Friday, May 29, 2009

Alkyne Reactions:



Using Lindlar's catalyst results in syn addition to form a cis alkene.


Using lithium in liquid ammonia yields a trans alkene



Ozonolysis


Preparation of alkynes

Diene Reactions:


Adding HBr to a diene forms a 1-2 adduct (top) or a 1-4 adduct (bottom)



Adding Br2 to a diene forms a 1-2 adduct (top) or a 1-4 adduct (bottom)

Alkene Reactions:



Creates a vicinal halide (anti addition)



syn addition


Creates an alkyl halide following Markovnikov's ru


Creates a Markovnikov alcohol


Creates an anti-Markovnikov alcohol


Alkenes react with peracides to form epoxodies/oxiranes through a syn addition


Yields two carbonyl compounds.




React with cold potassium permanganate to form vicinal diols through a syn addition.


Eliminates water to form the most stable alkene (use Zaitsev's rule)


Eliminates HX to form the most stable alkene (anti elimination) (use Zaitsev's rule)

How to Balance Redox Reactions - Balancing Redox Reactions

Balancing Redox 4 of 6 Reactions - Balance the Charge

Next, balance the charges in each half-reaction so that the reduction half-reaction consumes the same number of electrons as the oxidation half-reaction supplies. This is accomplished by adding electrons to the reactions:

2 I- → I2 + 2e-

5 e- + 8 H+ + MnO4- → Mn2+ + 4 H2O

Now multiple the oxidations numbers so that the two half-reactions will have the same number of electrons and can cancel each other out:

5(2I- → I2 +2e-)

2(5e- + 8H+ + MnO4- → Mn2+ + 4H2O)

Balancing Redox 5 of 6 Reactions - Add the Half-Reactions


Now add the two half-reactions:

10 I- → 5 I2 + 10 e-

16 H+ + 2 MnO4- + 10 e- → 2 Mn2+ + 8 H2O

This yields the following final equation:

10 I- + 10 e- + 16 H+ + 2 MnO4- → 5 I2 + 2 Mn2+ + 10 e- + 8 H2O

Get the overall equation by canceling out the electrons and H2O, H+, and OH- that may appear on both sides of the equation:

10 I- + 16 H+ + 2 MnO4- → 5 I2 + 2 Mn2+ + 8 H2O

Balancing Redox 6 of 6 Reactions - Check Your Work


Check your numbers to make certain that the mass and charge are balanced. In this example, the atoms are now stoichiometrically balanced with a +4 net charge on each side of the reaction.



How to Balance Redox Reactions - Balancing Redox Reactions

Balancing Redox Reactions 1 of 6- Half-Reaction Method


To balance redox reactions, assign oxidation numbers to the reactants and products to determine how many moles of each species are needed to conserve mass and charge. First, separate the equation into two half-reactions, the oxidation portion and the reduction portion. This is called the half-reaction method of balancing redox reactions or the ion-electron method. Each half-reaction is balanced separately and then the equations are added together to give a balanced overall reaction. We want the net charge and number of ions to be equal on both sides of the final balanced equation.

For this example, let's consider a redox reaction between KMnO4and HI in an acidic solution:

MnO4- + I- → I2 + Mn2+

Balancing Redox Reactions 2 of 6- Separate the Reactions

Separate the two half reactions:

I- → I2

MnO4- → Mn2+




Balancing Redox 3 of 6Reactions - Balance the Atoms


To balance the atoms of each half-reaction, first balance all of the atoms except H and O. For an acidic solution, next add H2O to balance the O atoms and H+ to balance the H atoms. In a basic solution, we would use OH- and H2O to balance the O and H.

Balance the iodine atoms:

2 I- → I2

The Mn in the permanganate reaction is already balanced, so let's balance the oxygen:

MnO4- → Mn2+ + 4 H2O

Add H+ to balance the 4 waters molecules:

MnO4- + 8 H+ → Mn2+ + 4 H2O

The two half-reactions are now balanced for atoms:

MnO4- + 8 H+ → Mn2+ + 4 H2O

Virtual Chemistry Text - Oxidation & Reduction Reactions

Virtual Chemistry Text Table of Contents

Oxidation and reduction reactions go hand in hand, which is why they are also called redox reactions. Acids and bases may be thought of as reactions involving hydrogen, or protons, while redox reactions tend to be concerned with electron gain and loss.

Introduction to Molecular Geometry

Three-Dimensional Arrangement of Atoms in a Molecule




There are two electron pairs around
the central atom in a molecule with
linear molecular geometry, 2 bonding
electron pairs and 0 lone pairs.
The ideal bond angle is 180°.Ben Mills




Molecular geometry or molecular structure is the three-dimensional arrangement of atoms within a molecule. It is important to be able to predict and understand the molecular structure of a molecule because many of the properties of a substance are determined by its geometry.

The Valence Shell, Bonding Pairs, and VSEPR Model

The outermost electrons of an atom are its valence electrons. The valence electrons are the electrons that are most often involved in forming bonds and making molecules.

Pairs of electrons are shared between atoms in a molecule and hold the atoms together. These pairs are called "bonding pairs".

One way to predict the way electrons within atoms will repel each other is to apply the VSEPR (valence-shell electron-pair repulsion) model. VSEPR can be used to determine a molecule's general geometry.

Predicting Molecular Geometry

Here is a chart that describes the usual geometry for molecules based on their bonding behavior. To use this key, first draw out the Lewis structure for a molecule. Count how many electron pairs are present, including both bonding pairs and lone pairs. Treat both double and triple bonds as if they were single electron pairs. A is used to represent the central atom. B indicates atoms surrounding A. E indicates the number of lone electron pairs. Bond angles are predicted in the following order:

lone pair versus lone pair repulsion > lone pair versus bonding pair repulsion > bonding pair versus bonding pair repulsion