Try it on paper first. A hint, the key idea and the worked solution are here when you need them.
A haloalkane has the general formula . It contains bromine by mass. Take C as , H as and Br as . Find the value of and the molar mass. Then count the structural isomers of this formula, and say how many of them have a carbon atom with four different groups.
A haloalkane is primary, secondary or tertiary by the number of carbon atoms joined to the carbon that holds the halogen. Isomers are counted by listing every carbon skeleton and every place for the halogen, then removing repeats. A carbon with four different groups is a stereocentre, and it doubles the count for its compound.
Have you tried it?
Work it out on paper first. It is fine to get stuck or make a mistake: that is how we learn. When you are ready, go through the solution one step at a time, and check each step against your own.
Given. Bromine is of the molar mass.
Step 1. Molar mass. .
Step 2. The value of n. , so and . The formula is .
Step 3. Isomers. Butane gives 1-bromobutane and 2-bromobutane. 2-Methylpropane gives 1-bromo-2-methylpropane and 2-bromo-2-methylpropane. That is 4.
Step 4. Stereocentre. Only 2-bromobutane has a carbon with four different groups (H, Br, CH3 and C2H5). So 1 isomer.
Answer: , , 4 structural isomers and 1 with a stereocentre.
Check. .
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