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What is a nucleophile?
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All Flashcards in Topic 6.4
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6.4.112 cards
What is a nucleophile?
An **electron-pair donor** — it has a lone pair and is attracted to a δ+ (electron-poor) carbon. Examples: OH⁻, CN⁻, NH_{3}.
Why is the carbon in a halogenoalkane δ+?
The C–halogen bond is **polar**: the more electronegative halogen pulls the bonding electrons, leaving carbon slightly positive (**δ+**).
What does a curly arrow show?
The movement of a **pair of electrons** — the tail is at the electrons that move, the head is where the pair ends up.
Describe the two curly arrows in nucleophilic substitution.
Arrow 1: the nucleophile's **lone pair → δ+ carbon** (new bond). Arrow 2: the **C–X bond → halogen**, which leaves as X⁻.
What is the leaving group?
The atom/ion that departs **with the bonding pair** — here the **halide ion, X⁻** (e.g. Br⁻, Cl⁻).
Product of a halogenoalkane + warm aqueous NaOH?
An **alcohol** (the –halogen is replaced by –OH), plus a halide ion.
Conditions for OH⁻ substitution?
**Warm** (gentle heat) and **aqueous** sodium or potassium hydroxide.
What is substitution?
A reaction in which **one group replaces another** on the carbon skeleton, which is otherwise unchanged.
Nucleophile vs electrophile?
Nucleophile = electron-pair **donor** (attacks δ+); electrophile = electron-pair **acceptor** (attacks δ−). Opposites.
Which C–halogen bond reacts fastest, and why?
**C–I** — it is the **weakest** bond, so it breaks most easily. C–F is strongest, so the fluoroalkane is slowest.
Product with cyanide, CN⁻?
A **nitrile** (–CN) — and the chain gains one carbon atom.
Product with ammonia, NH_{3}?
An **amine** (–NH_{2}), using excess ammonia.
6.4.212 cards
What is an electrophile?
An **electron-pair acceptor** — an electron-poor species (often δ+ or positive) attracted to electron-rich regions. Examples: Br_{2}, HBr, H⁺.
Why are alkenes reactive?
The **C=C double bond** has a **π bond** of high electron density that is easily attacked by **electrophiles**.
What is an addition reaction?
Two molecules combine into **one**: a group adds to **each** carbon and the **double bond becomes single** (nothing is left over).
What does 'unsaturated' mean?
The molecule contains a **C=C (or C≡C)** and can undergo **addition**; a saturated molecule has only single bonds.
Describe the two curly arrows in electrophilic addition.
Arrow 1: the **C=C π bond → the electrophile** (new bond). Arrow 2: the **X–X / H–X bond → the leaving atom**, which breaks heterolytically (e.g. as Br⁻).
Product of ethene + bromine?
**1,2-dibromoethane, CH_{2}BrCH_{2}Br** — a bromine atom adds to each carbon as the C=C opens.
Product of an alkene + a hydrogen halide (e.g. HBr)?
A **halogenoalkane** — H adds to one carbon and the halogen to the other (e.g. ethene + HBr → bromoethane).
Product of an alkene + steam (H_{2}O)?
An **alcohol** — using **steam with an H_{3}PO_{4} catalyst** (heat & pressure); –H and –OH add across the C=C.
What is the test for a C=C double bond?
Add **bromine water**: an alkene **decolourises** the orange bromine (it adds across the C=C).
Electrophile vs nucleophile?
Electrophile = electron-pair **acceptor** (electron-poor); nucleophile = electron-pair **donor** (electron-rich). Opposites.
Addition vs substitution — which for alkenes?
Alkenes (unsaturated) react by **addition** (C=C opens, nothing left over); alkanes (saturated) by **substitution** (an atom is replaced).
Why does the Br–Br bond break heterolytically here?
As Br_{2} meets the electron-rich C=C it becomes polarised (δ+/δ−); the far bromine leaves with **both** electrons as **Br⁻**.
6.4.312 cards
Why is an alkene's C=C reactive in electrophilic addition?
It is **electron-rich** — the exposed **π electrons** make it an electron-pair **donor** that attacks an electrophile.
What is an electrophile?
An **electron-pair acceptor** — an electron-poor (often positive) species attracted to an electron-rich centre such as a C=C (e.g. δ+ H of HBr, NO_{2}⁺).
What is a carbocation?
A carbon atom bearing a **positive charge** (only 3 bonds / 6 electrons) — the reactive **intermediate** in electrophilic addition.
Describe step 1 of HBr adding to an alkene.
The **C=C π electrons → the δ+ H** of HBr (arrow 1); the **H–Br bond → Br** (arrow 2), which leaves as **Br⁻**. A **carbocation** forms.
Describe step 2 of HBr adding to an alkene.
The **bromide ion (Br⁻) attacks the positive carbon** of the carbocation, forming the final C–Br bond — the reagent has now **added across** the C=C.
What is Markovnikov's rule (HL explanation)?
The major product forms via the **more stable carbocation**; equivalently, the **H of HX adds to the C with more H's**.
Order of carbocation stability?
**tertiary (3°) > secondary (2°) > primary (1°)** — more alkyl groups donate electron density and spread the positive charge.
Major product of propene + HBr, and why?
**2-bromopropane, CH_{3}CHBrCH_{3}** — via the more stable **secondary** carbocation (Markovnikov).
Major product of 2-methylpropene + HBr?
**2-bromo-2-methylpropane, (CH_{3})_{3}CBr** — via the most stable **tertiary** carbocation.
Why does benzene substitute rather than add?
Its **delocalised** ring is very **stable**; addition would destroy the delocalisation, while **substitution restores** the aromatic ring.
Electrophile, catalyst and product in benzene nitration?
Electrophile = **nitronium ion, NO_{2}⁺**; catalyst = conc. **H_{2}SO_{4}**; product = **nitrobenzene, C_{6}H_{5}NO_{2}** (plus H⁺ lost).
Addition vs substitution — the key difference?
**Addition**: reagent adds across C=C, nothing leaves (alkenes). **Substitution**: one group replaces a ring H, ring restored (benzene).
Topic 6.4 study notes
Full notes & explanations for Electron-pair sharing reactions
Chemistry exam skills
Paper structures, command terms & tips
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