CH3

CH3 CH3

Br

CH3

Br

H

CH3

H

CH3

H

Br

CH3

CH3 CH3 CH3

X

OH

CH3 CH3

BH2

CH3

OH

HH BH2

H

CH3 CH3 CH3

Hg(OAc)

OR

CH3

H

CH3

H

OH

X+

CH3 CH3 CH3

X

X

X+

Hg(OAc)

β-Mercury-substituted carbocation

H2SO4or

H3PO4or

NaHSO4

Which upon reaction with:

Reagent Yields

Halonium Ion

"X+"

Carbocation

Halonium Ion

What adds?

HOOH/NaOH

Hg(OAc)+

"X+"

To give the more stable:

Radical

Substrate

(NaBH4 reduction replaces Hg(OAc) with

H)

Radical Hydrohalogenation:

X2/H2O

Organoborane

Halogenation:

Hydration:

Halohydrin Formation:

Hydroboration/Oxidation:

Oxymercuration:

Summary of Alkene Reactions

Hydrohalogenation:

HBr/peroxides

Carbocation

B2H6

Br•

ROHHg(OAc)2

X–

Br–H+

H+

HBr

HBr

X2

H2O

H2O

ORGANIC CHEMISTRY

SYSTEMATIC NOMENCLATURE OF BICYCLIC COMPOUNDS

Bicyclic compounds require the breaking of two carbon-carbon bonds to convert them to openchained compounds--containing no rings.

Bicyclic compounds are named by prefixing bicyclo- to the name of the parent hydrocarbon.The name of the parent hydrocarbon is obtained by counting the total number of carbon atomsin all of the rings of the compound. Thus, because norbornane (below) contains seven carbonatoms in the two rings of the molecule it is a bicycloheptane. Carbon atoms shared by bothrings are referred to as bridgehead carbons.

1 2

3

45

6

7

1

2

3

4

5

67

norbornanebicyclo[2.2.1]heptane

The numbers of carbon atoms between bridgehead carbons in the molecule arespecified by counting from the bridgehead carbon and listing each of the numbers in bracketsin decreasing order prior to the name of the hydrocarbon. Thus, norbornane isbicyclo[2.2.1]heptane. The following are additional examples of bicycloheptanes:

12

3 4 5

6

7

1

2

3

4

5

6

7

bicyclo[3.1.1]heptane bicyclo[3.2.0]heptane

The carbon positions of a bicyclic hydrocarbon are numbered from the bridgeheadcarbon around the largest ring first. Substituents are located by the number assigned to thecarbon positions. For example:

H3C

CH3H3C

1,7,7-trimethylbicyclo[2.2.1]heptane

CH3

CH3

1,3-dimethylbicyclo[1.1.0]butane

bicyclo[2.2.2]octa-2,5,7-triene

H O

CO

H

The Curved Arrow Formalism

Curved arrows are used by chemists to indicate the flow of electrons in reactions.*

For each electron pair (either a bonding pair or lone pair) that changes position in a reaction, one arrow is required.

The tail of the arrow starts at the initial position of the electron pair (at an atom for a lone pair; at a bond for a bonding pair).

The head of the arrow points to the new position of the electron pair (to an atom for formation of a lone pair; midway between atoms for formation of a bond.)

CURVED ARROWS DO NOT REPRESENT THE MOVEMENT OF NUCLEI!!!

1.

2.

3.

4.

H O

CH3

H

H O H O

CH3

H

Ex. 1

H

H

H

C

H

O O

H

Ex. 2

Ex. 3

O H H

C

CH3

H3C

CH3

O

H

H

C

CH3

H3C

CH3

O

H

H

Note strict balance of charge and that only red electrons are moving (all other electrons and atomic positions remain the same). Keep careful track of formal charges.

*They are also sometimes used to indicate the interconversion of resonance forms. Do not confuse the two usages—interconversion of resonance forms is not a reaction!!!

Guide to Displacement ReactionsReactivity

Type of carbon SN1(carbocationintermediate)

E1(carbocationintermediate)

SN2(concerted)

E2(concerted)

1 ° X X• good LG• good Nu:• polar aproticsolvent

• strong, bulkybases

2 ° • good LG• poor Nu:• polar proticsolvent

• competeswith SN1when base ispresent

• good LG• good Nu:(weaker base thanHO–)

• polar aproticsolvent

• competes withSN2

• favored by bulkybases strongerthan HO–

3 ° • good LG• any Nu:• polar proticsolvent

• competeswith SN1when base ispresent

X• strong base

NucleophilicityClassification Nucleophiles Rel. React.

Excellent I–, HS–, RS– >105

Good Br–, HO–, RO–, CN–, N3– 104

Fair NH3, Cl–, F–, RCO2– 103

Weak H2O, ROH 1

Very Weak RCO2H 10–2

Leaving GroupsClassification Leaving Groups

Excellent ROSO2–, H2O, ROH, N2

Good I–, Br–, Cl–, NR3, RCO2–

Poor HO–, RO–, F–, CN–, NH2

HBr

CH3

Give all possible products, including stereochemistry!

H2SO4(dil.)

HBrperoxides

Br2CCl4

CH3

CH3

OH

Br

+ enantiomers

CH3

BrCH3

Br

CH3

Br

Br

+ enantiomer

OH

H

H

OH

+ enantiomer

+ enantiomerOH

H

H

OH

+ enantiomer

+ enantiomer

Br

H

H

Br

+ enantiomer

+ enantiomerBr

H

H

Br

+ enantiomer

+ enantiomer

Br

H

H

Br

+ enantiomer

+ enantiomerBr

H

H

Br

+ enantiomer

+ enantiomer

Br

Br

Br

Br

+ enantiomer

OH Br

Br

Br

Br

Br

Br

Br

Reagent

Substrate

CH3CH3

CH3CH3OH

Cl

Give all possible products, including stereochemistry!

Cl2H2O

1. Hg(OAc)22. NaBH4/OH–

1. BH32. H2O2/OH–

1. OsO42. H2O2/OH–

+ enantiomer

CH3

OH

OH

OH

OH

+ enantiomer + enantiomer

OH

Cl

+ enantiomer

OH

H

H

OH

+ enantiomer

+ enantiomer+ enantiomer

H OH

+ enantiomer

HO OH

OH

Cl

+ enantiomer

OH

H

H

OH

+ enantiomer

+ enantiomer+ enantiomer

H OH HO OH

Cl

OH

HO

ClOH

OH

OH

OH

OH

HO

OH

Reagent

Substrate

Analysis of Isomerism

Two molecules with the same

molecular formula

Molecules are Identical

Molecules are Constitutional

Isomers

Mirror Images?

Same Connectivity?

Superimposable?YES

YES

Molecules are EnantiomersMolecules are Diastereomers

YES

NO

NO

NO

C

0 Formal Charge

C

,

C

,

C

0 Formal Charge

C

Note: It should become second nature to recognize the common Lewis configurations for C, N, and O

C

+1 Formal Charge

O

–1 Formal Charge

O

NO!!! Violates Octet Rule

O

,

O

,

N

+1 Formal Charge

N,

N –1 Formal Charge

+1 Formal Charge

N N N

, 0 Formal Charge

–1 Formal Charge

LEWIS STRUCTURES OFTEN DO NOT ADEQUATELY REPRESENT ACTUAL STRUCTURES:

Cl NO

O

Remember...

Implies unequal bonding of N to two oxygens!Experimentally, they are the same length!

Linus Pauling (Feb. 28, 1901-Aug. 19, 1994) introduced the concept of...

Resonance: When a molecule can be represented by 2 or more Lewis structures where thearrangement of atoms is the same, but electrons are shifted

Resonance arrow (doubleheaded). Do not under anycircumstances confuse withan equilibrium arrow!

The small curved arrows are "electron pushing" arrows. They represent the figurative movement of an electron pair. This is for bookkeeping purposes only!

Cl NO

O

2

1

...RESONANCE HYBRID

2

Cl NO

O

22

1

THE ACTUAL MOLECULE IS MORE STABLE THAN ANY INDIVIDUAL RESONANCE FORM IMPLIES. IN FACT, THE MORE RESONANCE FORMS ONE CAN DRAW OF A SPECIES, THE MORE STABLE IT IS.

OZONE, O3, APPEARS TO HAVE A SEPARATION OF CHARGE, IMPLYING SEVERE INSTABILITY...

11

Neither Lewis structure by itself represents reality. The structures are not rapidly interconverting. Rather, the actual structure is a mix of the two (or more) resonance forms and is called a...

RESONANCE THEORY

...THE RESONANCE HYBRID SHOWS THAT BOTH TERMINAL OXYGENS SHARE THE NEGATIVE CHARGE, RESULTING IN GREATER STABILITY

RULES OF RESONANCE THEORY

Cl NO

O

3. RESONANCE FORMS FOR A GIVEN MOLECULE NEED NOT CONTRI- BUTE EQUALLY TO THE OVERALL STRUCTURE (THE RESONANCE HYBRID) SINCE THEY MAY DIFFER IN ENERGY

AND

1. ATOMIC POSITIONS REMAIN THE SAME

2. FOLLOW THE OCTET RULE

are not resonance forms!!!

...is not!

are resonance forms!!!

are valid resonance forms but...

OO

O

Separation of charge indicates a higher energy resonance form...

more stablecontributes more

less stablecontributes less

OO

O

more stable ( on O) less stable ( on C)

Formal negative charge on electropositive elements indicates a higherenergy resonance form, as well as positive charge on electronegative elements...

OOO

(kind of like blue and yellow mixing to make green)

4. INSURE THAT ALL RESONANCE FORMS HAVE THE SAME NET CHARGE, THE SAME NUMBER OF ELECTRONS, AND THE SAME NUMBER OF UNPAIRED ELECTRONS

N C O C N O

N C O N C O

N

O

C CH

HH H

H N

O

C CH

HH H

H

N

O

C CH

HH H

H

C CH

HH

HC C

H

HH

H

N

O

C CH

HH H

H N

O

C CH

HH H

H

A

B

A

B

Orbital HybridizationLinear Carbon (Atom)

• Only two other atoms or • One atom and one lone pair surrounding carbon (atom)

{one s + one p}two sp orbitals

ORIENT THEM: 180° apart

Two p orbitals left over

σ bonds are formed by thesp hybrids and π bonds by

the p orbitals

Example

Trigonal Planar Carbon (Atom)• Only three other atoms or• Two atoms and one lone pair or • One atom and two lone pairs (rare) surrounding carbon (atom)

H C N

{one s + two p}three sp2 orbitals

ORIENT THEM: 120° apart

Tetrahedral Carbon (Atom)• Four other atoms or• Three atoms and one lone pair or• Two atoms and one lone pair surrounding carbon (atom)

{one s + three p}four sp3 orbitals

ORIENT THEM: 109.5° apart

Example

Hydrogen cyanide

No p orbitals left over

σ bonds are formed by thesp3 hybrids—no π bonds

C C

Methane

(side view)

HH

HH

(top view)

Ethylene

HH

C

HH

One p orbital left over

σ bonds are formed by thesp2 hybrids and a π bond by

the p orbitalExample

C CH

H

H

H