- ABenedict's test
- ✓Iodoform test
- CTollen's reagent test
- DFehling's solution test
Febling-solution and Benedict solution. These reagents cannot be used to distinguish them
$C H_{3} C H O$ reacts with $N a O H$ and $I_{2}$ to form yellow crystalline iodoform while $C_{6} H_{5} C H_{2} C H O$
does not react with it. $\mathrm{CH}_{3} \mathrm{CHO}+3 \mathrm{I}_{2}+4 \mathrm{NaOH} \longrightarrow $$\mathrm{CHI}_{3}$ (Yellow crystalline ppt) $ +3 \mathrm{NaI}+3 \mathrm{H}_{2} \mathrm{O}$
$C_{6} H_{5} C H_{2} C H O+3 I_{2}+4 N a O H \rightarrow N$ oreaction
$\therefore$ lodoform test can be used to distinguish between $C H_{3} C H O$ and $C_{6} H_{5} C H_{2} C H O$
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Compound $(D)$ is sodium hexametaphosphate which is known as
$CoCl _{3} . xNH _{3}+ AgNO _{3}( aq ) \rightarrow$
If two equivalents of $AgCl$ precipitate out, then the value of $x$ will be $....$