Animated Solution for Chemistry - Organic Chemistry: $C{H}_{3}C{H}_2-\underset{Ph}{\stackrel{OH}{C}}-C{H}_3$ cannot be prepared by

Mastering Grignard Synthesis

Imagine you are an architect, but instead of buildings, you are constructing complex organic molecules. One of the most powerful tools in your molecular toolbox is the Grignard Reaction. It allows you to forge new carbon-carbon bonds with incredible precision. But what happens when you are given the final building and asked to figure out which bricks were used to build it? This is the essence of retrosynthetic analysis.

In this problem, we are presented with a specific target molecule: $C{H}_{3}C{H}_2-\underset{Ph}{\stackrel{OH}{C}}-C{H}_3$, which is 2-phenylbutan-2-ol. Our mission is to determine which combination of a carbonyl compound and a Grignard reagent cannot be used to synthesize this molecule.

Degree of Alcohols

Before we dive into the options, let's establish a fundamental rule of Grignard chemistry. When a Grignard reagent ($RMgX$) attacks a carbonyl group ($C=O$), the degree of the resulting alcohol depends entirely on the starting carbonyl compound:

1. Formaldehyde ($HCHO$) always yields a primary ($1^{\circ }$) alcohol. 2. Other Aldehydes ($RCHO$) yield secondary ($2^{\circ }$) alcohols. 3. Ketones ($RCO{R}^{\prime }$) yield tertiary ($3^{\circ }$) alcohols.

Take a close look at our target molecule. The carbon atom bonded to the $-OH$ group is also bonded to three other carbon groups: a methyl group ($-C{H}_3$), an ethyl group ($-C{H}_{2}C{H}_3$), and a phenyl group ($-Ph$). Therefore, our target is a tertiary alcohol.

To build a tertiary alcohol, we must start with a ketone. Because there are three different groups attached to the central carbon, there are exactly three ways to "disconnect" or break the molecule apart to find the starting materials.

Disconnection 1 (Option a): If we disconnect the phenyl group, we are left with butanone ($C{H}_{3}C{H}_{2}COC{H}_3$) and phenylmagnesium halide ($PhMgX$). When these react, the phenyl nucleophile attacks the ketone, perfectly forming our target tertiary alcohol.

Disconnection 2 (Option b): If we disconnect the ethyl group, the precursors are acetophenone ($PhCOC{H}_3$) and ethylmagnesium halide ($C{H}_{3}C{H}_{2}MgX$). This combination also flawlessly reconstructs our target molecule.

Disconnection 3 (Option c): If we disconnect the methyl group, we need propiophenone ($PhCOC{H}_{2}C{H}_3$) and methylmagnesium halide ($C{H}_{3}MgX$). Once again, this is a valid pathway to our tertiary alcohol.

Formaldehyde

Now, let's examine Option (d). It proposes reacting formaldehyde ($HCHO$) with a rather bulky Grignard reagent, $PhCH(C{H}_3)C{H}_{2}MgX$.

Remember our golden rule? Formaldehyde has two hydrogen atoms attached to its carbonyl carbon. No matter how massive or complex the Grignard reagent is, it will only add one carbon group to that central carbon. The resulting structure will be $R-C{H}_2-OH$.

In this specific case, the product would be $PhCH(C{H}_3)C{H}_{2}C{H}_{2}OH$, which is 3-phenylbutan-1-ol. This is a primary alcohol, not the tertiary alcohol we desire.

By simply identifying the degree of the target alcohol and recalling the fundamental rules of Grignard additions, we can instantly spot the imposter. Option (d) is the only combination that fails to produce a tertiary alcohol, making it the correct answer to our question. Always let the structure guide your logic!