A) sodium
B) sodium hydroxide
C) Tollen's reagent
D) alcohol
Correct Answer: C
Solution :
Key Idea: During distinction test both reactants should produce two different smells or colour or only one should react: With sodium: Both of them react with sodium to form salt and hydrogen \[2HCOOH+2Na\xrightarrow{{}}2HCOONa+{{H}_{2}}\] \[2C{{H}_{3}}COOH+2Na\xrightarrow[{}]{{}}2C{{H}_{3}}COONa+{{H}_{2}}\] So, it cannot be used as distinguishing test. With\[\mathbf{NaOH}\] \[HCOO{{H}_{4}}+NaOH\xrightarrow[{}]{{}}HCOONa+{{H}_{2}}O\] \[C{{H}_{3}}COOH+NaOH\xrightarrow{{}}C{{H}_{3}}COONa\] \[+{{H}_{2}}O\] \[\because \]both of them give same type of compounds. \[\therefore \]It cannot be used as distinguishing test. With alcohol: All acids react with alcohol to produce sweet smelling ester. \[\therefore \]It cannot be used as distinguishing test With Tollen's reagent: \[HCOOH\xrightarrow[{}]{Tollen's\text{ }reagent}silver\text{ }mirror\] (\[\because \] HCOOH reduces Tollen's reagent) \[C{{H}_{3}}COOH\xrightarrow[{}]{\text{ }Tollen's\,reagent}no\text{ }reaction.\] So, Tollen's reagent is used to distinguishing them
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