The reaction $Ar{-}\overset{+}{N}_2Cl^- \xrightarrow{Cu/HCl} ArCl + N_2 + CuCl$ is named as ____. (i) Sandmeyer reaction (ii) Gatterman reaction (iii) Claisen reaction (iv) Carbylamine reaction

The reaction Ar-overset+N₂Cl^- xrightarrowCu/HCl ArCl + N₂ + CuCl is named as __. (i) Sandmeyer reaction (ii)

NCERT Exemplar · Q20

Ar{-}\overset{+}{N}_2Cl^- \xrightarrow{Cu/HCl} ArCl + N_2 + CuCl

(i) Sandmeyer reaction

(ii) Gatterman reaction

(iii) Claisen reaction

(iv) Carbylamine reaction

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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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The reaction of an aryl diazonium chloride with Cu/HCl to form aryl chloride is the Sandmeyer reaction, where copper(I) chloride acts as a catalyst to replace the diazonium group with chlorine.

This is a classic naming reaction from organic chemistry, specifically from the chapter on amines and diazonium salts. The key is to recognise the reagent and the transformation.

1. What the reaction shows

You start with an aryl diazonium chloride ( $Ar{-}\overset{+}{N}_2Cl^-$ ) — a compound where a nitrogen triple-bonded to another nitrogen is attached to an aromatic ring, with a chloride counterion. When you treat it with copper(I) chloride in the presence of hydrochloric acid (written as Cu/HCl), you get an aryl chloride ( $ArCl$ ), nitrogen gas ( $N_2$ ), and copper(I) chloride is regenerated.

The copper(I) chloride here is a catalyst — it participates in the reaction but is re-formed at the end.

2. Why this is the Sandmeyer reaction

The Sandmeyer reaction is specifically the substitution of a diazonium group by a halogen (or other groups like CN, OH, etc.) using a copper(I) halide as the catalyst. The general form is:

ArN_2^+ X^- \xrightarrow{CuX} ArX + N_2

where $X$ is the halogen. In your question, $X = Cl$ , and the reagent is CuCl/HCl — exactly the Sandmeyer conditions.

Tip

A quick way to remember: Sandmeyer uses copper(I) salts (CuCl, CuBr, CuCN). Gatterman uses copper powder and HX — a different, less efficient method.

3. Distinguishing from the other options

Let's check each option:

(i) Sandmeyer reaction — Correct, as explained.
(ii) Gatterman reaction — This also converts diazonium salts to aryl halides, but uses copper powder (not CuCl) and HX. The reaction is $ArN_2^+Cl^- \xrightarrow{Cu, HCl} ArCl$ . It's a variation, but the reagent given is Cu/HCl, which is ambiguous — however, the standard name for CuCl/HCl is Sandmeyer.
(iii) Claisen reaction — This refers to the Claisen condensation (ester condensation) or the Claisen rearrangement (allyl phenyl ether rearrangement). Nothing to do with diazonium salts.
(iv) Carbylamine reaction — This is the test for primary amines: $RNH_2 + CHCl_3 + KOH \rightarrow RNC$ (isocyanide). Again, unrelated. …

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Concept: Diazonium Salt Substitution Reactions

This question tests your understanding of replacement reactions of diazonium salts — specifically, how the aryl diazonium group ( $-N_2^+$ ) can be replaced by a halogen using different copper-based reagents.

Method: Reagent-Product Matching

Steps:

Identify the reactant and product
- Reactant: Aryl diazonium chloride ( $Ar{-}\overset{+}{N}_2Cl^-$ )
- Product: Aryl chloride ( $ArCl$ ) + $N_2$ gas + $CuCl$
Recall the two key copper-catalyzed reactions that replace $-N_2^+$ with $-Cl$ :

Reaction Reagent Byproduct
Sandmeyer $CuCl$ (catalytic) + $HCl$ $CuCl$ is regenerated
Gatterman $Cu$ powder + $HCl$ $CuCl$ is formed as a product
Match the given equation
- The equation shows $Cu/HCl$ as reagent and $CuCl$ as a product (not regenerated). …

Reaction	Reagent	Byproduct
Sandmeyer	$CuCl$ (catalytic) + $HCl$	$CuCl$ is regenerated
Gatterman	$Cu$ powder + $HCl$	$CuCl$ is formed as a product

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The question tests your knowledge of named reactions in organic chemistry, specifically the conversion of diazonium salts to aryl halides.

Common Mistakes & How to Avoid Them

1. Confusing Sandmeyer and Gattermann Reactions

The Mistake: Choosing (A) Sandmeyer reaction because both use copper salts.
Why it happens: Students memorize "diazonium + Cu → aryl halide" without noting the catalyst form.
How to avoid:
- Sandmeyer: Uses CuCl or CuBr (stoichiometric copper(I) salt).
- Gattermann: Uses Cu powder (catalytic) + HCl (or HBr).
- Key difference: The given reaction uses Cu/HCl (copper metal + acid), not CuCl directly. This is the Gattermann reaction.

2. Misreading the Reagent

The Mistake: Overlooking the slash in Cu/HCl and assuming it means CuCl.
Why it happens: Slashes in reagents can be ambiguous (e.g., Cu/HCl means copper metal in hydrochloric acid).
How to avoid:
- Always parse the reagent carefully:
  - Cu/HCl = copper metal + HCl (Gattermann)
  - CuCl or CuBr = copper(I) salt (Sandmeyer)
- Write the reagent clearly in your notes.

3. Confusing with Claisen or Carbylamine Reactions

The Mistake: Selecting (C) or (D) because they sound similar or involve nitrogen.
Why it happens: Rote memorization without understanding reaction scope.
How to avoid:
- Claisen reaction: Condensation of esters or aldehydes (e.g., Claisen condensation).
- Carbylamine reaction: Primary amine + chloroform + base → isocyanide (foul smell).
- Neither involves diazonium salts or aryl halides.
- Memory aid:
  - Sandmeyer/Gattermann = diazonium → aryl halide …

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Q.The reaction Ar−N+2Cl−→Cu/HClArCl+N2+CuClAr{-}\overset{+}{N}_2Cl^- \xrightarrow{Cu/HCl} ArCl + N_2 + CuClAr−N+2​Cl−Cu/HCl​ArCl+N2​+CuCl is named as ____.

1. What the reaction shows

2. Why this is the Sandmeyer reaction

3. Distinguishing from the other options

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Q.The reaction $Ar{-}\overset{+}{N}_2Cl^- \xrightarrow{Cu/HCl} ArCl + N_2 + CuCl$ is named as ____.

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