3

HOW TO USE THIS BOOK - 5 -

UNIT 1 STRUCTURE, BONDING AND DRAWING - 7 - 1.1 Lewis Bonding and Formal Charges - 9 -

1.2 Drawing Condensed and Bond-Line Structures - 19 - 1.3 VSEPR Theory and Drawing Molecules in 3 Dimensions - 22 -

1.4 Molecular Orbital Theory: sp3, sp2 and sp Hybridization - 27 - 1.5 Constitutional Isomers and Units of Unsaturation - 39 -

UNIT 2 FUNCTIONAL GROUPS AND PHYSICAL PROPERTIES - 45 - 2.1 Functional Groups and Their Classification (1o, 2o Etc.) - 47 - 2.2 Intermolecular Forces, Boiling Points and Solubilities - 57 -

UNIT 3 NOMENCLATURE - 67 - 3.1 Nomenclature of Alkanes and Cycloalkanes - 69 - 3.2 Nomenclature of Alkenes and Alkynes - 75 - 3.3 Nomenclature of Alkyl Halides, Alcohols and Ethers - 79 -

UNIT 4 CONFORMATIONS - 85 - 4.1 Conformations of Ethane and Butane: Newman Projections - 87 - 4.2 Small Cycloalkanes (C3-C5): Ring Strain and Conformations - 91 - 4.3 Cyclohexane: Chair and Boat Conformations - 93 - 4.4 Substituted Cyclohexane Conformations - 97 -

UNIT 5 STEREOCHEMISTRY - 103 - 5.1 Chirality, Enantiomers and Chirality Centers - 105 - 5.2 R, S Configurations: The Cahn-Ingold-Prelog Rules - 111 - 5.3 Optical Activity and Racemic Mixtures - 115 - 5.4 Two Chirality Centers: Diastereomers and Meso Compounds - 120 - 5.5 3-D Bond-Line Structures and Fischer Projections - 124 - 5.6 Resolution of Enantiomers - 127 - 5.7Alkene Stereochemistry: Assigning E/Z and Cis/Trans - 128 - 5.8 Table of All Isomer Types Including Stereoisomers - 129 -

UNIT 6 HOW MOLECULES REACT: RESONANCE, ENERGETICS, - 131 - ACIDS AND BASES 6.1 Resonance and Electron-Pushing Rules - 133 - 6.2 Energetics: DGo, DG‡, Keq, Rates and Transition States - 140 - 6.3 Organic Acids and Bases: Brønsted-Lowry Theory, pKa and Equilibria - 146 - 6.4 Trends in Acidity: Electronegativity, Resonance and Inductive Effects - 157 - 6.5 Electrophiles, Nucleophiles and Mechanism Defined - 165 -

4

UNIT 7 ALKYL HALIDES: SN2/E2 AND SN1/E1 REACTIONS - 169 - 7.1Overview of All 4 Reactions - 171 - 7.2 The SN2 Reaction - 173 - 7.3 The E2 Reaction: Dehydrohalogenation - 184 - 7.4 The SN1/E1 Pair of Reactions - 196 - 7.5 Deciding Which Substitution or Elimination Happens - 202 - 7.6 List of Essential Alkyl Halide Reactions - 203 - 7.7 List of Reactions to Synthesize Alkyl Halides - 205 -

UNIT 8 ALKENES - 207 - 8.1 Unsymmetrical Additions: HX, H2O and Markovnikov’s Rule - 209 - 8.2 Symmetrical Additions: X2, H2, 1,2-Dihydroxylation and Epoxidation - 220 - 8.3 Oxidative Cleavages with Hot KMnO4 or O3: Carbonyl Products - 227 - 8.4 List of Essential Alkene Reactions - 231 - 8.5 List of Reactions to Synthesize Alkenes - 235 -

UNIT 9 ALKYNES - 237 - 9.1 Reactions of Alkynes - 239 - 9.2 List of Essential Alkyne Reactions - 247 - 9.3 A Synthesis of Alkynes - 251 -

UNIT 10 ALCOHOLS - 253 - 10.1 Acid-Base Reactions of Alcohols - 255 - 10.2 Conversion to Good Leaving Groups: Alkyl Halides and Sulfonates - 257 - 10.3 Substitution Reactions: Ether Synthesis and Protecting Groups - 261 - 10.4 Dehydration Reactions: Synthesis of Alkenes - 267 - 10.5 List of Essential Alcohol Reactions - 271 - 10.6 List of Reactions to Synthesize Alcohols - 275 -

UNIT 11 ETHERS AND EPOXIDES - 277 - 11.1 Reactions of Ethers and Epoxides - 279 - 11.2 List of Essential Ether and Epoxide Reactions - 286 - 11.3 List of Reactions to Synthesize Ethers and Epoxides - 288 -

UNIT 12: A SHORT INTRODUCTION TO MULTISTEP SYNTHESIS - 289 -

ANSWERS TO TEST YOUR KNOWLEDGE PROBLEMS - 299 -

ABOUT THE AUTHOR AND ACKNOWLEDGMENTS - 349 -

ABOUT RESCUE EDUCATION INC. - 350 -

85

UNIT 4 CONFORMATIONS

4.1 Conformations of Ethane and Butane: Newman Projections - 87 - A. Conformations of Ethane: Newman Projections - 87 - B. Conformations of Butane: Gauche, Anti and Steric Interactions - 88 - Test Your Knowledge - 90 -

4.2 Small Cycloalkanes (C3-C5): Ring Strain and Conformations - 91 - Test Your Knowledge - 92 -

4.3 Cyclohexane: Chair and Boat Conformations - 93 - A. How to Draw Chair Cyclohexane and “Flip” Forms - 93 - B. Boat Cyclohexane - 95 - Test Your Knowledge - 96 -

4.4 Substituted Cyclohexane Conformations - 97 - A. Monosubstituted Cyclohexanes - 97 - B. Disubstituted Cyclohexanes - 98 - C. More Substituted Cyclohexanes - 101 - Test Your Knowledge - 102 - -

93

4.3 CYCLOHEXANE: CHAIR AND BOAT CONFORMATIONS

A. How to Draw Chair Cyclohexane and “Flip” Forms • Flat cyclohexane would have bad C–C–C bond angles (120o instead of 109.5o) and terrible

torsional strain (all C–-H bonds eclipsing) • it would have very significant ring strain (4.2)

• Real cyclohexane exists mainly in the most stable chair conformation which is strainless • all bond angles are exactly 109.5o (no angle strain) and all C–H bonds on adjacent

carbons are perfectly staggered (no torsional strain)

• The chair conformation can be imagined to look somewhat like a patio “lounge” chair

• Drawing a credible chair cyclohexane is crucial and requires practice • building a correct 3-dimensional model at this point is INVALUABLE

• There is a step-wise procedure you can initially follow to understand how it should look • It is useful to know that the structure has 3 sets of parallel lines

• draw the two vertical parallel lines exactly the same length and slanted exactly the same amount to the right

• draw parallel lines (one connected to the left line and one connected to the right line already in place) and have them both slanted towards the center

• draw the remaining parallel lines to connect the carbons to complete the chair

• The attached hydrogens must be in either an axial or equatorial orientation

H H

H

H

H

H

H

H

H

H

H

H Often simplified to:

“Lounge” chairChair cyclohexane

H H

HH

H

H

Axial hydrogensHydrogens straight up on “up” carbons

Hydrogens straight down on “down” carbons

H H

HH

H

H

HH

H

H

HH

Equatorial hydrogensSticking out around periphery of ring

“Up” carbon

“Down” carbon

94

• axial hydrogens are drawn either straight up or down • on a carbon which is higher relative to its neighboring carbons (“up” carbon), the

axial hydrogen points straight up • on a carbon which is lower relative to its neighboring carbons (“down” carbon),

the axial hydrogen points straight down • equatorial hydrogens are more difficult to draw correctly but 2 observations will help

• the bond angles between the axial and equatorial hydrogens are ideally 109.50 • in practice, they can be drawn with a bond angle somewhat larger than 90o • when drawn correctly, each equatorial C–H bond will be parallel to both C–C

bonds 2 carbons away

• In reality there is usually no time on exams to carefully draw a chair conformation • learn to free-hand draw the chair skeleton itself • don’t draw in all hydrogens but only the groups, atoms or hydrogens necessary

• if you draw a substituent, it is customary to draw the hydrogen on that carbon too • draw in necessary axial hydrogen/substituents pointing straight up on “up” carbons and

pointing down on “down” carbons • then draw in necesssary equatorial hydrogens/substituents so they are somewhat more

than 900 but less than 120o bond angles with the attached axial hydrogen/substituent

• The first quickly-drawn chair of trans-1,2-dichlorocyclohexane (below) is less than perfect, but it is easy to see that it is a diequatorial dichloride

• the second attempted chair drawing is absolutely wrong for many reasons • the orientations of the substituents and many bond angles are wrong • it looks like and equatorial-axial dichloride which is incorrect

H

H

H

H

H

H

109.50

All H–C–H anglesideally 109.5o

HH H H

Equatorial bonds are parallel to C–C bonds 2 carbons away

109.50

Cl

H

H

Cl ClH

H

ClNOT:

95

• Cyclohexane rapidly “flips” back and forth between two equivalent chair conformations • if a particular hydrogen is labeled (circled), it is seen to rapidly interconvert between

axial and equatorial positions at room temperature (the same carbon is labeled x) • in the case of cyclohexane, the two flip forms exist as a 50:50 mixture

• To draw a flipped chair conformation properly (and to avoid mistakes) do the following: • draw the flipped chair in a mirror image chair orientation

• the result is that the same carbon (indicated by an x above) will be at a slightly different position relative to the viewer in the two flip forms

• do keep the same carbons as close to the same position in the two drawings • keep the observed group on the same face of the ring

• in this case, the circled hydrogen stays on the upper face of each ring

• Manipulating molecular models will make chair “flipping” crystal clear • models can be flipped by holding any two opposite carbons (4 away) and then torqueing

the “down” carbon “up” and the “up” carbon “down” • a colored group at an axial position should move to the equatorial position

• Chair cyclohexane is the most stable (lowest energy) conformation and this is how cyclohexane spends most of its time

• the carbon-carbon bonds in cyclohexane are rapidly rotating and so it is constantly changing conformations

• it spends time in other less stable conformations, but only briefly

B. Boat Cyclohexane • The boat conformation of cyclohexane is a less stable but important conformation

• it is a significant conformation but exists to a far lesser extent than the chair

H

H H

H

Ring flip drawing conventions:1.They have mirror image ring skeletons2.They are viewed from a slighly different angle3.The observed group stays on same face of ring (top-side in this case)

xx

In flip forms: All axial substituents become equatorial All equatorial substituents become axial

H

HH

H

How to Flip a Model:Hold any two opposite carbons and

torque the “down” carbon “up” and the “up” carbon “down”

Hold

Hold

96

• cyclohexane spends almost all of its time in the chair conformation and a small amount of time in the boat (and other less stable) conformations

• it is drawn as a rudimentary boat with two “prows” at left and right

• The boat conformation has two major sources of instability 1. The hydrogens at the prow have unfavorable steric interactions (i.e. they are bumping into

each other) • this is called a prow or masthead interaction

2. The hydrogens on the carbons at the bottom of the boat (front and back) are totally eclipsed and so have torsional instability

• there are 4 sets of these eclipsed hydrogens

TEST YOUR KNOWLEDGE 4.6 Using the “parallel lines” rules, carefully draw a chair cyclohexane with all of the hydrogens in

the correct axial and equatorial orientations. Circle all equatorial hydrogens. Quickly free-hand draw a chair cyclohexane and draw in all hydrogens.

4.7 Draw one chair with an axial bromine on the bottom face and then draw its correct flip form. 4.8 Draw a boat cyclohexane and identify and label its two sources of instability.

“Prows”

HH

H

H

HH

H

H

H

H

HH

1. Unfavorable prow or masthead interactions of hydrogens (a bad steric interaction)

2. There are 4 pairs of eclipsed H’s along the front and back bottom of the boat (bad torsional interaction)

Sources of instability in the boat conformation:

207

UNIT 8 ALKENES

8.1 Unsymmetrical Additions: HX, H2O and Markovnikov’s Rule - 209 - A. How Alkenes (and Alkynes) Usually React in Additions - 209 - B. HX: Markovnikov’s Rule - 210 - C. Carbocation Rearrangements - 212 - D. H2O, Catalytic Acid: Markovnikov Hydration - 214 - E. 1. BH3•THF 2. H2O2, NaOH: Anti-Markovnikov Hydration - 215 - Test Your Knowledge - 219 - 8.2 Symmetrical Additions: X2, H2, 1,2-Dihydroxylation and Epoxidation - 220 - A. Br2 or Cl2: A Stereospecific Anti Addition - 220 - B. H2 with Pt, Pd or Ni Catalyst: Catalytic Hydrogenation - 222 - C. Cold KMnO4 or OsO4: Syn 1,2-Dihydroxylation - 224 - D. RCO3H (Peroxyacids): Syn Epoxidation - 225 - Test Your Knowledge - 226 - 8.3 Oxidative Cleavage with Hot KMnO4 or O3: Carbonyl Products - 227 - A. Oxidation Levels of Carbon - 227 - B. Hot KMnO4: Most Highly Oxidized Carbonyl Products - 227 - C. O3 (Ozonolysis): Less Oxidized Ketone and Aldehyde Products - 228 - Test Your Knowledge - 230 - 8.4 List of Essential Alkene Reactions - 231 - Test Your Knowledge - 234 - 8.5 List of Reactions to Synthesize Alkenes - 235 -

209

8.1 UNSYMMETRICAL ADDITIONS: HX, H2O and MARKOVNIKOV’S RULE

A. How Alkenes (and Alkynes) Usually React in Additions • In many organic reactions, a nucleophile (electron-rich molecule or atom) reacts with an electrophile (electron-poor molecule or atom) (6.5)

• Recall that alkenes (and alkynes) have p bonds with loosely held electrons (1.4C, D) • alkene double bonds are electron-rich and are nucleophiles • they can react with electrophiles which are electron-deficient

• electrophiles have atoms with a partial (d+) or full + charge

• In the 1st step of the generic example below, the alkene is the nucleophile because it has an electron-rich p bond

• The unsymmetrical and polarized molecule Y-Z is the electrophile • it has a d+Y atom which is electron-deficient and the actual electrophilic center

• the whole molecule is called the electrophile even though only Y is the electrophilic atom

• In the 2nd step of this reaction the electron-deficient carbocation intermediate is the electrophile and electron-rich Z– is the nucleophile

• notice that the organic molecule has switched from being a nucleophile in the 1st step to an electrophile in the 2nd step

• Typical molecules that add to double bonds in this way are HX and H2O • Alkynes react in a similar way (9.1A) but since they have two p bonds they will usually

perform 2 addition reactions instead of 1 (next page) • sometimes the reaction stops halfway if 1 equivalent of reagent is added

δ -Z Y

δ+

YZ

NucleophileElectron-rich π bond

ElectrophileElectron-deficient

carbocation

Electrophilic center

Nucleophile

Y Z

Addition Product

Y Z

Electrophile

210

B. HX: Markovnikov’s Rule

• The generic reaction (top) shows that addition of H-X across double bonds yields alkyl halides

• In the example, the unsymmetrical alkene propene reacts with HBr, to produce one of the two possible products as practically the only product

• the major Markovnikov product has Br attached to the most substituted carbon and H attached to the least substituted carbon

• the very minor anti-Markovnikovproduct is usually not even drawn • this reaction is said to go with Markovnikov selectivity

• This preference was noted long ago and formulated into a rule by Markovnikov • this rule was based solely on empirical observations of the products formed

• you can still use it to correctly predict the major Markovnikov product

Original Markovnikov’s Rule HX adds to a double bond so that the H atom is attached

to the carbon that already has the most H atoms

• This is a specific example of a regioselective reaction • a regioselective reaction produces one of two possible constitutional isomers as

the major product • the competing products are called regioisomers

Two electron-rich π bonds

Reacts twice with Y-Z

Y Z Y Z

H H

Y ZH H

Y(Z) Z(Y)

1 eq.

1 eq. = 1 equivalent

Y Z1 eq.

δ–Z Y

δ+

H XH X

HX = HI, HBr, HCl

Markovnikov Selectivity

H3C CH CH2HBr H3C CH CH2

HBrMarkovnikov product

Major

+ H3C CH CH2BrH

Anti-Markovnikov productVery minor

Example

211

• in the example above, 1- and 2-bromopropane are regioisomers

• The basis of Markovnikov’s rule can be derived from the reaction mechanism • this is a very important mechanism to know and understand

• In the 1st step, the electron-rich p electrons react with the d+ hydrogen of HBr • The H can potentially attach to either of the carbons of the alkene

• the reaction proceeds to give predominantly the most stable 2o carbocation intermediate instead of the less stable 1o carbocation

• carbocation stabilities were discussed in 7.4B • In the 2nd step, nucleophilic Br– reacts with an electrophilic carbocation

(most of which is the 2o carbocation)

• The large preference for the formation of the most stable carbocation intermediate is the basis of a modern formulation of Markovnikov’s Rule

• this rule can be applied more broadly to many unsymmetrical additions

Modern Markovnikov’s Rule When unsymmetrical reagent Y-Z adds to an unsymmetrical double bond, the d+ atom attaches

to produce the most stable carbocation intermediate

SAMPLE PROBLEMS Give the major product or starting compound for the following reactions; whichever is indicated by the box.

H3C CH CH2

H3C CH CH2H

Most stable2o carbocation

Major

H3C CH CH2H

Less stable1o carbocation

Very minor

H Brδ+ δ –

H3C CH CH2HBr

H3C CH CH2BrH

Anti-Markovnikov productVery minor

Markovnikov productMajor

Br

Br

Usually notdrawn

a.HCl

b. c.HI HBr

Br

212

ANSWERS Draw H atoms on the alkene to help choose the correct Markovnikov product. Add the new H to the carbon with most H atoms already. In c., draw the two possible starting alkenes and subject them to the reaction conditions to determine which gives the correct major product.

• Even when a new chirality center (5.1) is formed in this reaction, there is no stereoselectivity • the product is a 1:1 mixture of enantiomers (i.e. a racemate) • reaction of butene with HBr results in a racemic mixture of chiral products

• the reaction goes through a planar carbocation and attack of the halide is equally likely from either side

• recall that when two achiral starting materials react to form a chiral product, the product must be a racemate (5.3B)

C. Carbocation Rearrangements • Any reaction which goes through a carbocation intermediate can potentially undergo a

carbocation rearrangement • the driving force for this rearrangement is almost always the shift of an atom or

group to form a more stable carbocation (7.4B) from a less stable carbocation

• The reaction of HBr with the following alkene gives both the expected Markovnikov product and a rearranged product in which the CH3 and Br appear to have switched

a.HCl

H

HClH

HH

H goes to carbon with most atoms already

(Least substituted carbon)

b.

HI

H HI

H

c.

HBrBr

Only alkenepossible

HBrBrH

This alkene givesthe wrong product

H

Major product

Major product

CH3CH2 CH CH2

H Br H2CH

CH3CH2 HBr Br

Planar carbocationEither attack equally likely

CCH3

CH2CH3HBr BrH

CH3CH2

CH3C+

Racemate1 : 1

* *

213

• The 1st step of the mechanism to form either product is the reaction of HBr to give the 2o carbocation (draw the rest of the mechanism for expected product yourself)

• In the 2nd step of the mechanism for the formation of the unexpected product, a methyl shift occurs (a shift of a CH3 with its electrons) from the 2nd carbon to the 3rd

• this shift converts the initial 2o carbocation into a more stable 3o carbocation on the adjacent carbon

• A similar substrate with one less CH3 will also react with HBr and rearrange but the type of shift is different

• in this case, the Br appears to “jump” to the next carbon

• The mechanism for formation of the unexpected product has the expected initial protonation of the alkene to form the 2o carbocation

• the 2nd step is instead a hydride shift (H moves with its electrons) to produce the more stable 3o carbocation form the less stable 2o carbocation

CH3 HBrCH3

BrBrCH3

Expected Markovnikov product

Unexpected product Br and CH3 “switched”

+

H BrCH3 CH3H

2o CarbocationLess stable

HCH3

3o CarbocationMore stable

CH3BrMethyl

shift

Br

Unexpectedproduct

HBrBr

Br

Expected Markovnikov product

Unexpected productBr “jumps” to next carbon

+

H BrH HH

2o CarbocationLess stable

H Br

3o CarbocationMore stable

Hydride shift

HBr

Unexpectedproduct

214

• When atoms and groups appear to “jump around”, a rearrangement of some sort is happening • either that or you drew something wrong!

• Although carbocation rearrangements can be rationalized after an unexpected product is isolated, it is very difficult to predict ahead of time that there will be one

D. H2O, Catalytic Acid: Markovnikov Hydration

• In the generic example, hydration of a double bond requires addition of H2O and a catalytic amount of strong acid

• with unsymmetrical alkenes, it will give Markovnikov selectivity

• Markovnikov hydration of 2-methylpropene yields the 3o alcohol as its only product • Direct addition of H2O to an alkene is impossible

• H2O is far too weak an acid to protonate the relatively unreactive alkene p bond • catalytic strong acid must be added to facilitate reaction of H2O with a p bond

• In the 1st step of this important mechanism a very strong acid (H2SO4) protonates H2O to form a hydronium (H3O+) which is still a strong acid

• recall H3O+ is the strongest acid allowable in H2O (6.3, the leveling effect)

H OH

cat. acid

Catalytic acid: H2SO4, HCl or other strong acid

H2Ocat. H2SO4

HOH

Example

Markovnikov SelectivityH-OH

Markovnikovproduct

H O HH O SO3H

H O HH Actual acid

Leveling effect

H O H

H

H

HO

OH H

OH+ H3O+

Strong acidreacts with π bond

Weak nucleophileH2O reacts

Protonated alcohol reacts with base (usually H2O)

RegeneratedHHO

H

215

• H3O+ is the acid that directly protonates the double bond • only a catalytic amount of H2SO4 is needed since H3O+ is regenerated in the last step

and can go back to protonate another double bond

• After formation of the 3o carbocation, the weak nucleophile H2O attacks it • carbocations are very powerful electrophiles and can react with weak (i.e. uncharged)

nucleophiles

• The protonated alcohol must be deprotonated by base to produce the final neutral alcohol • there is usually a large excess of H2O in this reaction so H2O is the likely base

• Hydration is an entirely reversible reaction • it is the reverse of dehydration of an alcohol to form an alkene (10.4A)

• hydration can be favored very easily by simply adding lots of water (because of Le Chetalier’s principle of equilibria)

• adding more catalyst will never shift an equilibrium • it will only cause the equilibrium to be achieved faster

• Since hydration goes through a carbocation mechanism, carbocation rearrangements are common with certain substrates such as 3-methyl-1-butene

E. 1. BH3•THF 2. H2O2, NaOH : Anti-Markovnikov Hydration • The previous two reactions, and in fact most unsymmetrical additions to unsymmetrical

alkenes, proceed with Markovnikov selectivity • Hydroboration/oxidation is the only reaction in this chapter to give anti-Markovnikov

selectivity • this is an extremely useful reaction because it gives exactly the opposite alcohol

regioisomer to that obtained with acid-catalyzed hydration (8.1D) • in this reaction, the least substituted alcohol is obtained

+ H2O OH

Hydration and dehydrationare in equilibrium

Favor alcohol by adding H2O

Hydration(cat. H2SO4+)

Dehydration

H2OOH

OH

Expected Markovnikov product

Unexpected productA hydride shift occurs

before H2O adds

+cat. H2SO4

H HH Carbocation

rearrangementsare common

216

• In the 1st step of this 2-step reaction, the molecule borane (BH3) is added across the double bond (“hydroboration”)

• we will discuss why there are two R groups on boron on the next page • In the 2nd step, the boron-containing group is replaced with a –OH (“oxidation”)

• this is accomplished with basic H2O2 • Notice in the example that the hydroxyl it attached to the less substituted carbon of the

double bond which means this reaction has anti-Markovnikov selectivity

• This reaction is also very powerful because it is selective for syn addition • in syn addition two atoms or groups add to the same face of a double bond • in anti addition two atoms or groups add to the opposite faces of a

double bond

• Let’s study the reaction of the simple substrate 2-methylpropene, in detail, to see why it has anti-Markovnikov selectivity

• after the 2-step reaction, the 1o alcohol is the major product

• Hydroboration-oxidation has 2 discreet reaction steps and reagents must be added sequentially and separately in the order indicated by the numbers 1 and 2

• adding both sets of reactants at once wrecks the reaction • The reagent for hydroboration is made by mixing diborane (B2H6) and THF (tetrahydrofuran)

• THF is a common solvent in organic chemistry but it is also used here to break apart the B2H6 to yield the more reactive BH3•THF complex

1. BH3 • THFHydroboration H BR2 H OH

2. H2O2, NaOHOxidation

Anti-Markovnikovand

syn selectivity

Example

CH3H

H OHH3C H

1. BH3 • THF2. H2O2, NaOH

Anti-Markovnikov productSyn (same side) addition product

X YX YSyn addition Anti addition

H H

X Y

H Y

X HH H

1. BH3 • THF

HBR2

2. H2O2, NaOH

HOH

217

• The addition of BH3 to the alkene is a concerted (“all at once”) addition • in concerted reactions, all bond-making and bond-breaking occurs at the same

time (or very close to it) • since the BH2 side of the reagent is much bigger than the H (hydride) side, the boron

goes to the least sterically hindered carbon

• The boron still has two B–H bonds left after the 1st addition • it can continue to add two more equivalents of alkene (3 equivalents in total) • the final product is therefore a trialkyl borane (i.e. R3B)

• all three R groups will turn into alcohols during the oxidation step

• The concerted nature of the reaction insures BH3 is added syn • in effect, BH3 “caps” the double bond with B and H on the same face • in a cyclic substrate, this syn addition is very easy to see

• Oxidation of the alkylborane with H2O2, NaOH has a complicated mechanism which we will not discuss

B2H6 + O

Diborane Tetrahydrofuran (THF) Common organic solvent

OBH

HH

BH3•THFMore reactive complex

H3CH3C

HH

H BHH

Concerted addition(“All at once”)

Syn addition H3CH3C H

H

H BH2

2 eq.

B

Larger BH2 adds to least hindered side of double bond

Small H adds to most hindered side

(R3B)

Syn addition

BH3•THF

BR2H

CH3 HH3C

BR2H

CH3 H2. H2O2, NaOH

OHH

CH3 H HO has exactly the same

stereochemistryas BR2

H and B mustattach to

the same faceHighly selective

Syn AND anti-Markovnikov

Addition

218

• all you should know is that –BR2 is replaced with –OH in exactly the same stereochemical position (if there is defined stereochemistry)

• syn addition of the borane leads to net syn addition of water • When the alkyl borane derived from 1-methylcyclohexene reacts, the final addition of water is

highly selective since it is both syn and anti-Markovnikov

SAMPLE PROBLEMS Give the major product, correct reagents or correct starting compound for the following reactions. Clearly show any stereochemistry in the product. Assume no carbocation rearrangements take place.

ANSWERS

a. CH31. BH3•THF2. H2O2, NaOH

b. CH3 CH3OH

c.

OH

1. BH3•THF2. H2O2, NaOH

Onlyd.

OH

H2O cat. H2SO4

a.CH3

OH H

CH3

OH

H Anti-Markovnikov and synCH3 and OH must be trans

You can draw this several waysor

b. CH3 CH3OH

H2O cat. acid

Markovnikov addition of H2ONo particular stereochemistry in product

You can say “acid” or specify (e.g. H2SO4)

c.

OH

NOT

OH

OHOnly

d.

Wrong product!

OHNOT

OH

+Somewhat symmetrical

alkene will give an unusable mixture of alcohols

Markovnikov

You must choose the alkene that will give the desired alcohol as

its only (or major) product

H2O cat. H2SO4

1. BH3•THF2. H2O2, NaOH

Roughly equal amounts

219

TEST YOUR KNOWLEDGE 8.1 For the following reactions, draw the major product and the complete step-by-step

mechanism including all correct electron-pushing arrows. What is the name for the selectivity in each of these reactions?

8.2 React each molecule with each reagent and draw the correct major product. If you expect a

particular stereochemistry in the product then show it.

Reagents: i. 1. BH3• THF 2. H2O2, NaOH ii. HCl iii. H2O, cat. H2SO4 8.3 Suggest how to make the following alcohols selectively from two different alkenes (if

possible). Draw the starting alkene and all necessary reagents. Do not attempt to use carbocation rearrangements.

8.4 In each reaction, an unexpected product is produced. Give a mechanism for formation of this

unexpected product.

a.HBr

b.H2O

cat. H2SO4

a. b. c. d. e. f.

OH OH OH OHa. b. c. d.

H2OOH

cat. H2SO4

HBrBr

BrHBr

a.

b.

c.

231

8.4 LIST OF ESSENTIAL ALKENE REACTIONS

UNSYMMETRICAL ADDITIONS

1. HX (8.1B) IMPORTANT MECHANISM

2. Carbocation Rearrangements (8.1C) IMPORTANT MECHANISMS

3. H2O/Catalytic Acid (8.1D) IMPORTANT MECHANISM

4. H2O: Hydroboration/Oxidation (8.2E)

H XH X

HX = HI, HBr, HCl

Markovnikov Selectivity

H3C CH CH2HBr H3C CH CH2

HBrMarkovnikov product

Major

+ H3C CH CH2BrH

Anti-Markovnikov productVery minor

Example

H OH

cat. acid

Catalytic acid: H2SO4, HCl or other strong acid

H2Ocat. H2SO4

HOH

Example

Markovnikov SelectivityH-OH

Markovnikovproduct

1. BH3 • THFHydroboration H BR2 H OH

2. H2O2, NaOHOxidation

Anti-Markovnikovand

syn selectivity

Example

CH3H

H OHH3C H

1. BH3 • THF2. H2O2, NaOH

Anti-Markovnikov productSyn (same side) addition product

232

SYMMETRICAL ADDITIONS

5. X2 (8.2A) IMPORTANT MECHANISM

6. H2: Catalytic Hydrogenation (8.2B)

7. Oxidation to 1,2-Diols (8.2C)

X

X(Cl2 or Br2)Anti addition

Br2 Trans product only

Example

X2

Br

Br

H2, cat. Pt, Pd or Ni

H HSyn addition

Example

CH3

CH3

H2, cat. Ni H

H

CH3

CH3

Cis product only

HO OH

Cold KMnO4 H2O, NaOH Syn addition

1,2-Diols

ExampleH

H

OH

OH

H

HCis product only

Cold KMnO4 H2O, NaOH

1. OsO42. NaHSO3, H2O

Syn dihydroxylation

or

233

8. Epoxidation (8.2D)

OXIDATIVE CLEAVAGES

9. Most highly oxidized products (8.3B)

10. Less highly oxidized products: Aldehydes and Ketones (8.3C)

O

RCO3HA peroxyacid

Syn addition

ExampleH

HO

H

H

mCPBACO

OOH

meta-Chloroperbenzoic acidmCPBA

Cl

An epoxide

Hot KMnO4NaOH , H2O

O O

As highly oxidized as possible

VERY oxidizing conditions

Examples

CH3CH2 CH3C

CHH

CH3CH2 CH

CCH3H

CH3CH2 CO

CCH3O

O O1. Hot KMnO4 NaOH , H2O

2. H3O+ CH3CH2 CHO

CCH3OH

O O

CH3CH2 CH3C

COHO

O O2. H3O+ CH3CH2 C

H3CCOHOH

O O

CO2+H2O

1. Hot KMnO4 NaOH , H2O

1. O32. Zn/HOAc or (CH3)2S

O O

ONLY exchange:

Less oxidizing conditions

Examples

CH3CH2 CH

CCH3H

CH3CH2 CH

CCH3H

O O

C C for 2 C O

1. O32. Zn/HOAc +

CH3CH2 CH3C

CHH

CH3CH2 CH3C

CHH

O O+

HOAc = acetic acid 1

1. O32. Zn/HOAc

Pages 3-4Page 85Pages 93-96Page 207Pages 209-219Pages 231-233