Oxidation of aniline with K$_2$Cr$_2$O$_7$/H$_2$SO$_4$ gives
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- phenylhydroxylamine
- p-benzoquinone
- nitrosobenzene
- nitrobenzene
The Correct Option is D
Solution and Explanation
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.
Top Questions on Amines
Amines have a lone pair of electrons on the nitrogen atom, due to which they behave as Lewis bases. Greater the value of \( K_b \) or smaller the value of \( pK_b \), stronger is the base. Amines are more basic than alcohols, ethers, esters, etc. The basic character of aliphatic amines should increase with the increase of alkyl substitution. However, it does not occur in a regular manner, as a secondary aliphatic amine is unexpectedly more basic than a tertiary amine in aqueous solutions. Aromatic amines are weaker bases than ammonia and aliphatic amines. Electron-releasing groups such as \( -CH_3 \), \( -NH_2 \), etc., increase the basicity, while electron-withdrawing substituents such as \( -NO_2 \), \( -CN \), halogens, etc., decrease the basicity of amines. The effect of these substituents is more pronounced at the para-position than at the meta-position.
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