Skip main navigation

Tools of the trade: understanding nucleophilic substitution reactions

understanding nucleophilic substitution reactions
Tools Of The Trade Understanding Nucleophilic Substitution Reactions

Let’s now return to the reaction of hydroxide ion with bromomethane (CH3Br). Here, the hydroxide ion attacks the carbon atom, not a hydrogen atom. The carbon atom is directly bonded to an electronegative bromine, which makes it an electron‐deficient or electron-poor site – more electron‐poor than the hydrogen atoms. A δ+ sign is used to indicate that the carbon atom has a partial positive charge.

nucleophilic substitutions

The hydroxide ion attacks the carbon atom to form a new C‐O bond and cleave the C‐Br bond, with loss of a bromide ion, which is called the leaving group (good leaving groups are relatively stable anions, such as Br, or neutral molecules). Two double-headed curly arrows show the mechanism of the reaction, which is called a nucleophilic substitution reaction. The electron‐rich hydroxide ion is called a nucleophile and the electron‐poor bromomethane is called an electrophile.


Most organic reactions involve a nucleophile donating a pair of electrons to an electrophile. So, it is very important to recognise which are the most electron‐rich (δ–), or nucleophilic sites within an organic compound, and which are the most electron‐poor (δ+), or electrophilic sites.

It is not fundamental to this course, but, for those of you interested in mastering the art of drawing sensible reaction mechanisms, you may find these guidelines useful. As with most things in life, from playing a piano to hitting a golf ball, practice makes perfect.

Andy’s 7-step guidelines for drawing reaction mechanisms

1. Draw the skeletal structure of the reactants showing lone pairs (••) on any oxygen or nitrogen atoms. (Halogens hold tightly onto their lone pairs and so they are rarely involved in forming bonds.) Also, show the polarities of polar bonds using δ+ and δ–.

2. Identify if the reaction starts with a proton transfer – look out for reagents that are acids (H+ donors) or bases (H+ acceptors).

3. Identify nucleophiles/nucleophilic sites (electron-rich sites) and electrophiles/electrophilic sites (electron-poor sites) in the reactants, and determine which reactant is the nucleophile and which is the electrophile. A charged reactant is usually more reactive than a neutral reactant, so a negatively charged reactant is normally the nucleophile and a positively charged reactant is normally the electrophile. In neutral compounds, carbons that are bonded to more electron-withdrawing atoms, or groups of atoms, are the stronger electrophiles.

4. Draw a double-headed curly arrow from the nucleophilic site (which can have a negative charge, lone pair or a multiple bond) in the nucleophile to the electrophilic site (which has a positive charge or a partial positive charge, δ+) in the electrophile.

5. If the nucleophilic site in the reactant was a negatively charged atom, this atom will be neutral in the product. If the nucleophilic site in the reactant was an uncharged atom, this atom will have a positive charge in the product.

6. If a new bond is made to a neutral atom in the electrophile, break one of the existing bonds and draw a second curly arrow (pointing in the same direction as the first arrow). Draw the second curly arrow so that the most electronegative atom accepts the electrons from the bond, and so acts as the leaving group.

7. Draw a skeletal structure of the product(s) including any charges. Check that the overall charge of the reactants is the same as the overall charge of the products. Finally, if the organic product is charged, and an uncharged product is required, then assess if loss of H+, addition of H+, or loss of a relatively stable negatively charged ion is required.

For practice, have a go at proposing mechanisms for these substitution reactions (answers are provided in the downloads section below):

(a) CH3CH2Br + NaCN –> CH3CH2CN + NaBr

(b) CH3I + N(CH3)3 –> (H3C)4N+ I

(c) CH3CH2ONa + CH3CH2Cl –> CH3CH2OCH2CH3 + NaCl

(d) CH3CH2OCH2CH3 + Hl –> CH3CH2OH + CH3CH2I (Hint: two steps)

If you would like some further information on drawing curly arrow reaction mechanisms you may find this YouTube clip of use.

This is an additional video, hosted on YouTube.

© University of York
This article is from the free online

Exploring Everyday Chemistry

Created by
FutureLearn - Learning For Life

Reach your personal and professional goals

Unlock access to hundreds of expert online courses and degrees from top universities and educators to gain accredited qualifications and professional CV-building certificates.

Join over 18 million learners to launch, switch or build upon your career, all at your own pace, across a wide range of topic areas.

Start Learning now