Planarity of given molecules in all conformations 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?
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!