10 min readPathway sheet1-mark MCQ on Zaitsev's rule and Grignard reagents; 2-mark equations for dehydrohalogenation and the Wurtz reaction; conversions in 3 and 5-mark answers

Elimination reactions and reaction with metals

Alcoholic KOH removes HX from a haloalkane to give an alkene; the more substituted alkene is the major product (Zaitsev). With Mg in dry ether, R-X forms a Grignard reagent.

β-EliminationH from the β-carbon and X from the α-carbon are removed; reagent: alcoholic KOH, heat

Zaitsev's (Saytzeff) ruleThe major alkene has more alkyl groups on the double-bond carbons

Grignard reagentR-MgX, from R-X + Mg in dry ether; the carbon of the C-Mg bond is partly negative

Wurtz reaction2R-X + 2Na → R-R + 2NaX, in dry ether

Dehydrohalogenation (β-elimination)

The carbon holding X is the **α-carbon**; the carbon next to it is the **β-carbon**. A base pulls an H off the β-carbon while X leaves from the α-carbon, forming a C=C bond.

1
CH₃CH₂Br + KOH → CH₂=CH₂ + KBr + H₂O
Alcoholic KOH, heat
2★
CH₃CH₂CH₂CH(Br)CH₃ + KOH → CH₃CH₂CH=CHCH₃ + KBr + H₂O
Alcoholic KOH, heat. Pent-2-ene is the major product (about 81%)
3
CH₃CH₂CH₂CH(Br)CH₃ + KOH → CH₃CH₂CH₂CH=CH₂ + KBr + H₂O
Pent-1-ene is the minor product (about 19%)
4★
Zaitsev's rule (also spelt Saytzeff): when more than one alkene can form, the major one has more alkyl groups on the double-bond carbons.

★ key step

Aqueous KOH vs alcoholic KOH

Aqueous KOHOH⁻ acts as a nucleophile: substitution. CH₃CH₂Br + KOH (aq) → CH₃CH₂OH + KBr.

Alcoholic KOHThe base is strong and the solvent is alcohol, so it pulls off a β-H: elimination. CH₃CH₂Br + KOH (alc) → CH₂=CH₂ + KBr + H₂O.

Substitution and elimination compete. A bulky or strong base and heat favour elimination. 1° halides prefer SN2; 3° halides prefer SN1 or elimination; 2° halides can go either way, depending on the base.

Grignard reagent

1
CH₃CH₂Br + Mg → CH₃CH₂MgBr
In dry ether. Victor Grignard won the Nobel Prize in Chemistry in 1912 for this discovery
2
The C-Mg bond is covalent but highly polar, with carbon partly negative; the Mg-X bond is mostly ionic. So the alkyl group acts as a strong nucleophile and base.
3★
CH₃CH₂MgBr + H₂O → CH₃CH₃ + Mg(OH)Br
Any proton source (water, alcohols, amines) turns it into the alkane, so even traces of moisture must be kept out

★ key step

Wurtz reaction

1★
2CH₃Br + 2Na → CH₃-CH₃ + 2NaBr
In dry ether. The alkane has double the number of carbon atoms of the halide
2
2CH₃CH₂Br + 2Na → CH₃CH₂CH₂CH₃ + 2NaBr
Bromoethane gives butane

★ key step

In real life

Grignard reagents are a workhorse for making medicines and fine chemicals in India's pharmaceutical industry. They are handled in dry, sealed vessels, because even moisture from humid air destroys them.

Exam angle

2 marks: “State Zaitsev's rule. Write the major product when 2-bromopentane is heated with alcoholic KOH.” The more substituted alkene is the major product. CH₃CH₂CH₂CH(Br)CH₃ → CH₃CH₂CH=CHCH₃ (pent-2-ene, major) + pent-1-ene (minor).

2 marks: “A student made ethylmagnesium bromide in a flask that was not fully dry and obtained ethane instead. Explain with an equation.” A Grignard reagent reacts with any proton source: CH₃CH₂MgBr + H₂O → CH₃CH₃ + Mg(OH)Br. So it must be made in dry ether with no moisture.

3 marks: “What happens when (i) bromoethane is heated with alcoholic KOH, (ii) 1-bromopropane is heated with sodium in dry ether, (iii) ethylmagnesium bromide is treated with water? Write equations.” (i) CH₃CH₂Br + KOH (alc) → CH₂=CH₂ + KBr + H₂O. (ii) 2CH₃CH₂CH₂Br + 2Na → CH₃(CH₂)₄CH₃ + 2NaBr (hexane). (iii) CH₃CH₂MgBr + H₂O → CH₃CH₃ + Mg(OH)Br.

Remember it

Aqueous: alcohol. Alcoholic: alkene. Zaitsev picks the alkene with the most alkyl friends.