(56) \ CH\_3MgCl + =O \longrightarrow 1:1 | Chemistry - Reactions of Ketones, Grignard Reagents | SolveeIt

Question

$(56) \ CH_3MgCl + =O \longrightarrow$ $1:1$

Answer

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

Explanation

Solution

The reaction involves the addition of a Grignard reagent, methylmagnesium chloride ( $CH_3MgCl$ ), to a ketone, cyclopropanone. Grignard reagents are strong nucleophiles and react with carbonyl compounds to form alcohols after hydrolysis.

Reaction Mechanism:

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)

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, which is a tertiary alcohol.

Explanation of the solution:

Methylmagnesium chloride ( $CH_3MgCl$ ) reacts with cyclopropanone via nucleophilic addition. The methyl group ( $CH_3^-$ ) attacks the carbonyl carbon, forming an alkoxide intermediate. Subsequent hydrolysis of this intermediate yields 1-methylcyclopropan-1-ol.

Question: $(56) \ CH_3MgCl + =O \longrightarrow$ $1:1$...

Solution