March 28, 2024, 10:17:42 PM
Forum Rules: Read This Before Posting


Topic: How to formally determine an ion stability?  (Read 9590 times)

0 Members and 1 Guest are viewing this topic.

Darkstar

  • Guest
How to formally determine an ion stability?
« on: September 12, 2004, 10:38:21 AM »
How do you formally determine an ion stability in a row? Is there a formal algorithm that allows to do that or is this rather intuitive and empirical?

For instance, which one is a stronger acid?
PhCH2COCH3
CH3COCH2COCH3

Which one is a stronger base?
(CH3)3N
p-(iPr)PhOMe

Offline Mitch

  • General Chemist
  • Administrator
  • Sr. Member
  • *
  • Posts: 5298
  • Mole Snacks: +376/-3
  • Gender: Male
  • "I bring you peace." -Mr. Burns
    • Chemistry Blog
Re:How to formally determine an ion stability?
« Reply #1 on: September 12, 2004, 03:53:13 PM »
For the first case you have a phenyl attached which greatly stabalizes the charge. "PhCH2COCH3"
Most Common Suggestions I Make on the Forums.
1. Start by writing a balanced chemical equation.
2. Don't confuse thermodynamic stability with chemical reactivity.
3. Forum Supports LaTex

Offline movies

  • Organic Minion
  • Retired Staff
  • Sr. Member
  • *
  • Posts: 1973
  • Mole Snacks: +222/-21
  • Gender: Male
  • Better living through chemistry!
Re:How to formally determine an ion stability?
« Reply #2 on: September 13, 2004, 11:42:49 AM »
I don't know that there is a formal algorithm, per se, but there is a lot of empirical data which has been rationalized by resonance, inductive, and steric effects.

What Mitch said is correct, the phenyl group will stabilize the charge if you form an enolate to the left side of the carbonyl (as you have it drawn).  Phenyl groups are essential neutral in terms of withdrawing/donating ability, that is to say they can stabilize a positive or negative charge about equally via resonance.

The second molecule you have drawn there is special, however.  If you were to deprotonate between the two carbonyls, you would have an anion that could be delocalized into either of the two carbonyls through resonance.  This effect makes this a very stable anion and therefore the compound is very acid (relatively).

In the bascitiy question you have two different types of functional groups.  You know that an amine is quite basic since it has a lone pair of electrons.  Ethers, however, are not nearly as acidic because oxygen is considerably more electronegative.  This ether is less basic than a normal ether because on side has an aromatic group which delocalizes the electrons from the oxygen into the ring (to a small extent).

Sponsored Links