The number of possible structural isomers for the compound $C_2H_3Cl_2Br$ is:

  • A
    $2$
  • B
    $3$
  • C
    $4$
  • D
    $5$

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The starting material is propene $(CH_3-CH=CH_2)$. The reaction with $H_3O^+$ (acid-catalyzed hydration) follows Markovnikov's rule to give $A$ (propan$-2-$ol,$CH_3-CH(OH)-CH_3$). The reaction with $(i) BH_3/THF$ followed by $(ii) H_2O_2/OH^-$ (hydroboration-oxidation) follows anti-Markovnikov's rule to give $B$ (propan$-1-$ol,$CH_3-CH_2-CH_2OH$). What is the relationship between products $A$ and $B$?

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