Ketone may react with nucleophilic hydride ion source (e.g. LiAIH4 or H-) and subsequently, a proton source (e.g. H2O or H+) to give a secondary alcohol: (a) For the first step of the reaction, highlight (in the structure below) the atom that is attacked by the hydride ion source. Highlight (in the structure below) the atom that will ultimately (i.e., in the second step of the reaction) be protonated by H+. Click on an atom to select it; click again to deselect. A selected atom will be colored green.
Here is the solution based on the principles of organic chemistry.
Correct Answer
(a) For the first step of the reaction, highlight (in the structure below) the atom that is attacked by the hydride ion source, H⁻.
In this step, the carbonyl carbon is attacked by the hydride.
(b) Highlight (in the structure below) the atom that will ultimately (i.e., in the second step of the reaction) be protonated by H⁺.
In this step, the carbonyl oxygen is protonated.
Explanation
The reaction shown is the reduction of a ketone to a secondary alcohol, a fundamental transformation in organic chemistry that proceeds via a two-step nucleophilic addition mechanism.
Part (a): The Site of Nucleophilic Attack
The first step of the reaction involves the attack of a nucleophile, the hydride ion (H⁻), on the ketone. To identify the site of attack, we must analyze the electronic properties of the ketone’s carbonyl group (C=O).
Oxygen is significantly more electronegative than carbon. This difference in electronegativity causes the electron density in the carbon-oxygen double bond to be polarized towards the oxygen atom. As a result, the carbonyl oxygen atom carries a partial negative charge (δ⁻), while the carbonyl carbon atom bears a partial positive charge (δ⁺).
This electron-deficient (δ⁺) carbonyl carbon is an electrophilic center, making it susceptible to attack by electron-rich species, known as nucleophiles. The hydride ion (H⁻), provided by sources like lithium aluminum hydride (LiAlH₄) or sodium borohydride (NaBH₄), is a potent nucleophile. It donates its pair of electrons to the electrophilic carbonyl carbon, forming a new carbon-hydrogen (C-H) bond. Simultaneously, to accommodate the new bond and maintain carbon’s octet, the weaker π-bond of the carbonyl group breaks, and its electron pair moves onto the electronegative oxygen atom. This step results in the formation of a tetrahedral alkoxide intermediate.
Part (b): The Protonation Step
The first step yields an alkoxide ion (R₂CHO⁻), an intermediate where the oxygen atom carries a full negative charge and has three lone pairs of electrons. This makes the oxygen atom highly basic and nucleophilic.
The second step of the reaction is an acid-base neutralization. A proton source, such as water (H₂O) or an acid (H⁺), is added to the reaction mixture. The negatively charged, basic oxygen atom of the alkoxide intermediate readily donates one of its lone pairs to a proton (H⁺) from the source. This forms a new oxygen-hydrogen (O-H) bond, neutralizing the charge and yielding the final product: a stable, neutral secondary alcohol. Therefore, the carbonyl oxygen of the initial ketone is the atom that is ultimately protonated in the second step.
