
Identify X,Y and Z.
\[
{C_6}{H_5}CON{H_2}\xrightarrow[{KOH}]{{B{r_2}}}X\xrightarrow[{HCL}]{{HN{O_2}}}Y\xrightarrow[{C{u^ + }}]{{NaN{O_2}}}Z. \\
\\
\]
Answer
560.7k+ views
Hint:In order to solve this question we need to apply first Hoffmann bromamide degradation by doing this we will get X. Now X is diazotized to form Y after which it will be treated with $NaN{O_2}/C{u^ + }$ and the product Z will also be found.
Complete step-by-step answer:As we have already know ${C_6}{H_5}CON{H_2}$ is amide which when reacts with bromine in aqueous or ethanolic solution of sodium or potassium hydroxide is known to be Hoffmann bromamide reaction so let's have a look it once:Hoffmann bromamide reaction: When amide is treated with bromine in an aqueous or ethanolic solution of sodium or potassium hydroxide, degradation of amide takes place which leads to the formation of primary amine. The primary amine which formed contains one carbon less than the number of carbon atoms in that amide.
$RCON{H_2} + B{r_2} + 4NaOH \to R - N{H_2} + N{a_2}C{O_3} + 2NaBr + 2{H_2}O$
Where; R is ${C_6}{H_5}$ is called Phenyl radical.So this is then a complete required reaction upto X and the product X is ${C_6}{H_5} - N{H_2}$ Now should be diazotized which will form benzene diazonium chloride.According to reaction:
${C_6}{H_5}N{H_2}\xrightarrow[{HCl}]{{HN{O_2}}}{C_6}{H_5}{N_2}Cl$
So now our Y is also found which is ${C_6}{H_5}{N_2}Cl$ this is known as diazonium salt.
Now diazonium salt is with $NaN{O_2}/C{u^ + }$ to give nitrobenzene.According to reaction:
${C_6}{H_5}{N_2}Cl\xrightarrow[{C{u^ + }}]{{NaN{O_2}}}{C_6}{H_5}N{O_2}$
Finally we get Z which is ${C_6}{H_5}N{O_2}$ which is nitrobenzene.
$
X = {C_6}{H_5}N{H_2} \\
Y = {C_6}{H_5}{N_2}Cl \\
Z = {C_6}{H_5}N{O_2} \\
$
These are our final products.
Note:In order to solve these types of questions we need to know the named reactions this will make the approach easy and complications will not occur.Balancing the reactions is not necessary because we only focus on main products.
Complete step-by-step answer:As we have already know ${C_6}{H_5}CON{H_2}$ is amide which when reacts with bromine in aqueous or ethanolic solution of sodium or potassium hydroxide is known to be Hoffmann bromamide reaction so let's have a look it once:Hoffmann bromamide reaction: When amide is treated with bromine in an aqueous or ethanolic solution of sodium or potassium hydroxide, degradation of amide takes place which leads to the formation of primary amine. The primary amine which formed contains one carbon less than the number of carbon atoms in that amide.
$RCON{H_2} + B{r_2} + 4NaOH \to R - N{H_2} + N{a_2}C{O_3} + 2NaBr + 2{H_2}O$
Where; R is ${C_6}{H_5}$ is called Phenyl radical.So this is then a complete required reaction upto X and the product X is ${C_6}{H_5} - N{H_2}$ Now should be diazotized which will form benzene diazonium chloride.According to reaction:
${C_6}{H_5}N{H_2}\xrightarrow[{HCl}]{{HN{O_2}}}{C_6}{H_5}{N_2}Cl$
So now our Y is also found which is ${C_6}{H_5}{N_2}Cl$ this is known as diazonium salt.
Now diazonium salt is with $NaN{O_2}/C{u^ + }$ to give nitrobenzene.According to reaction:
${C_6}{H_5}{N_2}Cl\xrightarrow[{C{u^ + }}]{{NaN{O_2}}}{C_6}{H_5}N{O_2}$
Finally we get Z which is ${C_6}{H_5}N{O_2}$ which is nitrobenzene.
$
X = {C_6}{H_5}N{H_2} \\
Y = {C_6}{H_5}{N_2}Cl \\
Z = {C_6}{H_5}N{O_2} \\
$
These are our final products.
Note:In order to solve these types of questions we need to know the named reactions this will make the approach easy and complications will not occur.Balancing the reactions is not necessary because we only focus on main products.
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