Why does the chlorination of indene occur with syn selectivity in heptane?
Key

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.

Background

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.]



Analysis

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.