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Diagonal Relationship (Group 1)

Which one of ...

Question

Which one of the following compounds liberates $C O_{2}$ from aqueous $N a H C O_{3}$ ?

Anilinium chloride (

C_{6} H_{8} C l N

)

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C H C l_{3}

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C C l_{4}

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C H_{3} C l

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Solution

The correct option is B Anilinium chloride ( $C_{6} H_{8} C l N$ )
Anilinium chloride is the hydrochloride salt of aniline. It reacts with aqueous sodium bicarbonate to liberate carbon dioxide. Chloroform, carbon tetrachloride and methyl chloride are solvents. They do not react with sodium bicarbonate.

$C_{6} H_{5} N H_{3}^{+} C l^{-} + N a H C O_{3} \to C_{6} H_{5} N H_{2} + N a C l + H_{2} O + C O_{2}$

Hence, the correct option is $A$

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Similar questions

C H_{4}, C H_{3} C l, C H_{2} C l_{2}, C H C l_{3}, C C l_{4}

C C l_{4}

C H_{3} C l

C H_{4} = C C l_{4} < C H C l_{3} < C H_{2} C l_{2} < C H_{3} C l

How many of the following compounds liberate $C O_{2}$ with $N a H C O_{3}$ ?

C O_{2}

N a H C O_{3}

C H_{3} C l

C H_{2} C l_{2}

C H C l_{3}

C C l_{4}

{C H}_{3} C l, {C H}_{2} {C l}_{2}, C H {C l}_{3}, C {C l}_{4}

{C H}_{3} C l > {C H}_{2} {C l}_{2} > C H {C l}_{3} > C {C l}_{4}

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