Number of compounds from the following which | vedclass

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Question

(A) Benedict's solution is used to detect reducing sugars. Reducing sugars contain a free aldehyde or ketone group that can reduce $Cu^{2+}$ ions to $

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(A) Benedict's solution is used to detect reducing sugars. Reducing sugars contain a free aldehyde or ketone group that can reduce $Cu^{2+}$ ions to $Cu^+$ ions,forming a red/orange precipitate of $Cu_2O$.Among the given compounds:$1$. Glucose: Reducing sugar (gives test).$2$. Maltose: Reducing sugar (gives test).$3$. Sucrose: Non-reducing sugar (does not give test).$4$. Ribose: Reducing sugar (gives test).$5$. $2$-deoxyribose: Reducing sugar (gives test).$6$. Amylose: Non-reducing polysaccharide (does not give test).$7$. Lactose: Reducing sugar (gives test).Therefore,only $2$ compounds (Sucrose and Amylose) will not produce an orange-red precipitate with Benedict's solution.

Item $I$ (compound)	Item $II$ (reagent)
$(a)$ Lysine	$(p)$ $1$-naphthol
$(b)$ Furfural	$(q)$ Ninhydrin
$(c)$ Benzyl alcohol	$(r)$ $KMnO_4$
$(d)$ Styrene	$(s)$ Ceric ammonium nitrate

Number of compounds from the following which will not produce orange red precipitate with Benedict's solution is $..............$. Glucose,maltose,sucrose,ribose,$2$-deoxyribose,amylose,lactose.

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The correct match between item $I$ and item $II$ is

Item $I$ (compound) Item $II$ (reagent)

$(a)$ Lysine $(p)$ $1$-naphthol

$(b)$ Furfural $(q)$ Ninhydrin

$(c)$ Benzyl alcohol $(r)$ $KMnO_4$

$(d)$ Styrene $(s)$ Ceric ammonium nitrate

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