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प्रश्न
An aromatic compound ‘A’ on treatment with aqueous ammonia and heating forms compound ‘B’ which on heating with Br2 and KOH forms a compound ‘C’ of molecular formula C6H7N. Write the structures and IUPAC names of compounds A, B and C.
उत्तर
It is given that compound ‘C’ having the molecular formula C6H7N is formed by heating compound ‘B’ with Br2 and KOH. This is a Hoffmann bromamide degradation reaction. Therefore, compound ‘B’ is an amide and compound ‘C’ is an amine. The only amine having the molecular formula C6H7N is aniline (C6H5NH2).
Therefore, compound ‘B’ (from which ‘C’ is formed) must be benzamide (C6H5CONH2).
Further, benzamide is formed by heating compound ‘A’ with aqueous ammonia. Therefore, compound ‘A’ must be benzoic acid.
The given reactions can be explained with the help of the following equations:
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संबंधित प्रश्न
Convert 3-Methylaniline into 3-nitrotoluene.
Write the IUPAC name of the following compound and classify it as primary, secondary and tertiary amine.
C6H5NHCH3
Give one chemical test to distinguish between the following pair of compounds.
Secondary and tertiary amines
Give one chemical test to distinguish between the following pair of compounds.
Ethylamine and aniline
Give one chemical test to distinguish between the following pair of compounds.
Aniline and benzylamine
Account for the following:
Ethylamine is soluble in water whereas aniline is not.
Account for the following:
Methylamine in water reacts with ferric chloride to precipitate hydrated ferric oxide.
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How will you convert ethanoic acid into propanoic acid?
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Accomplish the following conversion:
Chlorobenzene to p-chloroaniline
Complete the following reaction:
\[\ce{C6H5NH2 + Br2 (aq) ->}\]
Complete the following reaction:
\[\ce{C6H5N2Cl ->[(i) HBF4][(ii) NaNO2/Cu, \Delta]}\]
Do the following conversions in not more than two steps :
Ethyl benzene to Benzoic acid
Predict the product 'A' in the following reaction.
\[\ce{\underset{(Acetyl chloride)}{CH3COCl}->[H2][Pd-BaSO4]A + HCl}\]