9
$\begingroup$

Is there any trend here at all? This seems very chaotic as a trend.

$\endgroup$
3
  • 3
    $\begingroup$ This paper may help. $\endgroup$
    – DHMO
    Jan 18 '17 at 13:27
  • 3
    $\begingroup$ It has the common zig-zag practically all properties of alkylic homologues show, and there is a minimum at 5. Looks very regular to me. No.5 cannot decide if it's rather polar or nonpolar, and gets so confused it does not crystallise. $\endgroup$
    – Karl
    Jan 18 '17 at 18:02
  • 1
    $\begingroup$ More to the point the odd number carbons have one trend and the even another. $\endgroup$
    – MaxW
    Feb 18 '17 at 6:30
1
$\begingroup$

My guess without looking at any literature:

Higher acidity $\implies$ higher hydrogen bonding $\implies$ higher melting point.

Thus, changing from H and Me where there is virtually no +I effect from the chain, to Et, Pr, Bu where the +I effect of the chain decreases the melting point, as the hydrogen bonds get weaker, as the COOH groups are less acidic.

Further increasing the chain length after hexyl does not really lead to any additional +I effects on the COOH group that is now far away. Thus, the melting point increases as there are additional van der Waals interactions between the longer chains.

Then there seems to be a factor that makes crystals of odd chains more favorable.

$\endgroup$
6
  • $\begingroup$ So there are basicall two opposing trends: longer chains decrease the stability of the coo- anion but then after a certain chain length each methylene group adds additional vdw interactions. $\endgroup$ Jan 18 '17 at 14:02
  • $\begingroup$ Please, use proper punctuation, spacing, and capitalization, and do not use short forms unless in chemical formulas. $\endgroup$
    – DHMO
    Jan 18 '17 at 14:05
  • $\begingroup$ @DHMO Sorry, was kind of in a hurry ; ) $\endgroup$ Jan 18 '17 at 18:09
  • $\begingroup$ I'm afraid only last 2 sentences are OK to have them in answer to this question. $\endgroup$
    – Mithoron
    Jan 18 '17 at 19:37
  • $\begingroup$ @Mithoron Why exactly? $\endgroup$ Jan 18 '17 at 20:34
1
$\begingroup$

An obvious rule is that for even $n+1$ the melting point of the acid $\ce{C_{n+1}}$ is higher than the melting point of acid $\ce{C_n}$.

$\endgroup$
2
  • 1
    $\begingroup$ I don't find it obvious. Can you explain why? $\endgroup$
    – Yashas
    Mar 28 '17 at 12:58
  • $\begingroup$ The zigzag line in the above image shows that the melting point increases from 1 to 2, from 3 to 4, from 5 to 6, and so on. I'm quite sure that there's no exception from this rule for n-alkyl carboxylic acids. $\endgroup$
    – aventurin
    Mar 28 '17 at 20:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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