Question

Oxidation of aniline with K$_2$Cr$_2$O$_7$/H$_2$SO$_4$ gives

• A. phenylhydroxylamine
• B. p-benzoquinone
• C. nitrosobenzene
• D. nitrobenzene

Hint:

Oxidation of aniline with strong oxidizing agents like K\(_2\)Cr\(_2\)O\(_7\) leads to the formation of nitrobenzene.

Answer 1

The Correct Option is D

Step 1: Oxidation of Aniline
Aniline (C\(_6\)H\(_5\)NH\(_2\)) undergoes oxidation with potassium dichromate (K\(_2\)Cr\(_2\)O\(_7\)) in sulfuric acid (H\(_2\)SO\(_4\)) to form nitrobenzene (C\(_6\)H\(_5\)NO\(_2\)).
Step 2: Explanation of Other Options
Option (a) phenylhydroxylamine is incorrect because this is not the product of oxidation.
Option (b) p-benzoquinone is incorrect because quinones are not produced in this oxidation.

Option (c) nitrosobenzene is incorrect because nitroso compounds are not produced in this reaction.

Step 3: Conclusion Thus, the oxidation of aniline with K\(_2\)Cr\(_2\)O\(_7\)/H\(_2\)SO\(_4\) gives nitrobenzene.