6
$\begingroup$

From the Wikipedia article on Reimer-Tiemann reation:

given mechanism

In step 5, why doesn't the oxygen attack the carbene. The way I see it: upon attacking the carbene from ortho position, the mechanism proceeds via a non aromatic intermediate. Oxygen bearing a negative charge is a good nucleophile too.

So I proceeded with oxygen acting as an nucleophile and came up with this mechanism:

proposed mechanism

The final product is not a major product (I wasn't even able to find it listed as a product after a cursory google search), so why does phenoxide attack the carbene from ortho position.

EDIT

The product I'm asking about has not even been mentioned in the linked question.

Improve this question
$\endgroup$
2
  • 1
    $\begingroup$ Related: Products of Reimer–Tiemann reaction of 4-methylphenol $\endgroup$
    – Mithoron
    Commented Jun 12, 2018 at 17:42
  • $\begingroup$ Due to continuous conjugation of negative charge in phenoxide ion, the lone pair of electrons are present more at ortho position rather than at para position + carbene is highly electron deficient compound $\endgroup$
    – Rahul Verma
    Commented Jun 15, 2018 at 16:05

2 Answers 2

Reset to default
13
+50
$\begingroup$

First of all, in the phenoxide ion, the lone pair on the oxygen atom doesn't remain fully available, it remains in continuous conjugation with the double bond of the benzene ring. In the resonance hybrid, there is also electron density on the Carbon atom in the ortho and para position,and also definitely on the oxygen atom itself. But, Carbon being less electronegative acts as a better nucleophile in this case rather than Oxygen. So, attack from the Carbon atom rather than Oxygen atom is more probable.
More over, attack from the ortho position, though creates a chance of Steric crowding, but the intermidiate generated from the attack by ortho-position creates an ordered transition state.The metal ion(say, $\ce{Na+}$) can easily interact with the electronegative $\ce{O}$ and $\ce{Cl}$ atoms, and can increase the stability of the transition state. Here is a picture of the intermidiate and a 3D view of it,
enter image description here

This ordered transition state will not appear in case of attack by the Oxygen atom. This is also another reason why that transition state is more favoured, and the reaction rate through that step is higher than any other possible steps.

Improve this answer
$\endgroup$
2
  • $\begingroup$ what you have shown here needs a citation. Regardless I don't see why chlorine should have any interaction with the $\ce{Na+}$ ion. It doesn't have any charge on it. ("Backbonding" is not possible there is evidence that it forms a cyclopropyl dichloro ring) $\endgroup$
    – napstablook
    Commented Jul 22, 2021 at 2:13
  • $\begingroup$ @Soumik But the wikipedia states that ortho product is major even in case of KOH. Can you please shed some light on it? $\endgroup$
    – An_Elephant
    Commented Aug 24, 2022 at 14:23
4
$\begingroup$

Another way of rationalising is through the use of the Hard-Soft Acid-Base principle. The dichlorocarbene generated is a soft electrophilic species while the oxyanion form of the phenoxide is a hard nucleophilic species. The nucleophilic $\pi$ bond is much softer and thus, it is much more favourable for it to be the attacking nucleophile (soft-soft interaction) rather than the negatively-charged oxygen atom (hard-soft interaction).

Improve this answer
$\endgroup$

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.