A-Level Chemistry: Organic Mechanisms and Curly Arrows Explained

1. What a curly arrow actually means

Organic mechanisms look intimidating, but almost every mark comes down to one skill: drawing curly arrows correctly. A curly arrow is not decoration. It has a precise meaning, and examiners mark it precisely.

A full curly arrow (with a full arrowhead) shows the movement of a pair of electrons. It tells the story of the reaction: which electrons move, where they come from, and which new bond they make. Get the arrows right and the products follow automatically.

• An arrow always starts from a region of high electron density: either a covalent bond (a bonding pair) or a lone pair.
• An arrow always points to where the new bond forms: onto an atom, or into the gap between the two atoms that are joining up.
• A full arrowhead means a pair of electrons. A half arrowhead (a 'fish hook') means a single electron, and is only used in free-radical mechanisms.

Before you draw a single arrow, label the reactive parts. Find the nucleophile (an electron-pair donor that seeks a positive or slightly positive centre) and the electrophile (an electron-pair acceptor that seeks an electron-rich centre). Once you know which is which, the arrows have somewhere obvious to go.

2. The rules that stop you losing marks

Most dropped marks are not about understanding the chemistry, they are about arrow technique. Lock in these rules and you protect the easy marks.

1. 1Show the dipoles first: On any polar bond, mark the partial charges. In a C-Br bond, bromine is more electronegative than carbon, so carbon is delta+ and bromine is delta-. In H-Br, hydrogen is delta+ and bromine is delta-. These charges tell you where the nucleophile attacks and where the electrophile is.
2. 2Start the tail on electrons, never on an atom or a charge: The tail of the arrow sits on a lone pair or in the middle of a bond. It must not start from the atom symbol itself, and it must not start from the '-' sign of a negative charge. This is the single most common mark-losing error.
3. 3Point the head where the new bond forms: If a nucleophile is attacking a carbon, the head lands on that carbon. If a bond is breaking so an atom can leave with the electrons, the head lands on the leaving atom.
4. 4Show every charge on intermediates and products: A carbocation needs a '+' on the correct carbon. A leaving halide needs a '-'. Missing charges lose marks even when the arrows are right.

3. Nucleophilic substitution: haloalkane + hydroxide

This is the mechanism for the hydrolysis of a haloalkane. Worked example: bromoethane reacting with warm aqueous hydroxide ions to make ethanol. The overall equation is CH3CH2Br + OH- -> CH3CH2OH + Br-. It is one step and it needs exactly two curly arrows.

1. 1Spot the dipole: In CH3CH2Br, the C-Br bond is polar. Carbon is delta+ and bromine is delta-. That delta+ carbon is the target for the nucleophile.
2. 2Arrow 1: the nucleophile attacks: The hydroxide ion, OH-, is the nucleophile. Draw a curly arrow from a lone pair on its oxygen to the delta+ carbon. This forms the new C-O bond.
3. 3Arrow 2: the leaving group leaves: Draw a second curly arrow from the C-Br bond onto the bromine atom. Bromine leaves with both bonding electrons as a bromide ion, Br-.
4. 4Write the products: You get ethanol, CH3CH2OH, plus a bromide ion, Br-. Two arrows, one step, done.

4. SN1 vs SN2 at a high level

The two-arrow, one-step mechanism above is the SN2 pathway, and it is the one you draw for a primary haloalkane like bromoethane. Some boards ask you to go further and compare it with the SN1 pathway. Here is the high-level difference.

• SN2 (substitution, nucleophilic, bimolecular): one concerted step. The nucleophile attacks the delta+ carbon at the same moment the halogen leaves. Favoured by primary haloalkanes, where there is little steric crowding around the carbon. The rate depends on the concentration of both the haloalkane and the nucleophile.
• SN1 (substitution, nucleophilic, unimolecular): two steps. First the C-halogen bond breaks on its own to make a carbocation intermediate, then the nucleophile attacks that carbocation. Favoured by tertiary haloalkanes, because a tertiary carbocation is stabilised by the surrounding alkyl groups. The rate depends only on the concentration of the haloalkane.

The underlying idea worth remembering is that primary haloalkanes tend to react in one step, while tertiary ones react via a carbocation. Whether you need to name SN1 and SN2 depends on your exam board, so check your specification, as some require the labels and some do not.

5. Electrophilic addition: alkene + HBr

This is the mechanism for adding a hydrogen halide across a carbon-carbon double bond. Worked example: ethene reacting with hydrogen bromide to make bromoethane. The overall equation is CH2=CH2 + HBr -> CH3CH2Br. It runs in two steps through a carbocation intermediate, and it needs exactly three curly arrows.

1. 1Spot the electron-rich and electron-poor parts: The C=C double bond is a region of high electron density, so it acts as the nucleophile. In H-Br, hydrogen is delta+ and bromine is delta-, so the delta+ hydrogen is the electrophile.
2. 2Arrows 1 and 2: the first step: Arrow 1: from the C=C double bond to the delta+ hydrogen, forming a new C-H bond. Arrow 2: from the H-Br bond onto the bromine atom, which leaves as Br-. This gives a carbocation, CH3CH2+, and a bromide ion.
3. 3Arrow 3: the second step: Draw a curly arrow from a lone pair on the bromide ion, Br-, to the positively charged carbon. This forms the new C-Br bond.
4. 4Write the product: You get bromoethane, CH3CH2Br. Three arrows across two steps, with a carbocation intermediate in the middle.

Markovnikov's rule (sometimes written Markownikoff): ethene is symmetrical, so it does not matter which carbon the hydrogen adds to. For an unsymmetrical alkene like propene, CH3CH=CH2, it does matter. The hydrogen adds to the carbon that already has more hydrogens, and the bromine ends up on the more substituted carbon, so the major product is 2-bromopropane, CH3CHBrCH3.

The reason is carbocation stability. Adding H+ to the end carbon of propene gives a secondary carbocation, CH3CH+CH3, which is more stable than the primary carbocation you would get the other way round. Alkyl groups are electron-donating (a positive inductive effect), so they spread out and stabilise the positive charge. The more stable carbocation forms faster, so its product dominates.

6. Common mistakes and exam technique

Almost every lost mechanism mark falls into one of these traps. Run this checklist on any mechanism you draw.

• Arrow starting in the wrong place. The tail must sit on a bond or a lone pair, never on an atom symbol and never on a charge sign.
• Wrong number of arrows. Nucleophilic substitution needs two; electrophilic addition needs three. Count them before you move on. If your count is off, you have missed or invented an arrow.
• Arrow pointing to the wrong place. The head lands where the new bond forms, or on the atom that leaves with the electrons.
• Missing dipoles or charges. Mark delta+/delta- on the polar bond, a '+' on the carbocation, and a '-' on the ions.
• Mixing full and half arrowheads. A full head is an electron pair; a half head is a single electron and belongs only in free-radical mechanisms, such as the chlorination of methane.
• Not showing the lone pair you are drawing the arrow from. If the arrow starts from a lone pair, that lone pair should be visible on the structure.

Frequently asked questions