Compound A,C_5H_{10}O_5,gives a tetraacetate | vedclass

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(A) $(i)$ Formation of a tetraacetate with $Ac_2O$ indicates that compound $A$ contains four $-OH$ groups.$(ii)$ Oxidation of $A$ with $Br_2-H_2O$ yie

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(A) $(i)$ Formation of a tetraacetate with $Ac_2O$ indicates that compound $A$ contains four $-OH$ groups.$(ii)$ Oxidation of $A$ with $Br_2-H_2O$ yields an acid $C_5H_{10}O_6$,which confirms the presence of an aldehyde group $(-CHO)$ in $A$,as $Br_2-H_2O$ specifically oxidizes aldoses to aldonic acids.$(iii)$ Reduction of $A$ with $HI$ gives isopentane $(2-methylbutane)$,which indicates the carbon skeleton is branched (isopentane structure).$(iv)$ Combining these facts,the structure must be a branched pentose. The structure in option $A$ represents $2,3,4-trihydroxy-2-(hydroxymethyl)butanal$,which fits the molecular formula $C_5H_{10}O_5$ and the branched skeleton.

Compound $A$,$C_5H_{10}O_5$,gives a tetraacetate with $Ac_2O$ and oxidation of $A$ with $Br_2-H_2O$ gives an acid,$C_5H_{10}O_6$. Reduction of $A$ with $HI$ gives isopentane. The possible structure of $A$ is:

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