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.
Note
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.
Watch out
A common mistake: thinking the nucleophile attacks the oxygen. No — oxygen is already electron-rich; the nucleophile goes to the carbon because it is electron-deficient.
A Concrete Example – Addition of HCN to a Ketone
Take acetone (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. …
Methylamine (CH3NH2) reacts with nitrous acid (HNO2) via diazotization and subsequent decomposition to yield methanol (CH3OH) as the major product - option (iii).
The reaction of a primary aliphatic amine with nitrous acid proceeds via the nitrosonium ion (NO+), generated in situ from HNO2 under acidic conditions, attacking the amine nitrogen.
Formation of the nitrosonium ion: NaNO2 + HCl gives HNO2 + NaCl; HNO2 + H+ gives H2O + NO+.
Nucleophilic attack by methylamine: the amine's lone pair attacks NO+, forming an N-nitroso intermediate that loses a proton to give a nitrosamine-type species.
Tautomerisation and diazotisation: under acid, this tautomerises and loses water to form the diazonium ion, CH3N2+ - extremely unstable for an aliphatic system.
Watch out
Aromatic diazonium salts are stable at low temperatures (used in coupling reactions), but aliphatic diazonium salts decompose immediately, even at 0 C - a common exam pitfall.
Decomposition of the diazonium ion: CH3N2+ gives CH3+ + N2 (gas, visible effervescence). …
Common Mistakes Students Make on the Reaction of Methylamine with Nitrous Acid
The Correct Answer
Methylamine (CH3NH2) reacts with HNO2 to form methanol (CH3OH) — option (C).
The reaction is:
CH3NH2+HNO2→CH3OH+N2+H2O
Mistake #1: Confusing this with diazotisation of aromatic amines
What students do wrong:
They think HNO2 always gives a diazonium salt, so they write CH3N2+Cl− or similar.
Why it's wrong:
Diazotisation works only for aromatic primary amines (like aniline). Aliphatic primary amines like methylamine react differently — they undergo deamination (loss of NH2 as N2 gas), not diazotisation.
Diazotisation is specific to aryl amines at low temperature (0–5°C)
Mistake #2: Choosing CH3−O−N=O (option A)
What students do wrong:
They think the product is an alkyl nitrite, like methyl nitrite.
Why it's wrong:
Alkyl nitrites form from alcohols reacting with HNO2, not from amines. Methylamine gives N2 gas (bubbles) and an alcohol, not a nitrite ester.
How to avoid:
Track the nitrogen: methylamine has one N, HNO2 has one N — together they form N2 (gas), leaving the carbon skeleton as CH3OH
Alkyl nitrites require an OH group already present in the starting material
Mistake #3: Choosing CH3CHO (option D)
What students do wrong:
They confuse this with oxidation reactions (e.g., CH3NH2 oxidised to CH3CHO).
Why it's wrong:
HNO2 is not a simple oxidising agent here — it's a nitrosating agent. The reaction mechanism involves formation of an unstable diazonium intermediate that decomposes, not oxidation to an aldehyde.
How to avoid:
HNO2 with primary amines = deamination, not oxidation
Aldehydes come from oxidation of primary alcohols or from specific reactions like Rosenmund reduction — not from this reaction
Mistake #4: Forgetting the gas evolution
What students do wrong:
They memorise the product but don't realise N2 gas is evolved — so they miss a key identification clue.