this (https://www.thevespiary.org/rhodium/Rhodium/chemistry/phenyl-2-nitropropanol.html)
(https://www.thevespiary.org/rhodium/Rhodium/chemistry/phenyl-2-nitropropanol.html) route to the compound. The heavily substituted 2,4,5-benzaldehyde turned out to be very insoluble in methanol, and thus required higher temperatures. The Merck index, states that 4-fluorobenzaldehyde is soluble in the lighter alcohols and ethers. Thus a candidate for this route, could be a reaction run in cold methanol. Does this sound reasonable?One final remark, I guess the p-fluorophenyl-2-nitropropan-1-ol will dehydrate more easily to the 2-nitropropene than the non-substituted compound.
Why do you think that it will do that more readily? Flourine is very electronegative and will thus withdraw the electrons from the aromatic system inductively. The dehydration works AFAIK by creating an intermediate carbocation at the 1-ol oxygen, and later at the corresponding carbon atom. This would be destabilized from the eletronegative group, would it not? Thus dehydration should occur at a lower frequency? Maybe im missing something, i don't know :)
Regards
Bandil
Post 459064 (missing)
(methyl_ethyl: "The Pharmacology and Clinical Pharmacology of MDMA", General Discourse))