lodoform reaction is answered by all, except

For alcohols secondary and tertiary alcohols gives iodoform reaction and ethanol is the only primary alcohol which gives
iodoform reaction, therefore, propanol will not give iodoform reaction.
(1) \( \underset{\text { Acetaldehyde }}{\mathrm{C}_{3} \mathrm{CH} \mathrm{O}}+3 \mathrm{NaOI} \rightarrow \underset{\text { Yellow ppt }}{\mathrm{CH}}+ \) HCOON \( a+2 \mathrm{NaOH} \)
(2) \( \begin{array}{c}\text { Acetaldehyde } \\ \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{OH}\end{array}+\mathrm{NaOI}_{2}-\mathrm{NaOH}_{\mathrm{S}_{2}}> \) do not give iodoform reaction.
(3) \( \mathrm{CH}_{3}-\stackrel{\mathrm{OH}}{\mathrm{CH}}-\mathrm{CH}_{2} \mathrm{COOH} \) NaOI \( \underset{\text { Yellowppt. }}{\mathrm{CHI}_{3}}+\mathrm{NaOCCH}_{2} \mathrm{COOH} \)
\( \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}-\mathrm{NaOH}_{\mathrm{I}_{2}} \rightarrow \mathrm{CH}_{3} \mathrm{CHO}+\mathrm{NaI}+\mathrm{H}_{2} \mathrm{O} \)
\( \mathrm{CH}_{3} \mathrm{CHO}+3 \mathrm{NaOI} \rightarrow \mathrm{CHI}_{3}+\mathrm{HCOON} a+2 \mathrm{NaOH} \)