Reduction problem organic chemistry?
a: NaBH4 or LiAlH4, followed by an acidic workup: 2-hexanol b: H2 (g) and Pd/C, which shouldn't touch the ester group: ethyl 4-aminobenzoate c: NaBH4 or LiAlH4, followed by an acidic workup: hexanol
Why cant sodium borohydride (NaBH4) reduce carboxylic acids or esters?
For the best answers, search on this site https://shorturl.im/ayfGD The starting material that you've described is called and acid anhydride - it's the condensation product formed by dehydration of two carboxylic acids. The one you've described is cyclic so the two acids are attached to the same molecule. Anyway, sodium borohydride is a good reducing agent for aldehydes, ketones, and acid chlorides but it does not reduce esters or acid anhydrides. Therefore, no reaction will occur. If you used an excess of lithium aluminium hydride (LiAlH4) instead of NaBH4, then the diol would be the product formed. For instance, if maleic anhydride was allowed to react with excess LiAlH4, 1,4-butanediol would be the product. C-C(=O) I.....\ I......O (maleic anhydride) + LiAlH4 --> HO(CH2)4OH I....../ C-C(=O) Another possible product would be the cyclic ester (also known as a lactone) that would result from the hydride reduction of one of the two carbonyl carbons followed by a ring closure of the alkoxide oxygen onto the second carbonyl carbon. Sorry if I've confused you but it's a little complicated! Hope this helps!!!
What produces 4-ethyl-3-hexanol in the presence of a nickel catalust and hydrogen in an oxidation reduction?
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