Q.A carbonyl compound (A) has a carbonyl carbon that is doubly bonded to oxygen and also carries two different groups, a and b (both attached to the C=O carbon). A nucleophile Nu(-) adds to this carbonyl. Choose the correct representation(s) of the tetrahedral alkoxide intermediate that forms. (Two or more options may be correct.)

Nucleophilic Addition – From Intuition to Precision

Imagine you have a molecule with a carbon–oxygen double bond — a carbonyl group ($\ce{C=O}$). That oxygen is greedy for electrons; it pulls them away from carbon, leaving the carbon slightly positive ($\delta^+$) and the oxygen slightly negative ($\delta^-$). Now, if you bring a species that is rich in electrons (a nucleophile, meaning "nucleus-loving"), it will naturally be attracted to that electron-deficient carbon. The nucleophile attacks the carbon, the $\pi$ bond breaks, and the oxygen picks up a proton (or some other electrophile) to become stable. That, in a nutshell, is nucleophilic addition.

The Precise Statement

Nucleophilic addition is a reaction in which a nucleophile (an electron-rich species) forms a sigma bond with an electrophilic carbon atom of a polar multiple bond (typically a carbonyl group, $\ce{C=O}$, or a nitrile, $\ce{C≡N}$), while the $\pi$ bond breaks. The resulting intermediate then captures a proton (or another electrophile) to give a neutral product.

In general form:

The nucleophile ($\ce{Nu^-}$) attacks the carbonyl carbon; the oxygen becomes negatively charged; then a proton ($\ce{H^+}$) from the medium attaches to the oxygen, yielding an alcohol.

Why It Happens – The Driving Force

The carbonyl carbon is electrophilic because:

• Oxygen is more electronegative than carbon, so the $\ce{C=O}$ bond is polarised: $\ce{C^{\delta+}=O^{\delta-}}$.
• The $\pi$ bond is weaker than a $\sigma$ bond, so it can break relatively easily.

A nucleophile (like $\ce{OH^-}$, $\ce{CN^-}$, or $\ce{NH3}$) has a lone pair or a negative charge. It seeks positive centres. The attack forms a new $\sigma$ bond, and the $\pi$ electrons move entirely to oxygen, creating an alkoxide ion ($\ce{R2C-O^-}$). This intermediate is then quenched by a proton.

A Concrete Example – Addition of HCN to a Ketone

Take acetone ($\ce{CH3COCH3}$) and hydrogen cyanide ($\ce{HCN}$). In the presence of a base, $\ce{CN^-}$ (the nucleophile) attacks the carbonyl carbon:

The alkoxide intermediate then picks up a proton from $\ce{HCN}$ (or from water) to give a cyanohydrin:

The product is acetone cyanohydrin. Notice: two new sigma bonds formed ($\ce{C-CN}$ and $\ce{O-H}$), and the $\pi$ bond is gone.

What Makes a Good Nucleophile?

Strong nucleophiles are usually negatively charged or have lone pairs:

• $\ce{OH^-}$, $\ce{CN^-}$, $\ce{NH2^-}$, $\ce{CH3O^-}$, $\ce{H^-}$ (from hydride reagents like $\ce{NaBH4}$ or $\ce{LiAlH4}$)
• Neutral but polarisable: $\ce{NH3}$, $\ce{H2O}$ (weaker, but can add under acidic conditions) …