This is the question:


And this is my solution:


In the last step, I have tautomerised, and in the second last step, there is no double bond between carbon attached with Cl and OH (just some cutout work)

The answer given is B, but my answer comes out to be C.

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    $\begingroup$ You've already been told about improper formatting... $\endgroup$ – Mithoron Jun 20 '16 at 22:31
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    $\begingroup$ Arrows flow from negative to positive charges in general. In other words, nucleophiles attack electrophiles. The arrow going from the H+ is incorrect, all arrows involving H+ should point toward H+, as it can only be an electrophile. $\endgroup$ – Dissenter Jun 20 '16 at 22:41
  • $\begingroup$ @Mithoron, I really don't remember. :( $\endgroup$ – Reeshabh Ranjan Jun 21 '16 at 17:06
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    $\begingroup$ Doesn't it look better now? $\endgroup$ – Mithoron Jun 21 '16 at 17:15
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    $\begingroup$ I think your answer is correct, although I haven't seen this exact reaction reported in SciFinder. If this is from an instructor, ask how B can be produced. $\endgroup$ – jerepierre Jun 21 '16 at 18:15

The product of the addition of one chloride ion (C) is easily oxidized by atmospheric oxygen (radical mechanism). The obtained 2-chloro-1,4-benzoquinone is susceptible to conjugate addition of another chloride ion, this happens chemoselectively at the more electron deficient double bond which has already a chlorine attached to one carbon atom. enter image description here

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  • $\begingroup$ Which leads to the next question: Why is 2,3-dichlorohydroquinone not readily oxidised by atmospheric oxygen? $\endgroup$ – Jan Jun 26 '16 at 20:40
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    $\begingroup$ @Jan because it is already oxidized. Chlorine is a moderately electron-withdrawing group, and two chlorine atoms make ring electron-deficient enough to not being readily oxidized. Moving further this road, I point out that DDQ (dichloro-dicianot-quinone) is a rather strong oxidizer by organic chemistry standards. $\endgroup$ – permeakra Jun 26 '16 at 22:13
  • $\begingroup$ @permeakra Thanks, I suspected something like that but confirmation is always nice. $\endgroup$ – Jan Jun 26 '16 at 22:18
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    $\begingroup$ No, because the first step of the radical mechanism would be a single electron transfer from the aromatic ring to oxygen, producing a radical-cation. Cations are destabilized by electron-withdrawing groups. $\endgroup$ – EJC Jul 3 '16 at 19:25
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    $\begingroup$ @Marko Can you give a link to the radical mechanism you are talking about? $\endgroup$ – user14857 Sep 22 '16 at 15:11

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