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When drawing a skeletal formula, what is the difference between an angular version and a linear version?

I was asked to draw the Z isomer of Resveratrol:

(E)-Resveratrol

For which I drew:

My (Z)-Resveratrol

However the markscheme states that:

skeletal structure must be correct and angular not linear

I haven't come across the difference between the two before and can't find anything on google to suggest one. Their drawing of the correct answer is equivalent to mine, however I am concerned that I may have drawn the linear version as my benzene rings are in a line and theirs are not. What am I missing here?

Their version:

Their (Z)-Resveratrol

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enter image description here

Those are $sp^2$ carbons and should have a triangular planar 120° geometry. The angles you've drawn are clearly not 120° .

I've not exactly come across the terminology before, but I believe this is the distinctions:

  • In a "linear" formula, you disregard the actual bond angles and just try to keep everything in a line.
  • In an "angular" formula, you draw everything according to the bond angles.

Not sure of this, though.

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  • $\begingroup$ Are you getting the definitions from the question? If so, I would wonder that maybe @Rory's structure was simply counted wrong due to the bond angles, not because it's an example of the linear form of the structure. I've heard of "linear structural formulas" as simply the molecular formula written as the structure is bonded (i.e. writing CH3CH2OH instead of C2H6O). library.queensu.ca/research/guide/… uses this phrasing too. $\endgroup$
    – Emmie MC
    May 11, 2012 at 21:30
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    $\begingroup$ Rory - I would nominate that the person marking is potentially being a bit harsh, however the structure as you drew it does disguise the degree of steric collision that one would anticipate in (Z)-resveratrol and indeed in any (Z)-stilbenoid. There's going to be serious competition between the pressure to planar due to conjugation stabilisation and the fact that the rings would be virtually intersecting if the molecule were planar. $\endgroup$
    – Richard Terrett
    May 12, 2012 at 2:20
  • $\begingroup$ @EmmieMC possibly, however they asked for a skeletal formula. It's me marking a past paper so I was just keen to know if and where I'd gone wrong xD $\endgroup$
    – Rory
    May 12, 2012 at 12:51

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