Alkyl halides

MDCAT Chemistry · Chemistry · Alkyl halides. The idea in plain words, the rules to remember, the mistake to avoid and a worked example, written and checked by Mega Lecture.

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The idea

In an alkyl halide, R–X, the halogen pulls electrons away from carbon, so that carbon is slightly positive. Nucleophiles attack it and replace the halogen (nucleophilic substitution), or a base removes HX to give an alkene (elimination).

Classify the halide first: primary, secondary or tertiary, by the number of carbon atoms joined to the C–X carbon. Primary halides react by the one-step S_N2 path. Tertiary halides react by the two-step S_N1 path through a carbocation.

Reactivity depends on the strength of the C–X bond, not on its polarity. The C–I bond is the weakest, so iodoalkanes react fastest.

Rules to remember

Reactivity: R–I > R–Br > R–Cl > R–F (C–F is the strongest bond)
S_N2: one step, rate = k[RX][Nu⁻], attack from the back, inversion of configuration. Order: primary > secondary > tertiary
S_N1: two steps, slow step forms a carbocation, rate = k[RX], product is mostly racemic. Order: tertiary > secondary > primary
E2: one step, rate depends on both halide and base. E1: two steps through a carbocation, rate depends on the halide only
Aqueous KOH favours substitution (alcohol); hot concentrated alcoholic KOH favours elimination (alkene). Tertiary halides eliminate most easily
R–X + Mg in dry ether → R–Mg–X (Grignard reagent); with water it gives the alkane R–H. Wurtz: 2R–X + 2Na → R–R + 2NaX
Boiling points: R–I > R–Br > R–Cl for the same R, and branching lowers them. Freons (CFCs) were refrigerants and aerosol propellants; their Cl atoms destroy ozone

Common mistake

Students say fluoroalkanes react fastest because C–F is the most polar bond. Rate depends on how easily the C–X bond breaks, and C–F is the strongest bond, so fluoroalkanes are the slowest.

Worked example

1-Bromobutane is warmed with KCN in ethanol. Name the organic product and the mechanism.

1-Bromobutane, CH₃CH₂CH₂CH₂Br, is a primary halide, so there is little crowding and no stable carbocation: the path is S_N2.
CN⁻ is a good nucleophile; it attacks the C–Br carbon from the back as Br⁻ leaves, in one step.
The product is CH₃CH₂CH₂CH₂CN, which has five carbon atoms: the chain has grown by one.

Answer: Pentanenitrile, formed by the S_N2 mechanism

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