E-StoreOrganic reaction mechanisms.....
Free radical halogenation
Initiation Step:
The reaction begins with an initiation step, which is the separation of the halogen (X2) into two radicals (atoms with a single unpaired electron) by the addition of uv light. This is called the initiation step because it initiates the reaction.
Propogation Steps:
The initiation step, or the formation of the chlorine radicals, is immediately followed by the propogation steps--steps directly involved in the formation of the product. As an example, isobutane (C4H10) will be used in the chlorination reaction. The first step is the abstraction of the hydrogen atom from the tertiary carbon (a tertiary carbon is a carbon that is attached to three other carbon atoms) Note that these are not protons (H+ ions) that are being abstracted, but actual hydrogen atoms since each hydrogen has one electron. This first propogation step forms the tertiary radical.
In the last step, the tertiary radical then reacts with another one of the chlorine molecules to form the product. Notice that another chlorine radical is regenerated, so this reaction can, in theory, go on forever as long as there are reagents. This is called a chain reaction. 
A sidenote on free radical stabilities:
Hydrogens attached to more highly substituted carbons (ie. carbons with many carbons attached to them) are more reactive in free-radical halogenation reactions because the radical they form is stabilized by neighboring alkyl groups. These neighboring alkyl groups have the ability to donate some of their electron density to the electron-deficient radical carbon (a radical is short one electron of filling the atom's valence octet). Thus the hydrogen on the tertiary carbon here is abstracted in preference to the 9 other hydrogen atoms attached to a primary carbon (a carbon that is attached to only one other carbon atom) because it forms a more stable radical.
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| Here, the tertiary radical is stabilized by electron donation from neighboring alkyl groups. |
Selectivity of free-radical halogenation
A point of note about free radical processes is that the intermediates are so highly reactive and short lived that usually you obtain a mixture of products, even though there is preference for forming more highly substituted free radical intermediates. In this example with isobutane, for instance, there would certainly be some abstraction of hydrogens attached to the primary carbons, leading to a different product than the above product (can you draw it out?).
Bromine reacts exactly the same way as chlorine; however, it is far more selective. If propane (CH3CH2CH3), for example, was the substrate, 2-bromopropane would be the dominant product, and there would be only a small amount of 1-bromopropane. Free radical chlorination, though, would not be quite as selective, and there would be a greater amount of the chlorination of the primary carbon than in the bromination reaction.
Termination Steps:
Side reactions that can stop the chain reaction are called termination steps. These termination steps involve the destruction of the free-radical intermediates, typically by two of them coming together.
Other Halogens?
So why can't the other halogens such as fluorine or iodine be used? Iodine reacts endothermically (energetically uphill) and too slowly to be of much good in these free radical processes, while fluorine is at the other pole--it reacts too violently and too quickly to be selective, and can, if uncontrolled, even break carbon-carbon bonds. To understand why this is so, derive the DH's for the 4 reactions and compare them (you will find that flourination is highly exothermic, while iodonation is endothermic; chlorination and bromination, however, are right in the middle).
Hydroboration of alkenes
| Overview: The general form of the hydroboration of alkenes mechanism is as follows: First step is the attack of the alkene on BH3, which then forms a four membered ring intermediate of partial bonds. It is because of this intermediate that hydroboration forms the anti-Markovnikov product. The boron atom is highly electrophilic because of its empty p orbital (ie. it wants electrons), and forms a slight bonding interaction with the pi bond. Since some electron density from the double bond is going towards bonding with the boron, the carbon opposite the boron is slightly electron deficient, left with a slightly positive charge. Positive charges are best stabilized by more highly substituted carbons, so the carbon opposite the boron tends to be the most highly substituted. Once the transition state breaks down, BH2 is attached to the least substituted carbon.
Peroxide then removes the borane and replaces it with the alcohol to form the anti-markovnikov product.
An example of the hydroboration mechanism:
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Fischer esterification mechanism Overview: The general form of Fischer esterification mechanism is as follows: The first step involves protonation of the carbonyl oxygen, followed by the nucleophillic attack of the alcohol. Then a loss and regain of a proton, followed by loss of water as electrons from the alcohol oxygen kick down to form the double bond. Loss of a proton yields the ester. Example of fischer esterification:
Friedel-Crafts alkylation
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The general form of the Friedel-Crafts alkylation mechanism is as follows:
Adding an alkyl halide to the Lewis acid aluminum trichloride results in the formation of an organo-metallic complex. In this complex the carbon attached to the chlorine has a great deal of positive charge character (in fact, for practical purposes when dealing with this reaction, you can think of the partially positive charge as a carbocation).
The pi electrons in a benzene ring are mildly electrophilic, and can attack the partially positive carbon to create a non-aromatic intermediate (note that this intermediate has several resonance structures, so that it is not as unstable as it might appear). Elimination of a proton re-establishes the aromaticity of the ring, and the aluminum trichloride catalyst is regenerated along with a molecule of hydrochloric acid. A word of caution about this reaction: because the aluminum trichloride generates what can essentially be thought of as a carbocation, rearrangments can occur to produce a more highly-substituted carbocation. For example: Addition of 1-Chloro-2-Methylpropane to benzene with aluminum trichloride results in the rearranged product, t-butyl benzene, and not the product that you might initially expect (work out the mechanism if you cannot see how that product is attained).
An example of a Friedel-Crafts alkylation:
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Alcohol dehydration
| Overview: The general form of alcohol dehydrations is as follows:
The first step involves the protonation of the alcohol by an acid, followed by loss of water to give a carbocation. Elimination occurs when the acid conjugate base plucks off a hydrogen. Alcohol dehydrations generally go by the E1 mechanism. Example of alcohol dehydration:
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