15
$\begingroup$

Which of the contributing structures of the resonance below is more stable?

enter image description here

I'm watching a video lecture by a professor of my college where he puts this question to the class. The class unanimously says B. At first I thought they were wrong, but then the professor agreed with them.

Is this true? The reason he gave was that oxygen's octet would be complete in B, hence it is more stable. But I don't understand this.

1) Isn't the octet complete in A too? It has 2 covalent bonds - 4 electrons (2 from sharing) and 2 lone pairs. Adding up to 8.
2) Oxygen being more electronegative than carbon, would pull the electrons towards itself, leading me to think that it would be more comfortable with a lone pair rather than another bond. Also, won't it prefer to lose out that '+' charge which the bonding's burdened onto it?

Am I wrong?

Another thing which made me believe that Prof. may have made a mistake while talking about the more stable compound, is that when he added Br$^-$ to the compound, he added it to the compound A. And don't you continue reactions with the most stable products?


Answer

The presence of an extra bond in the canonical structure B, along with the completion of the carbon valency makes it more stable than A. Br$^-$ will attach to A itself, though, because the electrons on oxygen will repel the incoming nucleophile.
Thanks, everyone!

$\endgroup$
  • 1
    $\begingroup$ I guess the octet is complete in both the cases. But for the stability of these resonance structures, points to considered in decreasing order of importance are higher number of bonds; complete octet; charge separation; +charge on electropositive and -charge on electronegative elements. $\endgroup$ – Satwik Pasani Oct 25 '13 at 14:28
  • $\begingroup$ So you say that option (B) is more stable? $\endgroup$ – mikhailcazi Oct 25 '13 at 14:36
  • $\begingroup$ While adding Br$^-$ to the compound, why do we take the reaction further by adding it to compound (A), then? :/ $\endgroup$ – mikhailcazi Oct 25 '13 at 14:37
  • $\begingroup$ Between A and B, look at how the octet rule is satisfied for the carbon atom (and not the oxygen)… $\endgroup$ – F'x Oct 26 '13 at 13:32
  • 1
    $\begingroup$ @SatwikPasani okay, so option (B) is more stable because 1) extra bond 2) carbon octet complete. Thanks! This helped. $\endgroup$ – mikhailcazi Oct 29 '13 at 10:09
11
$\begingroup$

Which of the contributing structures of the resonance below is more stable?

Technically, neither. Both structures A and B are resonance contributors to the same true structure of the ion. Structures A and B, because they both represent the same species, cannot have different energies, and therefor they cannot have different stabilities.

What the question is asking is "Which contributor is more important in describing the structure and behavior of the hybrid?" Contributor B probably approximates the structure better (i.e. the positive charge is more on the oxygen than on the carbon), but structure A represents the reactivity better (i.e. nucleophiles want to attack the carbon and not the oxygen).

$\endgroup$
2
$\begingroup$

The difference between the two structures is just the amount of space available to the lone pair. If you think about it in terms of the particle in a box approximation, the path length for that lone pair is much longer when in the double bond state, so it will be a lower energy state.

$\endgroup$
1
$\begingroup$

Structure A does NOT have on octet on the carbon that has the plus charge. Structure B has an octet on every atom. Thus, structure be is more stable.

$\endgroup$
-1
$\begingroup$

RULE 1: resonating structure with more number of bonds wins and (B) has 1 more bond than (A) hence it wins

$\endgroup$
-2
$\begingroup$

Structure A is more stable. Reasons are:

1) The O has a complete octet with 2 electrons per bond around it(based off of octet rule, not formal charge or oxidation state) + 2 lone pairs.

2) The O is not positive. O prefers to be negative if charged at all because of its electronegativity being so high.

3) Carbon is not so determinant on charge. It is fine being positive, negative, or neutral.

4) Each carbon that isn't positively charged has its octet from C-C-C or O-C-C + 2 H's.

$\endgroup$

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.