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This question already has an answer here:

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?

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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
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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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