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(C) $NaBH_4$ reduces the aldehyde group of glucose to a primary alcohol group $(-CH_2OH)$,resulting in a single product,$D$-sorbitol (also known as gl

Vedclass · Accepted Answer

$1$ and $2$

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(C) $NaBH_4$ reduces the aldehyde group of glucose to a primary alcohol group $(-CH_2OH)$,resulting in a single product,$D$-sorbitol (also known as glucitol).$NaBH_4$ reduces the keto group of fructose to a secondary alcohol group $(-CHOH-)$. This reduction creates a new chiral center at the $C-2$ position,resulting in two epimeric products: $D$-sorbitol and $D$-mannitol.Therefore,the number of products for glucose and fructose are $1$ and $2$ respectively.

How many products are produced respectively when optically pure glucose and fructose are reduced one by one by $NaBH_4$?

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