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Question 44

Which of the following compounds will be suitable for Kjeldahl's method for nitrogen estimation?

Kjeldahl’s method works only when the nitrogen present in an organic compound is first converted to $$\left(NH_4\right)_2SO_4$$ by boiling the substance with excess conc. $$H_2SO_4$$. For this to happen, the nitrogen atom must be present as an -NH, -NH2, -CONH-, -CNH- etc., i.e. it must be bonded only to carbon and/or hydrogen (and, after protonation, be able to form $$NH_4^+$$).
The method fails whenever nitrogen is present in an oxidation state or environment that prevents the formation of ammonium ion. Thus compounds containing

  • a nitro group $$(-NO_2)$$,
  • azo / diazo linkages $$(-N=N-,\; -N_2^+Cl^-)$$,
  • nitrile $$(-C\equiv N)$$, isonitrile $$(-NC)$$,
  • nitroso $$(-NO)$$, or
  • ring nitrogen as in pyridine, quinoline etc.

are not estimated correctly by the Kjeldahl procedure.

Keeping this rule in mind, inspect the four choices (structures as given in the paper):

Case A: Nitro-compound - contains $$-NO_2$$. Not suitable.
Case B: Hetero-aromatic (pyridine-like) - ring nitrogen. Not suitable.
Case C: Acetamide $$CH_3CONH_2$$ - nitrogen is in an amide unit that is protonated by conc. $$H_2SO_4$$ to give $$NH_4^+$$. Suitable.
Case D: Azo compound $$Ar-N=N-Ar$$ - contains an azo linkage. Not suitable.

Therefore, the only compound that fulfils the requirement is the amide in Option C.

Final answer: Option C which is: acetamide $$\left(CH_3CONH_2\right)$$.

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