FORMAL CHARGEFORMAL CHARGE
Formal ChargeFormal Charge
N:... ..
N:....
H
H
NH2-
BondedUnbonded
Number of All One half of = valence electrons unshared + all shared in the neutral electrons electrons atom
( Formal Charge = 5 - 4 - 2 = -1 )
5e- 6e-
LEWIS DIAGRAMS SHOW IT ALL !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! )
When drawing a Lewis Diagram remember these rules.
Rumus KimiaRumus Kimia
Rumus empirikRumus MolekulRumus struktur
Rumus struktur lengkapRumus struktur panjang (expanded)Rumus struktur termampatkan (condensed)
Rumus Struktur pada Rumus Struktur pada senyawa siklis – sikloheksanasenyawa siklis – sikloheksana
Expanded formula
C
CC
C
CC
H H
H
H
H
HH
H
H HH
H
C6H12
Rumus Struktur pada Rumus Struktur pada senyawa siklis – sikloheksanasenyawa siklis – sikloheksana
Polygon formula(condensed formula)
C6H12
Rumus Struktur pada senyawa Rumus Struktur pada senyawa siklis – sikloheksanasiklis – sikloheksana
Condensed formula
C6H12
CH2
CH2CH2
CH2
CH2
CH2
Contoh Molekul siklisContoh Molekul siklis
C C
C
CH2 CH2
CH2
H H
H
HH
H
C
C C
C CH2
CH2 CH2
CH2
H
H
H
H
H H
HH
C C
C
C
C
CH2 CH2
CH2
CH2
CH2
H H
H
H
H
H
HH
HH
Cyclopropane C3H6
Cyclobutane C4H8
Cyclopentane C5H10
Beberapa cara penulisan strukturBeberapa cara penulisan struktur
H
O
citronellal(CH3)2C=CHCH2CH2CH(CH3)CH2CHO
C CH
H3C
H3C
CH2 CH2 C
CH3
H
CH2 CH
Oor
CC C
H
HH
H
HH
CC H
H
CH
H CH
C
CH H
H
C
HH
O
H
expanded
condensed
line
line structures are most compactand easy to read
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 (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,
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
O C O
Carbon dioxide = 0 D
B
F
F
F
Boron trifluoride = 0 D
C
Cl
ClClCl
Carbon tetrachloride = 0 D
Polar and Nonpolar Molecules• these molecules have polar bonds and
are polar moleculesN
HH
H
O
H H
Water = 1.85D
Ammonia = 1.47D
directionof dipolemoment
directionof dipolemoment
Polar and Nonpolar Molecules
– formaldehyde has polar bonds and is a polar molecule
Formaldehyde = 2.33 D
directionof dipolemoment H H
C
O
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 CH2 CH2 CH2 CH3
n-pentane, b.p. = 36°C
CH3 CH
CH3
CH2 CH3
isopentane, b.p. = 28°C
C
CH3
CH3
CH3
H3C
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
CH3 CH2 OH
ethanol, b.p. = 78°C
CH3 O CH3
dimethyl ether, b.p. = -25°C
trimethylamine, b.p. 3.5°C
N CH3H3C
CH3
propylamine, b.p. 49°C
CH3CH2CH2 N
H
H
ethylmethylamine, b.p. 37°C
N CH3CH3CH2
H
CH3 CH2 OH
ethanol, b.p. = 78°C ethyl amine, b.p. 17°C
CH3 CH2 NH2
ASAM DAN BASA
Brønsted-Lowry Theory of Acids and Bases
• Acid:Proton Donor
• Base: 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
Base(weaker base)
Acid(weaker acid)
Conjugate baseof CH3CO2H
(stronger base)
Conjugate acidof H2O
(stronger acid)
CH3COH + H2 O
+ H3 O+
CH3CO-
O
O
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 H3O++A
-
[H3O+][A
-]
[HA]Ka
+ H2O
= Keq[H2O]
pKa = - log Ka
Weak Acids and Bases
Acid Formula pKaConjugate Base
ethanol
water
bicarbonate ion
ammonium ion
carbonic acid
acetic acid
sulfuric acidhydrogen chloride
10.33
15.7
15.9
4.76
6.36
9.24
-5.2
-7
CH3CH2OH CH3CH2O-
H2O HO-
HCO3-
CO32-
NH4+ NH3
H2CO3 HCO3-
CH3CO2H CH3CO2-
H2SO4 HSO4-
HCl Cl -
Acidity Constant (Ka)
HA + H2OK
A- + H3O+
K =[A-] [H3O+]
[H2O][HA]
Ka =[A-] [H3O+]
[HA]K [H2O] =
pKa = - log Ka
pKa
pKa = - log Kaa
Strong acid = large Ka = small pKaa
Weak acid = Weak acid = small Ka = large pKaa
Relative Acid Strength
HClO4 ClO4
_stronger
weaker
ACIDSTRENGTH
weaker
stronger
BASESTRENGTH
ACID CONJ. BASE Ka pKa
10 -1010
1.8 x 10 4.8-5
1.0 x 10 10-10
10 50-50
OH O
CH3 C OH
O
CH3 C O
O
CH3 CH3CH3 CH2
_
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
++Stronger acid
Stronger base
Weaker base
Weaker acid
HA B-
A-
HB
(weaker acid)(stronger acid)pKa 9.24pKa 4.76
+ +
Acetic acid Ammonium ion
CH3CO2H NH3 CH3CO2-
NH4+
Position of equilibrium
• Stronger acid and stronger base react to give weaker acid and weaker base
Carbonic acidpKa 6.36
PhenolpKa 9.95
++C6 H5OH HCO3- C6 H5O
-H2 CO3
Bicarbonateion
Phenoxideion
+ +
Acetic acidBicarbonateion
Acetateion
Carbonic acidpKa 4.76 pKa 6.36
CH3 CO2H HCO3-
CH3 CO2-
H2 CO3