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
✓
Answer
Correct option: B.
Iodoform test
b
$\mathrm{CH}_{3} \mathrm{CHO}$ and $\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}_{2} \mathrm{CHO}$ are aliphatic aldehydes and hence react with Tollen's reagent,
$\mathrm{CH}_{3} \mathrm{CHO}$ and $\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}_{2} \mathrm{CHO}$ are aliphatic aldehydes and hence react with Tollen's reagent,
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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{C{H_3} - C - C{H_2} - C{H_2} - C{H_2} - C{H_2} - C - H}
\end{array}$ $\xrightarrow{{H{O^ - }/\Delta }}$ $\mathop {(A)}\limits_{(73\% )} $ Product $(A)$ is
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{C{H_3} - C - C{H_2} - C{H_2} - C{H_2} - C{H_2} - C - H}
\end{array}$ $\xrightarrow{{H{O^ - }/\Delta }}$ $\mathop {(A)}\limits_{(73\% )} $ Product $(A)$ is
But $-2-$ ene on reaction with alkaline $KMnO_4$ at elevated temperature followed by acidification will give
→$(A)\xrightarrow[dil.{{H}_{2}}S{{O}_{4}}]{HgS{{O}_{4}}}(B)\xrightarrow{LiAl{{H}_{4}}}\underset{recemic\,mixture}{\mathop{(C)}}\,$
→$\therefore$ reactant $(A)$ is