Skip to main content
Chemistry LibreTexts

3.4: Using Curved Arrows in Polar Reaction Mechanisms

  • Page ID
    205315
  • 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

    Understanding 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:

    cleavage of C-Br bond in tert-butylbromide to give tert-butyl cation and bromide ion; resonance in formaldhyde and hydrogen cyanide.svg

    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:

    formation of a bond between bromide ion and tert-butyl carbocation, resonance in formaldehyde and hydrogen cyanide.svg

    We always show electrons moving from electron rich to electron poor.

    Lesson 2

    reaction scheme for reaction of water and hydrochloric acid to generate hydronium and chloride ions.svg

    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:

    breakdown of bonds formed and broken in reaction between water and hydrochloric acid to generate hydronium and chloride ions.svg

    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 (H3O+) 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.

    mechanism of acid:base reaction between water and hydrochloric acid to generate hydronium and chloride ions.svg

    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:

    mechanism between chloride ion and hydronium ion to generate water and hydrochloric acid.svg

    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.

    generic reaction scheme for a substitution reaction.svg

    Let's consider the stepwise SN1 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.

    cleavage of C-Cl bond in (1-chloroethyl)benzene to form benzylic cation and chloride anion.svg

    In the second step, the electron-rich nucleophile donates electrons to form a new C-C bond with the electron-poor secondary carbocation.

    formation of C-C bond between cyanide anion and benzylic cation.svg

    In an SN2 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.

    mechanism of sn2 reaction between (1-chloroethyl)benzene and sodium cyanide.svg

    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 SN1 reaction of tert-butyl bromide with water.

    reaction equation for reaction of water and t-butyl bromide to give t-butyl alcohol and HBr.svg

    It can be helpful to take inventory of which bonds have been formed, and which bonds have been broken.

    breakdown of bonds formed and broken in reaction between t-butyl bromide and water to generate t-butanol and HBr.svg

    The curved arrows we draw must account for ALL of these bonding changes. Since we are dealing with an SN1 reaction process, the first step will be cleavage of the C-Br bond to give a carbocation and and a bromide anion.

    mechanism illustrating cleavage of C-Br bond in t-butyl bromide to give t-butyl cation and bromide anion.svg

    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.

    mechanism for bond formation between water and tert-butyl cation to generate an oxonium ion.svg

    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.

    mechanism illustrating deprotonation of the oxonium ion .svg

    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:

    overall mechanism for SN1 reaction between t-butyl bromide and water to generate t-butyl alcohol and HBr.svg

    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:

    breakdown of bonds formed and broken in reaction between t-butyl alcohol and HBr to generate t-butylbromide and water.svg

    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.

    mechanism for protonation of t-butyl alchol with HBr.svg

    mechanism showing C-O bond breaking to generate t-butyl cation and water.svg

    The bromide ion generated in the first step can then react with the t-butyl cation to generate t-butyl bromide.

    mechanism for bond formation between bromide ion and t-butyl cation to generate t-butyl bromide.svg

    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:

    overall mechanism for reaction between t-butyl alcohol and HBr to generate t-butyl bromide and water.svg

    Curved Arrow Summary double-headed mechanistic arrow.svg

    • Curved arrows double-headed mechanistic arrow.svg 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.

    Exercises

    Draw curved arrows to indicate mechanisms for the following reactions:

    6.6.png

    Solutions

    6.6sol.png

    Contributors and Attributions