- ABy the action of $HCl$
- ✓By the action of $I_2 + Na_2CO_3$
- CBy the action of $NH_3$
- DSolubility in water
Methanol and Ethanol can be distinguished by Iodoform test ,As Ethanol have one methyl group present on its alpha position, while methanol have no alpha methyl group present
$CH _3 OH + I _2+ NaOH _{\text {Iodoform }}^{\stackrel{\text { warm }}{\longrightarrow} \text { (yellow) }}$ No reaction
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$(I)$ Ortho -nitrophenol
$(II)$ Para -nitrophenol
$(III)$ Meta- nitrophenol
${P} \xrightarrow[\substack{\text { 2. } \mathrm{H}^{+}, \mathrm{H}_2 \mathrm{O} \\ \text { 3. } \mathrm{H}_2 \mathrm{SO}_4, \Delta}]{\text { 1. MeMgBr }}Q \xrightarrow[\text { 2. } \mathrm{Zn}, \mathrm{H}_2 \mathrm{O}]{1 . \mathrm{O}_3} {R} \xrightarrow[\text { 2. } \Delta]{1 . \mathrm{OH}^{-}} {S}$
$1.$ The structure of the carbonyl compound ${P}$ is
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$2.$ The structures of the products ${Q}$ and ${R}$, respectively, are
$Image$
$3.$ The structure of the product ${S}$ is
$Image$
Give hte answer question $1,2$ and $3.$
