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(B) The reaction is an $E2$ elimination reaction. The ethoxide ion $(C_2H_5O^-)$ acts as a strong base and abstracts a $\beta$-hydrogen. The hydrogen

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$C_6H_5-CH=C(CH_3)-CH_2-CH_3$

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(B) The reaction is an $E2$ elimination reaction. The ethoxide ion $(C_2H_5O^-)$ acts as a strong base and abstracts a $\beta$-hydrogen. The hydrogen on the carbon adjacent to the benzene ring is more acidic due to the electron-withdrawing effect of the phenyl group. Removal of this hydrogen leads to the formation of $C_6H_5-CH=C(CH_3)-CH_2-CH_3$. This product is the major product because the double bond is in conjugation with the benzene ring,which provides significant resonance stabilization.

The major product of the following reaction is:
$C_6H_5-CH_2-C(Br)(CH_3)-CH_2-CH_3 \xrightarrow[C_2H_5OH]{C_2H_5ONa}$

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The order of reactivity of the following halides towards $S_{N}1$ reaction is:
$(1) \ (CH_3)_3CBr$
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The major product of the following reaction is: $C_6H_5-CH_2-C(Br)(CH_3)-CH_2-CH_3 \xrightarrow[C_2H_5OH]{C_2H_5ONa}$