The boiling points of three unbranched bromoalkanes are 312 K for bromoethane, 344 K for…
Contributed by StemPy LabTry it on paper first. A hint, the key idea and the worked solution are here when you need them.
The boiling points of three unbranched bromoalkanes are for bromoethane, for 1-bromopropane and for 1-bromobutane. Assume that each extra carbon atom raises the boiling point by the same amount. Use the values for the two and the four carbon members to find that rise. Predict the boiling point of 1-bromopentane. The measured value is about . Find how far off the prediction is.
Need a reminder? The key ideaShow
The dipole moment of a bond is the charge multiplied by the bond length, . One debye is . Boiling points of an unbranched series rise by nearly the same amount for each extra carbon atom. Percent yield is .
The worked solution
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.
Step 1. The rise for one carbon atom. From 2 to 4 carbon atoms, in two steps. One step is .
Step 2. Prediction. .
Step 3. Difference. .
Answer: per carbon, predicted, too high.
Check. The real rises are and , and they shrink slowly as the chain lengthens, so the straight line runs slightly high.
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