Download Understanding the structure and bonding of organic molecules is

211 LSW Sp07
Exam 1
Organic Chemistry involves mostly CHNOPS and the halogens.
Organic compounds use valence shell electrons to bond.
Usually only in the s and p orbitals.
Orbital Shapes Carbon:
z
z
y
x
1s
n=1
l=0
ml=0
z
y
x
2s
n=2
l=0
ml=0
z
y
x
2pz
n=2
l=1
z
y
x
2px
n=2
l=1
x
2py
n=2
l=1
ml = 1,0,-1 arbitrary assignment
Electron configuration Carbon:
Full
Short form
Orbital energy diagram
Orbital energy levels diagram
1
y
211 LSW Sp07
Exam 1
Bonding:
•
Octet of electrons:
•
Ionic bonds:
•
Covalent bonds:
Example:
[CH3COO]
•
Non-polar bond:
•
Polar bond:
Examples:
_
Br2
Na +
HCl
CO
Formal Charges:
•
The charge on atoms within a neutral (or charged) molecule
•
Always label the formal charges on any atoms.
•
The location of formal charges often controls reactivity.
•
Equations:
Bonds
C
N
O
F
•
Recognition:
2
lone
pairs
211 LSW Sp07
Exam 1
Lewis Structures:
For NO3-:
1. Determine the total number of valence electrons in a molecule
2. Draw a skeleton.
a. Connects all atoms
b. Use only single bonds.
c. In simple molecules, the atom with the most available sites for bonding is usually
placed central.
d. Determine the number of bonding sites using the number of valence electrons
and the ability of an atom to expand its octet.
3. Place the remaining electrons to fill the octets of as many atoms as possible
a. Place the electrons on the most electronegative atoms first
b. Then proceed to the more electropositive atoms.
4. Fill any unfilled octets by making multiple bonds
5. Indicate the FORMAL CHARGES on all the atoms (keep them as low as possible)
O
N
O
6. Expanded octets occur only in atoms with n=3 (period) or greater.
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211 LSW Sp07
Exam 1
Molecular Shape: VSEPR
Electronic geometries:
Molecular geometries:
Electronic Geometry
Examples
Molecular Geometries
Linear
BeH2
Linear
Trigonal planar
BF3
Trigonal planar
SO2
Bent
CH4
Tetrahedral
NH3
Trigonal pyramidal
H2S
Bent
PF5
Trigonal bipyramidal
TeCl4
See-saw
BrF3
T-shaped
XeF2
Linear
SF6
Octahedral
IF5
Square pyramidal
XeF4
Square planar
Tetrahedral
Trigonal bipyramidal
Octahedral
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211 LSW Sp07
Exam 1
Rules for drawing molecules (3D) on paper (2D):
1.
2.
3.
4.
Place the largest number of atoms possible in the plane of the paper.
Unbroken lines
Hashed wedges
Solid wedges
H
Butane
C4H10
H
H
C
H
H
C
H
C
H
C
H
H
H
H
Ammonia
NH3
H
N
H
H
H
Methoxyethane
C3H8O
(Ethyl methyl ether)
H
N
H
H
H
C
O
H
C
H
-
You try CH3CO2 :
5
C
H
H
H
211 LSW Sp07
Exam 1
Molecular polarity:
You should be able to indicate a molecule's polarity.
Resonance structures:
Resonance hybrid:
LEWIS STRUCTURES
O
O
O
C
C
C
O
O
O
O
O
O
RESONANCE HYBRID
O
C
O
O
Rules to remember when considering resonance:
•
The bonding framework of a molecule remains intact.
•
You can only move electrons in multiple bonds or lone pairs.
•
The overall charge of the system remains the same.
Electron flow:
Curved arrows: indicate the movement of electrons not atoms,
the electrons move towards an electron deficient centers (+)
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211 LSW Sp07
Exam 1
Ranking the importance of resonance structures:
A) Satisfies the octet rule.
1) No charge separation are favored over
2) Charge separation as predicted by electronegtivity
3) Charge separation against that predicted by electronegativity.
B) Does not satisfies the octet rule.
Resonance structures for acetone:
O
CH3
O
CH3
I
CH3
O
CH3
CH3
O
CH3
CH3
II
III
What is the resonance hybrid of acetone?
What is the molecular polarity of acetone?
-
O
CH3
The enolate ion: [CH3CH-C(O)-CH3] (but-3-ene-2-olate)
Aromatic ring systems: p-methylaniline (CH3-Φ-NH2)
7
CH3
CH3
211 LSW Sp07
Exam 1
Hybridization:
Sigma, σ, overlap:
Pi, π, overlap:
Bond type
Overlaps
single
σ
double
σ and π
triple
σ and two π
VSEPR prediction is different from this prediction for CH4
Hybridization:
'hybrid' orbitals:
Name of hybrid:
8
211 LSW Sp07
Exam 1
You Try CO:
Name:
Lewis structure:
Predicted Shape:
Formal Charge on C:
Electronic Configuration:
C1-
E
And for Oxygen:
O1+ E
9
211 LSW Sp07
Exam 1
Y
X
CO
C O
Π BONDING
Y
Y
2p
C
C
2sp
C
Z
X
C
σ
O
X
O
O
O
Z
Z
BONDING
Draw the σ and π bonding for ethene, CH2CH2, and ethyne, CHCH.
ETHENE
H
H
C
H
C
H
H
H
C
C
H
H
Π BONDING
σ BONDING
ETHYNE
H
C
C
H
H
σ BONDING
C
C
Π BONDING
10
H
211 LSW Sp07
Exam 1
Acids and Bases:
Same chemical reaction happen simultaneously but looking at it from oppositely.
The base, B, removes a proton from the acid, H-A:
B
+H A
Acids
Bases
Arrenhius
BronstedLowry
Lewis
Acidity
Acid ionization:
H A
Ka
Ka =
pKa = - log10Ka
•
The more stable the conjugate base, A-, is then the more the equilibrium favors the
product side (Ka > 1)
•
The more the equilibrium favors products, the more H+ is present.
•
The more H+ there is then the stronger H-A is as an acid.
•
The larger Ka implies more dissociation of HA and so the stronger the acid.
•
The larger Ka is, the more negative the pKa so the lower the pKa, the stronger the acid.
11
211 LSW Sp07
Exam 1
Major factors that affect the stability of the conjugate base, A-.
Electronegativity. In the same row of periodic table:
HF H2O NH3 CH4
Size. In the same group of periodic table:
HI HBr HCl HF
Resonance. On the same atom:
RCO2H ROH
General acidity trend of common organic acids
OH
O
R OH
R NH2
R H
OH
R
CLASS
pKa
Basicity
Major factors that affect electron pair availability in a base, B
CH3- > NH2- > HO- > F-
Electronegativity. in the same row of periodic table:
F- > Cl- > Br- > I-
Size. in the same group of periodic table:
RO- > RCO2-
Resonance. On the same atom:
General acidity trend of some common organic bases:
R NH
CLASS
NH2
R O
OH
pKa
12
R NH2