Match List-I with List-II: List-I | vedclass

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(B) $1$. Glucose does not react with $NaHCO_3$ because it is not acidic enough to evolve $CO_2$ gas. Thus,$A-II$.$2$. Glucose on oxidation with concen

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$A-II, B-IV, C-III, D-I$

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(B) $1$. Glucose does not react with $NaHCO_3$ because it is not acidic enough to evolve $CO_2$ gas. Thus,$A-II$.$2$. Glucose on oxidation with concentrated $HNO_3$ gives saccharic acid (glucaric acid). Thus,$B-IV$.$3$. Glucose on prolonged heating with $HI$ gives $n$-hexane,indicating the presence of a straight chain of six carbon atoms. Thus,$C-III$.$4$. Glucose on oxidation with bromine water gives gluconic acid. Thus,$D-I$.Therefore,the correct matching is $A-II, B-IV, C-III, D-I$.

List-$I$	List-$II$
$A$. Glucose/$NaHCO_3$/$\Delta$	$I$. Gluconic acid
$B$. Glucose/$HNO_3$	$II$. No reaction
$C$. Glucose/$HI$/$\Delta$	$III$. $n$-hexane
$D$. Glucose/Bromine water	$IV$. Saccharic acid

Match List-$I$ with List-$II$:

List-$I$ List-$II$

$A$. Glucose/$NaHCO_3$/$\Delta$ $I$. Gluconic acid

$B$. Glucose/$HNO_3$ $II$. No reaction

$C$. Glucose/$HI$/$\Delta$ $III$. $n$-hexane

$D$. Glucose/Bromine water $IV$. Saccharic acid

Choose the correct answer from the options given below:

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Match List-$I$ with List-$II$: List-$I$ List-$II$ $A$. Glucose/$NaHCO_3$/$\Delta$ $I$. Gluconic acid $B$. Glucose/$HNO_3$ $II$. No reaction $C$. Glucose/$HI$/$\Delta$ $III$. $n$-hexane $D$. Glucose/Bromine water $IV$. Saccharic acid Choose the correct answer from the options given below: