Suppose the following reaction:CH_3-CH_2-OH x | vedclass

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(A) $1$. The reaction of $CH_3-CH_2-OH$ with $Cl_2/NaOH$ is a haloform reaction. Ethanol undergoes oxidation and chlorination to form $CHCl_3$ (chloro

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$2-$hydroxybenzaldehyde

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(A) $1$. The reaction of $CH_3-CH_2-OH$ with $Cl_2/NaOH$ is a haloform reaction. Ethanol undergoes oxidation and chlorination to form $CHCl_3$ (chloroform) and sodium formate $(HCOONa)$. Here,$A$ is $CHCl_3$.$2$. The reaction of phenol with $CHCl_3$ in the presence of aqueous $NaOH$ is the Reimer-Tiemann reaction.$3$. In the Reimer-Tiemann reaction,an aldehyde group $(-CHO)$ is introduced at the ortho position of the phenol ring.$4$. Therefore,the major product formed is $2-$hydroxybenzaldehyde (salicylaldehyde).

Suppose the following reaction:
$CH_3-CH_2-OH$ $\xrightarrow{Cl_2/NaOH} A$ $\xrightarrow{\text{Phenol}/OH^-} \text{Major Product}$
The structure of the Major Product will be:

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Suppose the following reaction:$CH_3-CH_2-OH$ $\xrightarrow{Cl_2/NaOH} A$ $\xrightarrow{\text{Phenol}/OH^-} \text{Major Product}$The structure of the Major Product will be: