A) inversion
B) esterification
C) saponification
D) hydration
Correct Answer: A
Solution :
(i) Esterification: The formation of ester by reaction of organic acid and alcohol in presence of mineral acid is called esterification. (ii) Saponification: The reaction of ester with alkali to give alcohol and salt of carboxylic acid is called saponification. \[\underset{ester}{\mathop{RCOOR}}\,+\underset{alkali}{\mathop{KOH}}\,\to \underset{salt\text{ }of\text{ }acid}{\mathop{RCOOK}}\,+\underset{alcohol}{\mathop{ROH}}\,\] (iii) Hydration: Addition of water molecule to organic reactant is called hydration. (iv) Inversion: \[\underset{\underset{optical\text{ }rotation+{{66.5}^{o}}}{\mathop{sucrose}}\,}{\mathop{{{C}_{12}}{{H}_{22}}{{O}_{11}}}}\,+{{H}_{2}}O\to \underset{\underset{+{{52.7}^{o}}}{\mathop{glucose}}\,}{\mathop{{{C}_{6}}{{H}_{12}}{{O}_{6}}}}\,+\underset{\underset{{{92.4}^{o}}}{\mathop{fructose}}\,}{\mathop{{{C}_{6}}{{H}_{12}}{{O}_{6}}}}\,\] The sucrose is dextrorotatory sugar which on hydrolysis gives glucose which is dextrorotatory and fructose which is laevo rotatory. The magnitude of laevo rotation is more than dextro rotation of sucrose. \[\because \] The sign of rotation has changed during the reaction. \[\therefore \] It is called inversion or inversion of sugar.
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