Neopentane (2,2-dimethylpropane, MW 72.15 g/mol) boils at 10 degree C, and tetra

Question

Neopentane (2,2-dimethylpropane, MW 72.15 g/mol) boils at 10 degree C, and tetrafluoromethane (MW 88.01 g/mol, in which each of the 4 methyl groups in neopentane has been replaced with a fluorine) boils at -128 degree C. Explain this large difference in b.p. by referring to all three types of IMF's. (b) What is your qualitative prediction for the boiling point of t-butyl fluoride the molecule in which only one methyl group has been replaced by a -F. Is it higher or lower than neopentane? Higher or lower than tetrafluoromethane? Explain your prediction, again making reference to the role played by all three types of non-covalent intermolecular forces. Which type of IMF to you expect to "dominate"? Make your prediction consistent with that expectation.

Explanation / Answer

(A) Neopentane molecule is bigger in size than tetrafluoromethane as methyl group is bigger in size than flourine. Hence neopetane has greater surface area than tetrafluoromethane. As a result the van Der Waals interactions are greater in neopentane resulting in greater boiling point than tetrafluoromethane.

(B) t-butyl fluoride has one methyl group less than neopentane. So neopentane has greater surface area than

t-butyl fluoride. This meight result in greater van Der Waals interactions in neopentane. Hence boiling point of neopentane meight be greater than t-butyl fluoride. From the same argument the boiling point of t-butyl fluoride meight be greater than tetrafluoromethane.

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