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KCET · Chemistry · Polymers

Match the reactant in Column - \( 1 \) with the reaction in Column - II
\begin{array}{|l|l|}
\hline I & II \\
\hline (i) Acetic acid & (a) Stephen \\
\hline (ii) Sodium phenate & (b) Friedel-Crafts \\
\hline (iii) Methyl cyanide & (c) HVZ \\
\hline (iv) Toluen & (d) Kolbe's \\
\hline
\end{array}

  1. A i-d , ii-b , iii-c, iv -a
  2. B i-c , ii-d , iii-a, iv -b
  3. C i-c, ii - a, iii - d, iv - b
  4. D i - b, ii - c, iii - a, iv - d
Verified Solution

Answer & Solution

Correct Answer

(B) i-c , ii-d , iii-a, iv -b

Step-by-step Solution

Detailed explanation

Acetic acid undergoes Hell-Volhard-Zelinsky reaction (HVZ)

(II) Sodium phenate undergoes Kolbe's electrolysis reaction.
(III) Methyl cyanide undergoes Stephen's reaction
\(
\begin{array}{l}
\mathrm{SnCl}_{2}+\mathrm{HCl} \rightarrow \mathrm{SnCl}_{4}+\underset{\text { Nascent }}{2[\mathrm{H}]} \\
\mathrm{CH}_{3} \mathrm{C} \equiv \mathrm{N}+2[\mathrm{H}]+2 \mathrm{HCl} \\
\downarrow \\
\mathrm{CH}_{3} \mathrm{CH}=\mathrm{NH}+\mathrm{HCl}_{-}+\underset{2}{\mathrm{~B}} \mathrm{O}_{3} \mathrm{Cl}
\end{array}
\)
(IV) Toluene undergoes Friedel Crafts reaction
reaction.
(III) Methyl cyanide undergoes Stephen's reaction
\(
\begin{array}{l}
\mathrm{SnCl}_{2}+\mathrm{HCl} \rightarrow \mathrm{SnCl}_{4}+\underset{\text { Nascent }}{2[\mathrm{H}]} \\
\mathrm{CH}_{3} \mathrm{C} \equiv \mathrm{N}+2[\mathrm{H}]+2 \mathrm{HCl} \\
\downarrow \\
\mathrm{CH}_{3} \mathrm{CH}=\mathrm{NH}+\mathrm{HCl}_{-}+\underset{2}{\mathrm{~B}} \mathrm{O}_{3} \mathrm{Cl}
\end{array}
\)
(IV) Toluene undergoes Friedel Crafts reaction