Mechanism of 2,6-dibromo-3-hydroxybenzaldehyde formation using chloroform

Hi,

Sorry, I probably asked question in incorrect way.

When 3-hydoxybenzaldehyde is brominated on mildly polar solvents like chloroform, phenoxy ion is not formed therefore arene ring is less activate (also aldehyde group is EWG). Solvent itself is not helpful  with stabilizing a (+)Br-Br(-) resonance, which again does not help bromination. There is a third factor: substitution of hydrogen with bromine makes arene ring even less active towards another substitution. In conclusion: bromination of 3-hydroxybenzaldehyde leads to di-bromo derivatives


Taking under consideration above facts, after reaction I should 2,4-dibromo-5-hydroxybenzaldehyde (less sterically hindered) or mixture of 2,4-dibromo-5-hydroxybenzaldehyde + 2,6-dibromo-3-hydroxybenzaldehyde.
But I have only 2,6-dibromo-3-hydroxybenzaldehyde (attached NMR of 2,6-dibromo-3-methoxybenzaldehyde- cant find hydroxy on my PC)


Does anyone know why do I have only one derivative which is sterically more hindered?!


thanks

B

It is a Forum Rule (see red link) that you must show your attempt before we can help you.

I think you may find it difficult to predict ortho-para ratios if you look at a lot of reactions. None the less, I did a quick search and I found the 2,4-dihalo-3-hydroxybenzaldehydes are commercially available. It would appear the bromination has a strong preference for ortho attack.