Aniline on direct nitration yields
Show Hint
- 51%-ortho, 47%-para, 2%-meta derivatives
- 51%-meta, 47%-ortho, 2%-para derivatives
- 51%-para, 47%-meta, 2%-ortho derivatives
- 51%-ortho, 47%-meta, 2%-para derivatives
The Correct Option is B
Solution and Explanation
Nitration is carried out in a strongly acidic medium (conc. \(HNO_3\) and \(H_2SO_4\)). In this medium aniline gets largely protonated to the anilinium ion \(C_6H_5NH_3^+\).
Step 1:
The \(-NH_2\) group is ortho/para directing and ring activating, but the protonated \(-NH_3^+\) group is meta directing and ring deactivating. Because a large fraction exists as the anilinium ion, a substantial amount of the meta product is formed along with ortho and para products.
Step 2:
The observed distribution is about 51% meta, 47% para and 2% ortho. Among the given choices the one giving meta as the major (51%) product is option (B).
Answer: Option (2) — 51%-meta is the only choice showing the characteristic large meta fraction obtained on direct nitration of aniline.
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Write IUPAC names of the following compounds and classify them into primary, secondary and tertiary amines.
(i) (CH3 )2CHNH2 (ii) CH3 (CH2 )2NH2 (iii) CH3NHCH(CH3 )2
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Write IUPAC names of the following compounds and classify them into primary, secondary and tertiary amines.
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(i) In decreasing order of the pKb values: C2H5NH2 , C6H5NHCH3 , (C2H5 )2NH and C6H5NH2
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How will you convert:
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(iii) Methanol to ethanoic acid
(iv) Ethanamine into methanamine
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(viii) Propanoic acid into ethanoic acid?
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