Wouldn't C1 and C2 in prop-2-enylbenzene be on the same plane as that of the phenyl ring, right? If the double bond connected to C2 was between it and C1, they would be, but is it correct that now they're both planar, but on different planes?


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    $\begingroup$ That's right, there is no conjugation, hence no reason for the whole thing to be planar. $\endgroup$ Feb 18 at 13:35
  • $\begingroup$ @Ivan Neretin; wait, ring planarity is due to conjugation? I thought planarity was a requirement for the latter. $\endgroup$
    – harry
    Feb 18 at 13:42
  • 2
    $\begingroup$ It is a mutual thing. Planarity is a requirement for conjugation, and conjugation enforces planarity. $\endgroup$ Feb 18 at 13:57
  • $\begingroup$ @HarryHolmes with reference to your last comment - to which Ivan Neretin already replied - I suggest that you read my answer to chemistry.stackexchange.com/questions/146178/… as it is strongly related as well as it clarifies some uses of terms. $\endgroup$
    – Alchimista
    Feb 18 at 16:09
  • $\begingroup$ Yes sp2 carbon is planar and sp3 isn't It depends if that is a transition state, a byproduct or the main product (probably not). To find the molecule with the lowest energy you have to do a theoretical calculation or measure the XRD in certain properties. $\endgroup$
    – deuti
    Feb 18 at 16:18

Probably not

Part of the problem with the question is geometry. Any two points always form a plane. What I suspect you mean is "is the allyl double bond in the same plane as the benzene ring?" in which case the answer is probably not.

To make the situation clearer here is a different numbering of the atoms in the allyl benzene molecule:

allyl benzene numbering

The question I think you intended is whether the 2* and 3* bond is in the same plane as the benzene ring.

There is no chemical or geometric reason why it would be. the 1-1* bond and the 1-2 bonds will freely rotate so there is no bond-related force driving the orientation of the 2-3 double bond. to illustrate here is a 3D view of one possible configuration:

allyl benzene 3D

I say one possible configuration as it is likely that the barrier to rotation of those two bonds is small and, at room temperature in the liquid, the molecule will explore all the possible configurations (it might be different in a solid crystal where packing forces between units might lead to one being favoured).

In short: there is no reason to suppose the double bond and the benzene ring will be in the same plane because the two single bonds connecting it to the ring will freely rotate.

  • $\begingroup$ I get this; it's just that I've heard of some sort of rule that states that a carbon attached to an $sp^2$ hybridised one is in the same plane as it. In that case(using your numbering), 1 and 1*, and 1* and 2*, would be coplanar. Could you clarify why this rule doesn't work? $\endgroup$
    – harry
    Feb 24 at 11:40
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    $\begingroup$ @HarryHolmes When sp2 carbons are attached to other sp2 carbons, then there are electronic reasons that might drive them to be coplanar (if your example where vinyl benzene then this might matter). The reason is that pi-bond systems will interact. But the allyl group isn't directly connected to the benzene so its pi-bonds can't interact with the bonds in the ring: there is a tetrahedral sp3 carbon in between. $\endgroup$
    – matt_black
    Feb 24 at 11:49
  • $\begingroup$ I think you mean conjugation. Yeah, that makes carbons coplanar, but this is something separate. $\endgroup$
    – harry
    Feb 24 at 12:12
  • $\begingroup$ @HarryHolmes To be even clearer, the 1* carbon is in the same plane as the ring (it has to be because the 1 carbon is sp2). But this does not constrain rotation about the 1-1* bond or the rotation about the 1*-2* bond. Perhaps that explains the confusion? $\endgroup$
    – matt_black
    Feb 24 at 14:04
  • $\begingroup$ If its being sp2, as you said just now, is why 1-carbon makes 1* carbon stay on the same plane as it, wouldn't the same apy for the sp2 carbon, 2*? Wouldn't that effectively make them all planar? $\endgroup$
    – harry
    Feb 24 at 14:11

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