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 
Statement II: R-CN hydrolyses under acidic medium to produce a compound which on treatment with $\text{SOCl}_2$, followed by the addition of $\text{NH}_3$ gives another compound(x). This compound (x) on treatment with $\text{NaOCl/NaOH}$ gives a product, that on treatment with $\text{CHCl}_3\text{/KOH/}\Delta$ produces R-NC
In the light of the above statements, choose the correct answer from the options given below
Let's analyze the statements one by one to identify which statement is correct.
Statement I: The dipole moment of R-CN is greater than R-NC and R-NC can undergo hydrolysis under acidic medium to produce \(R-\text{COOH}\).
Analysis:
Conclusion for Statement I: False, because R-NC does not produce carboxylic acid upon hydrolysis under acidic conditions.
Statement II: R-CN hydrolyses under acidic medium to produce a compound which on treatment with \(\text{SOCl}_2\), followed by the addition of \(\text{NH}_3\) gives another compound (x). This compound (x) on treatment with \(\text{NaOCl/NaOH}\) gives a product, that on treatment with \(\text{CHCl}_3/\text{KOH}/\Delta\) produces R-NC.
Analysis:
Conclusion for Statement II: True, as the procedure described is a viable synthetic pathway from nitrile to isonitrile.
Final Answer: Statement I is false but Statement II is true.
Given below are two statements :
Statement I : Compound (X), shown below, dissolves in $\text{NaHCO}_3$ solution and has two chiral carbon atoms 
Statement II : Compound (Y), shown below, has two carbons with $\text{sp}^3$ hybridization, one carbon with $\text{sp}^2$ and one carbon with $\text{sp}$ hybridization 
In the light of the above statements, choose the correct answer from the options given below :

Consider the above sequence of reactions. The number of bromine atom(s) in the final product (P) will be :