- A chemist plans to synthesize 3-phenyl-3-pentanol. Three of the proposed routes are viable. Which one of the following synthetic routes will FAIL to produce the desired tertiary alcohol?
Answer: 3-Oxo-3-phenylpropanoic acid treated with excess ethylmagnesium bromide, followed by workup.
The starting material in route C, 3-oxo-3-phenylpropanoic acid, contains a highly acidic carboxylic acid proton (pKa ≈ 4-5). The strongly basic Grignard reagent will be consumed by an irreversible acid-base reaction with this proton before it can act as a nucleophile at the ketone carbonyl. The other three routes are all valid disconnections for synthesizing the target tertiary alcohol.
- Both methyllithium and methylmagnesium bromide are effective for adding a methyl group to cyclohexanone. However, in which of the following reactions would methyllithium be expected to give a significantly different major product or reaction outcome compared to methylmagnesium bromide?
Answer: Reaction with 1-hexyne.
The primary difference between organolithium and Grignard reagents in an introductory context is basicity; organolithiums are substantially more basic. A terminal alkyne like 1-hexyne has a weakly acidic proton (pKa ≈ 25). While a Grignard reagent can deprotonate it, the reaction may be slow or incomplete. Methyllithium, a much stronger base, will rapidly and quantitatively deprotonate the alkyne in an acid-base reaction. In the other cases (A, B, D), both reagents act primarily as nucleophiles and give analogous addition products.
- A student attempts to prepare a Grignard reagent from 4-bromoaniline in anhydrous THF by reacting it with magnesium turnings. The reaction does not produce the expected Grignard reagent. What is the most likely reason for this failure?
Answer: Any Grignard reagent that forms is immediately destroyed by the acidic N-H protons of another 4-bromoaniline molecule.
This is an example of an incompatible functional group during the formation of a Grignard reagent. The amine (-NH₂) group has acidic protons. Grignard reagents are powerful bases. If any molecule of the 4-bromoaniline Grignard reagent were to form, it would immediately encounter another molecule of the starting material and be quenched in an acid-base reaction. This prevents any significant concentration of the desired reagent from accumulating.
- A student attempts to synthesize 4-phenyl-2-butanol from 4-phenyl-2-butanone using methylmagnesium bromide (CH₃MgBr). Prior to the reaction, the ketone is purified via column chromatography using a solvent system containing a trace amount of water. The collected fractions are immediately used. The reaction fails to yield any significant amount of the desired alcohol. What is the most likely cause for this failure?
Answer: Trace water from the chromatography quenched the Grignard reagent through an acid-base reaction.
Grignard reagents are extremely strong bases and are highly sensitive to protic substances. Any trace amount of water (pKa ≈ 15.7) will rapidly and irreversibly react with the Grignard reagent (the conjugate acid of which is methane, pKa ≈ 50) in an acid-base reaction, destroying the nucleophile before it can react with the ketone.
- In an attempt to synthesize triphenylmethanol, a student adds exactly 1.0 equivalent of phenylmagnesium bromide to a solution of ethyl benzoate. After aqueous workup, the student finds that the major product isolated is benzophenone, not the desired tertiary alcohol. What is the most plausible explanation for this result?
Answer: The stoichiometry was incorrect; two equivalents of Grignard reagent are needed to convert the ester to the tertiary alcohol.
The reaction of an ester with a Grignard reagent produces a ketone as an intermediate. This ketone is also reactive towards the Grignard reagent. To form the tertiary alcohol, two full equivalents of the Grignard reagent are required: one to convert the ester to the ketone, and a second to convert the ketone to the tertiary alcohol. Using only one equivalent will primarily result in the formation of the ketone intermediate.
- Which of the following reaction combinations is a valid and efficient synthesis of 3-ethyl-3-hexanol following an appropriate aqueous workup?
Answer: 3-hexanone with ethylmagnesium bromide
The target, 3-ethyl-3-hexanol, is a tertiary alcohol with two ethyl groups and one propyl group attached to the carbinol carbon. Retrosynthetic analysis shows two possible ketone/Grignard pairs: 3-hexanone (an ethyl group and a propyl group on the carbonyl) reacting with an ethyl Grignard, or diethyl ketone (two ethyl groups on the carbonyl) reacting with a propyl Grignard. Option A matches the first valid disconnection.
- The reaction of di-tert-butyl ketone with isopropylmagnesium bromide results in a very low yield of the expected tertiary alcohol. The major product is 2,2,4,4-tetramethyl-3-pentanol. This transformation is an example of the Grignard reagent playing an alternative mechanistic role. What is that role?
Answer: Acting as a hydride donor to cause reduction of the ketone.
When a sterically hindered ketone reacts with a Grignard reagent that possesses β-hydrogens (like isopropylmagnesium bromide), nucleophilic addition can be slow. A competing pathway, reduction, can become dominant. The Grignard reagent delivers a hydride (H⁻) from its β-position to the carbonyl carbon via a six-membered ring transition state, reducing the ketone to a secondary alcohol. The Grignard reagent is converted to propene.
- A sample of benzaldehyde is specially prepared such that its carbonyl oxygen is isotopically labeled with ¹⁸O. This labeled compound is treated with excess methylmagnesium bromide, followed by a workup with normal water (H₂¹⁶O). What is the location of the ¹⁸O label in the resulting 1-phenylethanol product?
Answer: The ¹⁸O label is found exclusively in the hydroxyl group of the 1-phenylethanol.
In a Grignard reaction, the nucleophile adds to the carbonyl carbon, and the C=O π-bond breaks, placing a negative charge on the oxygen. This oxygen atom, which was originally the carbonyl oxygen (¹⁸O), is then protonated during the workup step. The oxygen atom itself is never exchanged. The proton comes from the water, but the oxygen atom of the resulting hydroxyl group is the same oxygen atom that was in the starting carbonyl.
- When optically pure (R)-3-methyl-2-pentanone is treated with excess ethylmagnesium bromide followed by an aqueous workup, a new stereocenter is created at the carbonyl carbon. Which statement best describes the resulting product mixture?
Answer: A mixture of diastereomers is formed in unequal amounts.
The starting ketone is chiral. The nucleophilic attack of the Grignard reagent can occur from either the Re or Si face of the prochiral carbonyl, creating a new stereocenter. Because the existing stereocenter creates a chiral environment, the transition states for attack from the two faces are diastereomeric and have different energies. This results in a mixture of two diastereomeric products, formed in an unequal ratio.
- A Grignard reagent is prepared from 1-bromo-3-methylbenzene (m-bromotoluene) and magnesium metal. This reagent is then poured over crushed dry ice (solid CO₂), and the resulting mixture is acidified with dilute H₂SO₄. What is the final major organic product?
Answer: 3-Methylbenzoic acid
Grignard reagents act as strong nucleophiles and add to the electrophilic carbon of carbon dioxide. This forms a magnesium carboxylate salt. Subsequent acidification (workup) protonates the carboxylate to yield a carboxylic acid. In this case, the m-tolyl Grignard reagent forms m-toluic acid (3-methylbenzoic acid).
- A student is planning a synthesis that uses n-butyllithium (an organolithium reagent) as a strong base and nucleophile. Which of the following would be an inappropriate choice of solvent for this reaction?
Answer: Ethanol
Organolithium reagents, like Grignard reagents, are extremely strong bases. They will react destructively with any protic solvent. Ethanol (CH₃CH₂OH) has an acidic hydroxyl proton that will be readily abstracted by n-butyllithium, quenching the reagent and forming butane and lithium ethoxide. Aprotic solvents like alkanes (pentane) and ethers (diethyl ether, THF) are required.
- During the reaction of acetone with methylmagnesium bromide (CH₃MgBr) in anhydrous ether, a key intermediate is formed. Which structure represents this intermediate that exists in the solution immediately before the addition of aqueous acid?
Answer: The magnesium salt of tert-butoxide, (CH₃)₃C-O⁻ MgBr⁺
The nucleophilic carbon of the Grignard reagent attacks the electrophilic carbonyl carbon of acetone, pushing the pi electrons onto the oxygen. This forms a tetrahedral intermediate which is an alkoxide. In the ether solution, this alkoxide is stabilized as a magnesium salt, specifically a halomagnesium alkoxide. The final alcohol is not formed until the proton source (aqueous acid) is added in a separate workup step.
- The reaction of 4-tert-butylcyclohexanone with methylmagnesium bromide, followed by aqueous workup, yields two separable, stereoisomeric alcohol products. Which statement correctly describes the relationship between these two products?
Answer: They are diastereomers, resulting from nucleophilic attack on the two different faces of the carbonyl group.
The large tert-butyl group effectively locks the cyclohexane ring in a single chair conformation. The carbonyl group is planar, and the methyl Grignard can attack from either the axial face or the equatorial face. Axial attack leads to an equatorial alcohol, and equatorial attack leads to an axial alcohol. These two products have the same connectivity but different 3D arrangements of atoms, and they are not mirror images of each other. Therefore, they are diastereomers.
- Propylene oxide (methyloxirane) is treated with phenylmagnesium bromide in THF, followed by an acidic workup. What is the major product of this regioselective reaction?
Answer: 1-Phenyl-2-propanol
Under basic or neutral conditions (like with a Grignard reagent), the nucleophile attacks an epoxide at the less sterically hindered carbon atom in an Sₙ2-like fashion. In propylene oxide, the methyl group is on C2, so the less hindered carbon is C1. Attack by the phenyl nucleophile at C1, followed by protonation of the resulting alkoxide at O, yields 1-phenyl-2-propanol.
