315 SN E Syn & Chem Catalog

8/13/2019 315 SN E Syn & Chem Catalog

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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 electronsalways leave with X.SN2

E2

SN1

E1

XRBNu BHNuHCompetingReactions

CompetingReactions

CarbonGroup

LeavingGroup

Nu: / B: = is an electron pair donor to carbon (= nucleophile) orto 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 Ccarbon attached to the leaving group, X? Is it a

methyl, primary, secondary, tertiary, allylic, or benzylic carbon? What about any Ccarbon atoms? How

many additional carbon atoms are attached at a Cposition (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.


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

Na

O H

Na

S HO R

Na K

O

Na

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

Li

Na

cyanideacetylides

ester enolates

H2C

O

O

(make)

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

Na

N

N

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 H

N

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 H

NR2

= 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=Oto make alkenes

Ph

CrO3/ pyridine

pyridinium chlorochromatePCC

oxidizing E2 reaction

CrO3/ H2O

Jones reagentoxidizing E2 reaction

(buy)

(buy)


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

Br O

majorminorH

O

H

O

H

H

Br

Br

H

O

H

O

majorminor

H

Br

H

O

H

Rate = kSN2[HO ]

1

[RBr]

1

Rate = kE2[HO ]1

[RBr]1

Rate = kE1[RBr]1

Rate = 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

H

H

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 = anythingwith negative charge,and neutral sulfur,

phosphorous or nitrogen

SN2 > E2

(unless t-butoxide)

C

H3C

H3C

H3C

(also R2N = E2)

SN2 always

backside 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, 3oversions)

C C X

C-beta carbon(0-3 of these)

C-alpha carbon(only 1 of these)

benzylic

(1o, 2o, 3oversions)


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

XC

H

H

H

k 30methyl (unique)

XC

H

H3C

H

k 1ethyl (primary)

reference compound

XC

CH3

H3C

H

k 0.025isopropyl (secondary)

XC

CH3

H3C

CH3

k 0t-butyl (tertiary)

(very low)140

C

H

HH

X

methyl RX

Methyl has three easy paths ofapproach by the nucleophile. It isthe least sterically hindered carbonin SN2 reactions, but it is unique.

C

H

RH

X

primary RX

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

C

R

RH

X

secondary RX

Secondary substitution allows one easy pathof approach by the nucleophile. It reacts theslowest of the possible SN2 substitutionreactions.

tertiary RX

C

R

RR

X = C

C

X

C

H

H

H

H

HH

CH

H

H

Nu Nu

Tertiary substitution has no easy path of approachby the nucleophile from the backside. We do notpropose any SN2 reaction at tertiary RX centers.

When the Ccarbon is completelysubstituted the nucleophile cannotget close enough to make a bond

with the Ccarbon. Even C-Hsigma bonds block the nucleophilesapproach.

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

H2CC

H

H

H

k 1

ethylreference compound

H2CC

H

H3C

H

k 0.4

propyl

H2CC

CH3

H3C

H

k 0.03

2-methylpropyl

H2CC

CH3

H3C

CH3k 0.00001 0

2,2-dimethylpropyl(1oneopentyl)

X X XX

C

C

X

H

C

CH3CH3

H

H

H

H

Nu

A completely substituted Ccarbon atom also blocks theNu: approach to the backsideof the C-X bond. A large groupis always in the way at thebackside of the C-X bond.

C

C

X

H

H

CH3CH3

HNu

If even one bond at Chas a

hydrogen then approach byNu: to the backsideof the C-X bond is possibleand an SN2 reaction ispossible.


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sp2carbon transition state as carboninverts configuration forms a high PEcarbon with 10 electrons at carbon.This is a concerted, one-step reaction.

"Transition State"

C

H

H

H

BrOH

BrC

H

H H

HO

C

H

H

H

HO Br

reactants

products

Ea- this energy difference determines

how fast the reaction occurs: "kinetics".

Ea

GG - this energy difference determines theextent of the reaction; the ratio of products

versus reactants at equilibrium (whenkinetics allows the reaction to proceed.Thermodynamics is determined mostly by thestrengths of the bonds and solvation energies ofthe 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

H3C

CH3CH2

I

CH3C

O

O

a b

Problem 2 - Why might C3and C4rings 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 BrBr Br Br

BrBr

1.0 0.00001 0.008 1.6 0.01 0.97 0.22

Problem 3 - Why might C6rings 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)?


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H H

H

HH

H

H

H

H

HH

H

H

H

H

H

H

H

H

H

H

H

These two chairconformations

interconvert in avery 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

methoxide

t-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

H3C

O

O

O N

CH

S

H3C

S

N

N

N

?

?

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

1 3 4 6 7

X

X

X

?

?

Nu

A

B

C

D

25

8


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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.

N

NN

?

Br

problemsolution

NaN3

N3

2-azidopropane

TM = target molecule

N

O

O

O

TM-1 TM-2 TM-3

O

R

TM-4 TM-5 TM-6O

S

Ph Ph

CH3C

O

TM-7TM-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

S

R

D

TM-16TM-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 1oRCH2-X in SN2 reactions.

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

OR

Na

alkoxides

OR H

alcohols Na

H

Br R

O

SN2

ethers

Keq=Ka1

Ka2

10-16

10-37= = 10+21

H

Hacid/base


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Make potassium t-butoxide and use as sterically large, strong base at 1oRCH2-X, 2oR2CH-X and 3

oR3C-X in E2

reactions.

OK

K

O

H

t-butylalcohol potassiumt-butoxide

HBr

H

R E2

R

alkenes

H

H

Keq=Ka1

Ka2

10-19

10-37= = 10+18

potassium t-butoxide, sterically bulky basethat mostly does E2 reactions with RX

compounds (except SN2 with CH3-X).

acid/base

Make thiols using NaSH as the nucleophile at Me-X, 1oRCH2-X and 2oR2CH-X in SN2 reactions.

SH

Na

hydrogen sulfide

Br H

S

thiolSN2

Make thiolates and use as nucleophiles at Me-X, 1

o

RCH2-X and 2

o

R2CH-X in SN2 reactions.

acid/base

SR

Na

alkylthiolates

NaSR 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 thatcan be made into aldehydesand ketones

S

S

dithiane anion

LiS

S

dithiane

aldehydes

andketones(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 1oRCH2-X in SN2 reactions.

CCR

Na

terminalacetylides

CCR

terminalalkynes

H

NaNR2

H3C

Br

SN2

R

CH3alkynes

Keq=Ka1

Ka2

10-25

10-37= = 10+12

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

acid/base


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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 C-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

N

Li

Think - sterically bulky, very basic

that goes after weakly acidic protons.

LDA = lithium

diisopropyl amide

given

given

acid/base

React ketone enolate (nucleophile) with R-Br electrophile (Me, 1oand 2oRX compounds)

H2C

O Li

ketone enolates(resonance stabilized)

React ketone enolate with RX compounds (methyl, 1oand 2oRX)

Br

O

1oRX key bond

RS Larger ketone made

from smaller ketone.

SN2reactions

-78oC


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React ester enolate (nucleophile) with R-Br electrophile (Me, 1oand 2oRX compounds)

H2C

O

O

Li

ester enolates(resonance stabilized)

React ester enolate with RX compounds (methyl, 1oand 2oRX)

Br

O

O

1oRX key bondR

S Larger ester madefrom smaller ester.

R RSN2

reactions

-78oC

diphenylmethylsulfonium

bromide salt

S

Ph

Ph

CH2

sulfur ylid

to make epoxides

S

Ph

Ph

CH2make epoxides

from aldehydes

and ketones

H2C

Li

Keq =Ka1

Ka2

10-50

10-35= = 10+15

H

S

Ph

Ph

H3C

Br

diphenyl

sulfide

SN2

acid/base


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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 alwaysbasic (= electron pair donation by

hydride to a proton), nevera nucleophile. The conjugate acid of hydride is hydrogen gas (with a pKa= 37, H2can

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 saltscontaining 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

H

H Na Li

Al H

H

H

H

B D

D

D

D Na

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

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


MW = 41.8 g/mol

Aldrich, 2012$125 / 5 gramsMW = 42 g/mol

Li

Al D

D

D

D

The deuteride salts areway 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 wont be important for us to know until we study carbonyl reactions.


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

Li

Al D

D

D

D

C

Cl

CH3

H

+ ?

Na

B H

H

H

H

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 C-X. The C-H

proton has to be anti to the C-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= -2.3RTlog(kE2)

G = -2.3RT log(Keq)

OH

C

C

R1R2

Br

C

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> R2and R3> R4then this would be Zconfiguration, which is fixed by therequirement for anti C-H / C-Xelimination.

E2 mechanism depends on

steric factors and basicity ofthe electron pair donor. Moresteric hindrance and more basicfavors E2 over SN2

= negative(exergonic)

Keeping Track of the C Hydrogens in E2 reactions


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At least one Cposition, 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 Cpositions. In E2 reactions there can

be anywhere from one to three Ccarbons, and each Ccarbon 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 Cpositions

unique methyl,

only SN2 possible

one Cposition

1oRX, usually SN2 > E2

(except with t-butoxide)

two Cpositions

(2oRX, SN2 / E2

both competitive)

three Cpositions

(3oRX, 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 Lets

consider a moderately complicated example (next problem).

B

Either sketch will work for everypossibility above, IF you fill inthe blank positions correctly.

C

C

H

X

CH

C X?

B

?

horizontal perspectiveC-H / C-X

vertical perspectiveC-H / C-X

Problem 11 - How many total hydrogen atoms are on Ccarbons in the given RX compound? How many different

types of hydrogen atoms are on Ccarbons (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 allof the Chydrogen atoms. Notice the

base/nucleophile is strong, so SN2 or E2reaction are considered. The RX pattern istertiary, so no SN2 is possible. There arethree Ccarbon atoms and all of themhave hydrogen atoms on them. That leavesE2 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, Cand

C1, that make this structure morecomplicated 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.


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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

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?

Br

a.

H

O

H

b.

H3CO

H

c.

H3C

C

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).


38/65



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 Cpositions. After all, there can be either two or three Ccarbons 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 positions3o R-X has nine C positions

CC

C

H

CC

C

C

XH3CR

H2C

XNo Reaction

in our course.

NuH

NuHNuH

methyl primary

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

E1 Mechanism (unimolecular kinetics) loss of proton from any adjacent C-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.


39/65



SN1 / E1 possibilitiesextra complications at Cpositions, In this problem 2oRX, rearrangements are NOT considered

(H2O,ROH,RCO2H)


40/65


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42/65


43/65



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 Ccarbon 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 Cposition or even between

two positions in completely different molecules. However, these we will not emphasize such possibilities in ourcourse.

Transition state of a carbocation rearrangement

C

CH2

H3C

H

HC

X

CH2

SN1

E1

rearrangement

2ocarbocation

SN1 and E1 arepossible here

XH3C

1,2-hydride shift

CC

H

CH2CH2

H

transition state(no finite existance)

migrating group is positionedbetween two vicinal carbon atoms

(vicinity = neighbors, Latin)

CC

H

CH2

HH2C

H3C

3ocarbocation

CC

H2C

H3C

H

HH2C

R-X

TS1 TS2TS3

Ea

G

2oR3oR

SN1 & E1products

PE

Progress of Reaction (POR)

H: = hydride shift

R: = alkyl shift

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

SN1 and E1 arepossible 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 Catom to the Catom (the

carbocation center). These are the 1,2 hydride or 1,2 alkyl shifts mentioned above. The Catom 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 interpretinga

reaction mechanism involving rearrangements, you may have to consider both equal (3o3oR+) and more

stable (2o3oR+) carbocation possibilities. However in this book when you are asked topredictwhat mightbe possible, you usually only need to consider more stable carbocation possibilities (2o3oR+).


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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, 2

oand 3

ocarbocations. 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

slow

stepR.D.S.

R

O

H

CH2

HH

ab

c

CH2

H

H

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 tertiarycarbocation), 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/mole6 kcal/mole), resulting in an overall potential energy change of -5 kcal/mole.


45/65



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.

H

CH3

CH3

H CH3

CH3

major(85%)

minor(14%)

H CH3

CH3

HH

Path c leads toring expansion

rearrangement.2ocarbocation

3ocarbocation

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 3ocarbocation

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

HH

CH3

CH3

H

CH3

HO

H

R

CH3

H

CH3

CH3

CH3

HO

H

H

lanosterol precursor

lanosterol

B H

R

O protonatedsqualene

acidcatalysis

rearrangement1

R

CH3H

CH3

CH3

H

CH3

HO

H

H

rearrangement2

R

CH3

CH3

H

CH3

HO

H

H

H

rearrangement4

CH3

R

CH3

H

CH3

HO

H

H

HCH3

CH3

R

CH3

CH3

CH3

H

CH3

HO

H

H

H

rearrangement3

E1 reaction

19 moresteps

H3CH

CH3

CH3

HO

H

H H

H

cholesterol

other

bodysteroids

H

squalene

Requires 5arrows toshow thereaction.


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47/65



SN1 / E1 possibilitiesextra complications at Cpositions of 2oRX, rearrangement to more stable 3oR+considered


48/65


49/65



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. Waters conjugate acid is H3O+, (pKa= -2), while

hydroxides 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 ().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, 2oand 3oROH reacted with HX acids (HCl, HBr, HI) - usually SNchemistry

i. methyl alcohols (SN2 emphasized, no rearrangement)

H3C

O HBrH

methanol

H3C

O H

H

Br

H3C

O

H

H

Br

SN2

BrH

O

H

H H

bromomethane

Br

water is a goodleaving group

pKa= -9

ii. primary alcohols (SN2 emphasized, no rearrangement)

H3C CH2

O H

BrH

primary alcohol

H3

C CH2

O H

H

Br

H3C CH2

O

H

H

BrSN2

BrH

O

H

H

primarybromoalkane

Br

pKa= -9


iii. secondary alcohols (SN1 emphasized, rearrangements possible)


ClH

secondary alcohol (trans OH)

H2C H3C H

SN1

Cl

H3C

H

Cl

H3C

Cl

H

trans Cl

cis Cl

top and bottomattack

H3C

H

O

H

H

O

H

H

secondarychloroalkane

pKa= -7

O

H

H O

H

H H

ClH Cl


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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 bottomattack

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, 2oand 3oROH reacted with H2SO

4and 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

O

H OH SO3H

(heat)

O

H

H

C

H

H

H H

rearrangementH

H

O SO3H

E1bp = -4

distillsbp = +82oC

very difficult(high temperature)

OH SO3H O SO3H

alcohol alkene

Tbp= 129oC

pKa= -10

ii. secondary alcohols (E1 emphasized, rearrangements possible)

secondary alcohol

O

H

H O

H

H H

O

H OH SO3H

(heat)

O

H

H

O SO3H

E1

bp = +83oC

distills outbp = +161oC

OH SO3HO SO3H

H

H

H

alcohol alkene

Tbp= 78oC

pKa= -10 water is a goodleaving group


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iii. tertiary alcohols (E1 emphasized, rearrangements possible)

tertiary alcohol

O

H

H O

H

H H

OH SO3H

(heat)

O SO3H

bp = +33oC

distills out

bp = +102oC

OH SO3HO SO3H

O

H O

H

H

H

H H

E1

bp = +39o

distills ou

90% < alke(more substitu

minor alkene(less substituted)b

a

a b

alcohol alkene

Tbp63oC

pKa= -10water is a goodleaving group

We have some, limited control to direct the alcohol functionality toward SNor E choices. The conditions to

effect these different pathways are important, so you must be aware of the details mentioned above (halide acids =

SNreactions 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 (161oC)

(-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.


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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

ClO

O

N H

OS

O

O

Br

Na

separate

step

Br

HBr

rearrangement

and

R,S (racemic)

S

1. TsCl/pyridine

2. NaCl

(prevents R+ formation

and any rearrangement)

alkyl tosylate = RX compound

compare - a diff erent result

R

R RR

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 fromalcohols 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, 1oROH = acyl-like substitution at SOCl2, then SN2 at methyl and primary RX.

R O

H 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 andprimary alcohols

No rearrangement because no R+.

H

ClH

Cl

O

SCl

O

Hproduct

O

S

O

H

acyl-likesubstitution

Cl

Cl


53/65



Thionyl chloride with 2oand 3oROH = acyl substitution, then SN1 (there are various ways you can write this

mechanism)

O H O

S

O Br

Br

OS

O

H

O

SBr

O

HH

Br

SN1 at secondary andtertiary 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-likesubstitution

S

Br

O

Br

Br

Br

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

C

R

O

O

HC

R

O

O

H

S

O Cl

Cl

resonance

C

R

O

O

H

S

O

Cl

Cl

C

R

O

O

H

S

O

Cl

Cl

Base

C

R

O

O

S

O Cl

C

BaseHC

R

O

O

SCl

O

CR OCR O

resonance

C

R

Cl

O acylium ion

S

Cl

O

Cl

OS

O

Cl

Cl

resonance

Formation of esters from ROH + acid chlorides, amides from RNH2or R2NH + acid chlorides and anhydrides from

RCO2H + acid chlorides

OR

O

R

H

Cl

OCl O

H

O

Cl

H

O

O

O

R

R

There are many variations of ROH and

RCO2H joined together by oxygen.

ester synthesis from acid chloride and alcohols

R

O

H

R

O

H

H


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55/65


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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

H

O

H

SN1 E1 SN2 E2

O

O

O

H

O

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

H

O

2 6

1 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

H

O

H

D H D H


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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

H H

H H

C

C

Ha

Br

C

H

H3C

Hb H

H

Rotation of C-C1brings Haor Hbanti to C-Br, which allows SN2 andtwo different E2 possibilities: Ha(Z) and Hb(E). Since C2is a simplemethyl, there is no C2substituent to inhibit either of these reactions.

C

C

Hb

Br

C

H

Ha

H3C H

H

When methyl on C1 is antito C-Br, no SN2 is possible andno E2 is possible from C1.

SN2 possibleE2 from C1(2Z- butene)

E2 from C2(1-butene)

SN2 possible

E2 from C1(2E- butene)

E2 from C2(1-butene)

No SN2 possible and no

E2 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

H

HBr

H

H

H

H

H

No SN2 is possible (1,3 diaxialpositions block approach ofnucleophile), and no E2 is

possible because ring carbonsare anti.

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

SN2 possible if Cisnot tertiary and thereis no anti C"R" group.

E2 possible

with anti C-H.

Both SN2 and E2 are possiblein this conformation with leavinggroup in axial position.

H H

H

H

H HBr

H

C

H

H

H

HH

H

H

H

HBr

H

H

H3C

HH

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

No SN2 possible if thereis an anti C"R" group. E2 possible

with anti C-H.

Only E2 is possible in thisconformation. Leaving groupis in axial position.

H

H

H

full rotation at

C-Cis possiblein chain

only partialrotation is

possible in ring

only partialrotation is

possible in ring

H H H

equatorial leaving group

axialleavinggroup

No SN2 is possible (1,3 diaxialpositions block approach ofnucleophile), and no E2 is

possible because ring carbonsare anti.

No E2 possible,no anti C-H.

full rotation atC-Cis possiblein chain

alkene stabilitiestetrasubstituted > trisubstituted > trans-disubstituted > gem-disubstituted cis-disubstituted > monosubstitute


58/65


59/65



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?

H

O

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 6

SN1

X

X 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 carbocationrearrangements thatimmediately go to a more

stable carbocation (notequally stable carbocations= too many possibilities)

a.

b.

2 2

1 2

X

a.

b.

c.

d.

consider any 3oR+possibility

and consider stereoisomers

4 5

1 2

1 2

H

ONa

condition 1 condition 2


60/65



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)

SClO

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

N

pyridine =proton sponge

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

sodium borohydride(nucleophylic hydride)

S

S

dithiane Br2

Pd / H2

quinoline(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)

pyridineH2O / H3O

+

CO2carbon dioxidebromine chlorine

hydrogenchloride

hydrogenbromide

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

Na

sodiummetal

3 ozonereactions

1. O3, -78oC

2. CH3SCH3

1. O3, -78oC

2. NaBH4

1. O3, -78oC

2. H2O2, HO

1. Hg(OAc)2/H2O

2. NaBH4

1. Hg(OAc)2/ROH2. NaBH4

O

O O

H

Cl

meta chloroperbenzoic acid(mCPBA)

Na / NH3

in ammonia(Birch reagent)

= py

(PCC) (Jones)

N

O O

Na

Mn

O O

OO

Os

O O

OOK

O

O O Cr

O

OO

Cr

O

OO

B

H

H

H

H AlH

H

H

H

NaLi

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

Li

lithiummetal

Zn

zincmetal

Lewis aciAlBr3FeBr3BF3

SOCl2PBr3SO3etc.

Various metals

HgX2

(mecuric salts)

HOOH

H2

NNH2 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)

BH

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.


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Allowed starting structures our main sources of carbon 1. Free radical substitution of sp3C-H bonds to form sp

C-Br bonds at the weakest C-H position and 2. Anti-Markovnikov addition to alkenes makes 1oR-Br. From these two

reactions we can make 13 R-Br molecules below.

CH4

Br Br

BrWe need to make these 1oRBr

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

H3C

BrBr

Br

Br BrBr

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 RBr

compounds: This is just

free radical substitution

at allylic sp3C-H position

of an alkene.

Br

benzylic RBr,f rom above

Br

PhBr2/ h

2 eqs.

Ph

Br

Br

Br Br Br Br

E2 reaction(twice)

NaR2N

1. 3 eqs.

2. workup

Ph

a b c a ab

b

c

c

Br

bromobenzeneis given untilaromatic chemistryis 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


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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 sp3C-H bonds to form sp3C-Br bonds at weakest C-H

position

H3C

H2

CCH3 H3C

CHCH3

Br

BrBr

overall reactionh

BrH

1. initiation

BrBrh

BrBr H = 46 kcal/mole

weakest bond ruptures first

2b propagation

H3C

C

CH3

H

BrBr

H3C

C

CH3

H Br

BrH = -22 kcal/mole (overall)

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

2a propagation

H3C

C

CH3

H H

Br BrH

H3C

C

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

Br

H3C

C

CH3

H

H3C

C

CH3

H Br

H3CC

CH3

H

CH3

C

H3C

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)R

O

O

R hR

O

O

R

BrH

R

O R

O

H Br

H = 40 kcal/mole

H = -23 kcal/mole


(cat.)

reagent


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2a propagation

H3C

HC

CH2 Br

H3C

C

CH2

Br

H

H = -5 kcal/mole


H = -15

both steps(2a + 2b)

2b propagation

H3C

C

CH2

Br

H

BrH H3C

H2C

CH2

Br Br

H = -10 kcal/mole


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

1. PCC = pyridinium chlorochromate, (CrO3/pyridine), CrO3oxidations of alcohols (methyl, 1oand 2oROH)

without water. Steps are: 1. Cr=O addition, 2. acid/base and 3. E2 to form C=O (aldehydes and ketones).

H3C

CH2

C

H

O

H

H

Cr

O

OO H3C

CH2

C

H

O

H

H

Cr

O

O

O

H3C

CH2

C

H

O

H

Cr

O

O

O

NN H

N

H3C

CH2

C

H

O

Cr

O

O

ON H

PCC = pyridinium chlorochromate oxidationof primary alcohol to an aldehyde (no water tohydrate the carbonyl group)

primary alcohols

aldehydes

E2

CrO3oxidations of alcohols (methyl, 1oand 2oROH) without water = PCC, Cr=O addition, acid/base and E2 to

form C=O (aldehydes and ketones)

H3C

CH2

C

CH3

O

H

H

Cr

O

OO H3C

CH2

C

CH3

O

H

H

Cr

O

O

O

H3C

CH2

C

CH3

O

H

Cr

O

O

ON N H

N

H3C

CH2

C

CH3

O

Cr

O

O

O

N H

PCC = pyridinium chlorochromate oxidationof primary alcohol to an aldehyde (no water tohydrate the carbonyl group)

primary alcohols E2

ketones


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2. Jones reagent = CrO3/water/acid, CrO3oxidations of alcohols (methyl, 1oand 2oROH) 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 1oalcohol (forms carboxylic acids from primary alcohols and ketones

from secondary alcohols).

H3C

CH2

C

H

O

H

H

H3C

CH2

C

H

O

Jones = CrO3/ H2O / acidprimary alcohols oxidize to carboxylic acids

(water hydrates the carbonyl group,

which oxidizes a second time )

primary alcohols

aldehydes (cont. in water)

H

O

HH

OH

H

H3C

CH2

C

H

O

H

H3C

CH2

C

H

O

H

H

O

H

H3C

CH2

C

O

H

O

H

H

H

H

O

HH3C

CH2

C

O

O

H H

H

H

O

H

H

O

H

H3C

CH2

C

O

OH

hydration ofthe aldehyde

second oxidation of the carbonyl hydrate

H

O

H

resonance

carboxylic acids

Cr

O

OOH3C

CH2

C

H

O

H

H

Cr

O

O

O

H3C

CH2

C

H

O

H

Cr

O

O

O

Cr

O

O

O

H

O

H

H

H

O

H

H

Cr

O

OO H3C

CH2

C

O

O

H

Cr

O

O

O

H

H

H3C

CH2

C

O

O

H

Cr

O

O

O

H

H

O

H

H

Cr

O

O

O

H

O

H

H


65/65

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315 SN E Syn & Chem Catalog

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