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I am ond of those people who craves ice cream the most when its freezing outside, but I won't say no to it during the Summer,when most people appreciate it. You can apply freezing point depression or the thermal properties of materials to make ice cream yourself.

Reduction & Oxidatiom reactiom are action at a time in a reaction.

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

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)

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)

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 - → I 2 + 2e - 5 e - + 8 H + + MnO 4 - → Mn 2+ + 4 H 2 O 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 - → I 2 +2e - ) 2(5e - + 8H + + MnO 4 - → Mn 2+ + 4H 2 O) Balancing Redox 5 of 6 Reactions - Add the Half-Reactions Now add the two half-reactions: 10 I - → 5 I 2 + 10 e - 16 H + + 2 MnO 4 - + 10 e - → 2 Mn 2+ + 8 H 2 O This yields the following final equation: 10 I - + 10 e - + 16 H + + 2 MnO 4 - → 5 I 2 + 2 Mn 2+ + 10 e - + 8 H 2 O Get the overall equation by canceling out the electrons and H 2 O, H + , and OH - that may appear on both sides of the equation: 10 I - + 16 H + ...

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 KMnO 4 and HI in an acidic solution: MnO 4 - + I - → I 2 + Mn 2+ Balancing Redox Reactions 2 of 6- Separate the Reactions Separate the two half reactions: I - → I 2 MnO 4 - → Mn 2+ Balancing Redox 3 of 6Reactions - Balance the Atoms To balance the atoms of each half-r...