E2 Reaction Coordinate

Introduction

An alkyl halide can lose a hydrogen halide equivalent and form an alkene instead of receiving a new group. In an E2 reaction, a base removes a hydrogen from a carbon next to the leaving group while the leaving group departs. The C=C bond forms in the same elementary event. Its one-peak reaction coordinate resembles SN2 in being concerted, but the base attacks a proton and the product is an alkene.

Core explanation

Label the carbon bearing X as the alpha carbon and an adjacent carbon as a beta carbon . A beta hydrogen is required on a beta carbon. For a typical transformation, B⁻ + R–CH₂–CHX–R′ → BH + R–CH=CH–R′ + X⁻. The base's electron pair forms a bond to the beta hydrogen. The beta C–H bond electrons form the pi bond between alpha and beta carbons. At the same time, the alpha C–X bond electrons move onto X. Three curved arrows describe one step, not three successive intermediates.

The E2 transition state has partial B–H formation, partial C–H and C–X cleavage and partial C=C formation. On a free-energy versus reaction-progress plot, draw one maximum and no carbocation valley. The base and alkyl substrate both participate in the rate-determining event, so ideal E2 kinetics follow rate = k[alkyl substrate][base]. Doubling either concentration alone doubles the initial rate, while doubling both gives four times the rate under fixed conditions. A favourable alkene product still requires passage through an activation barrier.

Efficient E2 usually requires the breaking C–H and C–X bonds to be antiperiplanar : coplanar and pointing in opposite directions. This geometry lets the C–H bonding orbital overlap properly with the C–X antibonding orbital as the new pi bond forms. An acyclic chain can often rotate to reach this arrangement, but not every conformer reacts equally fast. In cyclohexane, the requirement becomes trans-diaxial, discussed separately. Because the geometry is constrained, E2 can be stereospecific.

Strong bases commonly promote E2, especially with secondary or tertiary substrates where SN2 is hindered. Hydroxide and alkoxides are typical; bulky bases such as tert-butoxide favour proton removal over backside carbon attack. A tertiary alkyl halide can undergo E2 even though normal SN2 is blocked. A primary halide can also eliminate with a sufficiently strong, bulky base, although an unhindered nucleophile often favours substitution. The availability of at least one beta hydrogen must always be checked; if none exists, this ordinary E2 route is impossible.

Which beta hydrogen is removed may affect the alkene product. If more than one beta carbon bears hydrogen, different positional isomers can form. Less hindered bases often favour a more substituted alkene when available, whereas a bulky base may remove the most accessible proton and favour a less substituted alkene. These are selectivity trends, not a substitute for drawing all accessible antiperiplanar conformations.

Step-by-step reasoning

Mark X, its alpha carbon and every adjacent beta carbon. Confirm that a beta hydrogen exists. Choose a conformation with a beta C–H bond opposite and coplanar to C–X. Draw three arrows simultaneously: base to H, C–H bond to the alpha–beta bond, and C–X bond to X. Replace the alpha–beta single bond with C=C, then check that atoms and charges balance.

Visual explanation

Draw a Newman projection looking down the alpha–beta bond. Place H and X 180 degrees apart, one at the front carbon and the other at the rear. Show a base approaching H from outside. Beside it, draw a single-hill energy plot with the transition-state peak labelled "C–H and C–X breaking; C=C forming." Do not insert a carbocation intermediate valley.

Real-world analogy

Imagine two people pulling opposite tabs while a hinged bar between them snaps into a straighter, locked position. Both tabs release during the same movement; neither release creates a stable waiting stage. The tabs represent H and X, while the locked bar represents the new pi bond. The analogy emphasises simultaneous change, not a literal mechanical force model.

Real-world example

Heating 2-bromobutane with an alkoxide base can produce butenes by E2. Removal of hydrogen from C1 gives but-1-ene, while removal from C3 gives but-2-ene; the latter may appear as E and Z stereoisomers. Product proportions depend on base size, solvent, temperature and accessible conformations. A chemist must therefore distinguish mechanism from regioselectivity and stereoselectivity.

Why?

Why do strong bases favour E2 with crowded substrates? They can reach a beta hydrogen more easily than they can attack the substituted alpha carbon from behind. At the transition state, proton removal, pi-bond formation and leaving-group departure cooperate. This route avoids forming an unstable free carbanion or carbocation while producing a stable alkene and the base's conjugate acid.

Common misconception

"E2 first makes a carbocation and then loses H." That is E1-like logic. E2 is concerted, with a one-step three-arrow mechanism and a one-peak energy profile. It is also not enough to find any hydrogen in the molecule; the hydrogen must be on an adjacent beta carbon and suitably aligned.

Worked example

Question: Predict the organic product when 2-bromo-2-methylpropane reacts with a strong base that removes a beta hydrogen, and state the shape of the ideal reaction-coordinate plot.

Reasoning: The brominated tertiary carbon is alpha. Each methyl group is a beta carbon with hydrogens. Base removes one such H while C–Br breaks and the alpha–beta pi bond forms. All three methyl groups are equivalent for this purpose.

Answer: The organic product is 2-methylpropene. The ideal E2 energy plot has one transition-state peak and no carbocation intermediate.

Quick check

1. How many curved arrows and how many elementary steps describe an ordinary E2 elimination? Answer: Three curved arrows show the coupled electron movement in one elementary step.

Exam focus

Identify alpha and beta carbons before drawing the mechanism. Start the base arrow at its lone pair, the pi-bond arrow at the beta C–H bond, and the leaving arrow at C–X. State the bimolecular rate law and antiperiplanar requirement. Check for a beta hydrogen and show all plausible alkene positions when asked.

Advanced insight

The transition state can be more carbanion-like or more carbocation-like depending on substrate, base and leaving group, even though it remains one elementary E2 event. This variation can influence isotope effects and selectivity. A primary kinetic isotope effect is often observed when beta H is replaced by D because the C–H bond is being broken in the rate-limiting transition state.

Summary

E2 is a concerted bimolecular elimination: base removes a beta hydrogen, the C=C bond forms, and the leaving group departs in one step. The reaction coordinate has one peak and no intermediate. It requires a beta hydrogen and usually favourable antiperiplanar geometry. Strong bases, especially bulky ones with crowded substrates, favour E2; base size and conformations influence which alkene forms.

Practice questions

1. What is the beta carbon in an alkyl halide? Answer: It is a carbon directly adjacent to the alpha carbon that bears the leaving group. 2. Write the ideal E2 rate law for an alkyl bromide reacting with ethoxide. Answer: Rate = k[alkyl bromide][ethoxide], first order in each reactant. 3. Why cannot an ordinary E2 elimination occur if there is no beta hydrogen? Answer: The base has no adjacent C–H bond to break while the alpha–beta pi bond forms. 4. What product results when tert-butyl bromide loses HBr by E2? Answer: 2-Methylpropene, formed from equivalent beta methyl groups.