Q.Write chemical equations for the following conversions:
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 (C=O). That oxygen is greedy for electrons; it pulls them away from carbon, leaving the carbon slightly positive (δ+) and the oxygen slightly negative (δ−). 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 π bond breaks, and the oxygen picks up a proton (or some other electrophile) to become stable. That, in a nutshell, is nucleophilic addition.
The key idea: a nucleophile adds across a polar multiple bond (usually C=O or C≡N), breaking the π 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, C=O, or a nitrile, C≡N), while the π bond breaks. The resulting intermediate then captures a proton (or another electrophile) to give a neutral product.
In general form:
RX2C=O+NuX−HX+RX2C(OH)Nu
The nucleophile (NuX−) attacks the carbonyl carbon; the oxygen becomes negatively charged; then a proton (HX+) 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 C=O bond is polarised: CXδ+=OXδ−.
- The π bond is weaker than a σ bond, so it can break relatively easily.
A nucleophile (like OHX−, CNX−, or NHX3) has a lone pair or a negative charge. It seeks positive centres. The attack forms a new σ bond, and the π electrons move entirely to oxygen, creating an alkoxide ion (RX2C−OX−). This intermediate is then quenched by a proton.
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 (CHX3COCHX3) and hydrogen cyanide (HCN). In the presence of a base, CNX− (the nucleophile) attacks the carbonyl carbon:
CHX3COCHX3+CNX−CHX3C(OX−)(CN)CHX3
The alkoxide intermediate then picks up a proton from HCN (or from water) to give a cyanohydrin:
CHX3C(OX−)(CN)CHX3+HX+CHX3C(OH)(CN)CHX3
The product is acetone cyanohydrin. Notice: two new sigma bonds formed (C−CN and O−H), and the π bond is gone.
What Makes a Good Nucleophile?
Strong nucleophiles are usually negatively charged or have lone pairs:
- OHX−, CNX−, NHX2X−, CHX3OX−, HX− (from hydride reagents like NaBHX4 or LiAlHX4)
- Neutral but polarisable: NHX3, HX2O (weaker, but can add under acidic conditions)
In exams, remember: hydride ion (HX−) is a nucleophile in reductions (e.g., NaBHX4 reduces aldehydes/ketones to alcohols via nucleophilic addition).
The Big Picture – Why This Matters
Nucleophilic addition is the fundamental reaction of carbonyl compounds. It is how:
- Aldehydes and ketones form alcohols (with NaBHX4 or LiAlHX4)
- Cyanohydrins are made (important in organic synthesis)
- Grignard reagents (RMgX) add to carbonyls to form new carbon–carbon bonds
- Hemiacetals and acetals form (key in carbohydrate chemistry)
Every time you see a C=O group, think: this carbon is a target for nucleophiles.
Final Answer
Nucleophilic addition is a reaction where an electron-rich nucleophile attacks the electrophilic carbon of a polar multiple bond (typically C=O or C≡N), breaking the π bond and forming two new sigma bonds — one to the nucleophile and one to a proton (or other electrophile). The driving force is the polarity of the C=O bond and the stability gained by forming stronger sigma bonds.
Nucleophilic addition reactions of carbonyl compounds form a central part of the NCERT Class 12 Chemistry chapter on aldehydes and ketones, and questions on cyanohydrin formation or the role of hydride nucleophiles like NaBH4 are common in CBSE boards and JEE Main. Students searching "nucleophilic addition mechanism class 12 chemistry important questions" will find this carbonyl-carbon-attack explanation is the standard NCERT approach.
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:
Rate=k[RX]
Why?
The reaction happens in two steps:
- Slow step (RDS): The C–X bond breaks spontaneously, forming a carbocation intermediate. Only the substrate (RX) is involved.
RXslowR++X−
- Fast step: The nucleophile (Nu−) attacks the carbocation.
R++Nu−fastRNu
Since the slow step depends only on the concentration of RX, the rate law has no dependence on [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:
Rate=k[RX][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 RX and 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:
Rate∝[RX]×[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.
For SN1, the transition state of the slow step involves only bond breaking (no bond formation yet). The carbocation stability (tertiary > secondary > primary) determines the activation energy — this is why SN1 is favored at tertiary carbons.
3. The Temperature Dependence: The Arrhenius Equation
Both rate constants k follow the Arrhenius equation:
k=Ae−Ea/RT
- A = frequency factor (how often collisions occur with correct orientation)
- Ea = activation energy (the energy barrier for the RDS)
- R = gas constant
- T = temperature
This is not a separate formula — it explains why the rate constants change with temperature. Higher T increases the fraction of molecules with energy ≥Ea, speeding up the reaction.
4. Summary: The "Why" in One Table
| Mechanism | Rate Law | Why? (Molecular Reason) |
|---|---|---|
| SN1 | Rate=k[RX] | Slow step involves only the substrate breaking apart. Nucleophile waits. |
| SN2 | Rate=k[RX][Nu−] | Both molecules must collide in the single, concerted step. |
Final takeaway: The formulas are not arbitrary — they are direct consequences of which molecules are present in the slowest step. Always ask: "What is happening in the rate-determining step?" The answer gives you the rate law.
Concept: Nucleophilic Substitution Reactions (SN2) — using cyanide ion as a nucleophile to extend the carbon chain by one carbon, followed by reduction of the nitrile to a primary amine.
Reasoning:
- Both starting materials are primary alkyl halides. Treat each with alcoholic KCN (or NaCN) to perform an SN2 attack, replacing the chlorine with a cyano group (−CN). This adds one carbon atom.
- The resulting nitrile (−CN) is then reduced to a primary amine (−CH2NH2). A common reducing agent for this step is LiAlH4 in dry ether, or catalytic hydrogenation (H2/Ni or H2/Pd).
Stepwise equations:
- CH3−CH2−ClKCN(alc.)CH3−CH2−CNLiAlH4/H2OCH3−CH2−CH2−NH2
- C6H5−CH2−ClKCN(alc.)C6H5−CH2−CNLiAlH4/H2OC6H5−CH2−CH2−NH2
✓Final answer
The conversions are achieved via SN2 with KCN followed by LiAlH4 reduction.
Both conversions are one-carbon chain elongations using the cyanide ion (CN−) as a nucleophile in an SN2 reaction, followed by reduction of the nitrile (−CN) to a primary amine (−CH2NH2). The final products are propan-1-amine and 2-phenylethan-1-amine, respectively.
The Core Idea: Nucleophilic Substitution + Reduction
You have an alkyl halide (a good electrophile) and you want a product whose carbon chain is one carbon longer, ending in CH2NH2. The way to do that is to replace the halogen with a carbon nucleophile that carries the nitrogen, then reduce.
The cyanide ion (CN−) is perfect: it's a strong nucleophile, attacks the carbon bearing the halogen in an SN2 reaction, and the resulting nitrile (R–CN) can be reduced to R–CH2NH2 — exactly the product you need, with the nitrile carbon supplying the extra CH2.
A common mistake is to reach for direct amination with NH3, or for the Gabriel phthalimide synthesis. Both of those put the nitrogen onto the same carbon skeleton — from CH3CH2Cl they give ethylamine (2 carbons), not the 3-carbon target propan-1-amine. Because each target here is one carbon longer than its halide, only a chain-extending route works, and the cyanide route is the standard one. Always count carbons before picking a method.
Step-by-Step Solution
1. First conversion: CH3CH2Cl→CH3CH2CH2NH2
Step 1a: Nucleophilic substitution with KCN (or NaCN)
The chlorine atom is a good leaving group. In ethanol (the NCERT solution writes "ethanolic NaCN" — ethanol is a polar protic solvent, and the reaction works well in it), the cyanide ion attacks the electrophilic carbon.
CH3CH2Cl+KCNethanolCH3CH2CN+KCl
This is an SN2 reaction — the cyanide approaches from the back, inverting the configuration (though here the carbon is not chiral, so no stereochemical consequence). The product is propanenitrile (ethyl cyanide).
Step 1b: Reduction of the nitrile to a primary amine
The nitrile group (−CN) can be reduced to a primary amine (−CH2NH2) using a strong reducing agent. The classic choice is lithium aluminium hydride (LiAlH4) in dry ether, followed by hydrolysis. Alternatively, catalytic hydrogenation (H2/Ni) works equally well — that is the reagent NCERT itself uses in part (ii).
CH3CH2CN1. LiAlH4/ether2. H2OCH3CH2CH2NH2
The reduction adds two hydrogen atoms to the carbon and one to the nitrogen, converting the triple bond into a single bond.
You can also use H2 / Raney Ni with ammonia to avoid coupling side-products (secondary amines). But LiAlH4 or plain H2/Ni is entirely acceptable in a typical exam context.
Overall equation for (i):
CH3CH2ClKCNCH3CH2CNLiAlH4/H2OCH3CH2CH2NH2
2. Second conversion: C6H5CH2Cl→C6H5CH2CH2NH2
Step 2a: Nucleophilic substitution with KCN
Benzyl chloride (C6H5CH2Cl) is even more reactive toward SN2 than a simple primary halide — the adjacent aromatic ring stabilises the transition state, so cyanide attack is fast.
C6H5CH2Cl+KCNethanolC6H5CH2CN+KCl
The product is phenylacetonitrile (phenylethanenitrile / benzyl cyanide).
Step 2b: Reduction of the nitrile
Same reduction as before (H2/Ni, as NCERT writes, or LiAlH4):
C6H5CH2CN1. LiAlH4/ether2. H2OC6H5CH2CH2NH2
The product is 2-phenylethan-1-amine (phenethylamine).
Phenethylamine is a naturally occurring compound (found in chocolate and some brain chemistry) — a nice real-world connection.
Overall equation for (ii):
C6H5CH2ClKCNC6H5CH2CNLiAlH4/H2OC6H5CH2CH2NH2
The required conversions are:
- CH3CH2ClKCNCH3CH2CNLiAlH4/H2OCH3CH2CH2NH2
- C6H5CH2ClKCNC6H5CH2CNLiAlH4/H2OC6H5CH2CH2NH2
Method: Nucleophilic Substitution via Alkyl Cyanide (Nitrile → Amine)
This is a two-step chain elongation method using cyanide ion (CN−) as a nucleophile, followed by reduction.
General Principle
- Step 1: Alkyl halide undergoes SN2 with KCN (or NaCN, in ethanol — NCERT writes "ethanolic NaCN") to form an alkyl cyanide (nitrile).
- Step 2: The nitrile is reduced (e.g., with LiAlH4 or catalytic hydrogenation, H2/Ni) to a primary amine with one extra carbon.
(i) CH3−CH2−Cl→CH3−CH2−CH2−NH2
Step 1 — Nucleophilic substitution:
CH3−CH2−Cl+KCNethanolΔCH3−CH2−CN+KCl
Step 2 — Reduction of nitrile:
CH3−CH2−CN+4[H]LiAlH4 or H2/NiCH3−CH2−CH2−NH2
Key point: The cyanide carbon becomes the extra CH2 group next to the amine.
(ii) C6H5−CH2−Cl→C6H5−CH2−CH2−NH2
Step 1 — Nucleophilic substitution:
C6H5−CH2−Cl+KCNethanolΔC6H5−CH2−CN+KCl
Step 2 — Reduction of nitrile:
C6H5−CH2−CN+4[H]LiAlH4 or H2/NiC6H5−CH2−CH2−NH2
Key point: Benzyl chloride (C6H5CH2Cl) is especially reactive in SN2 — the adjacent aromatic ring stabilises the transition state — so the substitution proceeds smoothly.
Summary of the Method
| Step | Reaction Type | Reagent | Product |
|---|---|---|---|
| 1 | SN2 | KCN (ethanolic) | Alkyl cyanide (nitrile) |
| 2 | Reduction | LiAlH4 or H2/Ni | Primary amine (+1 carbon) |
Final result: Both conversions increase the carbon chain by one and introduce a primary amine at the terminal position.
Here are the common mistakes students make when solving these nucleophilic substitution conversions, along with how to avoid each.
Mistake 1: Choosing a Route That Doesn't Change the Carbon Count
The Mistake:
Students reach for a standard amine preparation — direct ammonolysis or the Gabriel phthalimide synthesis — without counting carbons:
- CH3CH2Cl+NH3→CH3CH2NH2 ✗ (ethylamine — only 2 carbons)
- CH3CH2Cl + potassium phthalimide → N-ethylphthalimide → hydrolysis → CH3CH2NH2 ✗ (still ethylamine)
Why it's wrong:
Both routes attach nitrogen to the existing carbon skeleton. The target of (i) is CH3CH2CH2NH2 (propan-1-amine, 3 carbons) — one carbon longer than the starting halide — so any route that doesn't add a carbon cannot give it.
How to Avoid:
Count carbons first. A one-carbon extension means the cyanide route: the CN− nucleophile supplies the extra carbon, and reduction turns −C≡N into −CH2NH2.
✓ Correct approach for (i):
- CH3CH2Cl+KCNethanolCH3CH2CN+KCl
- Reduction (LiAlH4 or H2/Ni) → CH3CH2CH2NH2
Mistake 2: Forgetting the Carbon Chain Length in (ii)
The Mistake:
Students write the product as C6H5CH2NH2 (benzylamine) instead of C6H5CH2CH2NH2 (2-phenylethan-1-amine).
Why it's wrong:
The target has two carbons between the benzene ring and the amino group. The starting material has only one carbon. You must increase the chain length by one carbon.
How to Avoid:
Always count the carbon atoms in the product vs. starting material. If the product has one more carbon, you need a cyanide ion (CN−) as the nucleophile first, then reduce.
✓ Correct approach for (ii):
- C6H5CH2Cl + KCN (ethanolic) → C6H5CH2CN (benzyl cyanide)
- Reduction: H2/Ni or LiAlH₄ → C6H5CH2CH2NH2
Mistake 3: Inventing "Better" Solvent Conditions Than the Standard Ones
The Mistake:
Insisting the substitution must be run in an anhydrous polar aprotic solvent (acetone, DMF) and marking the ethanol route wrong.
Why it's wrong:
The standard (and NCERT's own printed) condition for this reaction is ethanolic NaCN/KCN — cyanide is a strong enough nucleophile that the SN2 displacement works well in ethanol. Aprotic solvents can accelerate SN2 reactions, but they are not required here, and "correcting" the printed conditions loses marks.
How to Avoid:
Write the reagent the syllabus uses: ethanolic KCN (or NaCN), with heating. Mention SN2 as the mechanism.
Mistake 4: Choosing a Reducing Agent That Doesn't Reduce Nitriles to Primary Amines
The Mistake:
Using NaBH4 (which does not reduce nitriles), or DIBAL-H (which stops at the aldehyde stage), and expecting a primary amine.
How to Avoid:
For R−CN→R−CH2NH2, use LiAlH4 in dry ether (then water) or catalytic hydrogenation (H2/Ni) — the reagent NCERT itself uses. (H2/Raney Ni with ammonia suppresses secondary-amine coupling by-products, a useful refinement but not required.)
✓ Correct reduction:
R−CNLiAlH4/ether, then H2O (or H2/Ni)R−CH2NH2
Mistake 5: Writing Incomplete or Unbalanced Equations
The Mistake:
Writing only the organic product and forgetting byproducts (like KCl) or not balancing atoms.
Example of wrong:
CH3CH2Cl+KCN→CH3CH2CN (missing KCl)
How to Avoid:
Always write complete, balanced equations with all products. Check that the number of atoms of each element is the same on both sides.
✓ Correct:
CH3CH2Cl+KCNethanolCH3CH2CN+KCl
Quick Summary Table
| Mistake | How to Avoid |
|---|---|
| Same-carbon-count route (direct NH3, Gabriel) | Count carbons — a +1 extension needs CN− then reduction |
| Forgetting chain extension in (ii) | Use CN− then reduce |
| "Correcting" the solvent | Ethanolic KCN/NaCN is the standard condition |
| Wrong reduction reagent | Use LiAlH4 or H2/Ni (not NaBH4/DIBAL-H) |
| Unbalanced equations | Always write complete products |
Final Tip: For primary amine preparation from alkyl halides, remember two standard routes:
- No chain extension → Gabriel phthalimide
- Chain extension by one carbon → KCN followed by reduction
Both targets in this question are one carbon longer than their halides — so both must go through the cyanide route.
Showing the 12 most recent of 13 on this concept.
- CBSE 2026Set 56/3/11 markMCQQ.Which of the following reagent is used to distinguish between (C2H5)2NH and (C2H5)3N ? (A) CHCl3+KOH (B) C6H5SO2Cl (C) Conc. HCl+ZnCl2 (D) NaOH+I2
›Reveal solutionSolution
Hinsberg's reagent (C6H5SO2Cl) is used to distinguish between secondary and tertiary amines because secondary amines react to form an alkali-insoluble sulfonamide, while tertiary amines do not react. The correct option is (B).
Amines are organic compounds derived from ammonia (NH3) where one or more hydrogen atoms are replaced by alkyl or aryl groups. They are classified as primary (1∘), secondary (2∘), or tertiary (3∘) based on the number of alkyl/aryl groups attached to the nitrogen atom. This structural difference, specifically the number of hydrogen atoms directly bonded to the nitrogen, dictates their chemical reactivity and forms the basis for distinguishing them.
In this problem, we need to differentiate between (C2H5)2NH and (C2H5)3N.
- (C2H5)2NH is diethylamine, a secondary amine, as the nitrogen atom is bonded to two ethyl groups and one hydrogen atom.
- (C2H5)3N is triethylamine, a tertiary amine, as the nitrogen atom is bonded to three ethyl groups and no hydrogen atoms.
The core idea for distinguishing these two lies in finding a reagent that reacts with the N-H bond present in the secondary amine but cannot react with the tertiary amine due to the absence of such a bond.
-
Analyze the given compounds:
- (C2H5)2NH is a secondary amine. It has one hydrogen atom directly attached to the nitrogen.
- (C2H5)3N is a tertiary amine. It has no hydrogen atoms directly attached to the nitrogen.
-
Evaluate option (A): CHCl3+KOH (Carbylamine reaction)
- The carbylamine reaction (also known as isocyanide test) is a characteristic reaction for primary amines (both aliphatic and aromatic).
- In this reaction, a primary amine reacts with chloroform (CHCl3) and alcoholic potassium hydroxide (KOH) to form an isocyanide (carbylamine), which has a highly unpleasant odor.
- Example: R−NH2+CHCl3+3KOHΔR−NC+3KCl+3H2O
- Secondary and tertiary amines do not give this test.
- Therefore, this reagent cannot distinguish between a secondary amine and a tertiary amine, as neither will give a positive test.
-
Evaluate option (B): C6H5SO2Cl (Hinsberg's reagent)
- C6H5SO2Cl is benzenesulfonyl chloride, commonly known as Hinsberg's reagent. This reagent is specifically used to distinguish between primary, secondary, and tertiary amines.
- Reaction with secondary amines: A secondary amine reacts with Hinsberg's reagent to form an N,N-dialkylbenzenesulfonamide.
(C2H5)2NH+C6H5SO2Cl⟶(C2H5)2N−SO2C6H5+HCl
The product, N,N-diethylbenzenesulfonamide, does not have any acidic hydrogen attached to the nitrogen atom. Therefore, it is insoluble in alkali (like $KOH$ or $NaOH$). * **Reaction with tertiary amines:** Tertiary amines do not have any hydrogen atoms attached to the nitrogen. Thus, they cannot undergo nucleophilic substitution with Hinsberg's reagent. They simply act as bases and may form a salt with the reagent if it's acidic, but no sulfonamide is formed.(C2H5)3N+C6H5SO2Cl⟶No reaction (no sulfonamide formed)
The tertiary amine remains unreacted and is insoluble in alkali. * **Distinction:** When $(C_2H_5)_2NH$ is treated with Hinsberg's reagent, an insoluble product (N,N-diethylbenzenesulfonamide) is formed. When $(C_2H_5)_3N$ is treated with Hinsberg's reagent, no reaction occurs, and the tertiary amine itself is insoluble in the aqueous layer. However, the key is the *formation of a new product* in the case of the secondary amine. The difference in reactivity (reaction vs. no reaction) allows for distinction. * This reagent can effectively distinguish between $(C_2H_5)_2NH$ and $(C_2H_5)_3N$. > [!IMPORTANT] > The Hinsberg test is a classic method for distinguishing primary, secondary, and tertiary amines based on the solubility of their sulfonamide products in alkali. > * Primary amines form N-alkylbenzenesulfonamides, which are soluble in alkali due to an acidic hydrogen on nitrogen. > * Secondary amines form N,N-dialkylbenzenesulfonamides, which are insoluble in alkali as they lack an acidic hydrogen. > * Tertiary amines do not react with Hinsberg's reagent.4. Evaluate option (C): Conc. HCl+ZnCl2 (Lucas reagent)
* Lucas reagent is a mixture of concentrated hydrochloric acid and anhydrous zinc chloride.
* It is primarily used to distinguish between primary, secondary, and tertiary alcohols based on the rate of formation of alkyl chlorides (turbidity).
* This reagent is not used for distinguishing amines.
* Therefore, this option is incorrect.
- Evaluate option (D): NaOH+I2 (Iodoform test)
- The iodoform test is used to detect the presence of a methyl ketone (CH3CO− group) or a secondary alcohol with a methyl group at the alpha position (CH3CH(OH)− group).
- Amines do not contain these specific functional groups and therefore do not give a positive iodoform test.
- Therefore, this option is incorrect.
Based on the analysis, Hinsberg's reagent (C6H5SO2Cl) is the correct choice for distinguishing between a secondary amine and a tertiary amine.
✓Final answerThe reagent used to distinguish between (C2H5)2NH and (C2H5)3N is (B) C6H5SO2Cl.
- CBSE 2025Set 56/6/11 markMCQQ.In the Hinsberg's method for separation of primary, secondary and tertiary amines, the reagent used is : (A) Nitrous acid (B) CHCl3 + aq. NaOH (C) C6H5SO2Cl (benzenesulphonyl chloride) (D) HCl/ZnCl2
›Reveal solutionSolution
Hinsberg's method separates amines based on their reactivity with benzenesulphonyl chloride (C6H5SO2Cl). Primary amines form a soluble salt, secondary amines form an insoluble solid, and tertiary amines do not react. The correct reagent is (C).
Why Hinsberg’s method works — the concept
The key idea is that amines differ in how many hydrogen atoms are attached to the nitrogen. A primary amine (RNH2) has two hydrogens, a secondary amine (R2NH) has one, and a tertiary amine (R3N) has none. Benzenesulphonyl chloride (C6H5SO2Cl) reacts with the N–H bond, replacing the hydrogen with a sulphonyl group. The product’s solubility in alkali depends on whether there is still an N–H hydrogen left to be removed by base.
This gives a clean, visual separation: one fraction dissolves in NaOH, another precipitates, and the third stays as an oily layer that doesn’t react at all.
Step-by-step reasoning
- What does Hinsberg’s reagent do? Benzenesulphonyl chloride (C6H5SO2Cl) is an electrophile. The nitrogen lone pair attacks the sulphur atom, displacing chloride. The product is a sulphonamide. The reaction is:
RNH2+C6H5SO2Cl→C6H5SO2NHR+HCl
- Primary amine — two N–H hydrogens The initial product C6H5SO2NHR still has one N–H hydrogen. This hydrogen is acidic enough to be removed by aqueous NaOH, forming a water-soluble sodium salt:
C6H5SO2NHR+NaOH→C6H5SO2N(R)Na++H2O
So the primary amine ends up dissolved in the alkaline layer.
-
Secondary amine — one N–H hydrogen
The product C6H5SO2NR2 has no N–H hydrogen left (both are replaced by R groups). It cannot be deprotonated by NaOH, so it remains as an insoluble solid or oil that can be filtered off.
-
Tertiary amine — no N–H hydrogen at all
Tertiary amines have no hydrogen on nitrogen. They cannot undergo the substitution reaction with C6H5SO2Cl at all (no N–H bond to attack). The amine remains unreacted and can be extracted as a separate layer.
-
Why not the other options?
- (A) Nitrous acid is used for diazotisation and test for primary vs secondary vs tertiary, but not for separation in Hinsberg’s method.
- (B) CHCl3 + aq. NaOH is the carbylamine reaction (isocyanide test) — specific for primary amines only, not a separation method.
- (D) HCl/ZnCl2 is Lucas reagent for alcohols, not amines.
Watch outA common mistake is to confuse Hinsberg’s reagent with nitrous acid. Both are used to distinguish amines, but Hinsberg’s method actually separates the mixture into three fractions, while nitrous acid gives different products (diazonium salts, nitrosoamines, etc.) without clean physical separation.
TipTo remember: Primary dissolves (soluble salt), Secondary stays solid (insoluble sulphonamide), Tertiary does nothing (no reaction). The mnemonic “PST” — Primary Soluble, Tertiary inert — helps.
✓Final answerThe correct option is (C) C6H5SO2Cl (benzenesulphonyl chloride).
- CBSE 2025Set X11 markMCQQ.Given below are two statements : Statement I : Ammonolysis of alkyl halides has the disadvantage of yielding a mixture of primary, secondary, tertiary amines and quaternary ammonium salt. Statement II : Tertiary amine is obtained as a major product by taking large excess of ammonia in ammonolysis of alkyl halides. In the light of the above statements, choose the appropriate answer from the options given below :(a) Statement I is incorrect but Statement II is correct(b) Both Statement I and Statement II are correct(c) Both Statement I and Statement II are incorrect(d) Statement I is correct but Statement II is incorrect
›Reveal solutionSolution
Ammonolysis genuinely gives a mixture (I correct), but a large excess of ammonia favours the PRIMARY amine (not tertiary), so II is incorrect → option (d).
Statement I — Ammonolysis of an alkyl halide with ammonia is a nucleophilic substitution in which the primary amine formed is itself a nucleophile and reacts further, giving a mixture of 1°, 2°, 3° amines and finally the quaternary ammonium salt. This is a well-known drawback of the method → correct.
R-XNH3RNH2R-XR2NHR-XR3NR-XR4N+X−
Statement II — Taking a large excess of ammonia makes it more likely that each alkyl halide meets an ammonia molecule rather than a partly-alkylated amine, so the reaction stops early and the primary amine is the major product — NOT the tertiary amine. Hence Statement II is incorrect.
Therefore: Statement I correct, Statement II incorrect → option (d).
✓Final answer(d) Statement I is correct but Statement II is incorrect
- CBSE 2024Set 56/3/11 markMCQQ.Which of the following compounds on treatment with benzene sulphonyl chloride forms an alkali-soluble precipitate ? (A) CH3CONH2 (B) (CH3)3N (C) (CH3)2NH (D) CH3CH2NH2
›Reveal solutionSolution
The Hinsberg test distinguishes amines by their reaction with benzene sulphonyl chloride: only primary amines form N-alkyl sulphonamides that are acidic enough to dissolve in alkali. The answer is (D) CH3CH2NH2.
The question tests the Hinsberg test, a classic method to distinguish between primary, secondary, and tertiary amines using benzene sulphonyl chloride (C6H5SO2Cl). The key insight is that different classes of amines react differently, and only one product has the right acidity to dissolve in base after initially precipitating.
When benzene sulphonyl chloride reacts with an amine, it acts as an electrophile. The nitrogen's lone pair attacks the sulphur, displacing chloride. But what happens next depends entirely on whether the nitrogen still has a hydrogen attached.
Why acidity matters
A sulphonamide with an N–H bond is surprisingly acidic (pKa ~ 10) because the negative charge on nitrogen, after deprotonation, is stabilized by resonance with the adjacent SO2 group. The sulphonyl group is strongly electron-withdrawing, delocalizing the negative charge onto the oxygens. This makes the conjugate base stable enough that aqueous alkali (NaOH) can deprotonate it, converting the precipitate into a soluble sodium salt.
Step-by-step analysis
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Option (A): CH3CONH2 (acetamide)
This is an amide, not an amine. Amides are extremely weak nucleophiles because the lone pair on nitrogen is delocalized into the carbonyl π∗ orbital. Benzene sulphonyl chloride won't react with it under normal Hinsberg conditions. No precipitate forms at all.
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Option (B): (CH3)3N (trimethylamine, tertiary)
Tertiary amines have no N–H bond. They can form an unstable ionic complex with the sulphonyl chloride, but they cannot form a stable sulphonamide (no hydrogen to lose as HCl). The product, if any, remains in solution or decomposes. No precipitate.
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Option (C): (CH3)2NH (dimethylamine, secondary)
Secondary amines react to form N,N-dialkyl sulphonamides:
(CH3)2NH+C6H5SO2Cl⟶C6H5SO2N(CH3)2+HCl
This product has no N–H bond, so it cannot be deprotonated. It precipitates but remains insoluble in alkali.
- Option (D): CH3CH2NH2 (ethylamine, primary) Primary amines form N-alkyl sulphonamides with one N–H bond remaining:
CH3CH2NH2+C6H5SO2Cl⟶C6H5SO2NHCH2CH3+HCl
The product precipitates initially. But in aqueous NaOH:
C6H5SO2NHCH2CH3+NaOH⟶C6H5SO2N−Na+CH2CH3+H2O
The sodium salt is water-soluble. The precipitate dissolves in alkali.
TipHinsberg test summary:
• Primary amine → precipitate that dissolves in alkali
• Secondary amine → precipitate insoluble in alkali
• Tertiary amine → no precipitate (or unstable complex)
Watch outDon't confuse amides with amines. Amides (like acetamide) are far less nucleophilic and won't undergo the Hinsberg reaction under standard conditions.
✓Final answer -
- CBSE 2024Set 56/2/11 markMCQQ.Anisole reacts with HI to give : (A) Phenol + CH3−I (B) Iodobenzene + CH3−OH (C) Benzyl alcohol + CH3−I (D) Benzyl iodide + CH3−OH
›Reveal solutionSolution
Anisole undergoes nucleophilic substitution with HI, where iodide ion attacks the less hindered methyl carbon (not the aromatic ring), cleaving the C−O bond to yield phenol and methyl iodide.
Understanding Ether Cleavage with Hydrogen Halides
Anisole is methoxybenzene, CX6HX5−O−CHX3, an aromatic ether. When ethers react with strong acids like HI, they undergo cleavage through nucleophilic substitution. The key is understanding where the bond breaks and why.
Hydrogen iodide is both a strong acid and an excellent nucleophile (iodide ion). The reaction proceeds in two conceptual stages: protonation followed by nucleophilic attack.
Step-by-Step Mechanism
- Protonation of the ether oxygen The lone pair on oxygen accepts a proton from HI, converting the ether into an oxonium ion:
CX6HX5−O−CHX3+HICX6HX5−O+H−CHX3+IX−
This protonation is crucial because it transforms oxygen from a poor leaving group (OX− would be terrible) into a good one (OH, a neutral molecule).
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Nucleophilic attack by iodide
Now the iodide ion must attack. But where? Two carbons are bonded to oxygen: the aromatic ring carbon and the methyl carbon. The iodide attacks the methyl carbon because:
- It's less sterically hindered (primary vs. aromatic)
- SN2 displacement at an sp3 carbon is facile
- Attack at the aromatic carbon would require breaking aromaticity, which is energetically prohibitive
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Bond cleavage and product formation
The C−O bond between methyl and oxygen breaks as iodide displaces the phenol:
CX6HX5−O+H−CHX3+IX−CX6HX5−OH+CHX3−I
The products are phenol (CX6HX5OH) and methyl iodide (CHX3I).
TipIn mixed ethers (alkyl-aryl ethers), HI always cleaves the bond between oxygen and the alkyl group, never the aromatic ring. The alkyl halide forms, and the phenol is liberated.
Watch outA common mistake is thinking iodobenzene forms. Aromatic rings are extraordinarily stable and resist nucleophilic substitution under these conditions. The sp2 carbon of benzene cannot undergo SN2, and SN1 would require a phenyl cation (impossibly unstable).
Why Not the Other Options?
- (B) Iodobenzene + CHX3OH: Would require breaking the aromatic C−O bond and forming a phenyl cation—energetically impossible.
- (C) & (D): Benzyl alcohol and benzyl iodide contain a −CHX2− group attached to benzene. Anisole has no such group; the methyl is directly on oxygen, not on the ring.
✓Final answerThe correct option is (A): Anisole reacts with HI to give phenol and methyl iodide.
- CBSE 2024Set 56/2/11 markMCQQ.Ethanol on heating with conc. H2SO4 at 413 K gives : (A) C2H5OSO3H (B) C2H5−O−CH3 (C) C2H5−O−C2H5 (D) CH2=CH2
›Reveal solutionSolution
At 413 K, concentrated sulfuric acid dehydrates ethanol to form diethyl ether via an intermolecular dehydration mechanism. The correct product is diethyl ether, option (C).
The Concept: Nucleophilic Substitution in Alcohol Dehydration
When ethanol is heated with concentrated sulfuric acid, the acid acts as both a catalyst and a dehydrating agent. The key is temperature control — the same reactants give different products at different temperatures. At 413 K (≈140 °C), the reaction favours intermolecular dehydration (between two ethanol molecules), producing an ether. At a higher temperature (443 K, ≈170 °C), intramolecular dehydration (within one molecule) dominates, giving ethene.
The mechanism is a classic nucleophilic substitution (SN2-like) where one ethanol molecule acts as the nucleophile and another, after protonation, becomes the electrophile.
Step-by-Step Reasoning
- Protonation of ethanol Concentrated H2SO4 donates a proton to the hydroxyl group of ethanol:
C2H5OH+H+⇌C2H5OH2+
This converts the poor leaving group (−OH) into a good one (−OH2+).
- Nucleophilic attack by a second ethanol molecule A second ethanol molecule (the nucleophile) attacks the electron-deficient carbon attached to the protonated hydroxyl:
C2H5OH+C2H5OH2+→[C2H5−O(H)−C2H5]++H2O
This is an SN2-like step — the oxygen lone pair of the attacking ethanol displaces water.
- Deprotonation to form the ether The oxonium ion intermediate loses a proton to a base (e.g., HSO4− or water):
[C2H5−O(H)−C2H5]+→C2H5−O−C2H5+H+
The proton is recycled, regenerating the acid catalyst.
Watch outA common mistake is to confuse the temperature conditions. At 413 K, the product is diethyl ether, not ethene. Ethene forms only at 443 K or above. Also, option (A) C2H5OSO3H (ethyl hydrogen sulfate) is an intermediate that forms at lower temperatures but is not the final product under these conditions.
- Why other options are incorrect
- (A) C2H5OSO3H: This forms when ethanol reacts with H2SO4 at room temperature or below 413 K. At 413 K, it further reacts with ethanol to give the ether.
- (B) C2H5−O−CH3: This would require a methyl group, which is not present in the reactants.
- (D) CH2=CH2: This requires intramolecular dehydration at a higher temperature (443 K). At 413 K, the energy is insufficient to break the C–O bond completely; instead, two molecules couple.
TipA quick memory aid: 413 → Ether (the digits "4-1-3" can be thought of as "for-one-three" → "for one ether"). 443 → Ethene (the digits "4-4-3" look like "ethene" if you squint — or just remember "higher temp, higher energy, double bond").
Final Answer
✓Final answerThe correct option is (C), diethyl ether (C2H5−O−C2H5).
- CBSE 2024Set 56/2/11 markMCQQ.Assertion (A) : Aliphatic primary amines can be prepared by Gabriel phthalimide synthesis. Reason (R) : Alkyl halides undergo nucleophilic substitution with anion formed by phthalimide. Select the correct answer from the codes given below : (A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A). (B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A). (C) Assertion (A) is true, but Reason (R) is false. (D) Assertion (A) is false, but Reason (R) is true.
›Reveal solutionSolution
Gabriel phthalimide synthesis is a highly effective method for preparing pure primary aliphatic amines because the phthalimide anion undergoes nucleophilic substitution with alkyl halides, and the subsequent hydrolysis yields only primary amines, preventing overalkylation. Both Assertion (A) and Reason (R) are true, and Reason (R) correctly explains Assertion (A).
Gabriel phthalimide synthesis is a classic and very important reaction in organic chemistry, specifically designed for the preparation of primary amines. The key challenge in synthesizing primary amines directly from ammonia and alkyl halides is that the primary amine formed can act as a nucleophile itself, reacting further to produce secondary and tertiary amines, and even quaternary ammonium salts. This leads to a mixture of products that is difficult to separate. Gabriel synthesis elegantly bypasses this problem.
The underlying principle of Gabriel synthesis relies on using a protected form of ammonia (phthalimide) that can only be alkylated once, followed by a reaction that releases the primary amine.
Here's a step-by-step breakdown of the process and the reasoning behind it:
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Formation of the Phthalimide Anion:
Phthalimide is an imide, meaning it has an −NH− group flanked by two carbonyl groups. The hydrogen atom attached to the nitrogen is acidic because the resulting anion (phthalimide anion) is resonance-stabilized by the two adjacent carbonyl groups.
When phthalimide is treated with a strong base, such as potassium hydroxide (KOH) or sodium ethoxide (NaOEt), it loses this acidic proton to form a stable, negatively charged phthalimide anion.
Phthalimide+KOH⟶Potassium phthalimide+H2O
The nitrogen atom in this anion carries a negative charge, making it a strong nucleophile.
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Nucleophilic Substitution Reaction:
The potassium phthalimide (or the phthalimide anion) then reacts with an alkyl halide (R−X, where R is an aliphatic alkyl group and X is a halogen like Cl, Br, or I). This is a classic SN2 (bimolecular nucleophilic substitution) reaction. The nucleophilic nitrogen of the phthalimide anion attacks the electrophilic carbon atom bearing the halogen in the alkyl halide, displacing the halide ion.
Potassium phthalimide+R−X⟶N-alkylphthalimide+KX
This step is precisely what Reason (R) describes: "Alkyl halides undergo nucleophilic substitution with anion formed by phthalimide." This reaction incorporates the desired alkyl group (R) onto the nitrogen atom.
Watch outThis SN2 reaction works best with primary alkyl halides. Secondary alkyl halides may undergo elimination (E2) reactions, and tertiary alkyl halides predominantly undergo elimination. Aryl halides (like bromobenzene) do not undergo this nucleophilic substitution reaction under these conditions because the carbon-halogen bond in aryl halides is much stronger and less susceptible to SN2 attack due to the sp2 hybridization of the carbon and resonance effects. This is why the assertion specifies "aliphatic primary amines."
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Hydrolysis to Yield Primary Amine:
The N-alkylphthalimide formed in the previous step is then hydrolyzed. This can be achieved by heating with an aqueous acid (like HCl) or a base (like NaOH), or more commonly and efficiently, by treating it with hydrazine (N2H4).
- Acidic/Basic Hydrolysis: This breaks the two amide bonds, releasing the primary amine (R−NH2) and phthalic acid (or its salt).
- Hydrazinolysis (Gabriel-Ing-Manske modification): This is often preferred because it's milder and gives a better yield. Hydrazine reacts with the N-alkylphthalimide to form a cyclic phthalhydrazide and the primary amine.
N-alkylphthalimide+N2H4⟶R−NH2+Phthalhydrazide
The crucial aspect here is that the nitrogen atom in the original phthalimide only had one hydrogen atom that could be replaced by an alkyl group. Once that alkyl group is attached, there are no more hydrogens on the nitrogen to be replaced, preventing further alkylation. Therefore, only a primary amine (R−NH2) is formed, free from secondary or tertiary amine impurities. This directly supports Assertion (A).
Conclusion:
- Assertion (A): Aliphatic primary amines can be prepared by Gabriel phthalimide synthesis. This is True. The synthesis is specifically designed to yield pure primary amines by preventing overalkylation.
- Reason (R): Alkyl halides undergo nucleophilic substitution with anion formed by phthalimide. This is also True. The SN2 reaction between the phthalimide anion and the alkyl halide is the central step that introduces the alkyl group into the molecule.
Furthermore, Reason (R) provides the mechanistic explanation for how the alkyl group is incorporated into the phthalimide structure, which subsequently leads to the formation of the primary amine upon hydrolysis. Without this nucleophilic substitution, the synthesis would not proceed. Therefore, Reason (R) is the correct explanation for Assertion (A).
✓Final answerThe correct option is (A).
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- CBSE 2024Set ANNUAL1 markQ.Why do amines act as nucleophiles?
›Reveal solutionSolution
Amines act as nucleophiles because the nitrogen atom carries a lone pair of electrons that it can readily donate to an electron-deficient (electrophilic) centre.
In an amine, R−N..H2, nitrogen is sp3 hybridized with three bond pairs (to R and two H, or the equivalent for secondary/tertiary amines) and one lone pair occupying the fourth sp3 orbital. This lone pair is:
- Not delocalized/tied up in any π-system (unlike, say, the nitrogen lone pair in an amide, which is drawn into conjugation with the carbonyl and is far less available).
- Available for donation — nitrogen's relatively low electronegativity and its non-bonding electron pair together make it a good electron-pair donor.
Because a nucleophile is, by definition, an electron-rich species that donates an electron pair to form a new bond with an electrophile, this lone pair is exactly what allows amines to attack electrophilic carbons (as in SN2 reactions with alkyl halides, or addition to carbonyl carbons) and to act as Lewis/Brønsted bases as well.
✓Final answerAmines are nucleophilic because the nitrogen atom has a lone pair of electrons freely available to donate to an electrophilic (electron-deficient) centre.
- CBSE 2023Set 56/1/11 markMCQQ.The synthesis of alkyl fluoride is best obtained from : (A) Free radicals (B) Swartz reaction (C) Sandmeyer reaction (D) Finkelstein reaction
›Reveal solutionSolution
The best method for synthesizing alkyl fluorides is the Swartz reaction, which uses Hg2F2 or CoF2 to replace chlorine/bromine with fluorine. The correct option is (B).
Why this question matters
Alkyl fluorides are the most stable of the alkyl halides due to the strong C–F bond, but they are also the hardest to make by simple nucleophilic substitution. Fluoride ion (F−) is a poor nucleophile in polar solvents because it is heavily solvated (small, high charge density) and also a strong base — so direct SN2 with F− often gives elimination instead. This is why special methods exist.
Let’s examine each option.
1. Free radicals (Option A)
Free radical halogenation of alkanes with fluorine is violently exothermic and uncontrollable — it typically explodes or gives polyfluorinated products. Even with careful conditions, selectivity is terrible. This is not a practical laboratory synthesis for a specific alkyl fluoride.
2. Swartz reaction (Option B)
This is the classic method. A silver or mercury fluoride (like AgF, Hg2F2, or CoF3) is used to replace a chlorine or bromine atom with fluorine:
R–Cl+Hg2F2→R–F+Hg2Cl2
The driving force is the precipitation of the metal halide (e.g., Hg2Cl2 is insoluble). This works cleanly for alkyl, allyl, and benzyl halides. It is the standard method for making alkyl fluorides in the lab.
TipSwartz reaction is to alkyl fluorides what the Finkelstein reaction is to alkyl iodides — a specific halide-exchange method that works because the byproduct is insoluble.
3. Sandmeyer reaction (Option C)
This is for converting aryl diazonium salts into aryl halides (Cl, Br, I, CN) using copper(I) salts. It does not give alkyl fluorides, and it does not work for fluorine (the fluoro analogue uses HBF4 — the Schiemann reaction, not Sandmeyer). So this is irrelevant here.
4. Finkelstein reaction (Option D)
This is an SN2 exchange using NaI in acetone to convert alkyl chlorides/bromides into alkyl iodides. The equilibrium is driven by the insolubility of NaCl/NaBr in acetone. But for fluorides, NaF is even less soluble in acetone than NaCl — so the reverse equilibrium is favoured. More importantly, F− is a poor nucleophile in acetone too. Finkelstein does not give alkyl fluorides.
Watch outA common mistake is to think Finkelstein works for all halogens. It works well for iodides (because I⁻ is a good nucleophile and NaI is soluble), but fails for fluorides because F⁻ is too small and strongly solvated.
Final comparison
Reaction Best for Works for alkyl fluorides? Free radical Chlorination/bromination No (too violent) Swartz Alkyl fluorides Yes Sandmeyer Aryl chlorides/bromides/iodides No (aryl only, not F) Finkelstein Alkyl iodides No (F⁻ too poor a nucleophile) ✓Final answerThe correct option is (B) Swartz reaction.
- CBSE 2023Set 56/2/11 markMCQQ.Given below are two statements labelled as Assertion (A) and Reason (R). Select the most appropriate answer from the options given below : Assertion (A) : Nucleophilic substitution of iodoethane is easier than chloroethane. Reason (R) : Bond enthalpy of C-I bond is less than that of C-Cl bond. (A) Both (A) and (R) are true and (R) is the correct explanation of (A). (B) Both (A) and (R) are true, but (R) is not the correct explanation of (A). (C) (A) is true, but (R) is false. (D) (A) is false, but (R) is true.
›Reveal solutionSolution
The ease of nucleophilic substitution depends on the leaving group's ability to depart. A weaker C–I bond (lower bond enthalpy) makes iodide a better leaving group than chloride, so both Assertion and Reason are true, and Reason correctly explains Assertion.
Concept first: what makes a good leaving group in nucleophilic substitution?
In an SN1 or SN2 reaction, the leaving group (halide ion) must break away from the carbon. The weaker the carbon–halogen bond, the easier it is to break — so the halide leaves more readily. Bond enthalpy (bond dissociation energy) is a direct measure of bond strength: lower bond enthalpy means a weaker bond.
Iodine is a larger atom than chlorine, so the C–I bond is longer and weaker. The C–I bond enthalpy is about 240 kJ/mol, while the C–Cl bond enthalpy is about 330 kJ/mol. That difference is the key.
Now let’s check each statement.
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Assertion (A): "Nucleophilic substitution of iodoethane is easier than chloroethane."
This is true. In both SN1 and SN2 mechanisms, the rate-determining step involves breaking the C–X bond (in SN1, it’s the first step; in SN2, it’s the concerted step where the leaving group departs). Since the C–I bond is weaker, iodoethane reacts faster than chloroethane under identical conditions. Iodide is a better leaving group than chloride.
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Reason (R): "Bond enthalpy of C–I bond is less than that of C–Cl bond."
This is also true. Bond enthalpy decreases down the halogen group: C–F > C–Cl > C–Br > C–I. The C–I bond is indeed weaker.
-
Does (R) correctly explain (A)?
Yes. The entire reason iodoethane undergoes nucleophilic substitution more easily is precisely because its C–I bond is weaker (lower bond enthalpy), making it easier to break. The Reason directly accounts for the Assertion.
Watch outA common mistake is to think that the leaving group ability depends only on the size or polarizability of the halide. While those factors matter, the fundamental reason is the bond strength — and bond enthalpy is the quantitative measure of that. Don’t confuse leaving group ability with nucleophilicity; they are opposite trends.
TipFor quick recall: leaving group ability in halogens follows the order I⁻ > Br⁻ > Cl⁻ > F⁻. This is exactly the reverse of bond enthalpy order. So whenever you see "easier substitution" for a heavier halogen, the bond enthalpy reason is almost always correct.
✓Final answerBoth (A) and (R) are true and (R) is the correct explanation of (A). The correct option is (A).
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- CBSE 2023Set 56/3/11 markMCQQ.Assertion (A): Nucleophilic substitution of iodoethane is easier than chloroethane. Reason (R): Bond energy of C-Cl bond is less than C-I bond. (A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A). (B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A). (C) Assertion (A) is true, but Reason (R) is false. (D) Assertion (A) is false, but Reason (R) is true.
›Reveal solutionSolution
The key idea is that nucleophilic substitution depends on the leaving group's ability to depart, which is governed by bond strength — but the C–I bond is actually weaker than C–Cl, so the reason given is factually wrong. The correct answer is (C).
Let’s start with the concept. In an SN1 or SN2 reaction, the leaving group (the atom or group that gets displaced) must break its bond to carbon. The easier that bond breaks, the faster the reaction. So the bond dissociation energy of the C–X bond is a direct measure of how good a leaving group X is: weaker bond → better leaving group → faster substitution.
Now, the assertion says iodoethane (CH3CH2I) undergoes nucleophilic substitution more easily than chloroethane (CH3CH2Cl). That is true — iodide is a much better leaving group than chloride because the C–I bond is weaker and the iodide ion is larger, more polarizable, and more stable in solution.
The reason claims that the C–Cl bond energy is less than the C–I bond energy. That is false. Let’s check the actual bond dissociation energies:
Bond Approximate bond energy (kJ/mol) C–F ~485 C–Cl ~339 C–Br ~285 C–I ~240 So C–I is actually weaker than C–Cl. The reason has the inequality backwards.
- Assertion (A) is true: iodoethane reacts faster in nucleophilic substitution than chloroethane.
- Reason (R) is false: the C–Cl bond energy is greater than C–I, not less.
- Since the reason is false, it cannot be the correct explanation — but even if it were true, the direction is wrong. The correct explanation would be that C–I is weaker, not stronger.
Watch outA common mistake is to assume that because iodine is larger, its bond to carbon must be stronger (more electrons, more attraction). In reality, bond strength depends on orbital overlap — the larger iodine atom has poorer overlap with carbon’s small sp3 orbital, making the bond longer and weaker.
✓Final answerThe correct option is (C) — Assertion (A) is true, but Reason (R) is false.
- CBSE 2021Set ANNUAL1 markQ.What is an ammonolysis?
›Reveal solutionSolution
Ammonolysis: NH3 acts as a nucleophile and displaces the halide from a haloalkane (SN2), giving an amine salt that is freed by excess ammonia.
R–X + NH₃(excess) → R–NH₂ + HX (the HX formed is neutralized by excess NH3 to NH4X). This is called ammonolysis because it is analogous to hydrolysis but uses ammonia instead of water as the nucleophile. It is carried out in a sealed tube (Hofmann's ammonolysis) using excess ammonia in ethanol. Since the primary amine formed is itself nucleophilic, it can further react with unreacted R–X to give secondary and tertiary amines and finally a quaternary ammonium salt — so ammonolysis of alkyl halides gives a mixture of amines unless ammonia is in large excess.
✓Final answerAmmonolysis = nucleophilic displacement of halogen from an alkyl halide by ammonia, giving a primary amine (R–X + NH₃ → R–NH₂ + HX).
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