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MHT CET · Chemistry · Hydrocarbons

Identify the product \(X\) obtained in following reaction \(\mathrm{CH}_{3}-\left(\mathrm{CH}_{2}\right)_{4}-\mathrm{CH}_{3} \quad \frac{\mathrm{Cr}_{2} \mathrm{O}_{3}}{773 \mathrm{~K}, 10-20 \mathrm{~atm}} \mathrm{X}\)

  1. A \(\mathrm{CH}_{3}-\mathrm{CH}=\mathrm{CH}-\mathrm{CH}_{3}+\mathrm{CH}_{2}=\mathrm{CH}_{2}\) \(+~\mathrm{H}_{2}\)
  2. B \(2 \mathrm{CH}_{3}-\mathrm{CH}=\mathrm{CH}_{2}+\mathrm{H}_{2}\)
  3. C \(\mathrm{C}_{6} \mathrm{H}_{6}+4 \mathrm{H}_{2}\)
  4. D \(\mathrm{CH}_{3}-\left(\mathrm{CH}_{2}\right)_{3}-\mathrm{CH}=\mathrm{CH}_{2}+\mathrm{H}_{2}\)
Verified Solution

Answer & Solution

Correct Answer

(C) \(\mathrm{C}_{6} \mathrm{H}_{6}+4 \mathrm{H}_{2}\)

Step-by-step Solution

Detailed explanation

Higher alkane with more than 5 carbon atom undergoes cyclization to form benzene in presence of \(\mathrm{Cr}_{2} \mathrm{O}_{3}\) under 10 to \(20 \mathrm{~atm}\) pressure at \(773 \mathrm{~K}\).