a. Tetrahydropyran has a boiling point of 88 C, and cyclopentanol has a boiling

Question

a. Tetrahydropyran has a boiling point of 88 C, and cyclopentanol has a boiling point of 141 C. These two isomers have a boiling point difference of 53 C. Explain why the two oxygen-containing isomers have a much larger boiling point difference than the two amine isomers.

b. N,N -Dimethylformamide has a boiling point of 150 C, and N-methylacetamide has a boiling point of 206 C, for a difference of 56 C. Explain why these two nitrogen-containing isomers have a much larger boiling point difference than the two amine isomers. Also explain why these two amides have higher boiling points than any of the other four compounds shown (two amines, an ether, and an alcohol).

Explanation / Answer

First problem, Intermolecular forces of attraction play a key role in boiling point. The higher the attractive force, the higher is boiling point of compound. In this case, each molecules has about the same Vanderwaals forces, cyclopentanol is capable of hydrogen bonding with itself, Tetrahydropyron is only capable of dipole dipole interaction. Therefore, Cyclopentanol is having higher boiling point than Tetrahydropyron because hydrogen bonding is much stronger than dipole interaction.

For the second question, The first structure has no hydrogen bonding because it has no O-H or N-H bond, but it is highly polar structure and has a large dipole moment, so it's dipole-dipole intermolecular forces is very high. The second structure has hydrogen bonding because of the N-H bond in addition to the strong dipole reflected in its higher boiling point.

Both amides have higher boiling point than any of the four compound because of the dipole dipole interactions that are accentuates by the resonating forms.

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