Question

Solve for the theoretical yield of this prelab

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This experiment involves three redox reactions. In the first, iodide ion and hypochlorite react in the presence of three protons to form iodine, which is the source of \(\mathrm{I}^{+}\). This leaves some \(\mathrm{I}\) ' behind, which combines with unused \(\mathrm{I}_{2}\) to form \(\mathrm{I}_{3}^{-} . \mathrm{I}_{3}^{-}\) is reduced to water-soluble I by thiosulfate ion, which stops the reaction.

The line from the benzene ring connecting to the I means that you do not know its exact location on the ring.

This reaction is thought to follow the classic mechanism for electrophilic aromatic substitution. The electrophile, in this case, is \(I^{+},\) which comes from \(I_{2}\). The nucleophilic electron pairs of the benzene ring are attracted to \(\mathrm{I}^{+}\), so the aromatic ring loses aromaticity to form a resonance stabilized carbocation intermediate. The ring is re-aromatized through deprotonation by a base. You will predict the product of the substitution using salicylamide as the starting material instead of benzene.

Answer 1

They are the same mechanism: an electrophilic aromatic substitution, but the intermediate structures differ due to electronic delocalization. When you have substituted benzene, the substituents direct the position of the entering group; in your example, you have a strong activating group (-OH) and a moderately deactivating group (-CONH₂) so, the iodine atom is more prone to attack at ortho and para positions to the -OH group (that's the reason of the line from benzene to iodine atom). If the iodine atom enters at the opera position to the -OH group, the mechanism is: