Chap 8 thiols and sulfides (1)

ThiolsThiols (R–S–HR–S–H) and sulfidessulfides (R–S–R’R–S–R’) are sulfur analogs of alcohols and ethers, respectively

Sulfur replaces oxygen

ThiolsThiols (RSHRSH), also known as mercaptans, are sulfur analogs of alcohols

They are named with the suffix ––thiolthiol SHSH group is called “mercapto groupmercapto group” (“capturer of

mercury”)

SulfidesSulfides (RSR’RSR’) are sulfur analogs of ethers

◦ They are named by rules used for ethers, with sulfidesulfide in place of etherether for simple compounds and alkylthioalkylthio in place of alkoxyalkoxy

Practice ProblemPractice Problem: Name the following compounds: : Name the following compounds:

Alkanes have only strong, nonpolar bonds

No reaction with nucleophiles or electrophiles

Not much reactivity - paraffins (little affinity)

© Prentice Hall 2001Chapter 8 6


Initiation: Homolytic cleavage


Cl Cl400o C

or h2 Cl

Br Br400o C

or h2 Br

radicals

Note that when an arrowhead with a single barb is used, it denotes movement of a single electron




The very reactive chlorine atom will have lower selectivity and attack pretty much any hydrogen available on an alkane

The less reactive bromine atom will be more selective and tends to react preferentially with the easy targets, i.e. tertiary hydrogens



Benzylic and allylic radicals are even more stable than tertiary alkyl radicals

It should be easy for a halogen radical to abstract a benzylic or allylic hydrogen


Problem is that for the allyl radical there is a greater likelihood that the halogen will add electrophilically to the adjacent double bond


Electrophilic addition can be minimized by maintaining the halogen at a very low concentration

Under these conditions, halogens can substitute for allylic and benzylic hydrogens


N-Bromosuccinimide (NBS) is a good reagent for supplying low concentrations of bromine radical


Bromine radical comes from the homolytic cleavage of the N–Br Bond

Low concentration of Br2 is generated by the reaction of NBS with HBr

Neither HBr nor Br2 accumulate, so electrophilic addition is slow

© Prentice Hall 2001 Chapter 8 18

+ Br + HBr

N

O

O

Br + HBr N

O

O

H + Br2

+ Br2

Br

+ Br

When a radical abstracts an allylic or benzylic hydrogen, a radical that is stabilized by resonance is obtained


If the resonance hybrid is not symmetrical, more than one product is obtained


CH3CHCH CH3CH CHCH2CH3CH2CH CH2Br CH2

Br2

CH3CHCHBr

CH3CH CHCH2BrCH2

+

3-bromo-1-butene 1-bromo-2-butene


If a chirality center already exists, it may affect the distribution of products

A pair of diastereomers will be formed, but in unequal proportions

© Prentice Hall 2001 Chapter 8 22


Cyclic alkanes react with halogens in much the same way as acyclic compounds

+ Cl2

Cl

+ HCl


Cyclopropane undergoes electrophilic addition much like an alkene


Ozone (O3) is a major constituent of smog In the stratosphere, a layer of ozone shields

the Earth from harmful solar radiation The ozone layer is thinnest at the equator

and thickest in polar regions


Ozone is formed in the stratosphere by interaction of short-wavelength ultraviolet light with oxygen


O2

O3

hO + O

O + O2

The stratospheric ozone layer acts as a filter for biologically harmful ultraviolet radiation

Scientists have noted a precipitous drop in the ozone concentrations over Antarctica since 1985

Circumstantial evidence links the depletion in ozone to synthetic chlorofluorocarbons (CFCs) - used as refrigerants


Chlorofluorocarbons (CFCs) are exceptionally stable, but under the intense ultraviolet radiation present in the stratosphere, they undergo a radical dissociation


C ClFCl

F

hCFCl

F+ Cl

The chlorine radicals are ozone-removing reagents

It has been estimated that each chlorine radical destroys 100,000 ozone molecules in a radical chain reaction


Cl + O3 ClO + O2

ClO + O3 Cl + 2 O2

2 O3 3 O2Overall

Production and use of CFCs has been slowed, but because these materials have a half-life of 70 - 120 years they will be around in the stratosphere for a long time

The ozone hole over Antarctica was observed in October 1999 to be a little smaller than in October 1998


Chapter 8 32

Electrons in pi bond are loosely held. Electrophiles are attracted to the pi

electrons. Carbocation intermediate forms. Nucleophile adds to the carbocation. Net result is addition to the double bond.

=>

Chapter 8 33

Step 1: Pi electrons attack the electrophile.

C C + E+C

E

C +

C

E

C + + Nuc:_

C

E

C

Nuc

=>

• Step 2: Nucleophile attacks the carbocation.

Chapter 8 34

=>

Chapter 8 35

Protonation of double bond yields the most stable carbocation. Positive charge goes to the carbon that was not protonated.

X =>

+ Br_

+

+CH3 C

CH3

CH CH3

H

CH3 C

CH3

CH CH3

H

H Br

CH3 C

CH3

CH CH3

Chapter 8 36

CH3 C

CH3

CH CH3

H Br

CH3 C

CH3

CH CH3

H+

+ Br_

CH3 C

CH3

CH CH3

H+

Br_

CH3 C

CH3

CH CH3

HBr

=>

Chapter 8 37

Markovnikov’s Rule: The proton of an acid adds to the carbon in the double bond that already has the most H’s. “Rich get richer.”

More general Markovnikov’s Rule: In an electrophilic addition to an alkene, the electrophile adds in such a way as to form the most stable intermediate.

HCl, HBr, and HI add to alkenes to form Markovnikov products. =>

Chapter 8 38

In the presence of peroxides, HBr adds to an alkene to form the “anti-Markovnikov” product.

Only HBr has the right bond energy. HCl bond is too strong. HI bond tends to break heterolytically to

form ions. =>

Chapter 8 39

Peroxide O-O bond breaks easily to form free radicals.

+R O H Br R O H + Br

O OR R +R O O Rheat

• Hydrogen is abstracted from HBr.

Electrophile =>

Chapter 8 40

Bromine adds to the double bond.

+C

Br

C H Br+ C

Br

C

HBr

Electrophile =>

C

Br

CC CBr +

• Hydrogen is abstracted from HBr.

Chapter 8 41

Tertiary radical is more stable, so that intermediate forms faster. =>

CH3 C

CH3

CH CH3 Br+

CH3 C

CH3

CH CH3

Br

CH3 C

CH3

CH CH3

Br

X

Chapter 8 42

Reverse of dehydration of alcohol Use very dilute solutions of H2SO4 or

H3PO4 to drive equilibrium toward hydration. =>

C C + H2OH+

C

H

C

OH

alkene alcohol

Chapter 8 43

+C

H

C+

H2O C

H

C

O H

H+

+ H2OC

H

C

O H

H+

C

H

C

OH

H3O++ =>

C C OH H

H

++ + H2OC

H

C+

Chapter 8 44

Markovnikov product is formed.

+CH3 C

CH3

CH CH3 OH H

H

++ H2O+

H

CH3CH

CH3

CCH3

H2OCH3 C

CH3

CH CH3

HOH H

+

H2OCH3 C

CH3

CH CH3

HOH

=>

Chapter 8 45

Oxymercuration-Demercuration◦ Markovnikov product formed◦ Anti addition of H-OH◦ No rearrangements

Hydroboration◦ Anti-Markovnikov product formed◦ Syn addition of H-OH

=>

Chapter 8 46

Reagent is mercury(II) acetate which dissociates slightly to form +Hg(OAc).

+Hg(OAc) is the electrophile that attacks the pi bond.

CH3 C

O

O Hg O C

O

CH3 CH3 C

O

O_

Hg O C

O

CH3+

=>

Chapter 8 47

The intermediate is a cyclic mercurinium ion, a three-membered ring with a positive charge.

C C +Hg(OAc) C CHg+OAc

=>

Chapter 8 48

Water approaches the mercurinium ion from the side opposite the ring (anti addition).

Water adds to the more substituted carbon to form the Markovnikov product.

C CHg+OAc

H2O

CO+

C

Hg

H

H

OAc

H2O

C

O

C

Hg

H

OAc

=>

Chapter 8 49

Sodium borohydride, a reducing agent, replaces the mercury with hydrogen.

C

O

C

Hg

H

OAc

4 4 C

O

C

H

H

+ NaBH4 + 4 OH_

+ NaB(OH)4

+ 4 Hg + 4 OAc_

=>

Chapter 8 50

Predict the product when the given alkene reacts with aqueous mercuric acetate, followed by reduction with sodium borohydride.

CH3

D

(1) Hg(OAc)2, H2O

(2) NaBH4

=>

OHCH3D

H

anti addition

Chapter 8 51

If the nucleophile is an alcohol, ROH, instead of water, HOH, the product is an ether.

C C(1) Hg(OAc)2, CH3OH

C

O

C

Hg(OAc)

H3C

(2) NaBH4C

O

C

H3C

H

=>

Chapter 8 52

Borane, BH3, adds a hydrogen to the most substituted carbon in the double bond.

The alkylborane is then oxidized to the alcohol which is the anti-Mark product.

C C(1) BH3

C

H

C

BH2

(2) H2O2, OH-

C

H

C

OH =>

Chapter 8 53

Borane exists as a dimer, B2H6, in equilibrium with its monomer.

Borane is a toxic, flammable, explosive gas. Safe when complexed with tetrahydrofuran.

THF THF . BH3

O B2H6 O+ B-

H

H

H+2 2 =>

Chapter 8 54

The electron-deficient borane adds to the least-substituted carbon.

The other carbon acquires a positive charge. H adds to adjacent C on same side (syn).

=>

Chapter 8 55

Borane prefers least-substituted carbon due to steric hindrance as well as charge distribution. =>

C CH3C

H3C

H

H+ BH3

B

CC H

CH3

H3C

H

H

C

CH

HH

CH3

CH3

C

C

HH

H3CCH3

H

3

Chapter 8 56

Oxidation of the alkyl borane with basic hydrogen peroxide produces the alcohol.

Orientation is anti-Markovnikov.

CH3 C

CH3

H

CH

HB

H2O2, NaOH

H2OCH3 C

CH3

H

CH

HOH

=>

Chapter 8 57

Predict the product when the given alkene reacts with borane in THF, followed by oxidation with basic hydrogen peroxide.

CH3

D

(1)

(2)

BH3, THF

H2O2, OH-

=>syn addition

HCH3

DOH

Chapter 8 58

Alkene + H2 Alkane Catalyst required, usually Pt, Pd, or Ni. Finely divided metal, heterogeneous Syn addition

=>

Chapter 8 59

Insertion of -CH2 group into a double bond produces a cyclopropane ring.

Three methods:◦ Diazomethane◦ Simmons-Smith: methylene iodide and Zn(Cu)◦ Alpha elimination, haloform

=>

Chapter 8 60

Extremely toxic and explosive. =>

N N CH2 N N CH2

diazomethane

N N CH2heat or uv light

N2 +

carbene

CH

H

CH

H

C

CC

CC

H

H

Chapter 8 61

Best method for preparing cyclopropanes.

CH2I2 + Zn(Cu) ICH2ZnIa carbenoid

CH2I2

Zn, CuCl =>

Chapter 8 62

Haloform reacts with base. H and X taken from same carbon

CHCl3 + KOH K+ -CCl3 + H2O

CCl

Cl

Cl Cl-+CCl

Cl

Cl

Cl

CHCl3

KOH, H2O =>

Chapter 8 63

Cis-trans isomerism maintained around carbons that were in the double bond.

C CH

CH3

H

H3C NaOH, H2O

CHBr3C C

H

CH3

H

H3CBrBr

=>

Chapter 8 64

Cl2, Br2, and sometimes I2 add to a double bond to form a vicinal dibromide.

Anti addition, so reaction is stereospecific.

CC + Br2 C C

Br

Br

=>

Chapter 8 65

Pi electrons attack the bromine molecule. A bromide ion splits off. Intermediate is a cyclic bromonium ion.

CC + Br Br CCBr

+ Br =>

Chapter 8 66

Halide ion approaches from side opposite the three-membered ring.

CCBr

BrCC

Br

Br

=>

Chapter 8 67

=>

Chapter 8 68

Add Br2 in CCl4 (dark, red-brown color) to an alkene in the presence of light.

The color quickly disappears as the bromine adds to the double bond.

“Decolorizing bromine” is the chemical test for the presence of a double bond.

=>

Chapter 8 69

If a halogen is added in the presence of water, a halohydrin is formed.

Water is the nucleophile, instead of halide.

Product is Markovnikov and anti.

CCBr

H2O

CC

Br

OH H

H2O

CC

Br

OH

+ H3O+

=>

Chapter 8 70

The most highly substituted carbon has the most positive charge, so nucleophile attacks there.

=>

Chapter 8 71

Predict the product when the given alkene reacts with chlorine in water.

CH3

D

Cl2, H2O

=>

OHCH3D

Cl

Chapter 8 72

Alkene reacts with a peroxyacid to form an epoxide (also called oxirane).

Usual reagent is peroxybenzoic acid.

CC + R C

O

O O H CCO

R C

O

O H+

=>

Chapter 8 73

One-step concerted reaction. Several bonds break and form simultaneously.

OC

O

R

H

C

C

OOH

OC

O

RC

C

+

=>

Chapter 8 74

Since there is no opportunity for rotation around the double-bonded carbons, cis or trans stereochemistry is maintained.

CCCH3 CH3

H H Ph C

O

O O HCC

CH3 CH3

H HO =>

Chapter 8 75

Acid catalyzed. Water attacks the protonated epoxide. Trans diol is formed.

CCO

H3O+

CCO

H

H2O

CC

O

OH

H H H2O

CC

O

OH

H

=>

Chapter 8 76

To synthesize the glycol without isolating the epoxide, use aqueous peroxyacetic acid or peroxyformic acid.

The reaction is stereospecific.

CH3COOH

O

OH

H

OH

H

=>

Chapter 8 77

Alkene is converted to a cis-1,2-diol, Two reagents:

◦ Osmium tetroxide (expensive!), followed by hydrogen peroxide or

◦ Cold, dilute aqueous potassium permanganate, followed by hydrolysis with base

=>

Chapter 8 78

Concerted syn addition of two oxygens to form a cyclic ester.

C

COs

O O

OO

C

CO O

OO

Os

C

C

OH

OH+ OsO4

H2O2 =>

Chapter 8 79

If a chiral carbon is formed, only one stereoisomer will be produced (or a pair of enantiomers).

C

C

CH2CH3

H CH2CH3

C

C

CH2CH3

CH2CH3

OH

OH

H

HH2O2

H

(2)

(1) OsO4

cis-3-hexene meso-3,4-hexanediol

=>

Chapter 8 80

Both the pi and sigma bonds break. C=C becomes C=O. Two methods:

◦ Warm or concentrated or acidic KMnO4.◦ Ozonolysis

Used to determine the position of a double bond in an unknown. =>

Chapter 8 81

Permanganate is a strong oxidizing agent.

Glycol initially formed is further oxidized. Disubstituted carbons become ketones. Monosubstituted carbons become

carboxylic acids. Terminal =CH2 becomes CO2.

=>

Chapter 8 82

CCCH3 CH3

H CH3 KMnO4

(warm, conc.)C C

CH3

CH3

OHOH

H3C

H

C

O

H3C

H

C

CH3

CH3

O

C

O

H3COH

+

=>

Chapter 8 83

Reaction with ozone forms an ozonide. Ozonides are not isolated, but are treated

with a mild reducing agent like Zn or dimethyl sulfide.

Milder oxidation than permanganate. Products formed are ketones or

aldehydes. =>

Chapter 8 84

CCCH3 CH3

H CH3 O3 CH3C

H

O OC

CH3

CH3

O

Ozonide

+(CH3)2S

CH3C

HO C

CH3

CH3

O CH3 S

O

CH3

DMSO

=>

Science

Chap 8 thiols and sulfides (1)