Post on 14-Mar-2018
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