The reagents that can be used to convert benzenediazonium chloride to benzene are __. (Two or more options may

NCERT Exemplar · Q62

Q. The reagents that can be used to convert benzenediazonium chloride to benzene are ____. (Two or more options may be correct.)

(i)

SnCl_2

/HCl

(ii)

CH_3CH_2OH

(iii)

H_3PO_2

(iv)

LiAlH_4

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Concept understanding — Nucleophilic Substitution Reactions

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.

Note

The key idea: a nucleophile adds across a polar multiple bond (usually $\ce{C=O}$ or $\ce{C≡N}$ ), breaking the $\pi$ bond and forming two new sigma bonds.

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:

\ce{R2C=O + Nu^- ->[H^+] R2C(OH)Nu}

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.

Watch out

A common mistake: thinking the nucleophile attacks the oxygen. No — oxygen is already electron-rich; the nucleophile goes to the carbon because it is electron-deficient.

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:

\ce{CH3COCH3 + CN^- -> CH3C(O^-)(CN)CH3}

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

\ce{CH3C(O^-)(CN)CH3 + H^+ -> CH3C(OH)(CN)CH3}

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) …

Why this formula?

Nucleophilic Substitution Reactions: Why the Key Formulas Hold

Nucleophilic substitution reactions are a cornerstone of organic chemistry. Instead of just memorizing the rate laws, let's understand why they arise from the molecular events.

1. The Two Main Mechanisms: A Tale of Timing

The key formulas (rate laws) for nucleophilic substitution come directly from how many molecules are involved in the rate-determining step (RDS) — the slowest step that controls the overall reaction speed.

SN1: Unimolecular — The Leaving Group Goes First

The Rate Law:

\text{Rate} = k[\text{RX}]

Why?

The reaction happens in two steps:

Slow step (RDS): The C–X bond breaks spontaneously, forming a carbocation intermediate. Only the substrate ( $\text{RX}$ ) is involved.

\text{RX} \xrightarrow{\text{slow}} \text{R}^+ + \text{X}^-

Fast step: The nucleophile ( $\text{Nu}^-$ ) attacks the carbocation.

\text{R}^+ + \text{Nu}^- \xrightarrow{\text{fast}} \text{RNu}

Since the slow step depends only on the concentration of $\text{RX}$ , the rate law has no dependence on $[\text{Nu}^-]$ . The nucleophile arrives after the carbocation is formed — it cannot affect the speed of the first step.

Key insight: The rate is determined by how easily the leaving group leaves, not by how fast the nucleophile attacks.

SN2: Bimolecular — Simultaneous Attack and Departure

The Rate Law:

\text{Rate} = k[\text{RX}][\text{Nu}^-]

Why?

The reaction occurs in one concerted step:

The nucleophile attacks the carbon from one side at the same time as the leaving group departs from the opposite side.
Both $\text{RX}$ and $\text{Nu}^-$ must collide with the correct orientation and sufficient energy.

The rate depends on the frequency of productive collisions between the two molecules. This is directly proportional to the product of their concentrations:

\text{Rate} \propto [\text{RX}] \times [\text{Nu}^-]

Key insight: Both partners are involved in the transition state simultaneously — if either is missing, the reaction cannot proceed.

2. The Transition State: Why the Formulas Are Not Just "Given"

For SN2, the transition state has a pentavalent carbon (five bonds partially formed/broken). The energy barrier depends on steric hindrance — bulkier groups around carbon make it harder for the nucleophile to approach, which is why SN2 is favored at primary carbons. …

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Benzenediazonium chloride is reduced to benzene by replacing the diazonium group ( $-N_2^+$ ) with a hydrogen atom. The reagents that can accomplish this are $CH_3CH_2OH$ (ethanol) and $H_3PO_2$ (hypophosphorous acid). So the correct options are (ii) and (iii).

The key to this question lies in understanding the chemistry of the diazonium group — specifically, how it can be replaced by a hydrogen atom. The diazonium group ( $-N_2^+$ ) is an excellent leaving group because nitrogen gas ( $N_2$ ) is extremely stable and escapes as a gas, driving the reaction forward. To get benzene, you need a reagent that can donate a hydride ion ( $H^-$ ) or a hydrogen atom to the carbon that loses the $N_2^+$ group.

Let’s examine each option carefully.

Option (i): $SnCl_2/HCl$

This is a classic reducing agent, but its specialty is reducing nitro groups ( $-NO_2$ ) to amino groups ( $-NH_2$ ). For example, it converts nitrobenzene to aniline. However, it does not replace the diazonium group with hydrogen. Instead, $SnCl_2/HCl$ can reduce the diazonium salt to phenylhydrazine ( $C_6H_5NHNH_2$ ) under certain conditions — not benzene. So this is incorrect for our purpose.
Option (ii): $CH_3CH_2OH$ (ethanol)

Ethanol is a mild reducing agent here. When benzenediazonium chloride is warmed with ethanol, the diazonium group is replaced by hydrogen, and ethanol itself gets oxidized to acetaldehyde. The reaction is:

C_6H_5N_2^+Cl^- + CH_3CH_2OH \rightarrow C_6H_6 + N_2 + CH_3CHO + HCl

This is a well-known method for converting diazonium salts to arenes. Ethanol donates a hydride-like species (or a hydrogen atom via a radical mechanism) to the aryl carbon. So this works.

Option (iii): $H_3PO_2$ (hypophosphorous acid) This is perhaps the most famous reagent for this specific transformation. Hypophosphorous acid is an excellent reducing agent for diazonium salts. The reaction proceeds as:

C_6H_5N_2^+Cl^- + H_3PO_2 + H_2O \rightarrow C_6H_6 + N_2 + H_3PO_3 + HCl

Here, $H_3PO_2$ acts as a source of hydride ion ( $H^-$ ), replacing the diazonium group cleanly. This is a standard method in organic synthesis to remove an amino group (via diazotization and reduction) from an aromatic ring. So this is correct.

Option (iv): $LiAlH_4$ (lithium aluminium hydride) …

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Common Mistakes & How to Avoid Them

Concept: Reduction of Benzenediazonium Chloride to Benzene

The key reaction is the replacement of the diazonium group (–N₂⁺) with a hydrogen atom (–H). This is a reductive deamination or dediazoniation reaction.

Mistake 1: Confusing Reduction with Diazonium Coupling

The error: Students think $SnCl_2/HCl$ (option A) reduces the diazonium group to benzene.

Why it's wrong: $SnCl_2/HCl$ is a reducing agent, but it reduces the nitro group (–NO₂) to an amino group (–NH₂) — not the diazonium group. It does not replace –N₂⁺ with –H.

How to avoid:

Remember: $SnCl_2/HCl$ is used to prepare aniline from nitrobenzene, not to remove a diazonium group.
For dediazoniation, you need reagents that donate a hydride (H⁻) or act as a source of hydrogen under mild conditions.

Mistake 2: Thinking $LiAlH_4$ (option D) Works Here

The error: Students assume a strong reducing agent like $LiAlH_4$ will reduce the diazonium salt to benzene.

Why it's wrong: $LiAlH_4$ is a powerful hydride donor, but it reacts violently with water and protic solvents. Diazonium salts are prepared and used in aqueous acidic medium. $LiAlH_4$ would decompose before reacting with the diazonium group.

How to avoid:

Check the reaction medium: Diazonium reactions are in water. $LiAlH_4$ is incompatible with water.
Use only reagents that are stable in aqueous acid for this conversion.

Mistake 3: Forgetting $H_3PO_2$ (Hypophosphorous Acid) is the Classic Reagent

The error: Students overlook option (C) $H_3PO_2$ because they don't recall its specific role.

Why it's correct: $H_3PO_2$ is the standard reagent for replacing –N₂⁺ with –H. It acts as a hydride donor in acidic medium.

How to avoid:

Memorise: $H_3PO_2$ + diazonium salt → benzene is a named reaction (often called dediazoniation).
It's a mild, water-compatible reducing agent — perfect for this job.

Mistake 4: Thinking Ethanol ( $CH_3CH_2OH$ , option B) Cannot Do This

The error: Students dismiss ethanol because it's not a "strong" reducing agent. …

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Q.The reagents that can be used to convert benzenediazonium chloride to benzene are ____.
(Two or more options may be correct.)

Concept understanding — Nucleophilic Substitution Reactions

Nucleophilic Addition – From Intuition to Precision

The Precise Statement

Why It Happens – The Driving Force

A Concrete Example – Addition of HCN to a Ketone

What Makes a Good Nucleophile?

Why this formula?

Nucleophilic Substitution Reactions: Why the Key Formulas Hold

1. The Two Main Mechanisms: A Tale of Timing

SN1: Unimolecular — The Leaving Group Goes First

SN2: Bimolecular — Simultaneous Attack and Departure

2. The Transition State: Why the Formulas Are Not Just "Given"

Concept: Reduction of Diazonium Salts to Arenes

Method: Reduction by Hypophosphorous Acid (or Ethanol)

Steps:

Common Mistakes & How to Avoid Them

Concept: Reduction of Benzenediazonium Chloride to Benzene

Mistake 1: Confusing Reduction with Diazonium Coupling

Mistake 2: Thinking $LiAlH_4$ (option D) Works Here

Mistake 3: Forgetting $H_3PO_2$ (Hypophosphorous Acid) is the Classic Reagent

Mistake 4: Thinking Ethanol ( $CH_3CH_2OH$ , option B) Cannot Do This

Q.The reagents that can be used to convert benzenediazonium chloride to benzene are ____. (Two or more options may be correct.)

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Q.The reagents that can be used to convert benzenediazonium chloride to benzene are ____.
(Two or more options may be correct.)

Unlock everything free for 14 days