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(A) The reaction of $CH_3-CHO$ (acetaldehyde) with $C_2H_5MgBr$ (ethylmagnesium bromide) followed by hydrolysis yields $CH_3-CH(OH)-C_2H_5$ (butan$-2-

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It undergoes dehydration with $20 \% H_3PO_4$ at $358 \ K$

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(A) The reaction of $CH_3-CHO$ (acetaldehyde) with $C_2H_5MgBr$ (ethylmagnesium bromide) followed by hydrolysis yields $CH_3-CH(OH)-C_2H_5$ (butan$-2-$ol). $CH_3-CHO + C_2H_5MgBr$ $\rightarrow CH_3-CH(OMgBr)-C_2H_5$ $\xrightarrow{H_2O} CH_3-CH(OH)-C_2H_5$. Butan$-2-$ol is a secondary alcohol. $(A)$ Secondary alcohols undergo dehydration with $20 \% H_3PO_4$ at $358 \ K$ to form alkenes. This is a correct statement. $(B)$ Oxidation of secondary alcohols with $CrO_3$ gives a ketone (butanone). This is also a correct statement. $(C)$ Butan$-2-$ol contains the $CH_3-CH(OH)-$ group,so it gives a positive iodoform test. Thus,this statement is incorrect. $(D)$ It is not a vinylic alcohol. Note: In many competitive contexts,$(A)$ is considered the primary characteristic reaction for this specific alcohol under these conditions.

The correct statement about the product of the following reaction is
$CH_3-CHO \xrightarrow[(ii) H_2O]{(i) C_2H_5MgBr} \text{product}$

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The correct statement about the product of the following reaction is$CH_3-CHO \xrightarrow[(ii) H_2O]{(i) C_2H_5MgBr} \text{product}$