Consider the above sequence of reactions. The number of bromine atom(s) in the final product (P) will be :
To determine the number of bromine atoms in the final product (P) from the given sequence of reactions, we need to analyze each step carefully:
Thus, the final product, compound P, contains 5 bromine atoms.
The correct answer is: \(5\).
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
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 :
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