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#1 2009-02-14 12:48:41


Chlorination of 3-methylhexane

Yahoo!Answers in chemistry wrote:

Our professor discussed two problems in class. He drew a line picture of hexane with a methyl group on the third carbon. He then asked these questions:
1) How many monohalosubstituted products can be obtained here?
2) How many kinds of hydrogen do you see?
3) Is there symmetry in the molecule?
His answer to 1) is 7, 2) is 7, and 3) is No.

1) Your professor erred here. He should have used 2-methylhexane. 3-Methylhexane has 10 monosubstituted products. There are 16 hydrogens (C7H16) and replacing each would give a monohalogenated product. However, replacing each hydrogen of the methyl group gives the same product. Replacing each hydrogen with another atom is an alternate method I use to identify the number of signals you may find in an NMR. If you replace a hydrogen with another atom, if any two substitutions give the same product, they are chemically equivalent. (If you replace any hydrogen of ethane, with chlorine they all give chloroethane. Therefore all six hydrogens are chemically equivalent.) Replacing the hydrogens of the methyl group gives two additional instances of the same product. There are three methyl groups so there are 10 possible products.

Because 3-methylhexane has a chirality center, the hydrogens of the CH2 groups are not equivalent. Each gives a separate product. If you start with R-3-methylhexane and replace the hydrogens at C2, the products are (2R,3R)- and (2S,3R)-2-chloro-3-methylhexane. If you adjust this to include all of the monohalogenated compounds and the fact that you are starting with a racemic mixture (R)- plus (S)-3-methylhexane, you will get 20 compounds or 10 racemic compounds.

If 2-methylhexane were used, it does not contain a chirality center, there would be seven compounds as suggested above.

2) For the kinds of hydrogens present, I again agree the professor erred. For alkanes, the rates of halogenation do not differ greatly. Thus, you may calculate the relative ratios of products based upon the kinds of hydrogens present, CH3, CH2, and CH. You may find examples and . Chlorination and bromination are similar reactions and differ in the ratios at which substitution of the three kinds of hydrogens are replaced.

The mechanism of a free radical substitution is also shown in .

3) Agreed



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