Given below are two statements.
Statement I: The nucleophilic addition of sodium hydrogen sulphite t

Q: Why does the proton transfer happen in the bisulphite addition?

The initial intermediate has a negatively charged oxygen (alkoxide) and an acidic proton on the sulphite group. The sulfonate ion ($\text{-SO}_3^-$) is highly resonance-stabilized compared to the alkoxide ion. Therefore, the proton rapidly transfers to the oxygen to form the more thermodynamically stable bisulphite adduct.

Q: Can the cyanohydrin be converted into an amine?

Yes, but not directly in this single addition step! If you take the cyanohydrin and reduce the nitrile group ($\text{-CN}$) using a strong reducing agent like $\text{LiAlH}_4$, it will convert into a primary amine ($\text{-CH}_2\text{NH}_2$). However, the addition reaction itself only yields the cyanohydrin.

Q: Why is the bisulphite addition reaction important in chemistry?

It's a classic purification technique! The bisulphite adduct is a crystalline solid that is water-soluble. You can react an impure aldehyde/ketone with $\text{NaHSO}_3$, crystallize the adduct to purify it, and then regenerate the original carbonyl compound by treating it with dilute acid or base.

Q: Do all ketones undergo the bisulphite addition?

No, steric hindrance plays a big role. While most aldehydes and methyl ketones readily form bisulphite adducts, bulky ketones (like diethyl ketone or benzophenone) do not react well because the bulky $\text{-SO}_3^-$ group cannot easily approach the crowded carbonyl carbon.

Q: Is the addition of HCN reversible?

Yes, cyanohydrin formation is reversible. In the presence of a strong base, the hydroxyl proton is removed, and the cyanide ion can be expelled, regenerating the original carbonyl compound. This reversibility is often utilized in synthetic organic chemistry.

Question

Given below are two statements.
Statement I The nucleophilic addition of sodium hydrogen sulphite to an aldehyde or a ketone involves proton transfer to form a stable ion.
Statement II The nucleophilic addition of hydrogen cyanide to an aldehyde or a ketone yields amine as final product.
In the light of the above statements, choose the most appropriate answer from the options given below.

Options:

(A) Both statement I and statement II are true.

(B) Statement I is false but statement II is true.

(C) Statement I is true but statement II is false.

(D) Both statement I and statement II are false.

Sigma Percentile · Accepted Answer

### The Magic of the Carbonyl Group

The carbonyl group ($>	ext{C=O}$) is one of the most fascinating and versatile functional groups in organic chemistry. Because oxygen is significantly more electronegative than carbon, the carbon-oxygen double bond is highly polarized. The oxygen atom pulls electron density towards itself, acquiring a partial negative charge ($\delta^-$), while leaving the carbon atom with a partial positive charge ($\delta^+$).

This electron deficiency makes the carbonyl carbon highly susceptible to attack by nucleophiles (electron-rich species). In this problem, we are evaluating two classic nucleophilic addition reactions of aldehydes and ketones to determine the validity of the given statements.

### Analyzing Statement I: The Bisulphite Addition

Statement I claims that the nucleophilic addition of sodium hydrogen sulphite ($	ext{NaHSO}_3$) to an aldehyde or ketone involves a proton transfer to form a stable ion. Let's break down the mechanism to see if this holds true.

When sodium bisulphite is added to a carbonyl compound, the active nucleophile is the bisulphite ion ($	ext{HSO}_3^-$). The sulfur atom, bearing a lone pair of electrons, attacks the electrophilic carbonyl carbon. Simultaneously, the $\pi$ electrons of the carbon-oxygen double bond shift entirely onto the oxygen atom.

This initial attack forms an intermediate where the oxygen atom carries a full negative charge (an alkoxide ion), and the sulfur atom is bonded to the carbon, still retaining its acidic proton ($	ext{-SO}_3	ext{H}$).

**The Crucial Proton Transfer**

Here is where thermodynamics takes the wheel. The intermediate we just formed has an alkoxide ion ($	ext{-O}^-$) and a sulfonic acid group ($	ext{-SO}_3	ext{H}$). Alkoxide ions are relatively strong bases, while sulfonic acids are strong acids.

To achieve a lower energy state, a rapid intramolecular proton transfer occurs. The acidic proton from the $	ext{-SO}_3	ext{H}$ group jumps to the negatively charged alkoxide oxygen.

Why does this happen? The resulting sulfonate ion ($	ext{-SO}_3^-$) is incredibly stable because its negative charge is delocalized over three highly electronegative oxygen atoms via resonance. The alkoxide ion, on the other hand, has its negative charge localized on a single oxygen atom. Nature always favors the more stable, resonance-stabilized structure.

Thus, the proton transfer is a vital step that drives the formation of the stable bisulphite adduct. **Statement I is absolutely true.**

### Analyzing Statement II: The Cyanide Addition

Statement II asserts that the nucleophilic addition of hydrogen cyanide ($	ext{HCN}$) to an aldehyde or ketone yields an amine as the final product. Let's trace the reaction pathway.

In the presence of a base catalyst (which generates the strong nucleophile $	ext{CN}^-$), the cyanide ion attacks the electrophilic carbonyl carbon. Just like in the previous reaction, the $\pi$ electrons shift to the oxygen, forming an alkoxide intermediate.

In the next step, this alkoxide intermediate abstracts a proton from the solvent or from an undissociated $	ext{HCN}$ molecule. The final product has a hydroxyl group ($	ext{-OH}$) and a cyano group ($	ext{-CN}$) attached to the exact same carbon atom.

**What is this product called?**

Compounds containing a hydroxyl group and a cyano group on the same carbon are known as **cyanohydrins**.

An amine, by definition, is an organic derivative of ammonia containing a basic nitrogen atom with a lone pair, typically represented as $	ext{-NH}_2$, $	ext{-NHR}$, or $	ext{-NR}_2$. Our product, the cyanohydrin, does not contain an amino group.

While it is true that the nitrile group ($	ext{-CN}$) of a cyanohydrin can be *subsequently* reduced using strong reducing agents (like $	ext{LiAlH}_4$) to form a primary amine, the direct nucleophilic addition of $	ext{HCN}$ strictly yields a cyanohydrin, not an amine. Therefore, **Statement II is completely false.**

### Final Conclusion

By carefully analyzing the reaction mechanisms, we have established that the bisulphite addition relies on a stabilizing proton transfer, making Statement I true. Conversely, the addition of $	ext{HCN}$ produces a cyanohydrin, not an amine, making Statement II false.

Matching our findings with the given options, the correct choice is **(c) Statement I is true but statement II is false.**

Select Answer:

Visualized Solution

Analyzing the Statements

Addition of $NaHSO_{3}$

Proton Transfer

Addition of $HCN$

Conclusion

The Sigma Insight: Carbonyl Compounds

The Magic of the Carbonyl Group

Analyzing Statement I

Analyzing Statement II

Final Conclusion

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