MCQ
$CH_3CHO$ and $C_6H_5CH_2CHO$ can be distinguished chemically by
- 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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($\Delta {H_{eg}} = e^-$ gain enthalpy)

| Column $I$ | Column $II$ |
| $(A)$ $\mathrm{Bi}^{3+} \longrightarrow(\mathrm{BiO})^{+}$ | $(P)$ Heat |
| $(B)$ $\left[\mathrm{AlO}_2\right]^{-} \longrightarrow \mathrm{Al}(\mathrm{OH})_3$ | $(Q)$ Hydrolysis |
| $(C)$ $\mathrm{SiO}_4^{4-} \longrightarrow \mathrm{Si}_2 \mathrm{O}_7^{6-}$ | $(R)$ Acidification |
| $(D)$ $\left(\mathrm{B}_4 \mathrm{O}_7^{2-}\right) \longrightarrow\left[\mathrm{B}(\mathrm{OH})_3\right]$ | $(S)$ Dilution by water |