Elimination Reactions and Reactions with Metals

Substitution is not the only fate of a haloalkane. When the right base comes along, the molecule can lose two atoms at once and produce an entirely new functional group: a carbon-carbon double bond. Beyond elimination, haloalkanes also open the door to a powerful class of reagents built on carbon-metal bonds. Let us explore both pathways.

How Elimination Works: Building a Double Bond

When a haloalkane that has at least one hydrogen on the carbon next to the halogen-bearing carbon is heated with alcoholic potassium hydroxide (alcoholic $KOH$), something different from substitution takes place. Instead of a nucleophile replacing the halogen, two groups leave the molecule simultaneously:

• A hydrogen atom is pulled off from the $\beta$-carbon (the carbon adjacent to the one holding the halogen).
• The halogen atom departs from the $\alpha$-carbon (the carbon directly bonded to the halogen).

The result is the formation of a carbon-carbon double bond, producing an alkene. Because the hydrogen that leaves comes specifically from the $\beta$-position, this process is called $\beta$-elimination (also known as dehydrohalogenation, since both a hydrogen and a halogen are removed).

Zaitsev’s Rule: Predicting the Major Alkene

Here is an interesting complication. If the haloalkane has $\beta$-hydrogens on more than one neighbouring carbon, more than one alkene could form. So which one wins?

The Russian chemist Alexander Zaitsev (also spelled Saytzeff) noticed a clear pattern in 1875 and formulated a rule:

In dehydrohalogenation reactions, the preferred product is the alkene that has the greater number of alkyl groups attached to the doubly bonded carbon atoms.

In simpler terms, the more substituted alkene is the major product because it is more stable.

Seeing Zaitsev’s rule in action

Take 2-bromopentane ($CH_3CH_2CH_2CHBrCH_3$). The bromine sits on carbon-2, and there are $\beta$-hydrogens available on both carbon-1 and carbon-3. Removing a hydrogen from carbon-3 gives pent-2-ene ($CH_3CH_2CH{=}CHCH_3$), where both double-bonded carbons carry alkyl groups. Removing a hydrogen from carbon-1 gives pent-1-ene ($CH_3CH_2CH_2CH{=}CH_2$), where only one double-bonded carbon has an alkyl substituent.

As Zaitsev’s rule predicts, pent-2-ene (the more substituted alkene) is the major product at 81%, while pent-1-ene forms only as the minor product at 19%.

The Big Competition: Elimination vs Substitution

Every time a haloalkane with $\beta$-hydrogens meets a base or nucleophile, two reaction pathways compete: substitution ($S_N1$ or $S_N2$) and elimination. The molecule does not “choose” consciously; the faster pathway wins. Several factors tip the balance:

• Size of the base or nucleophile: A bulky species (like tert-butoxide, $(CH_3)_3CO^-$) struggles to reach the crowded $\alpha$-carbon for substitution. It finds it much easier to pluck a proton from the more exposed $\beta$-carbon, so it acts as a base and drives elimination. A smaller nucleophile (like ethoxide, $CH_3CH_2O^-$) can approach the $\alpha$-carbon more readily and tends to carry out substitution.

• Class of the alkyl halide:

  • Primary halides prefer $S_N2$ substitution because the $\alpha$-carbon is relatively unhindered.
  • Secondary halides can go either way: $S_N2$ or elimination, depending on the strength and bulk of the base or nucleophile.
  • Tertiary halides cannot undergo $S_N2$ (too much steric crowding). They react through either $S_N1$ or elimination, depending on whether the carbocation gets captured by a nucleophile or whether the more substituted alkene can form.

• Base strength: A strong base generally pushes the reaction towards elimination, while a weaker nucleophile favours substitution.

Think of it as a race: substitution and elimination are always competing, and the conditions of the race (bulkiness, substrate class, base strength, temperature) decide the winner.

Reactions with Metals: Forming Carbon-Metal Bonds

Haloalkanes react with certain metals to form a completely different class of compounds, ones where carbon is directly bonded to a metal atom. Such compounds are called organometallic compounds (compounds containing carbon-metal bonds).

Grignard Reagents: A Chemist’s Best Friend

One of the most important families of organometallic compounds was discovered by the French chemist Victor Grignard in 1900. These are the Grignard reagents, with the general formula $RMgX$ (where $R$ is an alkyl group and $X$ is a halogen).

To prepare them, a haloalkane is treated with magnesium metal in dry ether as the solvent:

$CH_3CH_2Br + Mg \xrightarrow{\text{dry ether}} CH_3CH_2MgBr$

The product, ethylmagnesium bromide, is a typical Grignard reagent.

A note on Grignard’s journey: Victor Grignard had an unusual start for a chemist. He initially earned a degree in mathematics before moving to organic chemistry. While searching for better methylation catalysts, he wondered whether the magnesium-ether combination might work as well as zinc in diethyl ether. The resulting reagents, first reported in 1900, formed the basis of his doctoral thesis in 1901. He became a professor at the University of Nancy in 1910 and shared the 1912 Nobel Prize in Chemistry with Paul Sabatier, who had advanced nickel-catalysed hydrogenation.

Bonding in the Grignard Reagent

The bonding in $RMgX$ has two distinct parts:

• The $C{-}Mg$ bond is covalent but highly polar. Carbon is more electronegative than magnesium, so it pulls electron density towards itself. Carbon carries a partial negative charge ($\delta^-$) and magnesium a partial positive charge ($\delta^+$).
• The $Mg{-}X$ bond is essentially ionic.

$\overset{\delta^-}{R}{-}\overset{\delta^+}{Mg} \quad X^{\delta^-}$

This strong polarity is what makes Grignard reagents so reactive: the carbon end is electron-rich and behaves almost like a carbanion ($R^-$), ready to attack electrophilic centres.

Why Dry Ether is Non-Negotiable

Grignard reagents react with any source of protons, which means even tiny amounts of water will destroy them:

$RMgX + H_2O \longrightarrow RH + Mg(OH)X$

Water is not the only problem. Alcohols and amines are also acidic enough to donate a proton and convert the reagent into a plain hydrocarbon. This is why the reaction must be carried out in dry (anhydrous) ether: ether has no acidic hydrogens and cannot react with the Grignard reagent.

On the flip side, this sensitivity offers a useful synthetic trick. If you want to convert a haloalkane into the corresponding hydrocarbon, you can first make the Grignard reagent and then treat it with water. The halogen is effectively replaced by hydrogen.

The Wurtz Reaction: Doubling the Carbon Chain

There is another metal-mediated reaction of haloalkanes that builds bigger molecules. When an alkyl halide is heated with sodium in dry ether, two alkyl groups couple together to form a hydrocarbon with double the number of carbon atoms. This is the Wurtz reaction:

$2RX + 2Na \xrightarrow{\text{dry ether}} R{-}R + 2NaX$

For example, two molecules of bromoethane produce butane:

$2CH_3CH_2Br + 2Na \xrightarrow{\text{dry ether}} CH_3CH_2CH_2CH_3 + 2NaBr$

The sodium atoms strip the halogen from each molecule, and the two freed alkyl groups join together. This reaction is a straightforward way to extend a carbon chain, though it works best when both halide molecules are identical (using two different halides leads to a mixture of products).