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Standard XII

Chemistry

Preparation of Benzene and Homologues

An yellow col...

Question

An yellow coloured liquid (A) called as 'Oil of Mirbane' is reduced with Sn/HCl to give compound (B). Identify A and B and write the equation.

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Solution

A) (i) Acid medium reduction: When nitrobenzene is reduced with tin and hydrochloric acid aniline is formed.
$C_{6} H_{5} N O_{2} N i t r o b e n z e n e + 6 [H] S n / H C l - ---- \to C_{6} H_{5} N H_{2} A n i l i n e + 2 H_{2} O$
B) $A = C_{6} H_{5} N O_{2}$
$B = C_{6} H_{5} N H_{2}$

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An yellow coloured liquid (A) called as 'Oil of Mirbane' is reduced with Sn/HCl to give compound (B). Identify A and B and write the equation.

A) (i) Acid medium reduction: When nitrobenzene is reduced with tin and hydrochloric acid aniline is formed.C6H5NO2Nitrobenzene+6[H]Sn/HCl−−−−−→C6H5NH2Aniline+2H2OB) A=C6H5NO2B=C6H5NH2

A) (i) Acid medium reduction: When nitrobenzene is reduced with tin and hydrochloric acid aniline is formed.
$C_{6} H_{5} N O_{2} N i t r o b e n z e n e + 6 [H] S n / H C l - ---- \to C_{6} H_{5} N H_{2} A n i l i n e + 2 H_{2} O$
B) $A = C_{6} H_{5} N O_{2}$
$B = C_{6} H_{5} N H_{2}$