Question
How the conversion can be carried out :
Isopropyl alcohol to iodoform
Isopropyl alcohol to iodoform
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$\text{C}_6\text{H}_5\text{NH}_2+\text{CHCl}_3+\text{alc.KOH}\rightarrow$

$\text{CH}_3-\text{C}=\text{C}-\text{CH}_2\text{OH}\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ | \ \ \ \ \ \ \ |\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{CH}_3 \ \text{Br}$