

H3CH

CH2D

CH3

OH2

H2O

a

b

SN1 products (a. add from top and b. add from bottom)

CCDH HDC

H

O

HH

H3C H2C

CH3

C

DHC

HDHC

O

HH

H3CCH2

CH3

a

b

OH2

OH2

CCDH HDC

H

O

H

H3C H2C

CH3

C

DHC

HDHC

O

H

H3CCH2

CH3

(2S,4S) (2S,3R,4S)

(2S,3S,4S)

diastereomers

acid/baseproton transfer

acid/baseproton transfer

(2S,4S)(2S)

(2S)

(2S)

(2S,4S)

(2S,4S)

(2S,4S)

(2S)

rotate bond

rotate bond

rotate bond



Rearrangements of Carbocations

As we have seen, carbocations have very large potential energy differences. These differences provide a large driving force to form more stable carbocations from less stable carbocations, in the range of 15-20 kcal/mole in the gas phase for 1o to 2o and 2o to 3o choices. Rearrangements are a competitive pathway in any reaction where a carbocation is formed. A relatively simple example illustrates the necessity to be systematic in your approach to determine all of the varied possibilities. Consider the migration of every group on a C position, whether H or C. To keep our choices simpler (than they really are) we will only consider rearrangements of 2o to 3o and 3o to 3o carbocations.

C

CH2

H3C

H

HC

X

CH3

The RX compound must be

2o, 3o, allylic or benzylic to

form the initial carbocation.

SN1

E1redrawn All potentially migrating

bonds drawn with bold lines.

Consider all C -groups

(H and C).

SN1 and E1 arepossible here

2o carbocation

An adjacent C group migrates with its

electrons to the C carbocation position.

There are four possibilities in this problem.

C

CH2

H3C

H

C

H

CH3

C

CH2

H3C

H

C

H

CH

HH

d

a

b

dd

a H:hydridemigrates

H3C:methylmigrates

cb

C

CH2

H3C C

H

CH3 C

H

C

CH3

CH3C

CH2

H3C

H

C

H

CH2

HH H

1o carbocation

looks very poor2o carbocation

looks OK

3o carbocation

looks very good

We are not likelyto observe this

choice.SN1 and E1 arepossible here


a

c

H3C H3C

c

d

d

H3C

b

H3CH2C:ethylmigrates

CH3C

H

C

X

CH3

CH2

2o carbocation

looks OK


X X

CH2

H3C H3C H3CH3C

possiblerearrangements

H:hydridemigrates

H-Nu/H-B

Problem 25 – What are the likely SN1 and E1 products of the initial carbocation and the rearranged carbocations from “a”, “b” and “c”? Problem 26 – Write out your own mechanism for all reasonable products from the given R-X compound in water (2-halo-3-methylbutane).

C

CH3

H3C

H

HC

X

CH3 HO

H?

The most competitive rearrangement above will be to form the more stable tertiary carbocation from the initially formed secondary carbocation. It is also likely that at least some of the initially formed carbocation will react by the SN1 and E1 choices. However, those may be minor products when a much more stable carbocation can



form by rearrangement. In the end, SN1 and E1 possibilities are the ultimate fates of even the most stable carbocation that can form. Our goal at this point is to understand how rearrangements occur and what SN1 and E1 products are possible.

All groups on any Cβ carbon can potentially migrate to the adjacent carbocation carbon (also called a 1,2 shifts), if a similar or more stable carbocation can form. If hydrogen with its two electrons is the group migrating, the rearrangement is called a 1,2 hydride shift. If a carbon group migrates with its two electrons, the rearrangement if called a 1,2 alkyl shift. Hydride and alkyl shifts can occur from further away than a C position or even between two positions in completely different molecules. However, these we will not emphasize such possibilities in our course. Transition state of a carbocation rearrangement

C

CH2

H3C

H

HC

X

CH2

SN1

E1

rearrangement

2o carbocation

SN1 and E1 are possible here

XH3C

1,2-hydride shift

CC

H

CH2

CH2

H

transition state (no finite existance)

migrating group is positioned between two vicinal carbon atoms

(vicinity = neighbors, Latin)

CC

H

CH2

HH2C

H3C

3o carbocation

CC

H2C

H3C

H

HH2C

R-X

TS1 TS2TS3

Ea

G

2o R3o R

SN1 & E1products

PE

Progress of Reaction (POR)

H: = hydride shift

R: = alkyl shift

The migrating gropu is always attached to the carbon skeleton; it is never a free anion.

SN1 and E1 are possible here

CH3

CH3CH3

CH3

H3C

H3C

H3C H3C

Two main rules will help guide you in evaluating possible rearrangements. 1. Rearrangements usually occur so that the migrating groups moves from a C atom to the C atom (the

carbocation center). These are the 1,2 hydride or 1,2 alkyl shifts mentioned above. The C atom that gives up the migrating group becomes the new electron deficient carbocation center, often because it is a more stable carbocation site.

2. If a 1,2 shift of a hydrogen atom or an alkyl group can form a similar or more stable carbocation, then such a rearrangement is likely to be competitive with other reaction choices (SN1 and E1). When interpreting a reaction mechanism involving rearrangements, you may have to consider both equal (3o 3o R+) and more stable (2o 3o R+) carbocation possibilities. However in this book when you are asked to predict what might be possible, you usually only need to consider more stable carbocation possibilities (2o 3o R+).



Problem 27 - Consider all possible rearrangements from ionization of the following RX reactants. Which are reasonable? What are the possible SN1 and E1 products from the reasonable carbocation possibilities?

d. What would happen to the complexity of the above problems with a small change of an ethyl for a methyl? Use the key of “b” and “c” as a guide. This problem is a lot more messy than those above, (which is the point of asking it). There are too many possibilities to consider listing every answer.

There are other features that must also be considered in carbocation rearrangements in addition to the relative stabilities of 1o, 2o and 3o carbocations. One such feature that modifies the relative potential energies of the possible choices is strain energy. Consider the possible rearrangement choices available to the following tertiary carbocation in a polar ionizing solvent.

Br

slowstep

R.D.S.

RO

H

CH2

HH

ab

c

CH2

HH

CH3

H

CH3

H

H

CH3

CH3

a

b

c

a. A hydride migration makes a primary carbocation from a tertiary carbocation. This reaction

would increase the potential energy by about 35 kcal/mole and is not a realistic option.

b. At first this option (hydride shift) seems very reasonable (tertiary carbocation to tertiary carbocation), but there would be much additional ring strain energy because of bond angle changes in the small cyclobutane ring (109o = sp3 to 120o = sp2), while geometric shape in the ring is trying to be 90o. This would, therefore, not be a favorable option.

c. At first this looks like a very poor reaction (tertiary carbocation to secondary carbocation vial alkyl migration of a ring carbon) and would be uphill by about 15 kcal/mole based on carbocation stabilities. However, the reduction in ring strain would be downhill by about 20 kcal/mole (26 kcal/mole 6 kcal/mole), resulting in an overall potential energy change of -5 kcal/mole.



What is actually observed? (Only E1 reactions are shown, SN1 possibilities are not included.) Rearrangement ‘c’ occurs, followed by another rearrangement from a secondary to tertiary carbocation.

HCH3

CH3

H CH3

CH3

major(85%)

minor(14%)

H CH3

CH3

HH

Path c leads toring expansionrearrangement.

2o carbocation3o carbocation

E1

H CH3

CH2

veryminor (1%)

E1 products

Problem 28 – Lanosterol is the first steroid skeletal structure on the way to cholesterol and other steroids in our bodies. It is formed in a spectacular cyclization of protonated squalene oxide. The initially formed 3o carbocation rearranges 4 times before it undergoes an E1 reaction to form lanosterol. Add in the arrows and formal charge to show the rearrangements and the final E1 reaction.

R

CH3

H

H

CH3

CH3

H

CH3

HOH

R

CH3

H

CH3

CH3

CH3

HOH

H

lanosterol precursor

lanosterol

B H

R

O protonatedsqualene

acidcatalysis

rearrangement1

R

CH3H

CH3

CH3

H

CH3

HOH

H

rearrangement2

R

CH3

CH3

H

CH3

HOH

H

H

rearrangement4

CH3

R

CH3

H

CH3

HOH

H

HCH3

CH3

R

CH3

CH3

CH3

H

CH3

HOH

H

H

rearrangement3

E1 reaction

19 more steps

H3CH

CH3

CH3

HO

H

H H

H

cholesterol

otherbody

steroids

H

squalene

Requires 5 arrows to show the reaction.



Example SN1 / E1 Mechanisms with rearrangement

CH3H

H

12

34

56

2R-bromo-3R-methylhexane

H3CO

H

Br secondary RX (2o)

C

C

HBr

H

H

H

CHCH3

CH2CH2CH3

C

C

HO

C

HCH3

D

Ha CH2CH3

Hb

C

C

H

CH3

CH3

CH2CH2CH3

1. The first 2o R+ forms two SN1 products and three E1 products2. The rearranged 3o R+ forms two SN1 products and five E1 products

C

C

HH

H

H CCH3

CH2CH2CH3

H

a, b, c show attack on left face of carbocation,attack also occurs from the right side of R+.

H3CO

H

H3CO

H

b

a

c

C

C

H

H

H

C

CH3

H CH2CH2CH3

a

b c

E1 product after loss of beta protonfrom methyl (CH3)

E1 product after loss of beta protonfrom methine (CH) from either face

C

C

HO

C

HCH3

D

Ha CH2CH3

Hb

SN1 product from attack of left face

H

H3C

H3CO

H

C

C

HO

H

H

H

CHCH3

CH2CH2CH3

H

CH3

H3CO

H

C

C

HO

H

H

H

CHCH3

CH2CH2CH3

CH3

SN1 product from attack of right face

b, c

rearrangement(next page)

d

d

C

C

H

CH2CH2CH3

CH3

CH3

c

H3C

d

a

1. weak H-Nu:/H-B:2. 2o R-X

R+ reactions1. add H-Nu: (SN1)2. lose C-H (E1)3. rearrange (start over)

C

C

HH

H

H CCH3

CH2CH2CH3

H

carbocation continued with rearrangement)

C

CH2

CH2

O

H3C

H

CH3

rearrangement from 2oR+ to 3oR+,

redrawn with new C.

H3CO

H

CH3

OH

C

C

H3CH2C CH3

Same product forms from loss of methyl proton from either face (no E or Z).

H3C

O

H

CH3

O

H

CH2CH3H

Two possible products from loss of proton from left faceor right face on propyl branch.

C

C CH3

CH2CH2CH3

CH3

H

C

C

H

H

H3CH2C

CH2CH2CH3

Two possible products from loss of proton from left faceor right face on ethyl branch.

a

b

c

d e

a

b

c

c

a

H3CO

H

d

e

CH3

OH

H3C

C

CH

CH2CH3

H

CC H

HHH3C

H

H

redraw R+ to show SN1 reaction

a, b, c = E1 products(protons lost from

either face)

SN1 products

3E alkene

3Z alkene

2Z alkene

2E alkene

(3S)(3R)

C

H2C

H2C CH3

H3C

CH2CH3

C

H2C

H2C

O

CH3

H

H3CCH3

CH2CH3 C

H2C

H2C

O

CH3

H3CCH3

CH2CH3

H3CH2C

C

CH2

CH2

O

H3C

CH3H3C

H3CH2C

2o R+ from previous page

C

C CH3

CH2CH2CH3

H

H3C

C

C

H3CH2C CH3

HH3CH2C



SN1 / E1 possibilities –extra complications at Cβ positions of 2o RX, rearrangement to more stable 3o R+ considered



After rearrangement to 3o carbocation (R+) – We will skip rearrangements in Chem 314

E1 products from left C carbon atom (top and bottom, after rearrangement)

Redrawn (achiral)

H2O

CC

CH3

H2C

HH3C

2E-alkene2Z-alkene

3Z-alkene3E-alkene

OH2

C

C CH2

H2C

H3C

H

CH3

H3CC

CH2C

H

H3C

CH2

OH2

H2O

a

b

SN1 product (a. add from top and b. add from bottom), (after rearrangement)

C

O

HH

C

O

HH

a

b

OH2

OH2

(achiral) acid/base proton transfer

E1 product from right C carbon atom (top and bottom, after rearrangement)

(3R)

diastereomers

CH2C

H

C

H

H3C

CH2H3C

CH3

(4S)

initial 2o carbocation rearranges via hydride shift to a 3o carbocation

C

C

H3C

H

CH2

H3CH

C

CH3

4S stereochemistry is lost in new carbocation

add nuclephile (top/bottom) = SN1

lose -H (top/bottom) = E1

rearrangement to similar or more stable carbocation (R+)(none is possible here)

R+ reaction possibilities start all over again with new carbocation

H

H

C

C

H3C

H

CH2

H3CH

C

CH3

H

H

C

C

H3C

H

CH2

H3CH

C

CH3

H

H

"H" parallel to empty 2p orbital on top and bottom of right C position

CH2

H3C

CC

HH2C

CH3H3C

CH2

H3C

diastereomers

C

C

H3C

H

CH2

H3CH

C

CH3

H

H

H2C

CH3

CH2

H3C

CH2

H3C

CH2

CH3CH2H3C

H2C

H3C

acid/base proton transfer

C

O

H

CH2

CH3CH2H3C

H2C

H3C

C

O

H

H2C

CH3

CH2

H3C

CH2

H3C

(3S)

enantiomers

CH3

H3C

CH2

C

CH3

H2C

H2C 1-alkene(does not have E/Z stereochemistry)

OH2

E1 product from methyl C carbon atom (top and bottom, after rearrangement, only one product from the methyl)

C

C

H2C

H

CH2

H3CH

CH2

CH3CH2

H3C

H

(3R)

(3S)





Alcohols in strong acid = Protonated Alcohols - Water as a Good Leaving Group

We can make the hydroxyl group of an alcohol, OH, into a good leaving group in strong acid conditions (HCl, HBr, HI and H2SO4). Strong protic acids are used to extensively protonate the alcohol OH. When the alcohol OH is protonated, the leaving group is water, not hydroxide. Water’s conjugate acid is H3O

+, (pKa = -2), while hydroxide’s conjugate acid is H2O, (pKa = 16). If substitution is the desired goal, then the strong halide acids are normally used, HCl, HBr or HI. If elimination is the desired goal, then concentrated sulfuric acid (H2SO4) is used at an elevated temperature (Δ).

Using the hydrohalic acids (HCl, HBr or HI), very polar, strongly acidic conditions encourage SN1 reactions, and these are assumed to be operating at all tertiary and secondary alcohol (ROH) centers. Rearrangements are frequently observed under these conditions. The large energy expense of a methyl or primary carbocation prevents the escape of water on its own. The H2O at methyl and primary ROH2

+ is assumed to be pushed off by the halide (SN2) of the strong acid to form a methyl or primary haloalkane without rearrangement. a. 1o, 2o and 3o ROH reacted with HX acids (HCl, HBr, HI) - usually SN chemistry i. methyl alcohols (SN2 emphasized, no rearrangement)

H3C

O H BrH

methanol

H3C

O H

H

Br

H3C

O

H

H

Br

SN2BrH

O

H

H H

bromomethane

Br

water is a good leaving group

pKa = -9

ii. primary alcohols (SN2 emphasized, no rearrangement)

H3C CH2

O H

BrH

primary alcohol

H3C CH2

O H

H

Br

H3C CH2

O

H

H

BrSN2

BrH

O

H

H H

primary bromoalkane

Br

pKa = -9


iii. secondary alcohols (SN1 emphasized, rearrangements possible)


ClH

secondary alcohol (trans OH)

H2C H3C HSN1

Cl

H3CH

Cl

H3CCl

H

trans Cl

cis Cl

top and bottom attack

H3CH

O

H

H

O

H

H

secondary chloroalkane

pKa = -7

O

H

H O

H

H H

ClH Cl



iv. tertiary alcohols (SN1 emphasized, rearrangements possible)

IH

tertiary alcohol (trans OH)

H3C H3C CH3

O

H

H

SN1I

O

H

H H

I

I

trans I

cis I

top and bottom attack

H3CCH3

O

H

CH3

O

H

H

IH

pKa = -10

Problem 29 - Propose a synthesis of monodeuterated cyclohexane from cyclohexanol.

O H

H H

D

b. 1o, 2o and 3o ROH reacted with H2SO4 and high temperature = E1 chemistry

Using strongly acidic sulfuric acid, H2SO4, at elevated temperatures favors E1 reactions because lower boiling alkenes distill out and continually shift the equilibrium to make more alkene, which continues to distill out, until there is no more alcohol left in the reaction pot. We will assume that an E1 mechanism is operating in all of the reactions below (even the primary alcohol, an exception to our rule about no primary carbocations – the conditions are very harsh). Rearrangements are possible and observed.

i. primary alcohols (with high temperature E1-? - emphasized, rearrangements possible)


primary alcohol

O

H

H O

H

H H

OH OH SO3H

(heat)

OH

H

CH

H

H H

rearrangementH

H

O SO3H

E1 bp = -47oCdistills outbp = +82oC

very difficult(high temperature)

OH SO3HO SO3H

alcohol alkene Tbp = 129oC

pKa = -10

ii. secondary alcohols (E1 emphasized, rearrangements possible)

secondary alcohol

O

H

H O

H

H H

OH OH SO3H

(heat) O

H

H

O SO3H

E1

bp = +83oCdistills outbp = +161oC

OH SO3HO SO3H

H

H

H

alcohol alkene Tbp = 78oC

pKa = -10 water is a good leaving group



iii. tertiary alcohols (E1 emphasized, rearrangements possible)

tertiary alcohol

O

H

H O

H

H H

OH SO3H

(heat)

O SO3H

bp = +33oCdistills out

bp = +102oC

OH SO3HO SO3H

OH O

H

H

H

H H

E1

bp = +39oCdistills out

90% < alkene(more substituted)

minor alkene(less substituted)b

a

a b

alcohol alkene Tbp 63oC

pKa = -10water is a good leaving group

We have some, limited control to direct the alcohol functionality toward SN or E choices. The conditions to effect these different pathways are important, so you must be aware of the details mentioned above (halide acids = SN reactions and H2SO4/ = E1 reactions). Heat is a crucial aspect of the E1 reaction, since it allows the lower boiling alkene to escape from the reaction mixture by distillation, while the higher boiling alcohol or inorganic esters remains, in the reaction pot to reestablish equilibrium by forming more alkene, which distills......etc. The alkene boils much lower than the alcohol it comes from because it does not have an “OH” to form hydrogen bonds.

Examples of Boiling Point Differences Between Alcohols and Possible Alkene Products

boiling points of alcohols (oC)

boiling point of alkenes (oC)

OH H2C CH2

OH

OH

OH

(79 oC)

(82 oC)

(-104 oC)

(-47 oC)

(97 oC)

(100 oC)

OH (161 oC)

(-47 oC)

(-6 oC)

(1 oC)

(4 oC)

(83 oC)

DTbp

(183 oC)

(129 oC)

(144 oC)

(106 oC)

(99 oC)

(96 oC)

(78 oC)

There are important other ways to change ‘OH’ into ‘Br’. The OH group can be an alcohol or a carboxylic acid. Some possibilities are shown below.



1. Formation of tosylates from ROH + TsCl (toluenesulfonyl chloride = tosyl chloride), SN/E chemistry is possible without rearrangements.

ORORH

S

Cl

H

SCl

O

O

O

O

N

OS

Cl O

O

N H

OS

O

O

Br

Na

separatestep

Br

HBr

rearrangementand

R,S (racemic)

S

1. TsCl/pyridine

2. NaCl

(prevents R+ formation

and any rearrangement)

alkyl tosylate = RX compound

compare - a different result

RR R

R

RR

RBr

Br

SN1

toluene sulfonylchloride(tosyl chloride)

SN2

Problem 30 – We can now make the following molecules. Propose a synthesis for each. (Tosylates formed from alcohols and tosyl chloride/pyridine via acyl substitution reaction, convert “OH” from poor leaving group into a very good leaving group, similar to iodide)

2. Other acyl-like transformations include thionyl chloride (SOCl2) or thionyl bromide (SOBr2) with alcohols (makes R-Cl and R-Br) or carboxylic acids (makes acid chlorides, RCOCl). Acid chlorides formed can make esters, amides and anhydrides.

Thionyl chloride with methyl, 1o ROH = acyl-like substitution at SOCl2, then SN2 at methyl and primary RX.

R OH S

Cl

O

Cl

H2C

R O

S

O Cl

Cl

OS

O

H

H2C

R O

SCl

O

CH2R

synthesis of an alkyl chloride from an alcohol + thionyl chloride (SOCl2) [can also make RBr from SOBr2]

SN2 at methyl and primary alcohols

No rearrangement because no R+.

H

ClH

Cl

O

SCl

O

Hproduct

O

S

O

H

acyl-like substitution

Cl

Cl



Thionyl chloride with 2o and 3o ROH = acyl substitution, then SN1 (there are various ways you can write this mechanism)

OH O

S

O Br

Br

OS

O

H

O

SBr

O

HH

Br

SN1 at secondary and tertiary alcohols

synthesis of an alkyl chloride from an alcohol + thionyl chloride (SOCl2) [can also make RBr from SOBr2]

O

SBr

O

H

BrH

RR

R

RR/S

acyl-like substitution

SBr

O

Br

Br

Br

Synthesis of acid chlorides from acids + thionyl chloride (SOCl2), use the carbonyl oxygen instead of the OH.

CR

O

OH

CR

O

OH

S

O Cl

Cl

resonance

CR

O

OH

S

O Cl

Cl

CR

O

OH

S

O Cl

Cl

Base

CR

O

O

S

O Cl

Cl

BaseH

CR

O

O

SCl

O

CR OCR O

resonance

CR

Cl

Oacylium ion

SCl

O

Cl

OS

O

Cl

Cl

resonance

Formation of esters from ROH + acid chlorides, amides from RNH2 or R2NH + acid chlorides and anhydrides from RCO2H + acid chlorides

OR

O

R

H

Cl

O Cl O

H

O

Cl

H

O

O

O

R

R

There are many variations of ROH andRCO2H joined together by oxygen.

ester synthesis from acid chloride and alcohols

RO

H

RO

H

H



O

O

Cl

O Cl O

Cl

There are many variations of R1CO2H andR2CO2Hjoined together by oxygen.

anhydride synthesis from acid chloride and carboxylic acids

O

OH

H

O

O

H

O

O

O

O ClH

Phosphorous trichloride (PCl3) = SN2 of alcohol at phosphorous, then SN2 (at methyl and primary R(OH)PCl2)

OH

OP

H

Cl

ClCl

Cl

OP

H

Cl

Cl

reactstwicemore

P

Cl

Cl

Cl

SN2 SN2

Phosphorous tribromide (PBr3) = SN2 of ROH at phosphorous, then SN1 (at secondary, tertiary, allylic and benzylic R(OH)PBr2)

OH

S

Br

H

R,S

OH

R

P

Br

Br

Br

SN2

OP

H

Br

Br

Br

SN2

BrO

P

H

Br

Br

reacts twice more

P

Br

Br

BrO

P

H

Br

Br

Br

OP

H

Br

Br

reacts twice more

Br

SN2 SN1



Chart of SN and E Chemistry (note exceptions)

typical strong basenucleophiles are:(for our course)

HO

RO

R

O

O

HO

RO

O

OH HH

NC C

C

RHS

RS

NN

N

H3CX

CH2

XR

HC

XR

R

CXR

RR

methyl

primary

secondary

tertiary

only SN2 only SN2 only SN2 only SN2 only SN2 only SN2 only SN2 only SN2

SN2 > E2 SN2 > E2 SN2 > E2 SN2 > E2 SN2 > E2 SN2 > E2 SN2 > E2 SN2 > E2

E2 > SN2 E2 > SN2 E2 > SN2SN2 > E2 SN2 > E2 SN2 > E2 SN2 > E2 SN2 > E2

only E2 only E2 only E2 only E2 only E2 only E2 only E2 only E2

exception(too basic)

O

t-butoxide

only SN2

E2 > SN2exception

(bulky & basic)

E2 >> SN2

only E2

typical weak basenucleophiles are:(for our course)

H3CX

CH2

XR

HC

XR

R

CXR

RR

methyl

primary

secondary

noreaction

noreaction

noreaction

noreaction

noreaction

noreaction

SN1 > E1

SN1 > E1

SN1 > E1

SN1 > E1

SN1 > E1

SN1 > E1

BH

H

H

H

only SN2

SN2 > E2

SN2 > E2

NA

exception(bulky & basic)



tertiary

AlD

D

D

D

alcohol reactions instrong acid:

(for our course)

H3COH

CH2

OHR

HC

OHR

R

COHR

RR

methyl

primary

secondary

tertiary

HX

(X = Cl, Br or I)

SN1

SN1

E1

notdiscussed

H2SO4

E1

E1

SN2

SN2

SN1

SN1

SN2

SN2

SOCl2SOBr2

PCl3PBr3

1. TsCl/py2. NaBr

NA

SN2

SN2

SN2



Problem 31– Look back at the table of R-Br structures on page 2. Include stereoisomers together. Be able to list any relevant structures under each criteria below.

1. Isomers that can react fastest in SN2 reactions 2. Isomers that give E2 reaction but not SN2 with sodium methoxide 3. Isomers that react fastest in SN1 reactions 4. Isomers that can react by all four mechanisms, SN2, E2, SN1 and E1 (What are the necessary

conditions?) 5. Isomers that might rearrange to more stable carbocation in reactions with methanol. 6. Isomers that are completely unreactive with methoxide/methanol 7. Isomers that are completely unreactive with methanol, alone.

The number of each type of product (SN1, E1, SN2, E2) is listed after a reaction arrow for each starting structure (assuming I analyzed the possibilities accurately in my head, while sitting at the computer). Do you agree with these numbers? Can you draw a valid mechanism for each one?

Br H

H3C H

OH

O

OH

HO

H

SN1 E1 SN2 E2

O

O

O

HO

2 41 3

1 3

1 3

a.

2 5b.

2 4a.

2 5b.

2 4a.

2 5b.b. after rearrangement

a. initial carbocation

Br H

H3C H

Br H

H3C H

OH

O

OH

SN1 E1 SN2 E2

O

O

O

HO

2 61 3

1 3

1 3

a.

2 5b.

2 6a.

2 5b.

2 6a.

2 5b.b. after rearrangement

a. initial carbocation

Br H

H3C H

HO

H

D H D H



Similar patterns in cyclohexane RX structures. The leaving group has to be axial in SN2 and E2 reactions.

C

C

C

Br

C

H

Hb

Ha H

HH

H H

C

C

Ha

Br

C

H

H3CHb H

H

Rotation of C-C1 brings Ha or Hb anti to C-Br, which allows SN2 and two different E2 possibilities: Ha (Z) and Hb (E). Since C2 is a simple methyl, there is no C2 substituent to inhibit either of these reactions.

C

C

Hb

Br

C

H

Ha

H3C H

H

When methyl on C1 is anti to C-Br, no SN2 is possible and no E2 is possible from C1.

SN2 possibleE2 from C1 (2Z- butene)E2 from C2 (1-butene)

SN2 possibleE2 from C1 (2E- butene)E2 from C2 (1-butene)

No SN2 possible and noE2 from C1, but E2 from C2 (1-butene) is possible.

E2 possible here

Use these ideas to understand cyclohexane reactivity.

H H

H

H

H HBr

H

H

H

H

H

HH

H

H

HH

BrH

H

H

HH

No SN2 is possible (1,3 diaxial positions block approach of nucleophile), and no E2 is possible because ring carbons are anti.

No SN2 or E2 when "X" is in equatorial position.

SN2 possible if C is not tertiary and there is no anti C "R" group.

E2 possible with anti C-H.

Both SN2 and E2 are possible in this conformation with leaving group in axial position.

H H

H

H

H HBr

H

C

H

H

H

HH

H

H

HH

BrH

H

H3C

HH

No SN2 or E2 when "X" is in equatorial position.

No SN2 possible if there is an anti C "R" group.

E2 possible with anti C-H.

Only E2 is possible in this conformation. Leaving group is in axial position.

H

H

H

full rotation atC-C is possible in chain

only partial rotation is possible in ring

only partial rotation is possible in ring

H H H

equatorial leaving group

axial leaving group

No SN2 is possible (1,3 diaxial positions block approach of nucleophile), and no E2 is possible because ring carbons are anti.

No E2 possible, no anti C-H.

full rotation atC-C is possible in chain

alkene stabilities tetrasubstituted > trisubstituted > trans-disubstituted > gem-disubstituted cis-disubstituted > monosubstituted



Problem 32 – Predict possible products of a. water and b. ethanoate (acetate) with structures 2, 10 and 13. Only consider rearrangements to more stable carbocations where appropriate.

C

H3C

H3C

H3C

H

H

CCH3

CH3

H3CH

H

severe stericrepulsion

19,999

Keqt-butyl substituent locksin chair conformationwith equatorial t-butyl

X

X

X

X

1213

Cyclohexane structureshave two chair conformationspossible.

1. Which conformation is reactive?

2. Is SN2 possible? Requires an open approach

at C and C .

3. Is E2 possible? Requires anti C -H.

4. How many possible products are there?

5. What is the relationship among the starting

structures?

6. What is the relationship among the products?

7. Are any of the starting structures chiral?

8. Are any of the product structures chiral?

Cyclohexane Examples

1

X X

XX

X X

X

X XX X2

3

4

5

6

7

8

9

10

11

chair 1 chair 2

X = ClBrI

OTs

14 15

Mechanism predictions with:

OR H

R

O

OH

OH HSome examples

OH Na OR Na R

O

O

Na

O K

weaknucleophiles

strong nucleophile/bases + other anions shown above



The number of each type of product (SN1, E1, SN2, E2) is listed after a reaction arrow for each starting structure (assuming I analyzed the possibilities accurately in my head, while sitting at the computer). Only consider rearrangements in part 9. Do you agree with these numbers? Can you draw a valid mechanism for each one?

HO

H

2 3

chair 1 chair 2

SN1

XX X

a.

b.

SN2 E2

1 1

E1

1 1

1

SN1 SN2 E2

1 2

E1

2 2

SN1 SN2 E2

0 1

E1

5 6SN1

XX X

SN2 E2

1 2

E1

2 2

4

SN1 SN2 E2

1 2

E1

2 2

SN1 SN2 E2

1 2

E1

2 2

8 9

SN1

XX

SN2 E2

1 2

E1

2 2

7

SN1 SN2 E2

0 3

E1

2 3

SN1 SN2 E2

0 1

E1

2 1

1 2

Only consider carbocation rearrangements that immediately go to a more stable carbocation (not equally stable carbocations = too many possibilities)

a.

b.2 2

1 2

X

a.

b.c.

d.

consider any 3o R+ possibilityand consider stereoisomers

4 51 21 2

HO Na

condition 1 condition 2



Available chemicals from the catalog

Sources of carbon - you can invoke these whenever needed:

CH4

Br

Br2 Cl2

Na

C N

Na

O H

CH2

Li n-butyl lithium (very strong baseor nucleophile, useanytime)

Na H

N

Hdiisopropylamine

N

O

O

H

Na

S H

O

OH

H2O

O

sodium hydride (very strong base)

ketone(a carbonyl)

pre alds / ketsphthalimide (an imide)

ethanoic acida carboxylic acid)

pre alds / kets

Na

sodium amide (very strong base)

HCl HBr HI H2SO4

Commercially available chemicals and reagents - you can invoke these whenever you need them.

NOO

Br

NBS = N-bromosuccinimide(supplies free radical brominefor allylic & benzylic substitution)

SCl

O

O= Ts-Cl (tosyl chloride)makes ROH into tosylates

N

pyridine = proton sponge

lithium aluminumhydride = LAH(very strong nucleophilic hydride)(LiAlD4 too)

sodium borohydride (nucleophylic hydride)

S

S

dithiane Br2

Pd / H2quinoline

(Lindlar's cat)

Pd / H2

H3C Li

methyl lithium (very strong base)

CLi

phenyl lithium (very strong base)

Na

N N N sodium azide

(excellent nucleophile)

H3PO4 HNO3 H2O2

KH

OH

palladium &hydrogen

phosphoricacid

nitric acid

sodium nitrite

hydrogenperoxide

diboraneuse w/ alkenes

dialkylboraneuse w/ alkynes

t-butyl alcohol(use to maket-butoxide)

pyridine H2O / H3O+

CO2carbon dioxidebromine chlorine hydrogen

chloridehydrogenbromide

hydrogeniodide

sulfuricacid

water

sodiumhydroxide

sodiumhydrogen

sulfide

sodiumcyanide

NH3ammonia

potassium hydride (very strong base)

R = C or H

NHR2

HCCl3

HCBr3

CH2I2Zn / Cu

chloroform

bromoform

SimmonsSmith

reagent

potassium permanganateosmium tetroxide

Nasodiummetal

3 ozonereactions

1. O3, -78oC2. CH3SCH3

1. O3, -78oC2. NaBH4

1. O3, -78oC2. H2O2, HO

1. Hg(OAc)2/H2O2. NaBH4

1. Hg(OAc)2/ROH2. NaBH4

O

O O

H

Cl

meta chloroperbenzoic acid (mCPBA)

Na / NH3in ammonia

(Birch reagent)

= py

(PCC) (Jones)

NO O

Na

Mn

O O

OO

Os

O O

OOK

OO O Cr

O

OO

Cr

O

OO

B

H

H

H

H Al

H

H

H

HNa

Li

B

H

H

H

B

R

R

H

pre aldehydes/ketones

P

Ph

Ph

Phtriphenylphosphine

(to make Wittig salts)

Na Cl

Br

INa

NaBrCu

Salts / ionic substances

1.

2. H2O2/HO 2. Br2/CH3O

1.

2. H2O2/HO

diboraneuse w/ alkenes

B

H

H

H1.

Mgmagnesium

metal

Lilithiummetal

Znzinc

metal

Lewis acidsAlBr3FeBr3BF3

SOCl2PBr3SO3etc.

Various metals

HgX2(mecuric salts)

HOOH H2N

NH2 NH

ethyleneglycol

(protect C=O)

hydrazine(Wolff-Kishner)

pyrrolidine(enamines)

O

THP(protect O-H) SHO

O

O

toluene sulfonic acid = TsOH(very strong organic acid)

sodium cyanoborohydride (nucleophylic hydride)(NaBD4 too)

B

H

H

H

CNNa

cuprousbromide

For now, the structures below represent your hydrocarbon starting points to synthesize target molecules (TM) that are specified. We will only study two free radical reactions in our course, but they are very important reactions because they make versatile functionalized starting molecules for synthesis of all the other functional groups studied in this course.



Allowed starting structures – our main sources of carbon – 1. Free radical substitution of sp3 C-H bonds to form sp3 C-Br bonds at the weakest C-H position and 2. Anti-Markovnikov addition to alkenes makes 1o R-Br. From these two reactions we can make 13 R-Br molecules below.

CH4

Br Br

BrWe need to make these 1o RBr from anti-Markovnikov free radical addition of H-Br (ROOR) to alkenes.

Br2 / h Br2 / h Br2 / h Br2 / h Br2 / h Br2 / h Br2 / h

H3CBr

Br

BrBr Br

Br

Br

HBrH2O2 / h

HBrH2O2 / h

HBrH2O2 / h

E2 reaction

O

K

E2 reaction

O

K

E2 reaction

O

K

E2 reaction

O

K

E2 reaction

O

K

Br2 / h

Br2 / h

Br2 / h

Br

Br

Br

Examples of allylic RBrcompounds: This is just free radical substitution

at allylic sp3 C-H positionof an alkene.

Br

benzylic RBr, f rom above

Br

PhBr2 / h2 eqs.

Ph

Br

Br

Br Br Br Br

E2 reaction(twice)

NaR2N

1. 3 eqs.

2. workup

Ph

a b c aab

b

c

c

Br

bromobenzene is given until aromatic chemistry is covered in 316

You will need to propose a step-by-step synthesis for each target molecule from these given structures. Every step needs to show a reaction arrow with the appropriate reagent(s) above each arrow and the major product of each step. This is often accomplished by using retrosynthetic thinking. You start at the target molecule (the end) and work your



way backwards (towards the beginning), one step at a time until you reach an allowed starting material. The starting material of each step becomes the target molecule for the next step until you reach the beginning.

1. Mechanism for free radical substitution of alkane sp3 C-H bonds to form sp3 C-Br bonds at weakest C-H position

H3C

H2C

CH3 H3CCH

CH3

Br

BrBr

overall reactionh

BrH

1. initiation

BrBrh BrBr H = 46 kcal/mole

weakest bond ruptures first

2b propagation

H3CC

CH3

H

BrBrH3C

CCH3

H Br

BrH = -22 kcal/mole (overall)

BE = +46 kcal/moleBE = -68 kcal/mole

2a propagation

H3CC

CH3

H H

Br BrH

H3CC

CH3

H

H = +7 kcal/mole (overall)


H = -15

both steps

3. termination = combination of two free radicals - relatively rare because free radicals are at low concentrations

BrH3C

CCH3

H

H3CC

CH3

H Br

H3CC

CH3

H

CH3

CH3C

H

CH

CH3

H3C CHCH3

CH3

H = -68 kcal/mole

H = -80 kcal/mole

very minor product

2. Free radical addition mechanism of H-Br alkene pi bonds (alkenes can be made from E2 or E1 reactions at this point in course) (anti-Markovnikov addition to alkenes)

H3C

H2C

CH2

Br

H3C

HC

CH2

HBrR2O2 (cat.)

h

overall reaction

1. initiation (two steps)

RO

OR h R

OO

R

BrHR

O RO

H Br

H = 40 kcal/mole

H = -23 kcal/mole


(cat.)

reagent



2a propagation

H3C

HC

CH2Br

H3CC

CH2

Br

H

H = -5 kcal/mole

BE = +63 kcal/mole BE = -68 kcal/mole

H = -15

both steps(2a + 2b)

2b propagation

H3CC

CH2

Br

H

BrH H3C

H2C

CH2

BrBr

H = -10 kcal/mole

BE = +88 kcal/mole BE = -98 kcal/mole

Miscellaneous E2 mechanism not included above; An E2 reaction that makes carbonyl compounds (C=O)

1. PCC = pyridinium chlorochromate, (CrO3/pyridine), CrO3 oxidations of alcohols (methyl, 1o and 2o ROH) without water. Steps are: 1. Cr=O addition, 2. acid/base and 3. E2 to form C=O (aldehydes and ketones).

H3CCH2

CH

OH

H

Cr

O

OO H3CCH2

CH

OH

H

Cr

O

O

O

H3CCH2

CH

O

H

Cr

O

O

ON

N H

N

H3CCH2

CH

O

Cr

O

O

ON H

PCC = pyridinium chlorochromate oxidation of primary alcohol to an aldehyde (no water to hydrate the carbonyl group)

primary alcohols

aldehydes

E2

CrO3 oxidations of alcohols (methyl, 1o and 2o ROH) without water = PCC, Cr=O addition, acid/base and E2 to form C=O (aldehydes and ketones)

H3CCH2

CCH3

OH

H

Cr

O

OO H3CCH2

CCH3

OH

H

Cr

O

O

O

H3CCH2

CCH3

O

H

Cr

O

O

ON N H

N

H3CCH2

CCH3

O

Cr

O

O

O

N H

PCC = pyridinium chlorochromate oxidation of primary alcohol to an aldehyde (no water to hydrate the carbonyl group)

primary alcohols E2

ketones



2. Jones reagent = CrO3/water/acid, CrO3 oxidations of alcohols (methyl, 1o and 2o ROH) with water. Steps are: 1. Cr=O addition, 2. acid/base and 3. E2 to form C=O (aldehydes and ketones) 4. hydration of C=O and repeat reactions when the starting alcohol is a 1o alcohol (forms carboxylic acids from primary alcohols and ketones from secondary alcohols).

H3CCH2

CH

OH

H

H3CCH2

CH

O

Jones = CrO3 / H2O / acid primary alcohols oxidize to carboxylic acids

(water hydrates the carbonyl group, which oxidizes a second time )

primary alcohols

aldehydes (cont. in water)

HO

HH

OH

H

H3CCH2

CH

OH

H3CCH2

CH

OH

HO

H

H3CCH2

C

OH

O

H

H

H

HO

HH3C

CH2

C

O

O

H H

H

HO

H

HO

H

H3CCH2

C

O

OH

hydration of the aldehyde

second oxidation of the carbonyl hydrate

H

O

H

resonance

carboxylic acids

Cr

O

OOH3C

CH2

CH

OH

H

Cr

O

O

O

H3CCH2

CH

O

H

Cr

O

O

O

Cr

O

O

O

HO

H

H

HO

H

H

Cr

O

OO H3CCH2

CO

O

H

Cr

O

O

O

H

H

H3CCH2

CO

O

H

Cr

O

O

O

H

HO

H

H

Cr

O

O

O

HO

H

H



Problem 33 – We can now make the following molecules. Propose a synthesis for each from our starting materials.

C

O

H HC

O

H3C H

C

O

H3C CH3

C

O

CH2

H

C

O

CH3

H3C

O

C

O

HC H

H3C

CH3

C

O

HC

O

CH2

H

C

O

H OH

C

O

H OCH3

C

O

H3C OHC

O

CH2

OHH3C C

O

HC OH

H3C

CH3

C

O

OH

C

O

CH2

OH

C

O

H3C O

C

O

CH2

OH3C C

O

HC O

H3C

CH3

CH3CH3 CH3

C

O

O

C

O

CH2

O

CH3

CH3

aldehydes, carboxylic acids and esters

ketones

C

O

CH

HH2C C

O

CH

OH

H2C C

O

CH

O

H2C CH3

O

C

O

C HH2C

CH3

C

O

C OHH2C

CH3

C

O

C OH2C

CH3

CH3

conjugated aldehydes, carboxylic acids and esters

2 always leave with x. n1

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