5
$\begingroup$

enter image description here

When the above compound is treated with $\ce{Cl2}$ in the presence of sunlight ($h\nu$), what is the number of monochloro structural isomers formed?

I expected only 1 product becuase I don't believe that monochlorination can take place at the bridgehead or the bridge positions. And rest of the 4 positions, by symmetry, are equivalent. This is because chlorination takes place through free radical mechanism in which a $\ce{sp^2}$ hybridised carbon free radical is formed.

But the answer given in the test's answer key is $3$. I think they have considered the products of bridgehead and bridged positions too but I don't understand how that is possible.

Improve this question
$\endgroup$
2
  • $\begingroup$ You're getting 3 products as one derivative is formed on chlorination of the 5 membered ring and two enentiomers for the 6 membered ring. I'll draw them outand post them in jee chem by tomorrow $\endgroup$
    – Avyansh Katiyar
    Nov 13, 2018 at 16:36
  • $\begingroup$ The question in the test(this is an AITS problem) asked for number of monochloro 'structural' isomers. $\endgroup$
    – Akshat Joshi
    Nov 17, 2018 at 5:59

1 Answer 1

Reset to default
1
$\begingroup$

The "bridged position" is just another ring, when viewed differently.

Suppose you look at the molecule a little differently (rotate it around a bit), then you could have a five-membered main ring, with a two membered "bridge", as opposed to what we have now (6 membered main ring with 1 membered bridge).

It's definitely possible to have a free radical halogen attack on that position, and with two stereoisomers at the positions which you have identified, it becomes total 3 monochlorinated derivatives. enter image description here

(in case you're wondering how to check if the two stereoisomers are not identical, you could try to superimpose them, or go from one form to the other without breaking any bonds. In that case, you'll find that they're indeed different. You could even check for a plane/center of symmetry)

--EDIT--

My answer is wrong, as @Abcd and @AkshatJoshi pointed out, because I counted stereoisomers along with the required structural isomers.

I read about attacks at Bridgehead positions and there's NO place that mentions the existence of Bridgehead free radicals. Thus, there will only be 2 structural isomers, but 3 isomers overall.

So, I believe the answer ought to be 2.

Improve this answer
$\endgroup$
9
  • $\begingroup$ I don't understand your 3rd product's mechanism. $\endgroup$
    – Archer
    Nov 14, 2018 at 3:15
  • $\begingroup$ See it's like a 60 degree angle at the bridge head position. How on earth can it become 120 degrees? $\endgroup$
    – Archer
    Nov 14, 2018 at 3:44
  • $\begingroup$ @Abcd that's not a bridge head $\endgroup$
    – Avyansh Katiyar
    Nov 14, 2018 at 4:07
  • $\begingroup$ @AvnishKabaj Sorry I meant Bridged not bridge head $\endgroup$
    – Archer
    Nov 14, 2018 at 4:08
  • 2
    $\begingroup$ But monochlorination does take place on 5 membered rings and I don't think it's 72° it prolly exists in a conformer like cyclohexane $\endgroup$
    – Avyansh Katiyar
    Nov 14, 2018 at 4:24

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.