Iodination Reaction

I am supposed to perform the above shown reaction in the lab soon, but I don't really understand the mechanism behind it. I got told by my lab assistant that the mechanism is an electrophilic aromatic substitution in which resorcinol (benzin-1,3-diol) acts as the electrophile and iodine as nucleophile. I imagine the resorcinol also electron-poor because of the -I effect caused by the electronegative Hydroxide substitutes. But isn't the +M Effect stronger? Then why should resorcinol act electrophilic?

Why is Iodine nucleophilic when there is no Lewis-acid present?

And last but not least: why is the Product 2-iodo-1,3-diol and not 4-iodobenzen-1,3-diol?

  • 1
    $\begingroup$ Resorcinol is very electron-rich. Think of it as 1,3-dione. Iodine is a good electrohile as the I - I bond is weak. $\endgroup$
    – Waylander
    May 3, 2021 at 12:54
  • $\begingroup$ So you are suggesting that it is a nucleophilic aromatic substitution? But how does a weak bond alone make the iodine an electrophile? $\endgroup$
    – Anni.Lin
    May 3, 2021 at 14:14
  • 1
    $\begingroup$ The Iodine - Iodine bond polarises very readily giving the functional equivalent of I+ easily. The carbon between the two oxygens is an electron rich site - it is an enol both ways. $\endgroup$
    – Waylander
    May 3, 2021 at 14:31
  • 1
    $\begingroup$ The reaction is likely an electrophilic aromatic substitiution i.e. the ring acts as the nucleophile and iodine as the electrophile. $\endgroup$
    – S R Maiti
    May 3, 2021 at 16:22

2 Answers 2


Reading this question, I realized that OP is very new to organic chemistry, and in need for learning a lot about electrophilic aromatic substitution reactions. Thus, I recommend that OP should concentrate on the electrophilic aromatic substitution reactions and read the chapters of OP's textbook dedicated to that subject. Said that, I'm going to answer plethora of subjective questions, which should be answered and cleared to show the way to right direction.

I got told by my lab assistant that the mechanism is an electrophilic aromatic substitution in which the Resorcinol (Benzin-1,3-diol) acts as the electrophile and Iodine as nucleophile.

First, resorcinol is not called benzin-1,3-diol (may be in an old book). It's correct name is 1,3-dihydroxybenzene. It is true that the reaction is an electrophilic aromatic substitution. However, the electrophile here is polarizable $\ce{I2}$ molecule because it exsists in $\ce{I^{\delta +}-I^{\delta -}}$ form under reaction conditions (where $\ce{I^{\delta +}}$, which lacks electrons is the reactive center). Aromatic nucleus with two hydroxy group is electron rich (hydroxy groups activate the ring), hence acts as the nucleophile.

But isn't the +M Effect stronger?

Yes, mesomeric effect (old resonance effect) is stronger than inductive effect. Here you have two hydroxy groups to give strong $(+M)$-effect to activate the aromatic nucleus. Hence the reaction is o,p-directive. As a result, you can expect either 2-iodo-1,3-dihyroxybenzene or 4-iodo-1,3-dihyroxybenzene or 2,4-diiodo-1,3-dihyroxybenzene or etc. as your product or mixture of them based on the reaction condition used (Ref.1 & 2).

Why is the Product 2-Iodo-1,3-diol and not 4-Iodobenzen-1,3-diol?

As I explained before, the reaction conditions govern the product ratio (see the PDF of Ref.1 I attached below for better understanding). The iodination of resorcinol is not a straight forward reaction. Slight changes of reaction conditions gives you different product. For example, the major product given in the reaction using $\ce{I2/NaHCO3}/\pu{0 ^\circ C}$ condition is 2-iodo-1,3-dihydroxybenzene as indicated in the OP's question. In a large scale reaction, the yield obtained was only 66% (Ref.3). Using the same conditions, Ref.1 reported 77% isolated yield. Keep in mind that, some of 2-iodo-1,3-dihydroxybenzene isomerized to 4-iodo-1,3-dihydroxybenzene (major) and 4,6-diiodo-1,3-dihydroxybenzene in an acidic medium at high temperature (Ref.1). Thus, temperature plays an important role in percent yield in the reaction.

On the other hand, when the electrophile ($\ce{I+}$ source is changed to $\ce{ICl}$, the main product is 4-iodo-1,3-dihydroxybenzene (4-iodoresorcinol) with 4,6-diiodo-1,3-dihydroxybenzene (4,6-diiodoresorcinol) as minor product (Ref. 1 & 2). Ref.2 also stated that use of nascent iodine $(\ce{KIO3/KI/HCl})$ has resulted in the regiospecific 2,4-diiodonation of resorcinol at room temperature (which is a rapid reaction). The $\ce{^1H}$-NMR spectrum of this reaction mixture has revealed only two products, 2,4-diiodoresorcinol as major product (56% isolated yield) and 2,4,6-triiodoresorcinol as minor product in $89:11$ ratio.


  1. Ib Thomsen, and Kurt B. G. Torssell, "Iodination of Resorcinol, 5-Methoxyresorcinol, Phloroglucinol and Resorcyclic Acid," Acta Chemica Scandinavica 1991, 45, 539-542 (DOI: 10.3891/acta.chem.scand.45-0539)(PDF).
  2. Frederick L. Weitl, "Regiospecific 2,4-diiodination of resorcinol with nascent iodine," J. Org. Chem. 1976, 41(11), 2044–2045 (DOI: https://doi.org/10.1021/jo00873a040).
  3. Shin-ichiro Tsujiyama and Keisuke Suzuki (submitted); David B. Guthrie, Heather M. Gibney, and Dennis P. Curran (checked), "Preparation of Benzocyclobutenone Derivatives Based on an Efficient Ggeneration of Benzynes [Bicyclo[4.2.0]octa-1,3,5-trien-7-one, 5-(phenylmethoxy)-]," Org. Synth. 2007, 84, 272-284; Coll. Vol. 11 2009, p. 488-497 (DOI: 10.15227/orgsyn.084.0272).

An electrophilic aromatic substitution reaction is one in which the aromatic ring reacts (as a nucleophile) with an electrophilic reagent. In all of the reactions that follow this naming convention, the adjectives "nucleophilic" and "electrophilic" describe the reagents, not the substrate. For example, the following reaction is a nucleophilic substitution where benzyl chloride (the electrophilic substrate) reacts with a nucleophile.

Nucleophilic substitution reaction

So, in an electrophilic aromatic substitution reaction, the arene (a nucleophile) is reacting with an electrophile. In your case, resorcinol is the nucleophile and iodine is the electrophile.

The nucleophilicity of resorcinol is enhanced by the basic conditions (resorcinol's first $\mathrm{p}K_\mathrm{a}$ is $9.32$).

Iodine is a reasonable electrophile with a weak, highly polarizable single bond between the two iodine atoms.


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