Question

(64) $CH_3Mgcl + \bigtriangleup \longrightarrow$

Answer 1

The product of the reaction is 1-methylcyclopropan-1-ol.

Answer 2

Explanation of the solution:

The symbol $\bigtriangleup$ in the context of this reaction, especially when compared to the similar question, represents cyclopropanone. Grignard reagents like methylmagnesium chloride ($CH_3MgCl$) are strong nucleophiles that react with carbonyl compounds (aldehydes and ketones) via nucleophilic addition.

1. Nucleophilic Attack: The methyl group ($CH_3^-$) from $CH_3MgCl$ acts as a nucleophile and attacks the electrophilic carbon of the carbonyl group ($C=O$) in cyclopropanone. This causes the $\pi$-bond of the carbonyl group to break, and the electrons move to the oxygen atom, forming an alkoxide intermediate.

   \begin{center}
   $CH_3MgCl$ +
   \begin{tikzpicture}
     \draw (0,0) -- (1,0) -- (0.5,0.866) -- cycle;
     \node at (0.5,0.866) {C};
     \node at (0,0) {CH$_2$};
     \node at (1,0) {CH$_2$};
     \draw (0.5,0.866) -- (0.5,1.5);
     \node at (0.5,1.7) {O};
     \draw[dashed, ->] (0.5,1.5) -- (0.5,1.7);
     \node at (0.5,1.2) {\textbf{=}};
   \end{tikzpicture}
   $\longrightarrow$
   \begin{tikzpicture}
     \draw (0,0) -- (1,0) -- (0.5,0.866) -- cycle;
     \node at (0.5,0.866) {C};
     \node at (0,0) {CH$_2$};
     \node at (1,0) {CH$_2$};
     \draw (0.5,0.866) -- (0.5,1.5);
     \node at (0.5,1.7) {O$^-$Mg$^+$Cl};
     \draw (0.5,0.866) -- (1.2,0.866);
     \node at (1.4,0.866) {CH$_3$};
   \end{tikzpicture}
   \end{center}
   *(1-methylcyclopropyl)magnesium chloride (alkoxide intermediate)*
   
   

2. Hydrolysis: The alkoxide intermediate is then protonated upon hydrolysis (typically by adding water and acid, $H_3O^+$) to form the corresponding alcohol.

   \begin{center}
   \begin{tikzpicture}
     \draw (0,0) -- (1,0) -- (0.5,0.866) -- cycle;
     \node at (0.5,0.866) {C};
     \node at (0,0) {CH$_2$};
     \node at (1,0) {CH$_2$};
     \draw (0.5,0.866) -- (0.5,1.5);
     \node at (0.5,1.7) {O$^-$Mg$^+$Cl};
     \draw (0.5,0.866) -- (1.2,0.866);
     \node at (1.4,0.866) {CH$_3$};
   \end{tikzpicture}
   $+ H_2O/H^+$ $\longrightarrow$
   \begin{tikzpicture}
     \draw (0,0) -- (1,0) -- (0.5,0.866) -- cycle;
     \node at (0.5,0.866) {C};
     \node at (0,0) {CH$_2$};
     \node at (1,0) {CH$_2$};
     \draw (0.5,0.866) -- (0.5,1.5);
     \node at (0.5,1.7) {OH};
     \draw (0.5,0.866) -- (1.2,0.866);
     \node at (1.4,0.866) {CH$_3$};
   \end{tikzpicture}
   \end{center}
   *(1-methylcyclopropan-1-ol)*
   

The final product is 1-methylcyclopropan-1-ol.