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

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(A) The reactions of glucose are as follows:$A$. Glucose reacts with $HI$ to form $n$-hexane,which indicates the presence of a straight chain of six c

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) The reactions of glucose are as follows:$A$. Glucose reacts with $HI$ to form $n$-hexane,which indicates the presence of a straight chain of six carbon atoms. Thus,$A-IV$.$B$. Glucose reacts with $Br_{2}$ water (a mild oxidizing agent) to form gluconic acid,which indicates the presence of an aldehyde group. Thus,$B-I$.$C$. Glucose reacts with acetic anhydride to form glucose pentacetate,which indicates the presence of five $-OH$ groups. Thus,$C-II$.$D$. Glucose reacts with $HNO_{3}$ (a strong oxidizing agent) to form saccharic acid (glucaric acid),which indicates the presence of both a primary alcohol and an aldehyde group. Thus,$D-III$.Therefore,the correct matching is $A-IV, B-I, C-II, D-III$.

List-$I$	List-$II$
$A$. Glucose $+ HI$	$I$. Gluconic acid
$B$. Glucose $+ Br_{2}$ water	$II$. Glucose pentacetate
$C$. Glucose $+$ acetic anhydride	$III$. Saccharic acid
$D$. Glucose $+ HNO_{3}$	$IV$. Hexane

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

List-$I$ List-$II$

$A$. Glucose $+ HI$ $I$. Gluconic acid

$B$. Glucose $+ Br_{2}$ water $II$. Glucose pentacetate

$C$. Glucose $+$ acetic anhydride $III$. Saccharic acid

$D$. Glucose $+ HNO_{3}$ $IV$. Hexane

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 $+ HI$ $I$. Gluconic acid $B$. Glucose $+ Br_{2}$ water $II$. Glucose pentacetate $C$. Glucose $+$ acetic anhydride $III$. Saccharic acid $D$. Glucose $+ HNO_{3}$ $IV$. Hexane Choose the correct answer from the options given below.