Q.Write structures and IUPAC names of
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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. …
The key idea is the Hoffmann bromamide degradation, where a primary amide reacts with bromine and a strong base to give a primary amine with one fewer carbon atom.
Step 1: Identify the amide for propanamine.
Propanamine (CH3CH2CH2NH2) has 3 carbons. The amide must have one more carbon (4 carbons) because the reaction removes the carbonyl carbon. The amide is butanamide.
Step 2: Write the structure and IUPAC name for (i).
Structure: CH3CH2CH2CONH2
IUPAC name: Butanamide
Step 3: Identify the amine from benzamide. …
Hoffmann bromamide degradation converts an amide to a primary amine with one fewer carbon. (i) The amide that gives propanamine is butanamide.
(ii) The amine from benzamide is aniline.
The Concept: Hoffmann Bromamide Degradation
This reaction is a classic method to shorten a carbon chain by one while converting an amide into a primary amine. The key insight: the amide’s carbonyl carbon is removed as carbonate (NaX2COX3), so the amine that forms has one less carbon than the starting amide.
The reaction proceeds through a rearrangement — the alkyl group attached to the carbonyl carbon migrates to the nitrogen atom. This means the R group in the amide (R−CONHX2) becomes the R group in the amine (R−NHX2). The carbonyl carbon itself is ejected.
A common mistake is to think the amide and amine have the same number of carbons. They do not — the amide always has one more carbon than the resulting amine. Count carefully.
Step-by-step Solution
Part (i): Amide that gives propanamine
1. Identify the target amine.
Propanamine is CHX3CHX2CHX2NHX2. It has 3 carbon atoms.
2. Work backwards using the Hoffmann rule.
Since the amide loses one carbon during the reaction, the starting amide must have 4 carbon atoms. The amide’s general structure is R−CONHX2, where R is the group that will become the amine’s alkyl group.
For propanamine, the alkyl group is propyl (−CHX2CHX2CHX3). So the amide must be CHX3CHX2CHX2−CONHX2.
3. Name the amide.
The parent chain is butane (4 carbons). Replace the -e with -amide: butanamide.
The IUPAC name is butanamide (common name: butyramide).
To double-check: butanamide (CX4HX9NO) undergoes Hoffmann degradation to give propanamine (CX3HX9N) — carbon count drops from 4 to 3. Always verify the carbon count.
Part (ii): Amine from Hoffmann degradation of benzamide
1. Identify the starting amide. …
Here is the clear, concept-first solution for the given problem.
Method: Hoffmann Bromamide Degradation
This is a rearrangement reaction where a primary amide (RCONH2) is converted into a primary amine (RNH2) with one less carbon atom in the chain. The reaction uses bromine (Br2) in the presence of a strong base (like NaOH).
Key Concept: The carbonyl carbon (C=O) of the amide is lost as CO2 gas. Therefore, the amine produced has one carbon fewer than the starting amide.
(i) Amide which gives propanamine
Step 1: Identify the target amine.
- Propanamine is a primary amine with 3 carbon atoms.
- Structure: CH3CH2CH2NH2
Step 2: Work backwards (retrosynthetic analysis).
- Since Hoffmann degradation removes one carbon, the starting amide must have 4 carbon atoms.
- The amide group (-CONH2) is at the end of the chain. The alkyl group attached to the carbonyl must be the same as the alkyl group in the amine.
- For propanamine (C3H7NH2), the alkyl group is propyl (C3H7−).
- Therefore, the amide is Butanamide.
Step 3: Write the structure and IUPAC name.
- Structure: CH3CH2CH2CONH2
- IUPAC Name: Butanamide
Reaction check:
CH3CH2CH2CONH2+Br2+4NaOHΔCH3CH2CH2NH2+Na2CO3+2NaBr+2H2O
(ii) Amine from Hoffmann degradation of benzamide
Step 1: Identify the starting amide.
- Benzamide is an aromatic amide.
- Structure: C6H5CONH2 (benzene ring attached to -CONH2)
Step 2: Apply the Hoffmann degradation rule.
- The amide loses the carbonyl carbon (C=O) as CO2. …
Here are the common mistakes students make with this specific Hoffmann bromamide reaction question, and how to avoid each.
The Core Concept (Why it matters)
The Hoffmann bromamide reaction is a rearrangement where an amide (RCONH2) is treated with bromine (Br2) and a strong base (like NaOH) to give a primary amine with one less carbon atom in the chain.
- Reaction: RCONH2+Br2+4NaOH→RNH2+2NaBr+Na2CO3+2H2O
- Key result: The alkyl group (R) attached to the carbonyl (C=O) in the amide becomes the alkyl group attached to the NH2 in the amine. The carbonyl carbon is lost as CO2.
Mistake #1: Counting Carbons Incorrectly (The "One Carbon Less" Trap)
The Error:
For part (i), students see "propanamine" (3 carbons) and write the amide as propanamide (CH3CH2CONH2). They forget that the product has one less carbon than the starting amide.
Why it happens:
Students memorize the phrase "amide to amine" but ignore the loss of the carbonyl carbon. They think the carbon skeleton stays the same.
How to Avoid:
- Work backwards. The product is propanamine (CH3CH2CH2NH2). This has 3 carbons.
- The amide must have one more carbon than the amine. So the amide must have 4 carbons.
- The amide is butanamide (CH3CH2CH2CONH2).
Correct Answer (i):
- Structure: CH3CH2CH2CONH2
- IUPAC Name: Butanamide
Mistake #2: Confusing the Starting Material with the Product (Benzamide vs. Aniline)
The Error:
For part (ii), students see "benzamide" and think the product is benzylamine (C6H5CH2NH2) or benzamide itself. They forget that the benzene ring is directly attached to the carbonyl in benzamide.
Why it happens:
Students confuse the structure of benzamide (C6H5CONH2) with other aromatic amides. They also forget that the Hoffmann reaction removes the carbonyl carbon.
How to Avoid:
- Draw the structure of benzamide: C6H5CONH2. The benzene ring is the R group.
- Apply the reaction: RCONH2→RNH2. So C6H5CONH2→C6H5NH2.
- The product is aniline (aminobenzene), not benzylamine.
Correct Answer (ii):
- Structure: C6H5NH2
- IUPAC Name: Aniline (or Benzenamine)
Mistake #3: Writing the Wrong IUPAC Name (Common vs. IUPAC)
The Error:
Students write "propanamine" as "propylamine" or "n-propylamine". For aniline, they write "phenylamine" (which is accepted in some contexts but not the strict IUPAC name). …
- 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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