Does Tautomerization Regenerate Phenylethanone in NaH Reaction?

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Please post this type of questions in the homework section using the template.
The instructions only specified that phenylethanone would react with NaH followed by an acid quench. I know that NaH provides a hydride (H-), and this will "take" one of the hydrogens on the exposed methyl group.

I followed this by pushing electrons so that the negative charge was on the oxygen, rather than carbon, and the acid quench (H+) would form an enol. However, doesn't tautomerization to a ketone product just regenerate phenylethanone again, or am I missing something? Thank you for your time. No amounts for any reagent were specified.
 
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What happens if, instead of thinking of H- as a base, you think of it as a nucleophile?
 
Natalie456 said:
The instructions only specified that phenylethanone would react with NaH followed by an acid quench. I know that NaH provides a hydride (H-), and this will "take" one of the hydrogens on the exposed methyl group.

I followed this by pushing electrons so that the negative charge was on the oxygen, rather than carbon, and the acid quench (H+) would form an enol. However, doesn't tautomerization to a ketone product just regenerate phenylethanone again, or am I missing something? Thank you for your time.No amounts for any reagent were specified.

Does it? I’m rusty but I thought the H- went to the carbonyl C*, after which you can’t make that tautomerism work. Is NaH qualitatively different from other reducing agents?
Can you write out a scheme?

*Often pictured as δ+ due to electron withdrawal onto the more electronegative O pictured =O δ-.
 
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