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(B) $1$. Compound $A$ has the formula $C_5H_9Cl$. The degree of unsaturation is $5 - (9/2) + (1/2) + 1 = 2$. Since it does not react with $Br_2 / CCl_

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$1-$chloro$-2-$methylcyclobutane

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(B) $1$. Compound $A$ has the formula $C_5H_9Cl$. The degree of unsaturation is $5 - (9/2) + (1/2) + 1 = 2$. Since it does not react with $Br_2 / CCl_4$,it must be a bicyclic compound or have two rings. $2$. Reaction with a strong base gives $B$ $(C_5H_8)$,which reacts with $Br_2 / CCl_4$,indicating $B$ is an alkene. This implies $A$ is a cyclic alkyl halide that undergoes dehydrohalogenation to form a cyclic alkene. $3$. Ozonolysis of $B$ $(C_5H_8)$ gives $C_5H_8O_2$. This indicates the ring opened to form a dicarbonyl compound. $4$. Based on the options,$A$ is $1$-chloro-$2$-methylcyclobutane. Dehydrohalogenation gives $1$-methylcyclobutene $(B)$. Ozonolysis of $1$-methylcyclobutene gives $5$-oxopentanal $(C_5H_8O_2)$.

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