7. Tratamientos de lodos
7.1. Producción y características de los lodos
#19- If you don’t start with chirality, you can’t end with it.
This might be one of the finer points of the course, but it is an important one nonetheless. Some professors are sticklers about this point and frankly we think they should be. The rule (and make no mistake, it is a rule) is if you don’t start with chirality somewhere is your reaction, you cannot finish with it. A chiral center cannot just be created magically. Almost always when you create a chiral center without any starting chirality, you have also created an equal amount of its enantiomer, making it a racemic mixture.
Cl2 hv Cl Cl Cl
+
+
In the above reaction, the free radical halogenation of pentane produces three different products, none of which possess chirality. 1-chloropentane is not a chiral molecule for obvious reasons. 3-chloropentane is a meso compound and therefore also not chiral. 2- chloropentane contains a chiral center, but has its enantiomer present in equal amounts and therefore does not contribute to the overall chirality of the reaction mixture. This is because the chlorine radical has an equal chance of attacking from either face of the molecule, and therefore gives an equal mixture of each enantiomer. Because there is no chirality in the starting materials, the overall product mixture is also achiral.
The next logical question would be “how do you get chirality in your starting materials”? There are two methods for doing this: 1) chiral reactant, or 2) chiral catalyst. We will deal with chiral reactants, as chiral catalysts are a complex topic and will be discussed at a later time. CH3 CH3 CH3 CH3 H H H2, Pd-c Major product
In the above example, a chiral center is already set in the molecule, so when the molecule attaches to the Pd-C catalyst, hydrogen will only be added from one face of the molecule, on the face opposite of the isopropyl group. This means that the chiral center established by hydrogenation will have the methyl groups pointing in the same direction as the isopropyl group.
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#20- Markovnikov was a liar. A dirty, smelly, liar.
It is now time to step into the AceOrganicChem time machine. We travel back to Russia; the year is 1869. The czars are big, personal hygiene is not, and we find a young chemist named Vladimir Markovnikov working in his chemistry lab. Dr. Markovnikov has developed a “rule” for organic chemistry that says when adding a hydrogen halide (HX) to an alkene, the nucleophile (X-) will go to the more substituted carbon. The other way to think about this is to say that whichever carbon has more hydrogens, gets the H+ from HX. And thus Markovnikov’s rule was born. Since then, there have been a lot of ways to remember Markovnikov’s rule:
- “Them that has, gets” (hydrogens) - “The rich just get richer” (hydrogens)
- “The less substituted gets more lovin’” (hydrogens)
This was all good until the early 1930’s, when chemist Morris Kharash began to further investigate Markovnikov’s work and found numerous exceptions, or anti-Markovnikov reactions, that Markovnikov decide to ignore. Kharash concluded that Markovnikov’s rule was basically a guess and should be renamed from “Markovnikov’s rule” to “Markovnikov’s speculation”.
Nonetheless, we are less concerned with feuding dead chemists, and more concerned with how you use this on an exam. It is our opinion that you should not memorize which reactions are Markovnikov and which are anti-Markovnikov, but you should learn the theory and mechanism behind the reactions and be able to apply to any situation. The classic example here is HBr addition to propene, with and without peroxides.
HBr Br HBr Br HOOH via C + via C* Br
In the top reaction, we see Markovnikov addition of HBr across the double bond. This is because the first step of the reaction is the addition of H+ to the alkene, which when added to the carbon with more hydrogens, will provide a more stable secondary carbocation intermediate.
This is in contrast to the second reaction, which is called anti-Markovnikov addition. Here, the first step of the reaction is the addition of Br*. If added to the terminal carbon, the intermediate is the more stable secondary radical. Thus, this reaction provides 1-
Orgo I, Difficulty:
Bromopropane as the product, where as the reaction without peroxides gives 2- Bromopropane as the product.
Some other examples of Markovnikov vs. anti-Markovnikov reactions are:
Hg(OAc)2, H2O then NaBH4 OH H HBr Br hv H+ / H2O OH H2SO4 OSO2OH 1. 2. BH3-THF H2O2, NaOH OH H Anti-Markovnikov Anti-Markovnikov Markovnikov Anti-Markovnikov Markovnikov
A better way to think of would be if addition of H+ is the first step, it probably goes through the Markovnikov addition because the more stable intermediate would be a more highly substituted carbocation. If the first step of the reaction is addition of a radical or addition of an electrophilic metal, then think anti-Markovnikov, because the more stable
intermediate would be a more highly substituted carbon radical or an electrophilic metal on a more nucleophilic carbon. Either way, the key is to know which intermediate is more stable, then create a product from that.