A chemist has 4 samples of artificial sweetener A, B, C and D. To identify these samples, he performed certain experiments and noted the following observations :
(i) A and D both form blue-violet colour with ninhydrin.
(ii) Lassaigne extract of C gives positive AgNO₃ test and negative Fe₄[Fe(CN)₆]₃ test.
(iii) Lassaigne extract of B and D gives positive sodium nitroprusside test
Based on these observations which option is correct ?

Solution

Chemistry ma’am · 3 steps

Write down which elements each sweetener carries, then let each test pick its sample.

1. What each test detects. Ninhydrin gives blue-violet only with a free amino group, as in amino acids and peptides. A positive AgNO₃ test means halogen. Prussian blue, Fe₄[Fe(CN)₆]₃, means nitrogen, and sodium nitroprusside means sulphur.

   Sweetener	Free NH₂ (ninhydrin)	N	S (nitroprusside)	Cl (AgNO₃)
   Aspartame (Asp–Phe methyl ester)	yes	yes	no	no
   Alitame (Asp–Ala amide with a thietane ring)	yes	yes	yes	no
   Saccharin (cyclic sulphonyl imide)	no	yes	yes	no
   Sucralose (chlorinated sucrose)	no	no	no	yes

2. Match the observations. C has chlorine but no nitrogen, so C is sucralose. D must give both ninhydrin and the sulphur test, and only alitame does both, so D is alitame. The other ninhydrin-positive sample, A, is aspartame, and the other sulphur-containing one, B, is saccharin.
3. So. A is aspartame, B is saccharin, C is sucralose and D is alitame. That is option (c).

The idea to keep. For sweetener questions, remember the elements: aspartame (N, free NH₂), alitame (N and S, free NH₂), saccharin (N and S, no free NH₂), sucralose (Cl only).

Answer from the NTA final answer key. Solution by PrepWiser.