Which of these have all its atoms in one plane in all the possible conformations (if any):

A) buta-1,3-diene

B) but-1-en-3-yne

C) $\ce{CH2=C=O}$

D) $\ce{CH2=C=CH2}$

While I was attempting this question, I got stuck in option B! I was sure about option A that since it has conformational change possible between C2-C3 bond, atoms will be present in more than one single plane.

Next, I am also convinced that option D can't be the answer as allenes are non planar and they have H-C-H planes perpendicular to one another. Then, for option C also, I know that since all the atoms are $sp^2$ hybridised, it has to be a planar straight molecule.

But, I am having a doubt in option B; how should we determine if the molecule is planar or not and whether all of its atoms are present in a single plane?


  • 1
    $\begingroup$ Do you have a modeling kit to build it in 3D? $\endgroup$ Apr 26 '19 at 18:00
  • 2
    $\begingroup$ No, I don't have. $\endgroup$
    – Saumya
    Apr 26 '19 at 18:33
  • 1
    $\begingroup$ All the atoms includes hydrogens, consider the angles that the H atoms have wrt the plane of the c atoms around double and triple bonds. $\endgroup$
    – porphyrin
    Apr 26 '19 at 19:14
  • $\begingroup$ Perhaps I am still sleeping but if a molecules hss all its atoms there is no reason to say in all possible conformations. Question should be which molecules has just one confirmation, or even better which one doesn't have conformers. Not pedantic, it is really conceptual. ... $\endgroup$
    – Alchimista
    Apr 27 '19 at 6:45
  • $\begingroup$ Yes, it's obvious that a molecule has all its atoms...but the question is not just about whether it has a conformer or not; rather, it also questions if the atoms lie in a single plane provided, that molecule has conformers. $\endgroup$
    – Saumya
    Apr 27 '19 at 7:38

To answer this you need to consider the hybridization and geometry around each atom but also think about all of the single bonds present. Single bonds will (generally) allow rotation along their axes as this can happen to a sigma bond without breaking the orbital overlap. If rotation about a single bond moves other atoms in space then you must consider conformational changes that may take atoms out of the plane.

You are correct with A and D. Option C is a ketene and the central C is sp hybridized, it is very similar to the allene in D. However, the terminal oxgyen has no other atoms bonded to it so it doesn't suffer from atoms out of the plane like D. So, C is completely planar (so far as atoms are concerned!), but not for the reason you stated.

For B, which you have shown, it is planar as drawn. Now consider what would happen if you rotated any of the single bonds. Would you break planarity? If not, it is also completely planar. Interestingly, if you rotate the C2-C3 bond, you rotate the alkyne group compared to the alkene but, as the alkyne group has infinite rotational symmetry along its axis, you can't tell the difference, all atoms remain on the plane!

enter image description here

  • 2
    $\begingroup$ @Karsten Theis. Nice JSmol! $\endgroup$
    – Withnail
    Apr 26 '19 at 20:34

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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