D-(+)-glucose reacts with hydroxylamine and y | vedclass

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Question

(D) $D-(+)$-glucose contains an aldehyde group $(-CHO)$ at $C-1$. When it reacts with hydroxylamine $(NH_2OH)$,the carbonyl oxygen is replaced by $=NO

Vedclass · Accepted Answer

$CH=NOH-CH(OH)-CH(OH)-CH(OH)-CH(OH)-CH_2OH$

Vedclass · Answer

(D) $D-(+)$-glucose contains an aldehyde group $(-CHO)$ at $C-1$. When it reacts with hydroxylamine $(NH_2OH)$,the carbonyl oxygen is replaced by $=NOH$ to form an oxime $(-CH=NOH)$. The configuration of the remaining chiral centers ($C-2$ to $C-5$) remains unchanged from the original $D$-glucose structure. In the Fischer projection of $D$-glucose,the $OH$ groups are on the right at $C-2, C-4,$ and $C-5$,and on the left at $C-3$. Thus,the structure of the oxime is $CH=NOH-CH(OH)-CH(OH)-CH(OH)-CH(OH)-CH_2OH$.

$D-(+)$-glucose reacts with hydroxylamine and yields an oxime. The structure of the oxime would be:

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