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.
- AP EAPCET 2026Set eng-2026-05-12-FN1 markMCQQ.In which of the following set/s reactant and reagent are correctly matched to get n-propylamine as major product? I. CH3CH2CH2Cl (1-chloropropane) ----- AgNO2,Fe/HCl II. CH3CH2Cl (chloroethane) ----- NaCN,H2/Ni III. CH3CH2CONH2 (propanamide) ----- Br2/OH− Correct answer is (A) II only (B) I, II (C) II, III (D) I, III
›Reveal solutionSolution
Two independent routes correctly build n-propylamine (nitro reduction of 1-chloropropane, and nitrile chain-extension + reduction of chloroethane); the Hofmann degradation route instead shortens the chain by one carbon, giving ethylamine.
Concept and Intuition
Three very different strategies for making primary amines are being tested, and the key discipline is to count carbons through each transformation, since some routes preserve chain length and one (Hofmann bromamide degradation) deliberately removes a carbon.
- AgNO2 vs NaNO2: nitrite is an ambident nucleophile (can attack through O or N). With the softer, more covalent silver salt AgNO2, alkylation occurs preferentially through nitrogen, giving the nitroalkane as the major product (with NaNO2, the ionic character favours O-alkylation, giving the alkyl nitrite instead). Reducing a nitroalkane (Fe/HCl) gives the primary amine with the same carbon skeleton.
- The cyanide route (NaCN, then reduction) is a classic chain-extension: the nucleophilic CN− displaces the halide, adding one carbon to the chain as a nitrile, and reducing the nitrile (H2/Ni or LiAlH4) gives a primary amine that is one carbon longer than the starting halide.
- The Hofmann bromamide degradation (Br2/OH− on an amide) is the opposite: it removes the carbonyl carbon (as it leaves via an isocyanate/carbamate intermediate that loses CO2), giving a primary amine with one carbon fewer than the starting amide.
Step-by-Step Solution
- I: CH3CH2CH2Cl (3 carbons) + AgNO2 → CH3CH2CH2NO2 (1-nitropropane, N-alkylation favoured by the covalent Ag–N bond). + Fe/HCl (reduction) → CH3CH2CH2NH2 = n-propylamine. Matches.
- II: CH3CH2Cl (2 carbons) + NaCN → CH3CH2CN (propanenitrile, 3 carbons after chain extension). + H2/Ni → CH3CH2CH2NH2 = n-propylamine. Matches.
- III: CH3CH2CONH2 (propanamide, 3 carbons) + Br2/OH− (Hofmann degradation) → loses the carbonyl carbon as CO2, giving CH3CH2NH2 = ethylamine (2 carbons), not n-propylamine. Does not match.
- So only I and II correctly give n-propylamine.
Common Mistakes
- Assuming Hofmann degradation preserves carbon count — it always removes one carbon from the amide.
- Missing the Ag+ vs Na+ nitrite distinction and assuming route I gives the nitrite ester (which on reduction would not cleanly give the amine in one obvious step).
✓Final answerThe correct option is (B) — I, II.
ANSWER: B
- AP EAPCET 2025Set eng-2025-05-22-AN1 markMCQQ.The preferred reagent for the preparation of pure alkyl chloride from alcohol is (A) HCl+ZnCl2 (B) PCl5 (C) SOCl2 (D) PCl3
›Reveal solutionSolution
Tests why thionyl chloride, not the phosphorus halides, is the preferred reagent purely on the grounds of product purity.
Concept and Intuition
Several reagents convert R−OH→R−Cl, but "preferred for pure product" is a purity argument, not a yield argument. The reagent of choice is the one whose by-products leave the flask on their own (as gases), so no work-up/purification is needed to remove them from the alkyl chloride.
Step-by-Step Solution
- R−OH+HCl/ZnCl2→R−Cl+H2O (Lucas reagent) — works well only for tertiary/some secondary alcohols and leaves ZnCl2/water behind.
- R−OH+PCl5→R−Cl+POCl3+HCl — the by-product POCl3 is a liquid that must be removed by distillation, contaminating the crude product.
- R−OH+PCl3→R−Cl+H3PO3 — leaves phosphorous acid behind, again needing separation.
- 3R−OH+SOCl2→R−Cl+SO2↑+HCl↑ — both by-products are gases that simply bubble out, so the alkyl chloride is obtained directly in pure form with essentially no work-up.
- Hence SOCl2 is the textbook "preferred reagent for pure alkyl chloride."
Common Mistakes
- Picking PCl5 because it is the most commonly taught halogenating agent — but the question specifically asks about purity, which is SOCl2's distinguishing feature.
- Forgetting that this reaction is normally run with a trace of pyridine to mop up the HCl and drive the reaction, without changing the by-product argument.
✓Final answerThe correct option is (C) — SOCl2.
ANSWER: C
- AP EAPCET 2025Set eng-2025-05-23-FN1 markMCQQ.The reaction of benzene diazonium chloride with Cu and HCl is known as (A) Sandmeyer reaction (B) Etard reaction (C) Finkelstein reaction (D) Gattermann reaction
›Reveal solutionSolution
This distinguishes the Sandmeyer reaction (uses cuprous salts, e.g. Cu2Cl2/Cu2Br2) from the closely related Gattermann reaction (uses copper powder directly with HCl/HBr) for converting a diazonium salt to an aryl halide.
Concept and Intuition
Both reactions replace the diazonium group (−N2+) with a halogen via a radical/copper-mediated mechanism, and both give the same type of product (aryl halide + N2). The distinguishing feature examiners test is the reagent form: Sandmeyer's reaction historically uses a cuprous halide salt (e.g. Cu2Cl2) generated in situ or added directly, while the Gattermann reaction is the simplified variant that uses copper powder (metallic Cu) together with the corresponding hydrohalic acid (HCl/HBr) — no cuprous salt needed.
Step-by-Step Solution
- The reaction described is ArN2+Cl− treated with Cu (metal powder) and HCl.
- Because the copper source is elemental copper powder (not a cuprous salt like Cu2Cl2), this specific reagent combination is named the Gattermann reaction, distinguishing it from the Sandmeyer reaction (which specifically uses cuprous chloride/cuprous bromide).
- Both reactions replace −N2+ with −Cl to give chlorobenzene, but the naming convention in exams hinges on the copper reagent identity.
- Etard and Finkelstein reactions are unrelated: Etard converts toluene to benzaldehyde via CrO2Cl2; Finkelstein converts alkyl chlorides/bromides to alkyl iodides using NaI in acetone.
Common Mistakes
- Calling every diazonium-to-aryl-halide reaction 'Sandmeyer' regardless of whether Cu metal or a cuprous salt is used — the specific reagent (Cu powder here) is the Gattermann variant.
- Confusing this with Finkelstein (which is for alkyl, not aryl/diazonium, halide exchange).
✓Final answerThe correct option is (D) — Gattermann reaction.
ANSWER: D
- AP EAPCET 2023Set ap-2023-05-23-AN1 markMCQQ.Hydrolysis products of isobutyl isonitrile are (A) (CH3)2CHCH2NH2+HCOOH (B) CH3CH2CH(CH3)NH2+HCOOH (C) (CH3)2CHNH2+CH3COOH (D) (CH3)2CHCOOH+CH3NH2
›Reveal solutionSolution
Isonitrile hydrolysis always gives the corresponding primary amine (with the SAME alkyl group as the isonitrile) plus formic acid — the alkyl group itself is never altered.
Concept and Intuition
An isocyanide, R−N≡C, is hydrolysed under acidic aqueous conditions by nucleophilic attack of water at the terminal carbon, ultimately cleaving the C–N bond to release the alkyl group AS AN AMINE (with its nitrogen intact, unchanged connectivity to R) while the isocyanide carbon (originally bonded only to N and nothing else besides the lone pair/triple bond) ends up as formic acid, HCOOH. This is a key distinguishing test reaction of isocyanides vs their isomeric nitriles (which hydrolyse differently, to carboxylic acids + ammonia/amine on the OTHER side).
Step-by-Step Solution
- Isobutyl isonitrile: (CH3)2CHCH2−N≡C — the isobutyl group ((CH3)2CHCH2−) is attached to the isocyanide nitrogen.
- Acid-catalysed hydrolysis proceeds: R−NC+2H2OH+R−NH2+HCOOH.
- Applying this to isobutyl isonitrile: the alkyl group R = (CH3)2CHCH2− ends up as the primary amine (CH3)2CHCH2NH2 (isobutylamine), retaining the exact same carbon skeleton and nitrogen attachment.
- The isocyanide carbon (the one triple-bonded to N in R–N≡C) is hydrolysed to formic acid, HCOOH.
- So the hydrolysis products are (CH3)2CHCH2NH2 + HCOOH — option (A).
Common Mistakes
- Confusing isonitrile hydrolysis (gives R-NH₂ + HCOOH) with nitrile hydrolysis (R-CN gives R-COOH + NH₃) — these are structurally different functional groups (isocyanide vs nitrile) with different hydrolysis outcomes.
- Changing the alkyl group's carbon skeleton in the amine product — it must remain exactly the same as in the starting isonitrile (isobutyl throughout).
✓Final answerThe correct option is (A) — (CH3)2CHCH2NH2+HCOOH.
ANSWER: A
- AP EAPCET 2022Set eng-2022-07-04-FN1 markMCQQ.When 1-chloro butane is treated with aqueous KOH it gives P. However, when it is treated with alcoholic KOH it gives Q. Identify the products P and Q respectively (A) P = CH3CH2CH2CH2OH (1-butanol), Q = CH3CH2CH2CH2OH (1-butanol) (B) P = CH2=CHCH2CH3 (but-1-ene), Q = CH3CH=CHCH3 (but-2-ene) (C) P = CH2=CHCH2CH3 (but-1-ene), Q = CH3CH2CH2CH2OH (1-butanol) (D) P = CH3CH2CH2CH2OH (1-butanol), Q = CH2=CHCH2CH3 (but-1-ene)
›Reveal solutionSolution
Aqueous KOH gives substitution (1-butanol); alcoholic KOH gives elimination, and since only C2 has beta-hydrogens, the only possible alkene is but-1-ene.
Concept and Intuition
Alkyl halides react differently depending on the solvent/base used: aqueous KOH (with water as solvent) favours SN2 nucleophilic substitution because OH− acts predominantly as a nucleophile in the polar protic medium, while alcoholic KOH (KOH dissolved in ethanol, with much less free water) favours E2 elimination, as the base character of OH−/ethoxide dominates, abstracting a beta-hydrogen to form an alkene.
Step-by-Step Solution
- Substrate: 1-chlorobutane, CH3CH2CH2CH2Cl -- a primary alkyl halide, Cl on C1.
- Aqueous KOH: favours SN2 substitution (primary halides are ideal SN2 substrates -- unhindered backside attack). OH− displaces Cl− directly: product P = CH3CH2CH2CH2OH (1-butanol).
- Alcoholic KOH: favours E2 elimination. The base removes a beta-hydrogen (a hydrogen on the carbon adjacent to the one bearing Cl).
- In 1-chlorobutane, C1 bears Cl; the only carbon adjacent to C1 is C2 -- so the only beta-hydrogens available are those on C2. There is no other beta-carbon (C1 is a terminal/primary carbon), so there is only one possible elimination direction.
- Removing an H from C2 and Cl from C1 forms the double bond between C1-C2: CH2=CHCH2CH3, i.e. but-1-ene -- this is the only alkene that can form; but-2-ene is not possible from this substrate.
- So P = 1-butanol, Q = but-1-ene, matching option (C).
Common Mistakes
- Assuming Zaitsev's rule (more substituted alkene favoured) applies here to give but-2-ene -- Zaitsev's rule only matters when there is a choice of beta-hydrogens from different beta-carbons; here there is only one beta-carbon (C2), so only one alkene is geometrically possible.
✓Final answerThe correct option is (C) — P = 1-butanol, Q = but-1-ene.
ANSWER: C
- AP EAPCET 2021Set eng-2021-08-25-AN1 markMCQQ.Identify the compound (B) in given reaction. CH3ClKCN(A)H+/H2O(B) (A) CH3NH2 (B) HCOOH (C) CH3COOH (D) CH3COCH3
›Reveal solutionSolution
CH3Cl with KCN forms methyl cyanide, and acidic hydrolysis of that nitrile gives acetic acid.
Concept and Intuition
Potassium cyanide, being an ionic (predominantly carbon-nucleophilic) source of CN−, substitutes alkyl halides via their carbon end to give alkyl cyanides (nitriles), R–C≡N, rather than isocyanides. Nitriles are the functional equivalent of a "masked carboxylic acid" — acidic (or basic) hydrolysis converts the C≡N group all the way to −COOH (via an amide intermediate).
Step-by-Step Solution
- CH3Cl+KCN→CH3CN+KCl — this is (A), methyl cyanide (acetonitrile).
- Acidic hydrolysis of a nitrile: CH3CN+2H2OH+CH3COOH+NH3 (via the amide CH3CONH2 intermediate).
- So (B)=CH3COOH (acetic acid).
Common Mistakes
- Using AgCN instead of KCN mentally — AgCN (more covalent, nitrogen-nucleophilic) would give the isocyanide CH3NC instead, a completely different pathway.
- Stopping hydrolysis at the amide stage instead of carrying it through to the carboxylic acid under acidic/aqueous conditions.
✓Final answerThe correct option is (C) — CH3COOH.
ANSWER: C
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