In the chlorination of indene to form 1,2-dichloro-2,3-dihydro-1​H-indene in heptane as solvent, the syn dichloride is formed as the major product, as reported in J. Org. Chem. 1980, 45 (25), 5150–5155):

Reaction scheme

The explanation I was given for this was that "in a non-polar solvent, the ion pair collapses quickly". What does this mean, and why is the stereoselectivity reversed, as compared to typical halogenations which proceed via a halonium ion intermediate to give anti dihalides?


1 Answer 1



The reaction is run in heptane, a non-polar solvent. A non-polar solvent will not stabilize or support ions as well as a polar solvent.


When the $\ce{Cl2}$ begins to interact with the indene an initial ion pair is formed. In the case of chlorine, it is not always clear whether

  • a cyclic halonium ion (the 3-membered structure on the bottom left)
  • an open ion (the structure on the bottom right)
  • or an equilibrium between the two ions occurs

[Side Note: In electrophilic bromination reactions involving alkenes, bromonium ions are very common intermediates usually leading to trans addition. However, chlorine isn't as polarizable as bromine and in this particular case with indene, the open ion is extremely stable due to its benzylic nature. Hence the involvement of only a chloronium ion is questionable in this case.]

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As mentioned above, ions are not very well stabilized in the non-polar solvent. As a result

  • the two ions (one positive, one negative) will tend to stay close together (tight ion pair) and stabilize each other
  • and further due to this lack of stabilization they will have a short lifetime leading to quick collapse to products.

Both of these factors (chlorines located close to one another, short lifetime doesn't allow much movement of the second chlorine) favor syn collapse - addition of the second chlorine from the same side that the first chlorine is on.

If we were to repeat the reaction in a polar solvent that would allow the ions to separate and have a longer lifetime, more trans product from anti addition would be expected.


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