Unit 14 Biomolecules (Intext Questions-1)
Intext Questions
14.1 Glucose or sucrose are soluble in water but cyclohexane or benzene (simple six membered ring compounds) are insoluble in water. Explain.
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Answer
A glucose molecule contains five $-\mathrm{OH}$ groups while a sucrose molecule contains eight $-\mathrm{OH}$ groups. Thus, glucose and sucrose undergo extensive $\mathrm{H}$-bonding with water.
Hence, these are soluble in water.
But cyclohexane and benzene do not contain - $\mathrm{OH}$ groups. Hence, they cannot undergo $\mathrm{H}$-bonding with water and as a result, are insoluble in water.
14.2 What are the expected products of hydrolysis of lactose?
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Answer
Lactose is composed of $\beta-D$ galactose and $\beta-D$ glucose. Thus, on hydrolysis, it gives $\beta-D$ galactose and $\beta-D$ glucose.
$ \underset{\text { Lactose }}{\mathrm{C_{12}} \mathrm{H_22} \mathrm{O_{11}}}+\mathrm{H_2} \mathrm{O} \longrightarrow \underset{\text { D-(+)-Glucose }}{\mathrm{C_6} \mathrm{H_{12}} \mathrm{O_6}}+\underset{\text { D-(+)-Galactose }}{\mathrm{C_6} \mathrm{H_{12}} \mathrm{O_6}} $
14.3 How do you explain the absence of aldehyde group in the pentaacetate of D-glucose?
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Answer
D-glucose reacts with hydroxylamine $\left(\mathrm{NH_2} \mathrm{OH}\right)$ to form an oxime because of the presence of aldehydic (-CHO) group or carbonyl carbon. This happens as the cyclic structure of glucose forms an open chain structure in an aqueous medium, which then reacts with $\mathrm{NH_2} \mathrm{OH}$ to give an oxime.
But pentaacetate of $\mathrm{D}$-glucose does not react with $\mathrm{NH_2} \mathrm{OH}$. This is because pentaacetate does not form an open chain structure.
