1
$\begingroup$

I have a few questions about whether organic functional groups can be classified as another.

First, can carboxylic acids be also be considered alcohols (they have $\ce{-OH}$)? If not, can they at least behave like alcohols?

Similarly, can esters ($\ce{RCOO-R'}$) be considered ethers because the $\ce{R'}$ group is linked to the other half of the ester via an $\ce{-O -}$ linkage? In whichever case, what are some chemical characteristics that distinguish ethers and esters?

$\endgroup$
1
$\begingroup$

Carboxylic acids are not also alcohols.

In general carboxylic acids are just that acids. They are prone to ionizing in water. Alcohols in general don't ionize in aqueous solution, and they certainly don't ionize in water to give acidic solutions.

Esters are not also ethers.

Esters react with water to form a carboxylic acid and an alcohol. Ethers are relatively stable in water.

PS - It isn't a bad thought! Everyone who has taken organic chemistry has had to ponder the points raised by your question.

$\endgroup$
  • 3
    $\begingroup$ "Alcohols in general don't ionize." Not especially true -- they just require a considerably higher $\mathrm{pH}$ compared to carboxylic acids. Phenol's $\mathrm pK_\mathrm{a}$ is actually quite modest, only $\sim 10$. $\endgroup$ – hBy2Py Mar 17 '16 at 3:31
  • 1
    $\begingroup$ @Brian, ok I qualified it a bit more... $\endgroup$ – MaxW Mar 17 '16 at 3:43
4
$\begingroup$

Would it make sense to classify a pistol as a type of pipe? The barrel of a pistol is a cylindrical metal tube, after all.

Clearly, no. The other key defining components of a pistol (stock, trigger, etc.) combine with the cylindrical metal tube to radically change its function and behavior.

Similarly, even though a carboxyl group does contain an $\ce{-OH}$ group bonded to a carbon atom, its behavior is sufficiently different from that of an alcohol that it makes no sense to group them together. A parallel argument holds for ethers and esters as well, as ably noted by MaxW.

$\endgroup$

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.