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I faced a question

How many cyclic isomers are possible for $\ce{C4H6}$?

I and my friend found the following four. enter image description here

But the answer key says there are 5. So what's the other one?


marked as duplicate by Todd Minehardt, paracetamol, pentavalentcarbon, ron, airhuff May 25 '17 at 20:26

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  • $\begingroup$ Your formula and the figures you've drawn don't match, the formula contains $3$ carbons and each of your figures has $4$ carbons. $\endgroup$ – Berry Holmes May 25 '17 at 13:37
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    $\begingroup$ We prefer to not use MathJax in the title field due to issues it gives rise to; see here for details. $\endgroup$ – Berry Holmes May 25 '17 at 13:38
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    $\begingroup$ There's also bicyclobutane. $\endgroup$ – Mithoron May 25 '17 at 13:50
  • $\begingroup$ I find it easier to deal with such questions using the Degrees of Unsaturation... $\endgroup$ – Eashaan Godbole May 25 '17 at 13:55
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    $\begingroup$ @Mockingbird You'll find it easily enough on the internet. Check YouTube too. $\endgroup$ – Eashaan Godbole May 25 '17 at 16:20

You can answer such questions using D.U. (Degree of unsaturation)

The formula is $\mathrm{C + 1 - }\frac{H + X - N}{2}$

C = Carbon.
H = Hydrogen.
X = Halogen.
N = Nitrogen.

If you get D.U. to be one, then in the structure there could be:

  • 1 double bond.

  • 1 ring.

Let's take an example of $\ce{C4H8}$ which has D.U. equal to one.

  • It can have three structures of 1 double bond:

enter image description here

  • It can have two structures of 1 ring:

enter image description here

Now for D.U. equal to two, the possibilities are:

  • 2 double bond.

  • 2 ring.

  • 1 double bond and 1 ring.

  • 1 triple bond.

Let's take your example of $\ce{C4H6}$ which has D.U. equal to two:

  • It can have two structures of 2 double bond:

enter image description here

  • It can have one structure of 2 ring:

enter image description here

  • It can have four structures of 1 double bond + 1 ring:

enter image description here

  • It can have two structures of 1 triple bond:

enter image description here

So, the answer is 5 cyclic isomers as you can see above.

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    $\begingroup$ Drawing angled $sp$ carbons is a sin bordering on blasphemy. $\endgroup$ – Ivan Neretin May 25 '17 at 15:56
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    $\begingroup$ @Mockingbird That I have studied. But you can also make out like D.U. Is 2 means there should be 2 double bonds that are removing the hydrogen from the formula. And it can have 2 rings that all that. You will have two check what are the possibilities that can remove hydrogen from an alkane $\endgroup$ – user237650 May 25 '17 at 15:59
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    $\begingroup$ @DavePhD Thanks for the information. I have studied that cyclic alkenes does not show geometrical isomerism having carbon less than 8. $\endgroup$ – user237650 May 25 '17 at 16:02
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    $\begingroup$ @Mesentery well perhaps you will find this article interesting, concerning the limit supposedly being 8 pubs.acs.org/doi/abs/10.1021/ja055388i and also pubs.acs.org/doi/abs/10.1021/jo00389a067 $\endgroup$ – DavePhD May 25 '17 at 16:06
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    $\begingroup$ Now that's much better. But look, a carbon with two double bonds is also an $sp$ carbon. $\endgroup$ – Ivan Neretin May 25 '17 at 17:34

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