Question

The following synthesis can not be carried out by:

• A. $\underset{\left(1\right)}{\overset{C{l}_2}{\to }}$ $\underset{\left(2\right)}{\overset{ICl/C{H}_{3}COOH}{\to }}$ $\underset{\left(3\right)}{\overset{ICl/C{H}_{3}COOH}{\to }}$ $\underset{\left(4\right)}{\overset{Zndust}{\to }}$
• B. $\underset{\left(1\right)}{\overset{HOBr/H^{+}}{\to }}$ $\underset{\left(2\right)}{\overset{C{l}_2/Fe}{\to }}$ $\underset{\left(3\right)}{\overset{ICl/ZnC{l}_2\left(excess\right)}{\to }}$ $\underset{\left(4\right)}{\overset{NaOH,\Delta }{\to }}$
• C. $\underset{\left(1\right)}{\overset{HOBr/H^{+}}{\to }}$ $\underset{\left(2\right)}{\overset{C{l}_2/Fe}{\to }}$ $\underset{\left(3\right)}{\overset{ICl/ZnC{l}_2\left(excess\right)}{\to }}$ $\underset{\left(4\right)}{\overset{Zndust,C{H}_{2}COOH}{\to }}$
• D. $\underset{\left(1\right)}{\overset{B{r}_2/C{H}_{3}COOH}{\to }}$ $\underset{\left(2\right)}{\overset{C{l}_2/Fe}{\to }}$ $\underset{\left(3\right)}{\overset{ICl/C{H}_{3}COOH\left(excess\right)}{\to }}$ $\underset{\left(4\right)}{\overset{NaN{H}_2}{\to }}$

Answer 1

Answer: (A), (B), (D)

The correct options are
A $\underset{\left(1\right)}{\overset{C{l}_2}{\to }}$ $\underset{\left(2\right)}{\overset{ICl/C{H}_{3}COOH}{\to }}$ $\underset{\left(3\right)}{\overset{ICl/C{H}_{3}COOH}{\to }}$ $\underset{\left(4\right)}{\overset{Zndust}{\to }}$
B $\underset{\left(1\right)}{\overset{HOBr/H^{+}}{\to }}$ $\underset{\left(2\right)}{\overset{C{l}_2/Fe}{\to }}$ $\underset{\left(3\right)}{\overset{ICl/ZnC{l}_2\left(excess\right)}{\to }}$ $\underset{\left(4\right)}{\overset{NaOH,\Delta }{\to }}$
D $\underset{\left(1\right)}{\overset{B{r}_2/C{H}_{3}COOH}{\to }}$ $\underset{\left(2\right)}{\overset{C{l}_2/Fe}{\to }}$ $\underset{\left(3\right)}{\overset{ICl/C{H}_{3}COOH\left(excess\right)}{\to }}$ $\underset{\left(4\right)}{\overset{NaN{H}_2}{\to }}$

The given synthesis cannot be carried out by options $A,B$ and $D$. It can be carried out by option $C$.