The boiling points of $\ce{CH3COCH3}$, $\ce{CH3COC2H5}$, and $\ce{CH3COC3H7}$ are $56˚C$, $80 ˚C$, and $102 ˚C$, respectively. What factor(s) best account(s) for this?

I believe london dispersion forces is one of them. Is dipole-dipole interactions another one? It seems to me that there are dipole-dipole interactions since none of the structures of symmetrical, so can they account for this increase in boiling point?


1 Answer 1


You're definitely right on London. Between the three molecules, the only thing that's changing is the carbon chain on the right hand side of the Oxygen. That plus the corresponding increase in boiling temperatures together form a good indicator that London forces are at play here.

Dipole-dipole (or Hydrogen bonding for that matter) are non-significant in this case, because you're not changing the electronegativity of anything here, only adding more of the same repeating atoms.

Source: This was on the ACS 2011 local exam, question 55.

Further information here


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.