A-Level Chemistry mechanisms: how to draw curly arrows for full marks

Mechanism questions are some of the most reliable marks in A-Level Chemistry, because the rules for drawing them never change. Get the curly arrows right once and you can apply the same logic to every reaction on the specification. Get them slightly wrong and you can lose most of the marks on a question you understood.

What a curly arrow actually means

A curly arrow shows the movement of a pair of electrons. That gives you two rules:

1. The arrow starts where the electrons are: a lone pair or a bond.
2. The arrow ends where the electrons go: an atom (to form a new bond) or a bond (where it is heading to become a lone pair on an atom).

Most lost marks break one of these. An arrow that starts from a positive charge, from an atom with no lone pair drawn, or from empty space cannot be credited.

The mark points examiners look for

Whatever the mechanism, check your diagram for:

• Dipoles (δ+ and δ−) on polar bonds, such as C–Br or C=O.
• Lone pairs drawn on the nucleophile, such as the oxygen in hydroxide or the carbon in cyanide.
• Charges on ions and on any intermediate, such as a carbocation.
• Arrows that start and finish in the right place.
• Products, including the by-product, such as the bromide ion.

The core mechanisms, step by step

Nucleophilic substitution

A halogenoalkane reacting with hydroxide ions, for example bromoethane with aqueous sodium hydroxide, warmed.

• The C–Br bond is polar: carbon is δ+, bromine is δ−.
• An arrow goes from a lone pair on the oxygen of OH⁻ to the δ+ carbon.
• A second arrow goes from the C–Br bond to the bromine.
• Products: ethanol and Br⁻.

Practise it on halogenoalkane questions.

Elimination

The same halogenoalkane with hydroxide dissolved in ethanol, heated. Here hydroxide acts as a base, not a nucleophile.

• An arrow from a lone pair on OH⁻ to a hydrogen on the carbon next to the C–Br carbon.
• An arrow from that C–H bond into the C–C bond, forming the C=C double bond.
• An arrow from the C–Br bond to the bromine.
• Products: an alkene, water and Br⁻.

A common exam twist asks why both substitution and elimination happen. The answer is that hydroxide can act as either a nucleophile or a base, and the conditions decide which dominates.

Electrophilic addition

An alkene reacting with hydrogen bromide.

• An arrow from the C=C double bond to the δ+ hydrogen of H–Br.
• An arrow from the H–Br bond to the bromine.
• This gives a carbocation intermediate, with the positive charge clearly drawn, and Br⁻.
• An arrow from a lone pair on Br⁻ to the positive carbon.

For an unsymmetrical alkene such as propene, the major product comes from the more stable carbocation. Tertiary carbocations are more stable than secondary, which are more stable than primary, because of the electron-releasing effect of the alkyl groups around the positive carbon.

Practise it on alkene questions.

Nucleophilic addition

An aldehyde or ketone reacting with cyanide ions, from potassium cyanide followed by dilute acid.

• The C=O bond is polar: carbon is δ+, oxygen is δ−.
• An arrow from the lone pair on the carbon of CN⁻ to the carbonyl carbon.
• An arrow from the C=O bond to the oxygen, leaving a negatively charged oxygen.
• An arrow from a lone pair on that O⁻ to an H⁺, giving a hydroxynitrile.

Because the carbonyl group is planar, the cyanide can attack from either side. With an unsymmetrical carbonyl, that produces a racemic mixture of two optical isomers, which is a favourite follow-up question.

Practise it on aldehyde and ketone questions.

Electrophilic substitution

The nitration of benzene with concentrated nitric and sulfuric acids.

• First, show how the electrophile forms: the acids react to give the nitronium ion, NO₂⁺.
• An arrow from the delocalised ring to the nitrogen of NO₂⁺.
• The intermediate has a broken delocalised ring (drawn as a horseshoe, open towards the carbon that has been attacked) and a positive charge.
• An arrow from the C–H bond back into the ring, restoring the delocalised system and releasing H⁺.

Practise it on aromatic chemistry questions.

The slips that cost the most

• Drawing the nucleophile's arrow from the minus sign instead of from a drawn lone pair.
• Pointing an arrow at the wrong atom, such as at the bromine instead of the hydrogen in H–Br.
• Forgetting the positive charge on a carbocation or on the benzene intermediate.
• Writing the wrong conditions, for example aqueous instead of ethanolic hydroxide for elimination.
• Leaving out the by-product, such as Br⁻ or H⁺.

How to make mechanisms stick

Reading a mechanism and drawing one from memory are different skills. Cover the page, draw the whole mechanism on scrap paper, then check every arrow, dipole and charge against a mark scheme. Repeat until you can draw each one without hesitating.

Our chemistry questions give you the full mark scheme after every answer, so you can check each arrow against the points examiners credit.