Organic Chemistry · Part 4 of 9

Nucleophilic Substitution at Saturated Carbon

Reaction Mechanisms, Part 4 · 9 sections · about 18,169 words · CSIR-NET Chemical Sciences, GATE Chemistry & IIT-JAM

This is the most examined mechanism in organic chemistry and the one students think they already understand. The two-mechanism picture taught at undergraduate level is a simplification, and the exam knows it. This part builds the real picture: the continuous spectrum between the two extremes, the ion pairs that explain partial racemisation, the substrates that refuse to react by either route, and the experimental evidence — rate laws, stereochemistry, salt effects, isotope effects — that tells you which pathway you are looking at. Two layers on every section: a slow, hand-held beginner path and a research-grade advanced/reference path.

The 9 sections in Part 4

  • 1The SN2 transition state, backside attack and Walden inversion Free below
  • 2Substrate structure — steric control, and the carbons that never react
  • 3The nucleophile — nucleophilicity is not basicity
  • 4The SN1 mechanism, the rate law, and the price of a carbocation
  • 5Ion pairs — the mechanism between the two mechanisms
  • 6Solvolysis, the Winstein–Grunwald treatment, and the borderline region
  • 7Leaving-group ability and its correlation with conjugate-acid pKa
  • 8Solvent effects on both mechanisms — the Hughes–Ingold analysis
  • 9Ambident nucleophiles, allylic systems, and the decision procedure

The SN2 transition state, backside attack and Walden inversion

Free extract

Section D.1 of Part 4, reproduced in full from the book — figures and all. No sign-in, no paywall on this section.

Beginner layer

One step, one transition state, no intermediate — and a geometry that is forced on the reaction by the shape of an orbital.

Start from the observation, not the picture. When methyl bromide is treated with hydroxide in aqueous ethanol, doubling the concentration of methyl bromide doubles the rate; doubling the concentration of hydroxide also doubles the rate. The rate law is second order overall, first order in each partner.

rate = k2[R–X][Nu−]   (second order overall)

A rate law counts the species present in and before the rate-determining transition state. Both partners appear, so both are present in the transition state — hence bimolecular, the ‘2’ in SN2. The mechanism has exactly one step: bond making and bond breaking are concerted, and there is no intermediate of any lifetime.

Mechanism — SN2 — concerted bimolecular substitution
  1. The nucleophile approaches the carbon along the line of the C–X bond, from the face opposite the leaving group.
  2. In a single continuous motion the Nu–C bond forms while the C–X bond breaks; the transition state has partial bonds to both, and negative charge shared between Nu and X.
  3. The three spectator groups flatten into a plane containing the carbon and then continue through to the other side.
  4. Product and leaving group separate. Configuration at carbon is inverted.
The SN2 transition statenodeHHHNuCX‡δ−δ−180° — collinear Nu…C…Xfilled lobe = nucleophile HOMO and the large carbon lobe of σ*(C–X); they are in phasethree unchanged bonds lie in the plane perpendicular to the reaction axis; carbon is sp² at the saddle point
Read the geometry off the orbitals, not the other way round. The empty orbital the nucleophile must fill is σ*(C–X). That orbital has a node between C and X and its largest lobe sits on carbon, on the face opposite X. A nucleophile can only overlap it constructively by approaching along the extension of the X–C axis — that is backside attack, and the 180° requirement is a consequence of orbital shape, not an extra postulate. At the saddle point the carbon is sp²-hybridised with its remaining p orbital shared between Nu and X: five atoms, and a three-centre four-electron array along the Nu–C–X axis (the nucleophile’s lone pair plus the old C–X bonding pair). Negative charge is dispersed over Nu and X in the transition state — remember that when you get to solvent effects in D.8. Schematic, not to scale.

Why 180°? Not by decree. The nucleophile has to put electrons somewhere, and the only orbital available on the substrate that can take them and simultaneously weaken the C–X bond is the antibonding σ*(C–X). That orbital is built out of a carbon lobe and a halogen lobe of opposite phase, with a node between them, and the carbon lobe — the big one, because carbon is the less electronegative partner — points away from X. Overlap with it is best along the extension of the X–C axis. Approach from anywhere else means poorer overlap, and approach from the front means overlapping the small out-of-phase halogen lobe as well, which cancels part of the interaction.

The stereochemical consequence — Walden inversion

If the nucleophile always arrives on the face opposite the leaving group, then at a stereogenic carbon the product configuration is fixed by the mechanism, not by chance. The three spectator groups are pushed through the plane like an umbrella in a gale.

Walden inversion — the umbrella pictureCXRRRbeforeR groups lean away from XCNuXRRRtransition stateR groups coplanar with CCNuRRRafterthey have leaned through‡The umbrella turns inside out. Configuration is inverted whether or not the CIP letter changes.
Inversion is not a rule bolted on to the mechanism; it is the mechanism seen from the side. Because the nucleophile arrives 180° from the leaving group, the three spectator groups must flatten into a plane and then continue through to the opposite face, exactly as an umbrella turns inside out in the wind. Every S N2 reaction at a stereocentre gives 100% inversion of spatial configuration — there is no mechanism by which it could give anything else, because the product geometry is fixed the moment the trajectory is fixed. What is not guaranteed is that the descriptor R/S changes; that depends on CIP priorities, which is a naming question, not a chemical one (D.1, worked example 2).
Walden inversion: the inversion of spatial configuration at a stereogenic carbon that accompanies every SN2 substitution. A clean SN2 reaction on an enantiopure substrate gives an enantiopure product of inverted configuration — 100% inversion, 0% racemisation.

The name is historical, and the history is worth thirty seconds because it is examinable. In the 1890s Paul Walden found that malic acid could be converted into chlorosuccinic acid and back again, and that the sign of rotation you ended with depended on which reagents you used and in which order. Starting from a single enantiomer of malic acid, one two-step sequence returned that same enantiomer while a different two-step sequence returned its mirror image. Since the starting material was the same in both cases, at least one of the steps must have turned the molecule inside out. That was the first evidence that a substitution can change configuration at carbon, decades before anyone could draw a transition state.

The two ends of the Walden cycle. Replacing OH by Cl and Cl by OH takes you round; run the two replacements with different reagents and you can arrive at either enantiomer of the starting acid. RDKit-rendered from SMILES with CIP labels shown; the configuration drawn is (S) in both cases, which is not a claim about which reagent gives which sign.

The trap inside the definition: R/S is a name, not a geometry

Configuration inverts every time. The CIP descriptor may or may not change with it, because the descriptor depends on the priority ranking of four groups and you have just swapped one of them. Work through the mechanical rule once and you will never guess again:

Rule. Rank the four groups in the substrate and in the product. If the incoming nucleophile occupies the same priority rank in the product that the leaving group occupied in the substrate, the descriptor changes. If the rank is different, work out the parity of the permutation: an odd number of priority swaps cancels the inversion and the descriptor is retained.

(R)-2-Bromooctane is treated with sodium hydroxide in aqueous acetone. Give the product with its configuration. Easy

Substrate class. Secondary alkyl bromide; hydroxide is a strong, small, charged nucleophile; the solvent is only moderately ionising. Bimolecular attack wins. SN2.
Geometry. Hydroxide attacks the back face; the C–Br bond breaks in the same motion. Spatial configuration is inverted. The product is octan-2-ol, enantiopure.
Descriptor. In the bromide the priorities are Br > C6 chain > CH₃ > H. In the alcohol they are OH > C6 chain > CH₃ > H. The incoming group holds the same rank (first) that the leaving group held, so the descriptor changes with the geometry.
Answer: (S)-octan-2-ol. Confirmed by machine assignment of CIP labels to the two structures below.
(R) → (S): the everyday case, where the descriptor does track the inversion. CIP labels here are assigned by RDKit from the SMILES, not by hand.

Methyl (R)-2-bromopropanoate is treated (a) with NaCN and (b) with NaN3, both in DMF. Both reactions are clean SN2. Give both products with configurations. Medium

Both reactions invert. Same substrate, same mechanism, same backside trajectory. There is no stereochemical difference between the two reactions at all.
Substrate priorities. Br (35) > C of CO₂CH₃ > CH₃ > H. The leaving group is rank 1.
(a) With cyanide. The new group is C≡N. Compare its first atom, carbon, with the ester carbon: both are carbon, so go outwards. Nitrile carbon duplicates to (N,N,N); ester carbon is (O,O,O). Oxygen beats nitrogen, so CO₂CH₃ > C≡N. The nucleophile lands at rank 2, not rank 1, and the ester is promoted from 2 to 1. That is one swap — an odd permutation — which cancels the geometric inversion. The descriptor is retained: methyl (R)-2-cyanopropanoate.
(b) With azide. The new group’s first atom is nitrogen (7), which outranks the ester’s carbon (6) immediately. Azide lands at rank 1, the same rank Br held. No permutation, so the descriptor follows the geometry. The descriptor changes: methyl (S)-2-azidopropanoate.
The point. One substrate, one mechanism, one stereochemical outcome — and two opposite answers to ‘did R become S’. Anyone who has memorised ‘SN2 turns R into S’ gets exactly half of this question wrong. All four descriptors were checked with an independent CIP implementation.
Same substrate, same inversion, different labels. The two products are drawn with the same spatial arrangement at the stereocentre — look at the wedge pattern, not the letter.
Advanced / reference layer

The experimental case for the concerted mechanism, the isotope-effect evidence, and what the transition state looks like when it is not symmetrical.

Proving inversion without knowing any absolute configuration

The neatest experiment in the field, and a standing favourite in Part C, uses radioactive iodide and optically active 2-iodooctane. Treat the enantiopure iodide with 128I− (or any labelled iodide) in acetone and monitor two things independently: the rate at which label enters the organic compound, and the rate at which optical rotation decays.

If every substitution event inverts, then each act of exchange converts one molecule of the (R) form into one molecule of the (S) form. That destroys two units of optical activity per event — the molecule that inverted no longer contributes its rotation, and it now cancels a partner. Hence:

krac = 2 kexch   (complete inversion)
krac = kexch   (complete retention or attack on a symmetric intermediate)

The measured ratio is 2. Notice how much the experiment gets for free: it needs no knowledge of absolute configuration, no reference compound, and no assumption about which way round the molecule is. It is a relative measurement, and it is decisive.

The classic substrate for the exchange experiment: secondary, no rearrangement possible, and iodide is both a good nucleophile and a good leaving group, so the identity reaction can be followed with a label.

Kinetic isotope effects: measuring the crowding at the transition state

Replace the hydrogens on the reacting carbon by deuterium and the rate barely changes — but the direction and size of the small change is diagnostic. An α-deuterium secondary kinetic isotope effect reports on how the out-of-plane bending vibration of the C–H bond changes between the ground state and the transition state.

MechanismCoordination at the reacting carbon in the TSTypical kH/kD per α-D
SN2Five — more crowded than the ground state; the C–H bending mode stiffens≈ 0.95–1.06 (unity or slightly inverse)
SN1 (ionisation)Three — less crowded; the bending mode loosens as carbon flattens≈ 1.15–1.25 (clearly normal)
The physical picture is simply steric: going to a five-coordinate transition state tightens the C–H bends, going to a three-coordinate cation relaxes them, and deuterium is more sensitive than protium to a change in a bending force constant. The numbers are small, but they are measurable to better than 1% and they do not depend on any assumption about the rate law.

The transition state is not a fixed object

Textbook drawings show a symmetrical transition state with Nu–C and C–X equally formed. Real ones are almost never symmetrical, and where they sit is governed by the same Hammond-type reasoning developed in Part 2. Two independent coordinates matter: how far the C–X bond has broken and how far the Nu–C bond has formed. A transition state in which both are advanced is called tight (or associative); one in which C–X breaking runs ahead of Nu–C making, leaving substantial positive charge on carbon, is loose (or dissociative, or ‘exploded’).

Better nucleophiles and poorer leaving groups tighten the transition state; poorer nucleophiles, better leaving groups and cation-stabilising substituents loosen it. A loose SN2 transition state has a great deal in common with an SN1 ion pair, and that overlap — not any sharp boundary — is what the borderline region of D.6 is made of. The two-dimensional bookkeeping device for this is the More O’Ferrall–Jencks diagram, drawn in D.6.

Gas phase: the barrier you see in solution is mostly desolvation

Run the same reaction with no solvent at all, in a mass spectrometer, and the energy profile changes shape completely. An anion and a neutral molecule attract each other, so they first fall into an ion–dipole complex that lies below the separated reactants. From there the system climbs to a central barrier and falls into a second ion–dipole complex on the product side. The result is a double-well profile, and for many identity reactions the central barrier lies below the energy of the separated reactants — the reaction has a negative apparent activation energy measured from infinite separation.

Same reaction, with and without solventFree energy G →reaction coordinate →gas phasein waterVertical scales are not comparable between the two curves; the shapes are the point.
In the gas phase (red) the ion and the neutral substrate attract each other into a pre-reaction ion–dipole complex, so the profile has two wells and a central barrier that can lie below the separated reactants. In water (blue) the same anion arrives wrapped in a hydrogen-bonded shell that must be partly stripped before it can reach carbon, and the wells disappear into a single large barrier. The lesson for the exam: a large part of a solution-phase SN2 barrier is the cost of desolvating the nucleophile, not the cost of rearranging bonds. That single sentence explains almost everything in D.3 and D.8. Schematic profiles.
Easy
A student writes that SN2 attack on (S)-2-bromobutane by NaSH gives (S)-butane-2-thiol, ‘because sulfur and bromine are both in the same priority position’. Diagnose the error and give the right answer.
Show solution
The student has the priority analysis right and the conclusion backwards. In 2-bromobutane the ranking is Br > CH₂CH₃ > CH₃ > H; in butane-2-thiol it is SH > CH₂CH₃ > CH₃ > H. The incoming group takes the same rank the leaving group held, so there is no compensating permutation and the descriptor changes along with the geometry. Configuration inverts, so (S) becomes (R)-butane-2-thiol. The ‘same position’ observation is exactly the condition for the label to flip, not to stay.
Med
Optically active 2-iodooctane racemises in acetone containing NaI at a rate that is first order in the iodide concentration. What does the iodide dependence tell you, and what would you conclude if the racemisation rate were independent of [I−]?
Show solution
First order in iodide means iodide is present in the rate-determining transition state: this is a bimolecular displacement, and racemisation is the accumulated result of repeated inversions (each event inverting one molecule, so krac = 2kexch). If racemisation were zero order in iodide, iodide would not be in the rate-determining step: the substrate would be ionising unimolecularly and iodide would only be intercepting whatever it produced. That is the SN1 signature (D.4), and for a secondary substrate in acetone — a poorly ionising solvent — it would be surprising enough to demand another explanation. The concentration dependence of the rate, not the stereochemistry, is what distinguishes the mechanisms.
Med
Explain, using orbitals, why a nucleophile cannot displace chloride from chlorobenzene or from vinyl chloride by an SN2 pathway, even though both have a perfectly good C–Cl bond and a good leaving group.
Show solution
Two reasons, and both are structural rather than energetic. (i) The back face does not exist as an accessible region. The carbon bearing chlorine is sp²-hybridised and trigonal; the position 180° from the C–Cl bond is occupied by the rest of the ring or by the vinyl carbon, so there is no open trajectory. (ii) The transition state would be impossible geometrically. An SN2 transition state requires the reacting carbon to become sp²-hybridised with a p orbital along the Nu…C…X axis. That carbon is already sp², and its p orbital is committed to the π system perpendicular to the trajectory. Attack along the axis would have to use an in-plane orbital, which is the wrong symmetry. Add to that the shortened, strengthened C–Cl bond (partial double-bond character from lone-pair donation into the π system) and the reaction has no viable route. Aryl halides do substitute, but by addition–elimination or by benzyne — both are Part 6 material, and neither is SN2.

Read the rest of Part 4

The remaining 8 sections of this part — Substrate structure — steric control, and the carbons that never react, The nucleophile — nucleophilicity is not basicity, The SN1 mechanism, the rate law, and the price of a carbocation… — and all nine parts of Reaction Mechanisms are part of ChemVidya Full Access, along with the other books, 55 Study Notes and 6,000+ practice questions.

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