Consider the above reaction,the product $A$ and product $B$ respectively are ...

  • A
    $2,4,6$-tribromoaniline and $p$-bromoaniline
  • B
    $p$-bromoaniline and $2,4,6$-tribromoaniline
  • C
    $2,4,6$-tribromoaniline and $2,4,6$-tribromoaniline
  • D
    $p$-bromoaniline and $p$-bromoaniline

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What product is formed when ethylamine reacts with nitrous acid?

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$H_3CCONH_2 + Br_2 + 4NaOH \longrightarrow Y + Na_2CO_3 + 2NaBr + 2H_2O$
What is $Y$ in the above reaction?

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$A$ hydrocarbon $A$ $(C_{4}H_{8})$ on reaction with $HCl$ gives a compound $B$ $(C_{4}H_{9}Cl)$ which on reaction with $1 \ mol$ of $NH_{3}$ gives compound $C$ $(C_{4}H_{11}N)$. On reacting with $NaNO_{2}$ and $HCl$ followed by treatment with water,compound $C$ yields an optically active compound $D$. The compound $D$ is

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