Given below are two statements :
Statement I : The dipole moment of R-CN is greater than R-NC and R-NC can
undergo hydrolysis under acidic medium to produce R-C(∥)(O)-OH .
Statement II : R-CN hydrolyses under acidic medium to produce a compound which on treatment with SOCl₂, followed by the addition of NH₃ gives another compound (x). This compound (x) on treatment with NaOCl/NaOH gives a product, that on treatment with CHCl₃/KOH/Δ produces R-NC
In the light of the above statements, choose the correct answer from the options given below

Solution

Chemistry ma’am · 3 steps

An isocyanide hydrolyses to an amine and formic acid, never to RCOOH, so one wrong half sinks Statement I.

1. Check Statement I. The dipole comparison is fine: alkyl cyanides are more polar than alkyl isocyanides. But acidic hydrolysis of an isocyanide breaks the C=N link and gives R-NH₂ + HCOOH, not RCOOH. One false part makes the whole statement false.
2. Follow Statement II. Track each reagent in turn. NaOCl/NaOH is a hypohalite in alkali, so it does the Hofmann bromamide job on the amide. This step removes one carbon, which is exactly why the final R matches the starting R.

   Reagent	Product	Reaction
   H₃O⁺ on R-CN	R-COOH	nitrile hydrolysis
   SOCl₂, then NH₃	R-CONH₂ (x)	acid chloride, then amide
   NaOCl/NaOH	R-NH₂	Hofmann degradation, loses one C
   CHCl₃/KOH, heat	R-NC	carbylamine reaction

3. So. Statement I is false and Statement II is true, which is option (b).

The idea to keep. Hydrolysis keeps the atom bonded to R: cyanides keep C and give RCOOH, isocyanides keep N and give RNH₂. Amide plus hypohalite in alkali gives an amine with one carbon fewer.

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