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

Telangana TsbieMCQ· 1mImportance★★★★★

44% · 48/108 Questions

🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

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. …

Start your 14-day free trial to unlock the full solution →

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. …

Start your 14-day free trial to unlock the full solution →

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

Start your 14-day free trial to unlock the full solution →

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 …

Start your 14-day free trial to unlock the full solution →

Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

TG EAPCET 2026Set eng-2026-05-10-FN1 markMCQ
$\mathrm{C_6H_5N(CH_3)_2}$
›Reveal solutionSolution
The pair is a tertiary amine and a secondary amine. Only Hinsberg's reagent, $\mathrm{C_6H_5SO_2Cl}$ (option A), separates them: the secondary amine gives an alkali-insoluble sulphonamide, the tertiary amine does not react.

The concept first

Before choosing a reagent, classify the two compounds:

$\mathrm{C_6H_5N(CH_3)_2}$ ( $N,N$ -dimethylaniline): the nitrogen carries three carbon groups → tertiary amine, no N–H.

$\mathrm{(C_2H_5)_2NH}$ (diethylamine): the nitrogen carries two carbon groups and one H → secondary amine, one N–H.

So you need a test that distinguishes $2^\circ$ from $3^\circ$ — i.e. a test that keys off the presence of an N–H bond, not off basicity.

Hinsberg's test does exactly that. Benzenesulphonyl chloride acylates the nitrogen, but only if there is an N–H available:

$1^\circ$ amine → $\mathrm{C_6H_5SO_2NHR}$ . The N–H left over is made acidic by the two electron-withdrawing $\mathrm{S{=}O}$ groups → the sulphonamide dissolves in KOH/NaOH.

$2^\circ$ amine → $\mathrm{C_6H_5SO_2NR_2}$ . No N–H remains, so the sulphonamide is a solid that is insoluble in alkali.

$3^\circ$ amine → no reaction (no N–H to substitute); the amine remains as an oily layer, dissolving only on adding acid.

Step-by-step

Test option (D), $\mathrm{CHCl_3}$ + alc. KOH (carbylamine). It gives the foul-smelling isocyanide only with primary amines. Both compounds here are non-primary → both give a negative test → cannot distinguish. ✗

Test option (B), anhy. $\mathrm{ZnCl_2}$ + conc. HCl (Lucas reagent). This is a test for alcohols (distinguishing $1^\circ$ / $2^\circ$ / $3^\circ$ ROH by turbidity). Amines are not alcohols. ✗ …
Start your 14-day free trial to see the full solution →
TG EAPCET 2024Set ap-2024-05-08-FN1 markMCQ
$\mathrm{C_7H_9N}$
›Reveal solutionSolution
The alkali-insoluble benzenesulphonamide tells us Y must be a secondary amine, which fixes $\mathrm{Y}=\mathrm{C_6H_5NHCH_3}$ ; a secondary amine of this shape arises from catalytic reduction of an isocyanide, so $\mathrm{X}=\mathrm{H_2/Ni}$ . Option (D).

The concept: why alkali-solubility identifies the class of amine

When an amine reacts with a sulphonyl chloride (Hinsberg's reagent, here p-toluenesulphonyl chloride):

1° amine $\mathrm{RNH_2}$ gives $\mathrm{RNH{-}SO_2Ar}$ . The N–H left on nitrogen sits between an electron-withdrawing $\mathrm{SO_2}$ group and the ring, so it is acidic — the sulphonamide dissolves in NaOH forming a salt.

2° amine $\mathrm{R_2NH}$ gives $\mathrm{R_2N{-}SO_2Ar}$ . Nitrogen now has no hydrogen at all, nothing to ionise, so the product is insoluble in alkali.

3° amine does not react at all.

So "Z is insoluble in alkali" is a direct statement that Y is a secondary amine.

Step-by-step

Use the Hinsberg clue. Z insoluble in alkali $\Rightarrow$ Y has no N–H after sulphonylation $\Rightarrow$ Y is 2°.

Screen the options. $\mathrm{C_6H_5CH_2NH_2}$ (benzylamine) is a primary amine — its tosylamide $\mathrm{C_6H_5CH_2NH{-}SO_2C_6H_4CH_3}$ still has an acidic N–H and would dissolve in KOH. So options (A) and (C) are out. Y must be $\mathrm{C_6H_5NHCH_3}$ .

Now identify the reduction. A secondary amine bearing an $\mathrm{N{-}CH_3}$ group is the hallmark product of reducing an isocyanide (carbylamine): $\mathrm{C_6H_5{-}NC} + 4[\mathrm{H}] \xrightarrow{\ \mathrm{H_2/Ni}\ } \mathrm{C_6H_5{-}NH{-}CH_3}$ …
Start your 14-day free trial to see the full solution →
TG EAPCET 2024Set eng-2024-05-10-AN1 markMCQ
$\mathrm{NH_2OH}$
›Reveal solutionSolution
The question asks which pair of reagent (X) and product (Y) is correctly matched. The correct match is aniline ( $\mathrm{C_6H_5NH_2}$ ) with Schiff base, so option (C) is correct.

Concept & Intuition

This is a classic organic chemistry matching problem about carbonyl derivatives. Each reagent (X) reacts with a carbonyl compound (aldehyde or ketone) to give a specific nitrogen-containing derivative. The key is to recall the functional group of the product formed:

Oximes come from hydroxylamine ( $\mathrm{NH_2OH}$ ).

Hydrazones come from hydrazine ( $\mathrm{NH_2NH_2}$ ).

Semicarbazones come from semicarbazide ( $\mathrm{NH_2NHCONH_2}$ ).

Schiff bases (imines) come from primary amines ( $\mathrm{RNH_2}$ ), especially aromatic ones like aniline.

Let’s check each option step by step.

Option (A): $\mathrm{NH_2OH}$ → Hydrazone

Hydroxylamine ( $\mathrm{NH_2OH}$ ) reacts with a carbonyl to form an oxime (with a $\mathrm{C=NOH}$ group), not a hydrazone. Hydrazones come from hydrazine. So this is incorrect.

Option (B): $\mathrm{NH_2NH_2}$ → Semicarbazone

Hydrazine ( $\mathrm{NH_2NH_2}$ ) gives a hydrazone ( $\mathrm{C=NNH_2}$ ). Semicarbazones require semicarbazide ( $\mathrm{NH_2NHCONH_2}$ ). So this is incorrect.

Option (C): $\mathrm{C_6H_5NH_2}$ → Schiff base …
Start your 14-day free trial to see the full solution →
TG EAPCET 2023Set eng-2023-05-12-AN1 markMCQ
$\mathrm{CH_3{-}CO{-}NH_2} \xrightarrow[\mathrm{NaOH\ (aq)}]{\mathrm{Br_2}} \mathrm{X} \xrightarrow[\text{(ii) } \mathrm{H_2O}]{\text{(i) } \mathrm{NaNO_2 + HCl}} \mathrm{Y}$
›Reveal solutionSolution

Acetamide undergoes Hofmann bromamide degradation to methylamine (X); nitrous acid converts this primary aliphatic amine into an unstable diazonium ion that immediately loses $\ce{N2}$ and picks up water, giving methanol — option (C).

The concept first: why aliphatic diazonium salts die instantly

Diazotisation ( $\ce{NaNO2 + HCl}$ ) makes the same $\ce{-N2+}$ group from any primary amine. The difference is stability. In an aryl diazonium salt the $\ce{-N2+}$ is conjugated with the ring, so at 273–278 K it survives long enough to be used in coupling/Sandmeyer reactions. In an alkyl diazonium ion there is no such delocalisation, and $\ce{N2}$ is an outstanding leaving group — so the ion falls apart the moment it forms, giving a carbocation that the solvent (water) captures. That is why primary aliphatic amines simply effervesce and give alcohols with nitrous acid.

Step-by-step

Step 1 — Identify X. Acetamide with bromine in aqueous alkali is the textbook Hofmann bromamide degradation:

$\ce{CH3CONH2 + Br2 + 4NaOH -> CH3NH2 + Na2CO3 + 2NaBr + 2H2O}$ …
Start your 14-day free trial to see the full solution →
TG EAPCET 2023Set eng-2023-05-13-AN1 markMCQ
$\mathrm{CH_3CH_2CH(NH_2){-}COC_6H_5}$
›Reveal solutionSolution
A sulphonamide that is insoluble in alkali can only come from a secondary amine, so X is $\ce{CH3CH2NHCH3}$ ; benzoyl chloride then acylates its nitrogen to give $\ce{CH3CH2N(CH3)COC6H5}$ — option (D).

The concept first: why alkali-solubility fingerprints the amine class

p-Toluenesulphonyl chloride (a Hinsberg-type reagent) reacts with the amine's lone pair, replacing an $\ce{N-H}$ hydrogen with the bulky $\ce{-SO2Ar}$ group:

Primary amine $\ce{R-NH2}$ → $\ce{R-NH-SO2Ar}$ . One $\ce{N-H}$ remains. That hydrogen is acidic, because the resulting anion is stabilised by the strongly electron-withdrawing sulphonyl group. Hence the product dissolves in KOH/NaOH.

Secondary amine $\ce{R2NH}$ → $\ce{R2N-SO2Ar}$ . No $\ce{N-H}$ is left. There is no acidic proton to remove, so the product is insoluble in alkali. ✓

Tertiary amine $\ce{R3N}$ → no reaction (there is no $\ce{N-H}$ to substitute in the first place), so no product Y at all.

The stem says a product Y forms and is insoluble in alkali. Both facts together force X to be secondary.

Step-by-step

Step 1 — Classify X. Product formed ⇒ not tertiary. Product alkali-insoluble ⇒ not primary. Therefore X is a $2^\circ$ amine.

Step 2 — Benzoylation (Schotten–Baumann). Benzoyl chloride $\ce{C6H5COCl}$ acylates the amine nitrogen:
$\ce{R2NH + C6H5COCl -> R2N-CO-C6H5 + HCl}$
The amide nitrogen ends up carrying both original alkyl groups plus the benzoyl group — i.e. it is a tertiary (N,N-disubstituted) amide with no $\ce{N-H}$ .

Step 3 — Test each option against "benzoyl on a $2^\circ$ nitrogen".

(A) $\ce{CH3CH2CH(NH2)COC6H5}$ — the benzoyl is on carbon, and a free $\ce{-NH2}$ survives. Wrong on both counts. …
Start your 14-day free trial to see the full solution →
TG EAPCET 2022Set eng-2022-07-18-FN1 markMCQ
$\mathrm{Li_2O + NO_2}$
›Reveal solutionSolution
Lithium nitrate decomposes differently from other alkali metal nitrates because of the small size and high polarising power of Li⁺; it gives $\mathrm{Li_2O}$ , $\mathrm{NO_2}$ , and $\mathrm{O_2}$ , so the correct option is (B).

Concept & Intuition

Most alkali metal nitrates (like NaNO₃, KNO₃) decompose on strong heating to give the nitrite and oxygen:

$\mathrm{2\,MNO_3 \rightarrow 2\,MNO_2 + O_2}$ .

But lithium is an exception. Because Li⁺ is very small, it has a high charge density and strongly polarises the nitrate ion. This destabilises the nitrate, causing it to break down further — lithium nitrite (LiNO₂) itself is unstable at high temperature and decomposes to lithium oxide, nitrogen dioxide, and oxygen. So the final products are not simply nitrite + O₂, but oxide + NO₂ + O₂.

Step-by-step reasoning

General trend for alkali metal nitrates

For Na, K, Rb, Cs:

$\mathrm{2\,MNO_3 \xrightarrow{\Delta} 2\,MNO_2 + O_2}$

The nitrite is stable at the decomposition temperature.

Lithium’s anomaly

Li⁺ is much smaller than other alkali ions. Its high polarising power weakens the N–O bonds in the nitrate ion, so decomposition occurs at a lower temperature and proceeds further.

First step – formation of nitrite

Initially, LiNO₃ does form LiNO₂ and O₂:

$\mathrm{2\,LiNO_3 \rightarrow 2\,LiNO_2 + O_2}$

But LiNO₂ is not stable at the temperature needed for decomposition.

Second step – further decomposition of LiNO₂

Lithium nitrite decomposes to lithium oxide, nitrogen dioxide, and oxygen:

$\mathrm{2\,LiNO_2 \rightarrow Li_2O + NO_2 + NO}$

However, NO reacts immediately with O₂ to give NO₂: …
Start your 14-day free trial to see the full solution →

←Previous questionQ19 — Acid anhydrides on reaction with primary amines give __. (i) amide (ii…→Next questionQ21 — Best method for preparing primary amines from alkyl halides without ch…

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

🎓Unlock everything free for 14 days

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

Unlock everything free for 14 days