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 reacts with nitrous acid to give nitrogen gas (N2) as the gaseous product - option (ii).
Concept and Intuition
Primary aliphatic amines undergo diazotization with HNO2 (generated in situ from NaNO2 + HCl) to form a highly unstable aliphatic diazonium salt. Unlike aromatic diazonium salts (stable at low temperature), aliphatic diazonium salts decompose spontaneously, even at 0 C, releasing N2 gas and forming a carbocation that reacts further to give a mixture of products (alcohols, alkenes, etc.) - but the gas evolved is always N2.
Watch out
A common mistake is to think ammonia (NH3) is evolved. Ammonia is a base and would be protonated in the acidic medium, not released as gas.
Step-by-Step
N-nitrosation: CH3NH2 + HNO2 gives CH3NH2+NO (unstable), which loses a proton to CH3NHNO.
Tautomerisation and diazonium formation: CH3NHNO rearranges and loses water under acid to give CH3N2+.
Spontaneous decomposition: CH3N2+ gives CH3+ + N2 (gas) - the nitrogen bubbles out of solution; this is the gas evolved. …
Concept: Reaction of Primary Aliphatic Amines with Nitrous Acid (Diazotisation)
This is a classic reaction from amines chapter in organic chemistry. Methylamine (CH3NH2) is a primary aliphatic amine.
Method: Diazotisation & Decomposition Pathway
Step 1 — Formation of Diazonium Salt
Methylamine reacts with nitrous acid (HNO2, generated in situ from NaNO2+HCl) to form an aliphatic diazonium salt:
CH3NH2+HNO2+HCl→[CH3N2+]Cl−+2H2O
Step 2 — Immediate Decomposition
Unlike aromatic diazonium salts, aliphatic diazonium salts are unstable even at low temperatures. They spontaneously decompose to give a carbocation and nitrogen gas:
They confuse the reaction of an amine with nitrous acid (HNO2) with the reaction of an amide or a simple base–acid neutralisation. Methylamine (CH3NH2) is basic, so some students assume it simply accepts a proton to form CH3NH3+ and then releases NH3.
How to avoid:
Remember that primary aliphatic amines react with HNO2 to give diazonium salts, which are unstable and decompose to release nitrogen gas (N2). This is a classic test for primary amines. The reaction is:
CH3NH2+HNO2→CH3N2++2H2O→CH3OH+N2+H2O
The key: N2 gas is evolved, not NH3.
Mistake 2: Choosing hydrogen gas (H2)
Why students make this mistake:
They think the reaction is a simple displacement or reduction, or they confuse it with the reaction of a metal with an acid.
How to avoid:
Nitrous acid (HNO2) is an oxidising agent, not a reducing agent. It does not produce H2 with amines. The only gas formed from the decomposition of the diazonium intermediate is nitrogen (N2). No hydrogen gas is evolved here.
Mistake 3: Choosing ethane (C2H6)
Why students make this mistake:
They think the methyl group (CH3−) from methylamine couples to form a C–C bond, producing ethane. This is a confusion with the Kolbe electrolysis or Wurtz reaction, where alkyl halides couple.