Question

Acetone can be converted into 2-methylpropan-2-ol using

• A. Pd H\(_2\)
• B. B\(_2\)H\(_6\) H\(_2\)O\(_2\), NaOH
• C. CH\(_3\)MgI H\(_2\)O
• D. LiAlH\(_4\)
• E. NaBH\(_4\)

Hint:

Grignard reagents add to carbonyls. With formaldehyde, they give 1\(^\circ\) alcohols; with other aldehydes, 2\(^\circ\) alcohols; with ketones, 3\(^\circ\) alcohols.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
The question asks for the synthesis of a tertiary alcohol (2-methylpropan-2-ol) from a ketone (acetone). This transformation involves adding an alkyl group to the carbonyl carbon and converting the carbonyl group into a hydroxyl group. This is characteristic of a Grignard reaction.
Step 2: Detailed Explanation:
Let's analyze the reaction and the reagents.
Starting Material: Acetone (propan-2-one), a ketone with the structure CH\(_3\)-CO-CH\(_3\).
Product: 2-methylpropan-2-ol, a tertiary alcohol with the structure (CH\(_3\))\(_3\)COH.
The transformation involves changing a C=O bond to a C-OH bond and adding one methyl (-CH\(_3\)) group.
Let's evaluate the given reagents:

(A) Pd / H\(_2\): This is a catalytic hydrogenation reagent. It reduces ketones to secondary alcohols. Acetone would be reduced to propan-2-ol, not 2-methylpropan-2-ol.

(B) B\(_2\)H\(_6\) / H\(_2\)O\(_2\), NaOH: This is the hydroboration-oxidation reagent. It reduces ketones to secondary alcohols, similar to NaBH\(_4\) or LiAlH\(_4\). It would also produce propan-2-ol.

(C) CH\(_3\)MgI / H\(_2\)O: This is a Grignard reagent (methylmagnesium iodide) followed by an acidic workup (hydrolysis). The Grignard reagent acts as a nucleophile (CH\(_3^-\)).
Step 1: The nucleophilic methyl group from CH\(_3\)MgI attacks the electrophilic carbonyl carbon of acetone. The \(\pi\)-bond of the C=O group breaks, and the electrons move to the oxygen atom, forming an alkoxide intermediate.
\[ \text{CH}_3\text{COCH}_3 + \text{CH}_3\text{MgI} \rightarrow (\text{CH}_3)_3\text{COMgI} \]
Step 2: Hydrolysis (addition of H\(_2\)O or H\(_3\)O\(^+\)) protonates the alkoxide to yield the final tertiary alcohol product, 2-methylpropan-2-ol.
\[ (\text{CH}_3)_3\text{COMgI} + \text{H}_2\text{O} \rightarrow (\text{CH}_3)_3\text{COH} + \text{Mg(OH)I} \]
This sequence correctly produces the desired product.

(D) LiAlH\(_4\) and (E) NaBH\(_4\): These are strong and mild reducing agents, respectively. They reduce ketones by adding a hydride ion (H\(^-\)), converting them to secondary alcohols. They would also convert acetone to propan-2-ol.

Step 3: Final Answer:
The correct reagent to convert acetone to 2-methylpropan-2-ol is the Grignard reagent CH\(_3\)MgI followed by hydrolysis. This corresponds to option (C).