Q.Write chemical reaction of aniline with benzoyl chloride and write the name of the product obtained.
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🔒 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 (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) …
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. …
Concept: Nucleophilic substitution on an acyl chloride (benzoylation).
Reasoning:
- Aniline (CX6HX5NHX2) acts as a nucleophile via its lone pair on nitrogen.
- Benzoyl chloride (CX6HX5COCl) has an electrophilic carbonyl carbon, which is attacked by aniline’s nitrogen. …
Aniline undergoes nucleophilic substitution with benzoyl chloride to form N-phenylbenzamide (also called benzanilide) via an addition-elimination mechanism, with HCl as the byproduct.
The Concept: Why This Reaction Works
Aniline (CX6HX5NHX2) is a primary aromatic amine. Its nitrogen atom carries a lone pair of electrons, making it a good nucleophile. Benzoyl chloride (CX6HX5COCl) is an acyl chloride — one of the most reactive carboxylic acid derivatives. The carbonyl carbon in benzoyl chloride is strongly electrophilic because:
- The electronegative oxygen pulls electron density away via the C=O double bond.
- The chlorine atom is both electron-withdrawing (inductive effect) and a good leaving group.
So when aniline meets benzoyl chloride, the nucleophilic nitrogen attacks the electrophilic carbonyl carbon. This is a classic nucleophilic acyl substitution reaction — not a simple SN1 or SN2, but a two-step addition-elimination that is characteristic of acyl compounds.
A common mistake is to think this is an electrophilic substitution on the benzene ring. It is not — the reaction occurs at the carbonyl carbon of benzoyl chloride, not on the aromatic ring of aniline.
Step-by-Step Mechanism
-
Nucleophilic attack
The lone pair on the nitrogen of aniline attacks the electrophilic carbonyl carbon of benzoyl chloride. This forms a tetrahedral intermediate where the oxygen now bears a negative charge (an alkoxide ion).
-
Elimination of the leaving group
The tetrahedral intermediate collapses. The C=O bond reforms, and the chlorine atom leaves as chloride ion (ClX−). The chloride ion then abstracts a proton from the positively charged nitrogen (which had become −NHX2X+ after attack), regenerating the neutral amide.
-
Overall reaction
The net result is that the −Cl group of benzoyl chloride is replaced by the −NH−CX6HX5 group from aniline. The byproduct is hydrogen chloride (HCl). …
Method: Nucleophilic Acyl Substitution (Addition-Elimination Mechanism)
This reaction follows the standard mechanism for nucleophilic substitution at an acyl carbon (carbonyl group).
Steps of the Mechanism
-
Nucleophilic Attack
The lone pair on the nitrogen atom of aniline (CX6HX5NHX2) attacks the electrophilic carbonyl carbon of benzoyl chloride (CX6HX5COCl).
- This forms a tetrahedral intermediate.
-
Elimination of Leaving Group
The tetrahedral intermediate collapses, expelling the chloride ion (ClX−) as a leaving group.
- The carbonyl group is reformed.
-
Deprotonation
The intermediate loses a proton (from the nitrogen) to give the final neutral product.
Chemical Reaction …
Here are the common mistakes students make when writing the reaction of aniline with benzoyl chloride, along with how to avoid each one.
1. Writing the wrong product (confusing with other acylations)
The Mistake:
Students often write the product as a secondary amide (like C6H5NHCOCH3) or confuse it with acetylation (using acetyl chloride). They forget that benzoyl chloride has a benzene ring attached to the carbonyl group.
How to Avoid:
- Always identify the acyl group: benzoyl chloride = C6H5COCl.
- The product is N-phenylbenzamide (or benzanilide).
- Write the structure clearly:
C6H5NH2+C6H5COClpyridineC6H5NHCOC6H5+HCl
Key result: The product is N-phenylbenzamide (common name: benzanilide).
2. Forgetting the role of a base (pyridine or NaOH)
The Mistake:
Students write the reaction without a base, showing only aniline + benzoyl chloride → product + HCl. They don’t realise that the HCl produced will protonate aniline (a base) and stop the reaction.
How to Avoid:
- Remember: aniline is a weak base; HCl will form anilinium chloride (C6H5NH3+Cl−), which is unreactive.
- Always include a base (usually pyridine or dilute NaOH) to neutralise the HCl.
- Write the balanced equation with the base shown.
3. Naming the product incorrectly
The Mistake:
Students give names like “benzoyl aniline” or “phenyl benzamide” — these are not IUPAC or common names accepted in exams.
How to Avoid:
- Learn the correct names:
- IUPAC: N-phenylbenzamide
- Common name: Benzanilide
- The “N-” prefix is crucial because the benzoyl group is attached to the nitrogen of aniline.
4. Writing the wrong mechanism (SN1 vs SN2 confusion)
The Mistake:
Students treat this as an SN1 or SN2 reaction on the carbon of benzoyl chloride. In reality, it’s a nucleophilic acyl substitution — the mechanism is different (addition-elimination via a tetrahedral intermediate).
How to Avoid:
- Understand that the carbonyl carbon is electrophilic, not the carbon attached to Cl.
- The nucleophile (aniline) attacks the carbonyl carbon, not the C–Cl bond.
- Draw the mechanism step-by-step:
- Nucleophilic attack by aniline’s lone pair on C=O
- Tetrahedral intermediate forms
- Elimination of Cl−
- Deprotonation gives the amide
5. Ignoring the effect of substituents on reactivity
The Mistake: …
- 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. …
- 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. …
- 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). …
- 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. …
- 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. …
- 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 …
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