FORMAL CHARGE
Unbonded
Formal Charge =
NH2-
Number of
valence electrons
in the neutral
atom
.
.N :
.
Bonded
All
One half of
unshared + all shared
electrons
electrons
5e-
( Formal Charge = 5 - 4 - 2 = -1 )
..
H . . N:
..
H
6e-
When drawing a Lewis Diagram remember these rules.
LEWIS DIAGRAMS SHOW IT ALL !
- all atoms including hydrogens
- all bonds (lines not dots )
- all unshared pairs ( dots )
- all formal charges
- all atoms with octets ( except H )
- the correct number of electrons ( count! )
Rumus Kimia

Rumus empirik
 Rumus Molekul
 Rumus struktur
Rumus struktur lengkap
Rumus struktur panjang (expanded)
Rumus struktur termampatkan (condensed)
Rumus Struktur pada
senyawa siklis – sikloheksana

Expanded formula
H
H
H
H
H
H
C
C
C
C
C
C
H
H
H
H
H
H
C 6 H 12
Rumus Struktur pada
senyawa siklis – sikloheksana

Polygon formula
(condensed formula)
C 6 H 12
Rumus Struktur pada senyawa
siklis – sikloheksana

Condensed formula
CH 2
CH 2
CH 2
CH 2
CH 2
CH 2
C 6 H 12
Contoh Molekul siklis
H
H
CH2
C
H
H
C
CH2 CH2
C
H
H
C yc lo p ro p a n e
H
H
H
C 3H 6
H
C
C
C
C
H
H
CH2 CH2
H
C y clo b u ta n e
H
C 4H 8
H
C
H
CH2 CH2
H
CH2
H
C
C
CH2
H
CH2
H
C
H
C
H H
CH2 CH2
H
C y c lo p e n ta n e
C 5H 10
Beberapa cara penulisan struktur
c itro n e lla l
(CH 3 ) 2 C=C HCH 2 CH 2 CH(CH
3 )CH 2 CH
O
or
H 3C
C
CH
CH 2
H 3C
CH 2
CH 3
O
C
CH
CH 2
H
condensed
H
H
H
C
O
O
H
H
C
C
H
H
C
H
H
H
H
lin e s tru c tu re s a re m o s t c o m p a c t
a n d e a s y to re a d
H
C
H
H
C
C
C
H
C
C
H
H
H
expanded
lin e
Molekul polar dan Nonpolar
• To determine if a molecule is polar, we
need to determine
– if the molecule has polar bonds
– the arrangement of these bonds in space
• Molecular dipole moment (): the vector
sum of the individual bond dipole moments
in a molecule
– reported in debyes (D)
Bond Dipole Moments
• are due to differences in electronegativity.
• depend on the amount of charge and
distance of separation.
• In debyes,
 = 4.8 x  (electron charge) x d(angstroms)
Molecular Dipole Moments
• Depend on bond polarity and bond angles.
• Vector sum of the bond dipole moments.
• Lone pairs of electrons contribute to the
dipole moment.
Polar and Nonpolar Molecules
• these molecules have polar bonds, but
each has a zero dipole moment
Cl
F
O
C
B
O
F
F
Carbon dioxide
 =0D
Boron trifluoride
 =0D
C
Cl
Cl
Cl
Carbon tetrachloride
 =0D
Polar and Nonpolar Molecules
• these molecules have polar bonds and
are polar molecules
d i re cti o n
of d i p ol e
mo m ent
N
O
H
H
H
H
H
W at e r
 = 1 .8 5 D
A m mo nia
 = 1.47D
d i re cti o n
of d i p ol e
mo m ent
Polar and Nonpolar Molecules
– formaldehyde has polar bonds and is a
polar molecule
d i re cti o n
of d i p ol e
mo m ent
O
H
C
H
Form al d e h y d e
 = 2.33 D
Intermolecular Forces
• Strength of attractions between molecules
influence m.p., b.p., and solubility; esp. for
solids and liquids.
• Classification depends on structure.
– Dipole-dipole interactions
– London dispersions
– Hydrogen bonding
Dipole-Dipole
=>
Dipole-Dipole Forces
• Between polar molecules
• Positive end of one molecule aligns with
negative end of another molecule.
• Lower energy than repulsions, so net
force is attractive.
• Larger dipoles cause higher boiling points
and higher heats of vaporization.
London Dispersions
•
•
•
•
Between nonpolar molecules
Temporary dipole-dipole interactions
Larger atoms are more polarizable.
Branching lowers b.p. because of
decreased surface contact between
molecules.
CH3
CH3
CH3
CH2
CH2
CH2
n-pentane, b.p. = 36°C
CH3
CH3
CH
CH2
CH3
isopentane, b.p. = 28°C
H 3C
C
CH3
CH3
neopentane, b.p. = 10°C
=>
Dispersions
=>
Hydrogen Bonding
• Strong dipole-dipole attraction
• Organic molecule must have N-H or O-H.
• The hydrogen from one molecule is
strongly attracted to a lone pair of
electrons on the other molecule.
• O-H more polar than N-H, so stronger
hydrogen bonding
H Bonds
Boiling Points and
Intermolecular Forces
CH 3
CH 2
ethanol, b.p. = 78°C
H 3C
N
O
CH 3
dimethyl ether, b.p. = -25°C
C H 3C H 2
CH3
N
CH3
C H 3C H 2C H 2
trimethylamine, b.p. 3.5°C
CH 2
ethylmethylamine, b.p. 37°C
OH
ethanol, b.p. = 78°C
N
H
H
H
CH3
CH 3
CH 3
OH
CH 3
propylamine, b.p. 49°C
CH 2
NH 2
ethyl amine, b.p. 17°C
ASAM DAN BASA
Brønsted-Lowry Theory of
Acids and Bases
• Acid:
• Base:
Proton Donor
Proton Acceptor
Conjugate Acid: Base + Proton
Conjugate Base: Acid - Proton
Strong Acids and Bases
• Strong acid - completely ionized in
aqueous solution. Examples are:
– HCl, HBr, HI, HNO3, HClO4, and H2SO4
• Strong base - completely ionized in
aqueous solution. Examples are:
– LiOH, NaOH, KOH, Ca(OH)2, and Ba(OH)2
Weak Acids and Bases
• Acetic acid is a weak acid
– it is incompletely ionized in aqueous
solution
O
CH3 COH
Acid
(weaker acid)
+
H2 O
Base
(weaker base)
O
CH3 CO-
Conjugate base
of CH 3 CO 2 H
(stronger base)
+
H3 O
+
Conjugate acid
of H 2 O
(stronger acid)
Lewis Theory of Acids and
Bases
• Acid: Electron-Pair Acceptor
–
Electrophile
• Base: Electron-Pair Donor
–
Nucleophile
Weak Acids and Bases
• The equation for the ionization of a weak
acid, HA, in water and the acid
ionization constant, Ka, for this
equilibrium are
HA
+
Ka =
H2 O
K eq [H
pK a = - log K a
A
2 O] =
-
[H
+
+
3
H3 O
O ][A
[HA]
-
]
+
Weak Acids and Bases
CH 3 CH 2 OH
pK a
15.9
Conjugate
Base
CH 3 CH 2 O
water
H2 O
15.7
HO
bicarbonate ion
HCO 3
ammonium ion
Acid
Formula
etha nol
carbonic acid
acetic acid
-
2-
10.33
CO 3
NH 4
H 2 CO 3
9.24
NH 3
HCO 3
CH 3 CO 2 H
4.76
+
6.36
CH 3 CO 2
sulfuric acid
H 2 SO 4
-5.2
HSO 4
hydrogen chloride
HCl
-7
Cl
-
-
-
Acidity Constant (Ka)
K
HA
+
H 2O
A
-
-
+
+
[A ] [H 3 O ]
K =
[H A ] [H 2 O ]
-
K [H 2 O ]
=
Ka
=
+
[A ] [H 3 O ]
[H A ]
pKa =
- lo g K a
H 3O
+
pKa
pKa = - log Ka
Strong acid = large Ka = small pKa
Weak acid = small Ka = large pKa
Relative Acid Strength
AC ID
C O N J . B AS E
s tro n g e r
C lO
4
AC ID
STRENGTH
4
10
10
-1 0
O
O
C H3 C
pKa
w eaker
_
H C lO
Ka
C H3 C
OH
-5
1 .8 x 1 0 4 .8
O
OH
O
B AS E
STRENGTH
-1 0
1 .0 x 1 0
10
_
C H3 C H3
w eaker
-5 0
10
C H3 C H2
s tro n g e r
50
Acid Strength
• Strong Acid
– Conjugate base is weak
– pKa is small
• Weak Acid
–
Conjugate base is strong
– pKa is large
Base Strength
• Strong Base
–
Conjugate acid is weak
– pKa is large
• Weak Base
–
Conjugate acid is strong
– pKa is small
Position of equilibrium
• Favors reaction of the stronger acid and
stronger base to give the weaker acid
and weaker base
+
HA
Stronger
acid
CH 3 CO 2 H
Acetic acid
pK
a
4.76
(stronger acid)
B
-
Stronger
base
+
NH
3
A
-
+
Weaker
base
CH 3 CO 2
-
HB
Weaker
acid
+
NH
+
4
Ammonium ion
pK
a
9.24
(weaker acid)
Position of equilibrium
• Stronger acid and stronger base react to
give weaker acid and weaker base
C H 3 C O2 H
Acetic acid
pK a 4.76
C 6 H 5 OH
Phenol
pK a 9.95
+
+
H C O3
-
Bicarbonate
ion
H C O3
-
Bicarbonate
ion
C H 3 C O2
Acetate
ion
C6 H5 O
Phenoxide
ion
-
-
+
H 2 C O3
Carbonic acid
pK a 6.36
+
H 2 C O3
Carbonic acid
pK a 6.36