Predicting the product in a series of reactions
You seem very sure about your intermediate B; we’ll come back to that in a second.

For now, let’s concentrate on the reagents for your final step.

$$\ce{B -&gt;[MnO2] C}$$

Manganese dioxide is the product you get from heating any manganese salts in the bunsen burner until red-hot; thus, one might call it the most stable or thermodynamic oxide. It seems very unreactive and in fact I only know of one type of reaction that it is able to perform on organic molecules: allylic and benzylic alcohols are oxidised to the corresponding carbonyl compounds while fully saturated (i.e. not allylic) alcohols are untouched.

Since you have that reagent, it might be helpful to go backwards; assuming that the final step is an oxidation of a benzylic or allylic alcohol to a corresponding carbonyl. Thus, $\ce{C}$ would either be a ketone or an aldehyde and $\ce{B}$ would have to be an alcohol. But your proposed structure $\ce{B}$ does not include an alcohol! Therefore, your proposed reaction of the second step is very likely wrong.

$$\ce{A -&gt;[Ph-CHO][H2O] B}$$

Indeed, here you assumed that the anion $\ce{A}$ you proposed reacts with the phenyl ring of benzaldehyde in an electrophilic aromatic substitution. Take a good look at the two emboldened words and consider whether an anion can really be electrophilic. Then ask yourself what anions are in philicity terms. Once you have done that, attempt to locate the position of benzaldehyde with the corresponding electronic properties.

The solution is given in the spoiler below; please try to find it yourself, though; it is a much better exercise than just looking things up.


   Anions are negatively charged as are electrons. Thus, anions are practically never drawn towards negative charges — remember that like and like repulse each other. Instead, anions are drawn to positive charges: they are nucleophilic.

 Nucleophiles are drawn to the most positively charged bit of a molecule they are attacking. In the case of benzaldehyde, the carbon side of the $\ce{C=O}$ bond is most positively charged due to the resonance depiction we can draw below:

 $$\ce{C=O &lt;-&gt; \overset{+}{C}-\overset{-}{O}}$$

 Therefore, the acetylide anion will attack the carbonyl carbon of benzaldehyde nucleophilicly, creating a benzylic alcohol. It is now no secret that this is cleanly oxidised to the corresponding alkynophenone.


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   $$\ce{Bu-C#C^- (A) -&gt;[Ph-CHO][H2O] Ph-CHOH-C#C-Bu (B) -&gt;[MnO2] Ph-CO-C#C-Bu (C)}$$