- A

- B

- C

- D

View written solutionFree
Correct answer: C
-
Interpret each test
We are given three isomeric amines with molecular formula .
We must identify:
- : can be prepared by Gabriel phthalimide synthesis
- : with Hinsberg reagent gives a solid insoluble in NaOH
- : with followed by -naphthol in NaOH gives red dye
-
Meaning of each test
(i) Gabriel phthalimide synthesis
Gabriel synthesis prepares only primary alkyl amines effectively, not aryl amines, not secondary/tertiary amines.
So must be a primary amine.
(ii) Hinsberg test
- Primary amine: forms sulphonamide containing N–H, soluble in NaOH.
- Secondary amine: forms sulphonamide without N–H, hence insoluble in NaOH.
- Tertiary amine: generally does not form sulphonamide.
Therefore must be a secondary amine.
(iii) Reaction with , then coupling with -naphthol
This is the diazotization + azo coupling test, given by primary aromatic amines.
Therefore must be a primary aromatic amine.
-
Find isomeric amines of formula
Let us list common isomers fitting this formula:
- Primary aliphatic amine: (phenethylamine)
- Primary aromatic amine: i.e. toluidines (-, -, -toluidine)
- Secondary amine: (N-methylaniline)
- Another secondary amine: would be ? Check: benzyl methyl amine = yes, but it is not aromatic primary.
-
Assign identities using tests
For
Since it gives diazotization followed by azo dye formation, it must be a primary aromatic amine.
Hence is toluidine (//-toluidine).
For
It gives Hinsberg product insoluble in NaOH, so it must be a secondary amine.
The suitable aromatic secondary amine is N-methylaniline.
For
It can be prepared by Gabriel synthesis, so it should be a primary aliphatic amine.
Hence is phenethylamine .
-
Final identification
Therefore,
-
Match with options
This corresponds to Option C.
-
Comparison with stored answer
Stored correct answer: C
Our derived answer: C
So they agree.
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