Counting Halomethanes ~arthasarathyNambi Veterans Administration Medical Center Kansas City, MO 64128
Undergraduate students learning organic chemistry are often asked to enumerate the possible isomers of simple alkanes. However, very little emphasis is given to illustrate the range of molecular species that can be derived by substituting the hydrogens of an alkane, for example, by halogens. The simple case is to substitute the hydrogens of methane with the halogens.
How many halomethanes are there? Acceptable Solutions The obvious solution is to write out all the possible halomethanes. One does arrive at the correct answer of 70 halomethanes, including methane. However, there is an easier way, which is also a simple example of the application of combinatorics in chemistry. We consider the four hydrogens of methane as "boxes". We have five "objects" (H, F, Cl, Br, and I) to use and place in these boxes. We need to remember that repetitions are possible, that is, two boxes can have objects of the same kind. Case 1: We can have all the boxes containing the same
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kind of objects, that is halomethanes of the type CX4.Since we have five different objects it is obvious we have 5 halomethanes of this type. Case 2: Three boxes can have objects of the same kind, and the fourth can be different, that is, CXsY. In this case for X we have five choices and for Y four choices. Therefore, there are 5 X 4 = 20 halomethanes of the type CXiY. Case 3: In this case two boxes have objects of the same kind, and the other two boxes have different objects. Now, we have two possibilities with the halomethanes of the type CX2Y2and CX2YZ.Let us first consider the case of CX2Y2.For X we have five choices and for Y four choices, but we have to remember that CX2Y2= CY2X2.Hence, there are (5 X 4)12 = 10 halomethanes of the type CX2Y2.For the case of halomethanes of the type CX2YZ,note that for X we have five choices and for Y and Z there are four and three choices, respectively. Since CXgYZ = CX2ZY,we must divide by 2; that is, 5 X (4 X 312) = 30 halomethanes of the type CX2YZ. Case 4: In this case all the boxes contain different objects, that is, halomethanes of the type CWXYZ. It is a simple matter to find out that there are five halomethanes of this type. Adding all the above cases we find that there are 70 halomethanes. Moreover, there is an equation that can be used to arrive at the answer in one step! The number of combinations of n objects taken k at a time with the possibility of repetition is given by the equation:
In our case n = 5 and k = 4. Substituting these values for n and k we find that C n k= 70.
