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NaBH4 and LiAlH4 will be a source for the hydride ion and reaction will begin with addition of hydride by displacing the carbonyl group.

But I am not sure will it attack both the carbonyl or only one. Moreover, Will it displace the chlorine at the bottom carbon?

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    $\begingroup$ There are three carbonyls here, not two. Both of those reagents will actually react with all three of the carbonyls in this particular case. $\endgroup$ – SendersReagent Oct 12 '18 at 13:53
  • $\begingroup$ In the book, when the compound reacted with NaBH4, it reduce only the first carbonyl and the bottom one to -OH. Whereas when it reacted with LiAlH4, it reduce all the carbonyls to -OH $\endgroup$ – kiv Oct 12 '18 at 13:58
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    $\begingroup$ This is a β-ketoester, though, and that changes things. I don't have time to draw out the reaction right now, but I wanted to make sure that if you get an answer, the person answering takes this into account. $\endgroup$ – SendersReagent Oct 12 '18 at 14:13
  • $\begingroup$ What solvent is this in? $\endgroup$ – Waylander Oct 12 '18 at 14:23
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    $\begingroup$ LiAlH4/ether or THF will reduce the three functional groups forming a tetraol. NaBH4 in hydroxylic solvent will behave the same way. Normally, esters reduce slowly with sodium borohydride such that the acyl chloride and ketone can be reduced selectively. I have found beta-ketoesters to reduce not only the ketone rapidly, but also the ester group. Presumably, by initial reduction of the ketone and then the ester via a neighboring group participation. $\endgroup$ – user55119 Oct 12 '18 at 14:52

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