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Consider the molecular formula $\ce{CH2=CH-CH=CH-OH}$ $(\ce{C4H5OH})$. Tell the types of bond, hybridizations, bond angles and geometry of this compound.

I didn't have any trouble with the first 3 topics. Carbon hybridizes in $\mathrm{sp^2}$ and oxygen hybridizes in $\mathrm{sp^3}$. Each carbon makes 3 $\sigma$ bonds and 1 $\pi$ bond and the oxygen makes 2 $\sigma$ bonds. The angles of the bonds $\ce{H-C-H}$, $\ce{H-C-C}$, $\ce{C-C-C}$ are 120 degrees and the angle of bond $\ce{H-O-C}$ is 109 degrees.

My problem is with the geometry. How can I deduce it? It doesn't seem to be one of the standard geometries.

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    $\begingroup$ Well, there are not awfully many possible options. Just draw the molecule like that, planar (because of conjugation) and with the angles like you said. (In fact, they may be slightly different; also, this whole thing does not quite exist, but who cares.) $\endgroup$
    – Ivan Neretin
    Jun 20, 2016 at 15:49

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Let’s start off with the things you thought you got right: Oxygen will not be $\mathrm{sp^3}$ hybridised but somewhat closer to $\mathrm{sp^2}$. This is because that is how the oxygen lone pair in a p-type orbital can take part in π conjugation. I’m not sure what $\angle(\ce{C-O-H})$ I would expect but it’ll be somewhere between $109^\circ$ and $120^\circ$, that’s for sure.

To the entire molecule’s structure: Yes, at first glance it seems like a potato. However, there is still something that you can say: for proper conjugation of the double bonds (usually favourable), you need a planar structure. Thus, the entire molecule will be planar with all atoms in the same plane. That is about all that you can safely derive, since there can still be rotation around the $\ce{C-C}$ single bond.

Also, as Ivan pointed out, the molecule is likely to prefer the keto-tautomer rather than the enol one. The conjugation argument may apply, but it is rather weak.

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The molecule is a 1,3-butadiene derivative. The butadiene has a planar structure as conjugation from the pi bonds spreads out along the carbon backbone so the carbon carbon angles would be close to 120 degrees, $\mathrm{sp^2}$ hybridization.
butadiene


In your structure the $\ce{-OH}$ group replaces just one of the $\ce{H}$ atoms at the end of the molecule. Thus there are two isomeric structures depending upon which $\ce{H}$ atom is replaced. I have very crudely placed these onto the structure close to the $\ce{H}$ atoms but the positions should be clear. (Sorry, I don't have software at hand just now to make proper pictures). The rotation about the 2-3 (middle) bond will be very, very slow because of the pi conjugation holding the structure planar so the two isomers $(E)$ & $(Z)$ will be distinct. The $\ce{C-C-O}$ bond angle will be close to 120 degrees. Try and look up the crystal structure then you will be able to get accurate angles and bond lengths.

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