Organic Chemistry · Reactions of Alkyl Halides: Nucleophilic Substitutions and Eliminations

The E2 Reaction and the Deuterium Isotope Effect

8 min read

On this page 8 sections

In 30 seconds
Why this matters
The college version
Eli explains
Key takeaway
Check yourself
Study tools
Sources & references

In 30 seconds

The is a bimolecular elimination: in one concerted step, a base removes a β-hydrogen while the leaving group departs from the α-carbon — no intermediate forms. The rate depends on the concentrations of both the alkyl halide and the base:

$rate = k[RX][base]$

The reaction is second order overall; the "2" in E2 records that bimolecularity, just as it does in S_N2. Two structural requirements control E2: the β-hydrogen and the leaving group must be (roughly opposite each other across the C–C bond), and the base must be able to reach that β-hydrogen.

The deuterium isotope effect is the experimental tool that reveals how the β C–H bond participates. Because deuterium (²H, D) is twice as heavy as protium (¹H), C–D bonds vibrate more slowly and sit in a lower level; they are harder to break. If replacing a β-H with D slows the reaction substantially, the C–H bond is breaking in the — exactly what E2 predicts.

Why this matters

E2 is the most common elimination mechanism for alkyl halides, the standard laboratory route to alkenes, and a direct competitor of S_N2 for the same substrates and bases. The isotope effect is a mechanistic probe: a large (KIE) means a C–H bond breaks in the slow step, distinguishing E2 from E1 and E1cB; it is used throughout physical organic chemistry and enzymology to assign mechanisms. E2's geometry also predicts alkene stereochemistry, relevant to drug synthesis and to enzyme-catalyzed dehydrations that form trans alkenes.

The college version

Core Concepts

One step, two events: the concerted transition state

In E2 there is no carbocation and no carbanion. In the single transition state, four events happen together: the base begins pulling the β-hydrogen, the C–H bond starts to break, the C–C bond takes on double-bond character, and the C–X bond begins to break as the leaving group departs. In arrow-pushing language: the base's electron pair forms the new H–base bond; the C–H bonding pair shifts to become the C=C π bond; and the C–X pair moves onto the departing leaving group. Because everything happens in one step, E2 is favored by strong bases and by good leaving groups.

Rate law and the meaning of bimolecular

Doubling either the base or the alkyl halide concentration doubles the rate. The units of the second-order rate constant follow from the rate law; rate has units of concentration per time, M s^-1:

$[rate] = M s -1 = [k] \cdot M \cdot M$

Solving for the units of k:

$[k] = M s -1 M \cdot M = M -1 s -1$

A first-order (E1) rate constant, by contrast, has units of s^-1 — a useful check when comparing mechanisms.

The anti-periplanar geometry requirement

For the π bond to form, the β-H and the leaving group must be anti-periplanar: the C–H and C–X bonds should lie in the same plane, pointing in opposite directions (roughly 180° apart). In acyclic molecules this is achieved in a staggered conformation. This requirement makes E2 stereospecific: the substrate's stereochemistry determines the alkene's, and simple acyclic substrates generally give the more stable trans (E) alkene.

The deuterium isotope effect: how heavy hydrogen reports on mechanism

Deuterium and protium have identical electronic structures, so C–D and C–H bonds have the same intrinsic strength — but not the same zero-point energy. A vibrating bond has a lowest energy level E₀ = 12hν, and the vibrational frequency depends on the reduced mass μ of the two atoms:

$ν = 1 2 π \sqrt k force μ$

The C–D pair has a larger reduced mass than C–H, so C–D vibrates at roughly 1/1.36 the frequency and sits about 1.2 kcal/mol lower in zero-point energy. Breaking the bond therefore costs more energy for C–D, which slows the reaction when that bond breaks in the rate-determining step. The measured ratio

$k H k D \approx 5 -- 10$

is a primary kinetic isotope effect. A ratio near 1–2 (a secondary effect) means the C–H bond was not broken in the slow step. E2 shows a primary KIE on the β-hydrogen because β C–H cleavage is part of the rate-determining step — strong evidence against E1, whose slow step is only loss of the leaving group.

How It Works / Step-by-Step Process

Identify the substrate and base; confirm that a β-H exists.
Verify the E2 rate law: second order, one in each reactant.
Place the β-H and LG anti-periplanar; expect the trans alkene in acyclic cases.
Interpret kinetics: a primary KIE on the β-H supports E2; a near-1 ratio points to a stepwise mechanism.

Worked Example: Second-Order Rate Constant and Units

Suppose a student measures the E2 reaction of 2-bromopropane with ethoxide at 25 °C: with [2-bromopropane] = [EtO^-] = 0.10 M, the initial rate is 1.2 × 10^-3 M s^-1.

Step 1 — write the rate law. E2 is second order:

$rate = k[CH 3 CHBrCH 3][EtO -]$

Step 2 — substitute and solve for k:

$1.2 \times 10 -3 M s -1 = k (0.10 M)(0.10 M)$

$k = 1.2 \times 10 -3 M s -1 1.0 \times 10 -2 M 2 = 1.2 \times 10 -1 M -1 s -1$

Step 3 — check units. The M in the numerator cancels one M in the denominator, leaving M^-1s^-1, exactly what a second-order constant requires.

Worked Example: What the Deuterium Isotope Effect Tells You

Now suppose the same reaction is run with the deuterated substrate CD₃CHBrCD₃, and the student measures k_H/k_D = 6.0 under identical conditions.

Interpret the number. A ratio of 6.0 sits in the primary-isotope-effect range (5–10): replacing the β-H with D slowed the reaction sixfold, so the β C–H(D) bond breaks in the rate-determining step — consistent with the concerted E2 transition state and inconsistent with E1, whose slow step (C–Br ionization) does not touch the β-hydrogens.

Quantify the slowdown. Using k_H = 1.2 × 10^-1 M^-1s^-1 from the example above:

$k D = k H k H /k D = 1.2 \times 10 -1 M -1 s -1 6.0 = 2.0 \times 10 -2 M -1 s -1$

The deuterated substrate reacts six times more slowly — the kinetic consequence of the ~1.2 kcal/mol higher barrier for breaking the C–D bond. (These are practice-exercise numbers illustrating the method, not a report of a specific published experiment.)

Common Confusions

Do not confuse	With	Difference
E2 (one step)	E1 (two steps)	E2 is concerted, second order, needs a strong base; E1 goes through a carbocation and is first order
E2	S_N2	Both are bimolecular, but S_N2 attacks the carbon and substitutes; E2 removes a β-H and forms a π bond
Primary KIE (≈5–10)	Secondary KIE (≈1–2)	Primary: the labeled C–H bond breaks in the slow step; secondary: it does not
“Deuterium is radioactive”	Deuterium	It is a stable isotope; effects arise from vibration and zero-point energy
“E2 needs only a weak base”	E2 conditions	E2 needs a strong base; weak bases under polar protic conditions favor E1/SN1
Anti-periplanar means “same direction”	Anti geometry	Anti means opposite directions (~180°); syn (same side) is the forbidden geometry

Eli explains

The same idea, in plain words

Explain it like I’m 10

E2 is like two kids letting go of a rope at the same time so a flag snaps up between them — the base pulls one end while the leaving group drops the other, in one motion. Swap a normal hydrogen for a heavy one and the rope is harder to pull, so the trick slows down a lot. That slowdown tells you the hydrogen really was being pulled in the slow step.

Key takeaways

E2 is one step, bimolecular: rate = k[RX][base]; k has units M^-1s^-1.
The base removes the β-H as the leaving group departs — no intermediate forms.
β-H and leaving group must be anti-periplanar (coplanar, ~180° apart); acyclic E2 usually gives the more stable trans alkene.
Strong bases (OH⁻, RO⁻) favor E2; the leaving group must accept the electron pair.
Primary KIE: k_H/k_D ≈ 5--10 when the β C–H bond breaks in the rate-determining step — observed for E2.
A small KIE (near 1) means the C–H bond breaks after the slow step, as in E1.
Deuterium is a stable isotope; isotope effects come from zero-point energy differences, not radioactivity.

Check yourself

6 review questions from the chapter. Try each one, then open the answer.

What does the “2” in E2 mean, and what is the E2 rate law?
Show answer
The reaction is bimolecular: base and substrate meet in the rate-determining step. Rate = k[RX][base].
What geometric relationship must hold between the β-H and the leaving group?
Show answer
They must be anti-periplanar — coplanar and pointing in opposite directions (~180° apart).
What are the units of a second-order rate constant, and why?
Show answer
M^-1s^-1, from M s^-1 = [k] · M · M.
What is a primary kinetic isotope effect, and what range does it occupy?
Show answer
A large slowdown (typically k_H/k_D ≈ 5--10) caused by substituting D for H at a bond that breaks in the rate-determining step.
Why does replacing a β-H with deuterium slow an E2 reaction?
Show answer
The C–D bond has a lower zero-point energy (higher effective barrier, ~1.2 kcal/mol more), so bond breaking is slower.
If you measured k_H/k_D ≈ 1.1 for an elimination, what would that suggest?
Show answer
The C–H bond is not broken in the slow step — consistent with a stepwise mechanism such as E1 (slow ionization first) rather than E2.

Keep learning

Ready to build on this? Continue to the next lesson.

Study tools & related lessonsKey vocabulary · Related

Key vocabulary

E2 reaction: A one-step elimination whose rate depends on both base and substrate
concerted mechanism: A reaction with a single transition state and no intermediate
anti-periplanar: Geometry in which the β-H and leaving group lie in one plane, pointing opposite ways
rate-determining step: The slowest step, which sets the overall rate
primary kinetic isotope effect: A large k_H/k_D (≈5–10) when the labeled bond breaks in the slow step
zero-point energy: The lowest vibrational energy of a bond, E₀ = 12hν

Sources & references

openstax.org — Organic Chemistry

This lesson was adapted from the open educational references above; their licenses and attributions are preserved. See Copyright & Licensing.

Educational content only. It is not medical, legal or professional advice. Found an error? Tell us.