Chapter 1 Organic Chem
-
Upload
elfisusanti -
Category
Education
-
view
2.325 -
download
7
Transcript of Chapter 1 Organic Chem
Elfi Susanti VH
SBI CLASSChemistry Department
FKIP UNS
Advices for studying organic chemistry
1. Keep up with your studying day to day –– never let yourself
get behind, or better yet, be a little ahead of your instructor.
Organic chemistry is a course in which one idea almost always
builds on another that has gone before.
2. Study materials in small units, and be sure that you
understand each new section before you go on to the next.
3. Work all of the in-chapter and assigned problems.
4. Write when you study. Write the reactions, mechanisms,
structures, and so on, over and over again.
5. Learning by teaching and explaining. Study with your student peers and practice explaining concepts and mechanisms to each other.
6. Use the introductory material in the Study Guide
entitled “Solving the puzzle ––or –– Structure is
everything (Almost)” as a bridge from general
chemistry to your beginning study of organic chemistry.
Once you have a firm understanding of structure,
the puzzle of organic chemistry can become one of
very manageable size and comprehensible pieces. 7. Use molecular models when you study.
Organic Chemistry:
Study carbon compounds
Material: 1. Structure of organic compound2. Alkanes & sicloalkanes3. Alkenes & Alkynes4. Aromatic 5. Alkyl halide6. Alcohol, ether7. Aldehyde & ketone8. Carboxylic acid9. Amine
Electron Configurations in the Periodic Table
1A 2A 3A 4A 5A 6A 7A 8A
1 H 1s1
2
He 1s2
3 Li 1s2 2s1
4 Be 1s2 2s2
5 B
1s2 2s22p1
6 C
1s2 2s22p2
7 N
1s2 2s22p3
8 O 1s2
2s22p4
9 F
1s2 2s22p5
10 Ne 1s2
2s22p6
11 Na
[Ne] 3s1
12 Mg [Ne] 3s2
13 Al
[Ne] 3s23p1
14 Si
[Ne] 3s23p2
15 P
[Ne] 3s23p3
16 S
[Ne] 3s23p4
17 Cl
[Ne] 3s23p5
18 Ar
[Ne] 3s23p6
H 1 1s1 [CORE] VALENCE SHELL
He 2 1s2
Li 3 1s22s1 [1s2]2s1
Be 4 1s22s2 [1s2]2s2
B 5 1s22s22p1 [1s2]2s22px1
C 6 1s22s22p2 [1s2]2s22px12py
1
N 7 1s22s22p3
O 8 1s22s22p4
F 9 1s22s22p5
Ne 10 1s22s22p6
Na 11 1s22s22p63s1 [1s22s22p6]3s1
1s
2s
2p
3s
3p
4s
3d ENERGY
Be [1s2]2s2
Be ..
nuclues
Valence Electron
LEWIS DOT SYMBOLS FOR COMMON ELEMENTS
GROUP I II III IV V VI VII VIII
H He
Li Be B C N O F Ne
Na Mg Al Si P S Cl Ar
K Ca Ge As Se Br Kr
Sn Sb Te I Xe
: : : : :
: : : : :
: :: ::
: : : : :
..
..
..
..
..
..
..
..
..
..
..
..
:
:
:
:
:
:
:
:
.
.......
.
.
.
.
.
.
.
. .
. .
. .
. ..
. ..
...... ..
....
....
.
.
.
.
.
.
.
.
.
.
.
.
.
Table 1-2 in textThe A groups are regular.
Structure formula
Common Name Molecular Formula
Lewis Formula Kekulé Formula
Methane CH4
Ammonia NH3
Ethane C2H6
Methyl Alcohol CH4O
Ethylene C2H4
Formaldehyde CH2O
Acetylene C2H2
Hydrogen Cyanide CHN
Structural Formulas for C4H10O Isomers
Kekulé Formula Condensed Formula Shorthand Formula
Covalen bonding
Polar Covalen bonding
To determine stability of compound as ion or neutral
N:... ..
N:....
H
H
NH2-
( Formal Charge = 5 - 4 - 2 = -1 )
5e-
azide anion has two negative-charged nitrogens and one positive-charged nitrogen, the total charge being minus one.
Ozone, the central oxygen atom has three bonds and a full positive charge while the right hand oxygen has a single bond and is negatively charged. The overall charge of the ozone molecule is therefore zero.
nitromethane has a positive-charged nitrogen and a negative-charged oxygen, the total molecular charge again being zero.
SO O
O
O....
:
..::
..
..:
..::
_
_
_
_
+2
2 -
Determine formal charges !!!
net ioniccharge
Group Formula
Class NameSpecific Example
IUPAC Name
Common Name
Alkene H2C=CH2 Ethene Ethylene
Alkyne HC≡CH Ethyne Acetylene
Arene C6H6 Benzene Benzene
Group Formula
Class Name
Specific Example
IUPAC Name
Common Name
Halide H3C-I Iodomethane Methyl iodide
Alcohol CH3CH2OH Ethanol Ethyl alcohol
Ether CH3CH2OCH2CH3 Diethyl ether Ether
Amine H3C-NH2 Aminomethane Methylamine
Nitro Compound
H3C-NO2 Nitromethane
Thiol H3C-SH MethanethiolMethyl
mercaptan
Sulfide H3C-S-CH3Dimethyl
sulfide
Group Formula Class Name Specific Example IUPAC Name Common Name
Nitrile H3C-CN Ethanenitrile Acetonitrile
Aldehyde H3CCHO Ethanal Acetaldehyde
Ketone H3CCOCH3 Propanone Acetone
Carboxylic Acid H3CCO2H Ethanoic Acid Acetic acid
Ester H3CCO2CH2CH3 Ethyl ethanoate Ethyl acetate
Acid Halide H3CCOCl Ethanoyl chloride Acetyl chloride
Amide H3CCON(CH3)2N,N-
DimethylethanamideN,N-
Dimethylacetamide
Acid Anhydride (H3CCO)2O Ethanoic anhydride Acetic anhydride
The Shape of Molecules
Methane
Ammonia
Water
Configuration Bonding Partners
Bond Angles
Example
Tetrahedral 4 109.5º
Trigonal 3 120º
Linear 2 180º
Distinguishing Carbon Atoms
Resonance
1) sulfur dioxide
2) nitric acid
3) formaldehyde
This averaging of electron distribution over two or more hypothetical contributing structures (canonical forms) to produce a hybrid electronic
structure
4) carbon monoxide
5) azide anion
evaluating the contribution each of these canonical structures makes to
the actual molecule:
1.The number of covalent bonds in a structure.
(The greater the bonding, the more important and stable the
contributing structure.)
2. Formal charge separation.
(Other factors aside, charge separation decreases the stability
and importance of the contributing structure.)
3. Electronegativity of charge bearing atoms and charge density.
(High charge density is destabilizing. Positive charge is best
accommodated on atoms of low electronegativity, and negative charge
on high electronegative atoms.)
The stability of a resonance hybrid is always greater than the stability of any canonical contributor
Atomic and Molecular Orbitals
structure of methane (CH4),
the 2s and three 2p orbitals must be converted to four equivalent hybrid
atomic orbitals, each having 25% s and 75% p character, and designated
sp3. These hybrid orbitals have a specific orientation, and the four are
naturally oriented in a tetrahedral fashion.
Molecular Orbitals
In general, this mixing of n atomic orbitals always generates n molecular orbitals.
Intermolecular Forces
van der Waals attraction Hydrogen Bonding
This attractive force has its origin in the electrostatic attraction
of the electrons of one molecule or atom for the nuclei of
another. If there were no van der Waals forces, all matter
would exist in a gaseous state, and life as we know it would
not be possible. It should be noted that there are also smaller
repulsive forces between molecules that increase rapidly at
very small intermolecular distances.
some of the factors that influence the strength of intermolecular attractions:
The formula of each entry is followed by its formula weight in parentheses
and the boiling point in degrees Celsius.
1.Molecular size.
Large molecules have more electrons and nuclei that create van
der Waals attractive forces, so their compounds usually have higher boiling
points than similar compounds made up of smaller molecules.
2.Molecular shape
3.Molecular dipoles generated by polar covalent bonds
Boiling Points (ºC) of Selected Elements and Compounds
Increasing Size
Atomic Ar (40) -186 Kr (83) -153 Xe (131) -109
Molecular CH4 (16) -161 (CH3)4C (72) 9.5 (CH3)4Si (88) 27 CCl4 (154) 77
Molecular Shape
Spherical: (CH3)4C (72) 9.5 (CH3)2CCl2 (113) 69 (CH3)3CC(CH3)3 (114) 106
Linear: CH3(CH2)3CH3 (72) 36 Cl(CH2)3Cl (113) 121 CH3(CH2)6CH3 (114) 126
Molecular Polarity
Non-polar: H2C=CH2 (28) -104 F2 (38) -188 CH3C≡CCH3 (54) -32 CF4 (88) -130
Polar: H2C=O (30) -21 CH3CH=O (44) 20 (CH3)3N (59) 3.5 (CH3)2C=O (58) 56
HC≡N (27) 26 CH3C≡N (41) 82 (CH2)3O (58) 50 CH3NO2 (61) 101
Hydrogen Bonding
The most powerful intermolecular force
influencing neutral (uncharged) molecules
Hydrogen forms polar covalent bonds to more
electronegative atoms such as oxygen, and
because a hydrogen atom is quite small, the
positive end of the bond dipole (the hydrogen)
can approach neighboring nucleophilic or basic
sites more closely than can other polar bonds.
The molecule providing a polar hydrogen for a
hydrogen bond is called a donor.
The molecule that provides the electron rich site to
which the hydrogen is attracted is called an acceptor.
Water and alcohols may serve as both donors and
acceptors, whereas ethers, aldehydes, ketones and
esters can function only as acceptors.
Primary and secondary amines are both donors and
acceptors, but tertiary amines function only as
acceptors.
Compound FormulaMol. Wt.
Boiling Point Melting Point
dimethyl ether CH3OCH3 46 –24ºC –138ºC
ethanol CH3CH2OH 46 78ºC –130ºC
propanol CH3(CH2)2OH 60 98ºC –127ºC
diethyl ether (CH3CH2)2O 74 34ºC –116ºC
propyl amine CH3(CH2)2NH2 59 48ºC –83ºC
methylaminoethane
CH3CH2NHCH
3
59 37ºC
trimethylamine (CH3)3N 59 3ºC –117ºC
ethylene glycol HOCH2CH2OH 62 197ºC –13ºC
acetic acid CH3CO2H 60 118ºC 17ºC
ethylene diamineH2NCH2CH2N
H2
60 118ºC 8.5ºC
All atoms and molecules have a weak attraction for one another, known as van der Waals attraction. This attractive force has its origin in the electrostatic attraction of the electrons of one molecule or atom for the nuclei of another, and has been called London dispersion force.
In general, larger molecules have higher boiling points than smaller molecules of the same kind, indicating that dispersion forces increase with mass, number of electrons, number of atoms or some combination thereof. The following table lists the boiling points of an assortment of elements and covalent compounds composed of molecules lacking a permanent dipole. The number of electrons in each species is noted in the first column, and the mass of each is given as a superscript number preceding the formula.