Haloalkanes and Haloarenes · Lesson 7 of 16
Nucleophiles and the SN2 Mechanism
“Predict substitution products and explain the concerted backside pathway through electron flow, rate and steric access.”
• Identify nucleophiles, electrophilic carbon and leaving groups in haloalkane substitution. • Predict products from common and ambident nucleophiles with carbon-count checks. • Explain the one-step SN2 pathway and distinguish a transition state from an intermediate. • Use a simple bimolecular rate law and steric reasoning under comparable conditions.
Follow the electron donor, not just the reagent bottle
In CH₃Br+HO⁻→CH₃OH+Br⁻, oxygen donates a lone pair to the electrophilic carbon while the C–Br electron pair leaves with bromine. The entering species is the nucleophile, the haloalkane is the substrate and bromide is the leaving group. A nucleophile can be negatively charged, such as OH⁻, or neutral, such as NH₃ or H₂O. Its defining role is electron-pair donation to an electron-deficient reaction centre.
| Reagent / attacking group | Formal organic product | Product family or qualification |
|---|---|---|
| Aqueous OH⁻ | ROH | Alcohol |
| Water in suitable solvolysis | ROH after proton transfer | Alcohol; neutral attack first gives oxonium |
| R′O⁻ | ROR′ | Ether; efficient SN2 usually needs a suitable unhindered alkyl partner |
| I⁻ | RI | Halide exchange |
| NH₃, then proton transfer | RNH₂ | Primary amine; excess ammonia can limit further alkylation |
| R′NH₂ or R′R″NH | RNHR′ or RNR′R″ after proton transfer | Further amine alkylation; mixtures possible |
| CN⁻ from KCN | R–C≡N | Nitrile; adds one carbon |
| AgCN in the stated comparison | R–N≡C | Isocyanide, better represented R–N⁺≡C⁻ |
| Nitrite with KNO₂ | R–O–N=O | Alkyl nitrite in the stated comparison |
| AgNO₂ | R–NO₂ | Nitroalkane in the stated comparison |
| Carboxylate, R′COO⁻ | R′COOR | Ester |
| LiAlH₄ as a reducing reagent | RH | Reduction outcome, not a universal identical SN2 mechanism |
| Suitable carbon nucleophile / organometallic | R–R′ | C–C formation depends strongly on reagent/substrate |
A table of formal products does not make every reaction equally selective or give it the same mechanism. Neutral nucleophile attack often gives a positively charged intermediate that must lose a proton. For ammonia a balanced net equation is RX+2NH₃→RNH₂+NH₄X, and the amine product can itself react again. In an ether synthesis, phenoxide plus a primary ethyl halide can give an aryl–O–ethyl ether; this is attack at the ethyl carbon, not ordinary displacement on the aromatic ring.
Ambident nucleophiles offer two attachment atoms
Cyanide can attack through carbon or nitrogen. In the common KCN comparison, carbon attack predominantly forms nitrile R–C≡N, whereas AgCN commonly favours isocyanide R–N≡C under the stated preparation conditions. Silver bonding and the availability of the two sites help change the outcome. It is too crude to explain every such selectivity solely by saying one final bond is “always stronger”; counterion, solvent and kinetic pathway matter. Nitrite similarly can attach through O to give an alkyl nitrite or through N to give a nitroalkane.
Problem
Predict the product of 1-bromobutane with KCN in the stated ethanol/water substitution conditions.
- 1.The attacking atom is the carbon end of cyanide. It bonds to the terminal carbon that originally bore Br.
- 2.The product is CH₃CH₂CH₂CH₂C≡N, pentanenitrile, plus bromide with its counterion. A four-carbon butyl group becomes a five-carbon nitrile.
- 3.Do not call the product butanenitrile or butyl isocyanide. The first loses a carbon and the second uses the wrong attachment atom for this stated comparison.
One concerted step, with two bond changes
In the ordinary SN2 model, the nucleophile approaches from the side opposite the C–X bond. Donation into an accepting antibonding orbital allows C–Nu formation and C–X cleavage in the same elementary event. Full-headed curved arrows begin at the nucleophile’s lone pair and at the breaking C–X bond, ending at carbon and at X respectively. A curved arrow is an electron-flow description; it is not a trajectory of the whole molecule or the same symbol as the overall reaction arrow.
At the transition state, carbon has partial bonds to both entering and leaving groups. The three remaining substituent bonds are approximately coplanar in the idealised picture, while the entering and leaving groups lie approximately opposite each other. This is not an isolable five-full-bond carbon intermediate. There is one barrier and no separate carbocation minimum in this pathway. Progress through the barrier inverts the three-dimensional arrangement at the reacting tetrahedral carbon; the stereochemical meaning is developed in the chirality lesson.
Problem
For a model SN2 system, k=0.20 L mol⁻¹ s⁻¹, [RX]=0.10 mol L⁻¹ and [Nu⁻]=0.20 mol L⁻¹. Find v and the effect of doubling both concentrations.
- 1.Substitute into the product law: v=0.20×0.10×0.20=0.0040 mol L⁻¹ s⁻¹. The L/mol in k cancels one concentration factor.
- 2.Doubling RX alone doubles rate and doubling nucleophile alone doubles rate. Doubling both multiplies rate by 2×2=4.
- 3.The new rate is 0.016 mol L⁻¹ s⁻¹. These are illustrative model data, not measured universal constants for all alkyl halides.
Steric access and leaving-group ability work together
For comparable simple alkyl substrates with the same leaving group and conditions, the usual SN2 order is methyl>primary>secondary≫tertiary. Extra groups obstruct backside approach and raise the barrier. The nearby carbon also matters: neopentyl bromide, BrCH₂C(CH₃)₃, is primary but very hindered by the adjacent quaternary carbon. It can be slower than an ordinary secondary substrate; “all primary halides are fast” is not a valid universal rule.
Problem
Choose faster SN2 partners: cyclohexylmethyl chloride versus chlorocyclohexane; 1-iodobutane versus 1-chlorobutane.
- 1.Cyclohexylmethyl chloride has a primary CH₂Cl outside the ring; chlorocyclohexane has a secondary ring reaction centre. The primary side-chain chloride is ordinarily more accessible to backside attack.
- 2.For the same butyl group, iodide is a better leaving group than chloride and the C–I bond is easier to cleave, so 1-iodobutane reacts faster in the intended comparable SN2 conditions.
- 3.The two comparisons isolate steric environment and leaving-group ability respectively. Do not change solvent and nucleophile simultaneously and attribute every rate difference to only one structural feature.
The familiar alkyl leaving-group order is I>Br>Cl≫F under comparable substitution conditions. A polar aprotic solvent often helps anionic nucleophiles by not strongly hydrogen-bonding around them, whereas protic solvents can stabilise and restrain the attacking anion. Nucleophilicity also depends on the solvent and on steric bulk; a strong base is not automatically the best nucleophile for an inaccessible carbon. Allylic and benzylic substrates can be especially reactive, but direct vinylic and aryl C–X bonds do not follow this ordinary saturated-carbon SN2 scheme.
Compare 1-bromo-3-methylbutane, 2-bromo-3-methylbutane and 2-bromo-2-methylbutane with the same nucleophile and solvent. The intended ordinary SN2 order is primary > secondary ≫ tertiary. The secondary member is additionally crowded by nearby branching. Together with the separate primary-halide comparison, this shows why both the degree of the reacting carbon and the position of branches must be inspected.
Quick Quiz
Quiz
Where does the electron-flow arrow for nucleophile attack begin?
Explanation: Curved arrows track electrons from a source to an accepting site.
What is the principal KCN product from C₂H₅Br?
Explanation: Carbon-end attack adds the cyanide carbon, giving propanenitrile.
What is present in a simple SN2 transition state?
Explanation: Bond formation and cleavage are coupled in one event.
What happens to the model rate if both reactant concentrations double?
Explanation: v=k[RX][Nu] contains both concentration factors.
Why is neopentyl bromide an exception to “primary means fast”?
Explanation: The primary CH₂ centre is sterically crowded by the neighbouring C(CH₃)₃ group.
Practice Problems
- Give products for bromoethane with aqueous OH⁻, ethoxide, KCN and AgNO₂ under appropriate substitution conditions.
- Describe the electron flow in 1-bromobutane plus cyanide.
- Rank 1-bromopentane, 2-bromopentane and 2-bromo-2-methylbutane for ordinary SN2.
- Rank 1-bromobutane, 1-bromo-3-methylbutane, 1-bromo-2-methylbutane and neopentyl bromide qualitatively for SN2.
- A simple SN2 system has k=0.50 L mol⁻¹ s⁻¹, [RX]=0.040 M and [Nu]=0.10 M. Find v.
- Explain KNO₂ versus AgNO₂ product linkage without confusing nitrite and nitro groups.
- Why can sodium phenoxide and ethyl chloride make phenetole without proving aryl SN2?
The formal products are ethanol, diethyl ether, propanenitrile and nitroethane respectively. OH and ethoxide attack through O, cyanide mainly through C in the KCN route, and the stated silver nitrite comparison favours N-linked nitro product. Conditions and competing elimination must still be checked.
Key Takeaways
• Nucleophiles donate electron pairs and leaving groups retain the broken bond pair. • Product identity depends on the attacking atom, including ambident linkage choices. • Cyanide substitution adds one carbon when it forms a nitrile. • The ordinary SN2 pathway is concerted and has no carbocation intermediate. • Backside attack produces geometric inversion at the reaction centre. • The simple bimolecular rate law depends on both reactants. • Nearby branching, solvent and leaving-group identity qualify simple rate rankings.