random Ochem Q...

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I think A is correct. The base removes an alpha hydrogen and forms an enolate. The alpha carbon then attacks the carbonyl of the second one and forms an alkoxide. Hope that helps cant draw it here.
(I think the problem should have showed acid workup in a second step to protonate the alkoxide)
 
awwwwwwwww yes....once again i shot myself in the foot with not paying attention to the detials.

Thank you both.
 
C is wrond becase it is an aldol reaction. The alpha carbon of one molecule attacks the carbonyl and forms a beta hydroxy ketone. For choice C to be correct you would have to switch the place of the double bonded Oxygen with the place of the alcohol. When the alpha carbon attacks it keeps its carbonyl. The carbionyl that is being attacked loses the carbonyl and forms an alcohol. The correct answer is A. Furthermore choice C has the carbon that has 5 bonds which is impossible.
 
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