CH_3-CH_2-CH_2-Br + NaOH xrightarrow{C_2H_5OH | vedclass

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(B) $1$. The reaction of $CH_3-CH_2-CH_2-Br$ with $NaOH$ in $C_2H_5OH$ is a dehydrohalogenation reaction (elimination),which produces propene $(CH_3-C

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$B=2	ext{-Propanol}, C=1	ext{-Propanol}$

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(B) $1$. The reaction of $CH_3-CH_2-CH_2-Br$ with $NaOH$ in $C_2H_5OH$ is a dehydrohalogenation reaction (elimination),which produces propene $(CH_3-CH=CH_2)$ as Product $A$.$2$. The hydration of propene in the presence of acid $(H_2O/H^+)$ follows Markovnikov's rule,leading to the formation of $2	ext{-Propanol}$ as Product $B$.$3$. The hydroboration-oxidation of propene $(B_2H_6, H_2O_2/OH^-)$ follows anti-Markovnikov's rule,leading to the formation of $1	ext{-Propanol}$ as Product $C$.$4$. Therefore,$B = 2	ext{-Propanol}$ and $C = 1	ext{-Propanol}$.

$CH_3-CH_2-CH_2-Br + NaOH \xrightarrow{C_2H_5OH} \text{Product } A$. Consider the above reactions,identify product $B$ and product $C$.

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