Try it on paper first. A hint, the key idea and the worked solution are here when you need them.
Arrange benzoyl chloride, ethyl benzoate, benzamide, benzaldehyde and acetophenone in order of decreasing reactivity towards a nucleophile at the carbonyl carbon. For each, name the group that raises or lowers the positive charge on the carbon. Then say which of the five sodium borohydride reduces, and which it leaves.
Reactivity towards nucleophiles: acyl chloride > aldehyde > ketone > ester > acid > amide. NaBH reduces aldehydes and ketones, LiAlH every carbonyl group, and cold DIBAL-H takes an ester or a nitrile to the aldehyde. A Grignard reagent uses 1 equivalent on an aldehyde, ketone or nitrile and 2 on an ester or acyl chloride. One LiAlH holds 4 hydrides, and a ketone takes 1 and an ester 2. An acetal protects a carbonyl group.
Have you tried it?
Work it out on paper first. It is fine to get stuck or make a mistake: that is how we learn. When you are ready, go through the solution one step at a time, and check each step against your own.
Step 1. The order. Benzoyl chloride, then benzaldehyde, then acetophenone, then ethyl benzoate, then benzamide.
Step 2. The reasons. The chlorine of the acyl chloride withdraws electrons, so the carbon is very positive. The aldehyde has a small hydrogen and one aryl group. The ketone has two groups that donate and crowd the carbon. The ester oxygen donates by resonance. The amide nitrogen donates most, so the amide is the least reactive.
Step 3. Sodium borohydride. It reduces the acyl chloride (which also reacts with the solvent), the aldehyde and the ketone. It leaves the ester and the amide.
Answer: benzoyl chloride, benzaldehyde, acetophenone, ethyl benzoate, benzamide.
Check. The order of the five follows the ladder: acyl chloride, aldehyde, ketone, ester, amide.
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