1
$\begingroup$

The order I was provided with is(comparing molecules of similar molecular mass from each);

alkanes< ethers< esters< aldehydes< ketones<< alcohols<< carboxylic acids<< amides

Where << is supposed to signify a considerable hike.

  1. Here, how do you compare esters and ethers?
  2. I assume the higher position of the carbonyls with respect to those are due to the keto-enol tautomerism they have(with the enol forms having hydrogen bonding).But how are ketones placed below alcohols?
  3. Why are amides at the very top? Nitrogen being less electronegative than oxygen, I would've thought the hydrogen bonding in those wouldn't be as strong as in the carboxylic acids.
$\endgroup$
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  • 2
    $\begingroup$ 1. Much can be attributed to dipole moments. Ethers typically have smaller ones. 2. The enol form is typically negligible. Again you can think about dipole moments (polarity). Alcohols can H-bond 3. Amides have large dipole moments (polarity) and can H-bond. See a solvent polarity table eg people.chem.umass.edu/xray/solvent.html $\endgroup$
    – Buck Thorn
    May 1, 2021 at 10:12
  • $\begingroup$ The values are okay, but it's the explanation I needed. I'll take your word for it, thanks! $\endgroup$
    – harry
    May 1, 2021 at 10:45
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    $\begingroup$ You may also want to refer to a the relative order of the strength of different intermolecular interactions: induced-dipole-induced-dipole<dipole-induced-dipole<dipole-dipole<H-bond $\endgroup$
    – Buck Thorn
    May 1, 2021 at 12:15
  • $\begingroup$ Substituted amides can H-bond less of course. Compare formamide with DMF and you see the effect of methyl-substitution (less available sites for H-bonding) on the bp. $\endgroup$
    – Buck Thorn
    May 1, 2021 at 12:53

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