27
$\begingroup$

When there are $2$ hydroxy groups on same carbon, they tend to undergo dehydration ( of course with exceptions like chloral where $\ce H$ bonding prevails).

Why does this happen? We are breaking a sigma bond in order to get a pi bond between $\ce C$ and $\ce O$ ?

Improve this question
$\endgroup$
1
  • $\begingroup$ "chloral where H bonding prevails" that's not the reason why chloral is mostly hydrated. $\endgroup$
    – Mithoron
    Commented Aug 16, 2023 at 15:08

1 Answer 1

Reset to default
31
$\begingroup$

Bond stability is the answer.

The energy required to break a $\ce{C-O}$ bond is $\pu{85.5 kcal/m}$; the energy given off when a $\ce{C=O}$ bond is formed is around $\pu{178 kcal/m}$ (reference). These bond energies suggest that a carbonyl will generally be a few $\pu{kcal/mol}$ more stable than the corresponding gem-diol $[(2 \cdot 85.5) - 178 = \pu{-7 kcal/mol}]$.

[Entropic effects also play a role (see @Martins comment), but while bond enthalpies will change within a series of compounds the entropic effects should be similar and roughly cancel out when we compare compounds.]

However, the $\pu{7 kcal/mol}$ difference noted above isn't large, so there is a delicate balance in the equilibrium between a carbonyl compound and its corresponding gem-diol. Since $\pu{1.4 kcal/mol}$ will shift an equilibrium by roughly a power of 10, small factors like dipole moments, inductive effects, etc., are enough to stabilize one side of the equilibrium more than the other and significantly shift the equilibrium from one side to the other.

If water is present, then an equilibrium will be established between a carbonyl compound and its corresponding gem-diol. Below is an interesting comparison of acetone and formaldehyde. A seemingly small change (changing $\ce{H}$ to $\ce{CH3}$) shifts the equilibrium from one side to the other (by 6 powers of 10!).

carbonyl-diol equilibria of Acetone and Formaldehyde

Here is a link to a Wikipedia article that has a few more interesting examples of carbonyl-diol equilibria.

In the case of chloral ($\ce{CCl3CHO}$) it is the strong electron withdrawing inductive effect of the $\ce{CCl3}$ group that destabilizes the already positively polarized carbonyl carbon driving the equilibrium to the diol side.

As an aside, if you're not familiar with the history of chloral (or as it is often called, chloral hydrate) google "Mickey Finn" or "knock out drops."

As I've mentioned above a change of just a few $\pu{kcal/mol}$ can markedly shift the equilibrium between a carbonyl and the corresponding gem-diol. Lest the reader think that gem-diols are a rarity, here are a number of examples of compounds that exist predominately in the gem-diol form:

  • ninhydrin (because of the H-bond)
  • $\ce{R-(CO)-(CO)-(CO)-R'}$ (middle ketone houses the gem-diol; see detailed explanation)
  • 1,1-dihydroxycyclopropane ($\mathrm{sp^3}$ carbon relieves ring strain)
  • floral hydrate (similar to chloral hydrate)
Improve this answer
$\endgroup$
6
  • 5
    $\begingroup$ I think we should also consider a small entropic effect, we are making one molecule from two after all. $\endgroup$
    – Martin - マーチン ♦
    Commented Mar 16, 2015 at 14:31
  • 2
    $\begingroup$ The entropy term must be important. E.g, cyclic hemiacetals predominate over the open forms. I don't know how reliable this is, but ChemDraw predicts the enthalpy of hydration of acetone to be ~15 kcal toward the hydrate side. $\endgroup$
    – jerepierre
    Commented Mar 16, 2015 at 15:55
  • 5
    $\begingroup$ @Martin and jerepierre, I understand that entropy plays a role here, but given the shift in equilibria of 10 6 between acetone and formaldehyde, just how "important" is it? I think the key is that as long as we compare a series of compounds, the entropic effects will be similar and cancel allowing us make reasonable predictions using bond enthalpies. I'll edit my answer to make this point. $\endgroup$
    – ron
    Commented Mar 16, 2015 at 17:53
  • $\begingroup$ @jerepierre see my comment to Martin $\endgroup$
    – ron
    Commented Mar 16, 2015 at 17:54
  • 3
    $\begingroup$ @Dissenter ?? The equilibrium can occur under acid or base conditions (just add water). How do you see kinetics playing into a thermodynamic (equilibrium, $\ce{Delta G}$, enthalpy, entropy) problem. $\endgroup$
    – ron
    Commented Mar 17, 2015 at 0:05

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.