What are the products formed when phenol reacts with iodomethane in presence of dilute alkali? Also, a reaction with a mechanism would be helpful.

EDIT: Phenol reacts with NaOH to give a salt, sodium phenoxide. And yes, I was thinking about addition of iodine to the ring, but I don't know how.


A good approach to this type of problem is to break it down into smaller problems.

We know we have phenol $\ce{PhOH}$, methyl iodide $\ce{CH3I}$, and base, let's say $\ce{NaOH(aq)}$. Instead of trying to figure out what these three things do all together, can we determine what might happen with just two of them?

Phenol and methyl iodide

Since this is the main thrust of the question, we probably cannot decide what happens here. If we knew this one, then we would know the answer.

Phenol and NaoH

Phenol is acidic, more acidic than aliphatic alcohols. NaOH is a base and probably reacts with phenol. What would the product of this acid base reaction be?

$$\ce{PhOH + NaOH -> ??}$$

NaOH and methyl iodide

The reaction between these two is probably unproductive, since it would happen without the phenol present.

If you can figure out what happens when phenol reacts with NaOH, then can you figure out what one of the products of that reaction might do when it encounters $\ce{CH3I}$?

  • $\begingroup$ I edited my question. Phenol with NaOH give Sodium Phenoxide. I don't think that is relevant here, because it is just due to its acidic nature $\endgroup$ – Shubham Dec 20 '13 at 16:56
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    $\begingroup$ Yes, but sodium phenoxide will react with methyl iodide. $\endgroup$ – Ben Norris Dec 20 '13 at 20:00
  • $\begingroup$ @Shubham, between methyl iodide and sodium phenoxide, decide which is nucleophilic and which is electrophilic. Remember that nucleophiles are electron-rich (i.e., have lone electron pairs or reactive $\pi$-bonds, typically on atoms with $\delta^{-}$ or full negative charge) and electrophiles are electron-poor (i.e., typically have $\delta^{+}$ or full positive charge). If you can decide that, the course of the reaction should hopefully become clear. $\endgroup$ – Greg E. Dec 21 '13 at 2:16
  • $\begingroup$ @BenNorris, or some canonical form of phenoxide ion. Thanks! $\endgroup$ – Shubham Dec 21 '13 at 12:42

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