Toluene is nitrated and the resulting product is reduced with tin and hydrochloric acid. The product so obtained is diazotised and then heated with cuprous bromide. The reaction mixture so formed contains

mixture of o− and p−bromotoluenes

mixture of o− and p−dibromobenzenes

mixture of o− and p−bromoanilines

mixture of o− and m−bromotoluenes

Hydrocarbons PYQ — JEE Main 2008 · Shift 0 · Q2 with Solution

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Correct answer: A

Nitration of toluene

Toluene is an activating, ortho/para-directing substituent because of the methyl group $(-CH_3)$ .

So, on nitration:

\text{Toluene} \xrightarrow{HNO_3/H_2SO_4} \text{mixture of } o\text{-nitrotoluene and } p\text{-nitrotoluene}

Reduction of nitro group

Reduction with tin and hydrochloric acid converts the nitro group $(-NO_2)$ into an amino group $(-NH_2)$ :

o\text{-nitrotoluene} \to o\text{-toluidine}

p\text{-nitrotoluene} \to p\text{-toluidine}

Thus, we get a mixture of $o$ - and $p$ -toluidines.

Diazotisation

Aromatic amines on diazotisation form diazonium salts:

ArNH_2 \xrightarrow{NaNO_2/HCl,\ 0-5^\circ C} ArN_2^+Cl^-

Therefore:

o\text{-toluidine} \to o\text{-methylbenzenediazonium chloride}

p\text{-toluidine} \to p\text{-methylbenzenediazonium chloride}

Sandmeyer reaction with cuprous bromide

On heating diazonium salts with cuprous bromide, the diazonium group is replaced by bromine:

ArN_2^+Cl^- \xrightarrow{CuBr} ArBr

So,

$o$ -toluidine gives $o$ -bromotoluene
$p$ -toluidine gives $p$ -bromotoluene

Hence the final reaction mixture contains:

\boxed{\text{mixture of } o\text{- and } p\text{-bromotoluenes}}

Option check

A: mixture of $o$ - and $p$ -bromotoluenes — Correct
B: mixture of $o$ - and $p$ -dibromobenzenes — Incorrect
C: mixture of $o$ - and $p$ -bromoanilines — Incorrect
D: mixture of $o$ - and $m$ -bromotoluenes — Incorrect

Therefore, the correct answer is A.

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View written solutionFree

Correct answer: A

Nitration of toluene

Toluene is an activating, ortho/para-directing substituent because of the methyl group $(-CH_3)$ .

So, on nitration:

\text{Toluene} \xrightarrow{HNO_3/H_2SO_4} \text{mixture of } o\text{-nitrotoluene and } p\text{-nitrotoluene}

Reduction of nitro group

Reduction with tin and hydrochloric acid converts the nitro group $(-NO_2)$ into an amino group $(-NH_2)$ :

o\text{-nitrotoluene} \to o\text{-toluidine}

p\text{-nitrotoluene} \to p\text{-toluidine}

Thus, we get a mixture of $o$ - and $p$ -toluidines.

Diazotisation

Aromatic amines on diazotisation form diazonium salts:

ArNH_2 \xrightarrow{NaNO_2/HCl,\ 0-5^\circ C} ArN_2^+Cl^-

Therefore:

o\text{-toluidine} \to o\text{-methylbenzenediazonium chloride}

p\text{-toluidine} \to p\text{-methylbenzenediazonium chloride}

Sandmeyer reaction with cuprous bromide

On heating diazonium salts with cuprous bromide, the diazonium group is replaced by bromine:

ArN_2^+Cl^- \xrightarrow{CuBr} ArBr

So,

$o$ -toluidine gives $o$ -bromotoluene
$p$ -toluidine gives $p$ -bromotoluene

Hence the final reaction mixture contains:

\boxed{\text{mixture of } o\text{- and } p\text{-bromotoluenes}}

Option check

A: mixture of $o$ - and $p$ -bromotoluenes — Correct
B: mixture of $o$ - and $p$ -dibromobenzenes — Incorrect
C: mixture of $o$ - and $p$ -bromoanilines — Incorrect
D: mixture of $o$ - and $m$ -bromotoluenes — Incorrect

Therefore, the correct answer is A.

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