A chemist has $4$ samples of artificial sweetener $\mathrm{A},\mathrm{}\mathrm{B},\mathrm{}\mathrm{C}$ and $\mathrm{D}$ . To identify these samples, he performed certain experiments and noted the following observations:
(i) $\mathrm{A}$ and $\mathrm{D}$ both form blue-violet colour with ninhydrin.
(ii) Lassaigne extract of $\mathrm{C}$ gives positive ${\mathrm{A}\mathrm{g}\mathrm{N}\mathrm{O}}_3$ test and negative $\mathrm{F}{\mathrm{e}}_4{\left[\mathrm{F}\mathrm{e}{\left(\mathrm{C}\mathrm{N}\right)}_6\right]}_3$ test.
(iii) Lassaigne extract of $\mathrm{B}$ and $\mathrm{D}$ gives positive sodium nitroprusside test.
Based on these observations which option is correct?

• Correct answer
  A

  A : Aspartame; B : Saccharin; C : Sucralose; D : Alitame
• B

  A : Alitame; B : Saccharin; C : Aspartame; D : Sucralose
• C

  A : Saccharin; B : Alitame; C : Sucralose; D : Aspartame
• D

  A : Aspartame; B : Alitame; C : Saccharin; D : Sucralose

Solution

$A-$ Aspartame

$B-$ Saccharine

$C-$ Sucralose

$D-$ Alitame

(i) $\mathrm{A }\& D$ give positive test with ninhydrin because both have free carboxylic and amine groups.
(ii) $C$ form precipitate with $AgN{O}_3$ in the Lassaigne extract of the sugar because it has chlorine atoms.
(iii) $\mathrm{B }\& D$ give positive test with sodium nitroprusside because both have sulphur atoms.

Hence, the correct answer is option: 'A'.