Ph-CH_2-CH(Br)-CH_2-CH_3 xrightarrow{NaNH_2} | vedclass

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(B) The reaction is a dehydrohalogenation ($E2$ elimination) using a strong base $NaNH_2$. The reactant is $Ph-CH_2-CH(Br)-CH_2-CH_3$. There are two t

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$4$

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(B) The reaction is a dehydrohalogenation ($E2$ elimination) using a strong base $NaNH_2$. The reactant is $Ph-CH_2-CH(Br)-CH_2-CH_3$. There are two types of $\beta$-hydrogens available for elimination: $1.$ Elimination involving $\beta$-hydrogen from the $CH_2$ group adjacent to the phenyl ring $(C1)$ gives $Ph-CH=CH-CH_2-CH_3$ ($1$-phenylbut-$1$-ene). This alkene can exist as two stereoisomers: $E$ (trans) and $Z$ (cis). $2.$ Elimination involving $\beta$-hydrogen from the $CH_2$ group on the other side $(C3)$ gives $Ph-CH_2-CH=CH-CH_3$ ($1$-phenylbut-$2$-ene). This alkene can also exist as two stereoisomers: $E$ (trans) and $Z$ (cis). Therefore,the total number of elimination products including stereoisomers is $2 + 2 = 4$.

$Ph-CH_2-CH(Br)-CH_2-CH_3 \xrightarrow{NaNH_2} (A)$
Total number of elimination products are (includes stereoisomers):

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$Ph-CH_2-CH(Br)-CH_2-CH_3 \xrightarrow{NaNH_2} (A)$ Total number of elimination products are (includes stereoisomers):