Objective After completing this section, you should be able to use curved (curly) arrows, in conjunction with a chemical equation, to show the movement of electron pairs in a simple polar reaction, such as electrophilic addition. Key Terms Make certain that you can define, and use in context, the key terms below. electrophlic nucleophlic Pushing Electrons and Curved Arrows U nderstanding the location of electrons and being able to draw the curved arrows that depict the mechanisms by which the reactions occur is one of the most critical tools for learning organic chemistry since they allow you to understand what controls reactions, and how reactions proceed. Before you can do this you need to understand that a bond is due to a pair of electrons shared between atoms. When asked to draw a mechanism, curved arrows should be used to show all the bonding changes that occur. A few simple lessons that illustrate these concepts can be found below. Lesson 1 If we remove the pair of electrons in a bond, then we BREAK that bond. This is true for single and multiple bonds as shown below: Notice that since the starting materials were neutral , the products are also neutral . In general terms, the sum of the charges on the starting materials MUST equal the sum of the charges on the products since we have the same number of electrons . The first example is a REACTION since we broke a sigma bond. In the second two examples, we moved pi electrons into long pairs. This is RESONANCE. If we move electrons between two atoms, then we MAKE a new bond: We always show electrons moving from electron rich to electron poor . Lesson 2 This is a simple acid/base reaction, showing the formation of the hydronium ion produced when hydrochloric acid is dissolved in water. It is useful to analyze the bond changes that are occurring. Water is functioning as a base and hydrochloric acid as an acid . Consider the differences in bonding between the starting materials and the products: One of the lone pairs on the oxygen atom of water was used to form a bond to a hydrogen atom, creating the hydronium ion (H 3 O + ) seen in the products. The hydrogen-chlorine bond of HCl was broken, and the electrons in this bond became a lone pair on the chlorine atom, thus generating a chloride ion. We can illustrate these changes in bonding using the curved arrows shown below. Note that in this diagram, the overall charge of the reactants is the same as the overall charge of the products. We can also show the curved arrows for the reverse reaction: This shows the formation of the new H-Cl bond by using a lone pair of electrons from the electron-rich chloride ion to form a bond to an electron poor hydrogen atom of the hydronium ion. Because hydrogen can only form one bond, the oxygen-hydrogen bond is broken and its electrons become a lone pair on the electron-poor oxygen atom. Notice that the charges balance! Lesson 3 In this section, we will look at the curved arrows for some nucleophilic substitution reactions. Overall, the processes involved are similar to those for the acid/base reactions described above. In a nucleophilic substitution reaction, an electron-rich nucleophile (Nu) becomes bonded to an electron-poor carbon atom, and a leaving group (LG) is displaced. In bonding terms, we must make a Nu-C bond and break a C-LG bond. Let's consider the stepwise S N 1 reaction between (1-chloroethyl)benzene and sodium cyanide. The first step of this process is breaking the C-Cl bond, where the electrons in that bond become a lone pair on the chlorine atom. The carbon atom has lost electrons and therefore becomes positive, generating a secondary carbocation. Because the chlorine atom gained an additional lone pair of electrons, it becomes a negatively charged chloride ion. In the second step, the electron-rich nucleophile donates electrons to form a new C-C bond with the electron-poor secondary carbocation. In an S N 2 reaction, the bond forming and breaking processes occur simultaneously. The scheme below shows the Nu donating electrons to form a new C-C bond at the same time that the C-Cl bond is breaking. The electrons in the C-Cl bond become a long pair on the chlorine atom, generating a chloride ion. Forming and breaking the bonds simultaneously allows carbon to obey the octet rule throughout this process. Notice that in all steps for the processes above, the overall charges of the starting materials match those of the products. Lesson 4 This section will dissect another substitution reaction, although it is more involved. Let's consider the S N 1 reaction of tert -butyl bromide with water. It can be helpful to take inventory of which bonds have been formed, and which bonds have been broken. The curved arrows we draw must account for ALL of these bonding changes. Since we are dealing with an S N 1 reaction process, the first step will be cleavage of the C-Br bond to give a carbocation and and a bromide anion. Water then acts as a nucleophile, using one of its lone pairs to form a bond to the electron-poor t -butyl cation. This generates an oxonium ion, where oxygen has three bonds and a positive formal charge. The final step is an acid/base reaction between the bromide anion generated in step 1 and the oxonium product of step 2. The bromide anion acts as a base, using a lone pair to form a bond to one of the hydrogen atoms. The O-H bond then breaks, and its electrons become a lone pair on oxygen. This gives the final products of HBr and t -butyl alcohol. Notice that in each of the mechanistic steps above, the overall charge of the reactant side balances with the overall charge of the product side. While the above process was broken down into distinct steps, however it is important to note that mechanisms are almost always shown as a continuous process. The overall mechanism for this processes can be found below: Now consider the reverse reaction, i.e. the reaction of t -butyl alcohol with hydrobromic acid to generate t -butyl bromide and water. The scheme is shown below, along with an analysis of the bonds formed and broken in this process: The mechanism must occur via the same pathway as shown above (Law of Macroscopic Reversibility), however this mechanism can still be deduced without knowing that. First, it is known that HBr is a strong acid and can donate a proton to a base. The most basic sites in the whole system are the lone pairs on the oxygen atom of t -butanol. Since the lone pairs are the electron-rich area of the molecule, the arrow starts at a lone pair and ends at the proton of HBr. The H-Br bond breaks, pushing its electrons onto the bromine atom and generating a bromide ion. The bromide ion generated in the first step can then react with the t -butyl cation to generate t -butyl bromide. Once again, the above the overall process is broken down into individual steps, however it is more common to illustrate this as one overall process: Curved Arrow Summary Curved arrows flow from electron rich to electron poor. Therefore they start from lone pairs or bonds. The charges in any particular step should always be balanced. Remember to obey the rules of valence ( eg . octet rule for C,N,O,F etc .) If electrons are taken out of a bond, then that bond is broken. If electrons are placed between two atoms then it implies a bond is being made. Exercise \(\PageIndex{1}\) Answer