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.