Haloalkanes and Haloarenes

Chemistry · Class 12

Lesson 10 of 13 · 5 min

Elimination and reactions with metals

NCERT §6.7.1

Kavya heats 2-bromopentane with KOH dissolved in ethanol instead of water. No alcohol forms; an alkene does, and two different ones are possible.

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In short

The carbon bearing the halogen is the α-carbon and its neighbour is the β-carbon. Heating a haloalkane that has a β-hydrogen with alcoholic KOH removes H from the β-carbon and X from the α-carbon to give an alkene: β-elimination, or dehydrohalogenation.

Zaitsev (Saytzeff) rule (1875): when more than one alkene can form, the major product is the one with more alkyl groups on the doubly bonded carbons. 2-Bromopentane gives mainly pent-2-ene.

Substitution and elimination compete. A bulky nucleophile tends to act as a base and pull off a proton. Primary halides favour SN2; secondary halides go SN2 or elimination depending on the strength of the base or nucleophile; tertiary halides go SN1 or elimination depending on the stability of the carbocation or of the more substituted alkene.

Aqueous KOH mainly substitutes (alcohol); alcoholic KOH mainly eliminates (alkene).

With certain metals, most organic chlorides, bromides and iodides give organometallic compounds, which hold a carbon-metal bond.

Grignard reagents, RMgX, form from a haloalkane and magnesium in dry ether; Victor Grignard reported them in 1900 and shared the 1912 Nobel Prize in Chemistry with Paul Sabatier.

In RMgX the C–Mg bond is covalent but highly polar, with carbon drawing electrons from magnesium; the Mg–X bond is essentially ionic.

Grignard reagents react with any proton source to give hydrocarbons; even water, alcohols and amines are acidic enough (RMgX + H₂O → RH + Mg(OH)X). Moisture must therefore be kept out, which is why dry ether is used; the reaction is also a way to turn a halide into a hydrocarbon.

Wurtz reaction: an alkyl halide with sodium in dry ether gives an alkane with twice the number of carbon atoms (2RX + 2Na → R–R + 2NaX).

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