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9476 / 2027
Organic Chemistry

Lesson 11 of 34

SN2 forms and breaks bonds in one concerted step

Backside approach makes steric access and inversion central to the mechanism.

A-Level 9476 (2026-2027)

  • I can explain both nucleophilic-substitution mechanisms.Syllabus

    Syllabus 9476, 11.3(n). Explain both nucleophilic-substitution mechanisms.

  • I can explain substitution stereochemistry for optically active substrates.Syllabus

    Syllabus 9476, 11.5(b). Explain substitution stereochemistry for optically active substrates.

SN2 forms and breaks bonds in one concerted step

Backside approach makes steric access and inversion central to the mechanism.

In SN2, a nucleophile gives a pair of electrons to the carbon that carries the leaving group. At the same time, the C-X bond pair moves onto X. Both changes happen together, in one step with one transition state. No carbocation forms. The nucleophile comes in from the side opposite the leaving group, so bulky groups around that carbon get in the way and slow the reaction a lot.

SN2: electron movements and the inverted product

Hydroxide attacks 2-bromobutane from the left, opposite the carbon-bromine bond on the right. One curly arrow runs from an oxygen lone pair to carbon and the second from C-Br to Br. The product is drawn below with OH on the left and all three retained groups turned to the right: methyl remains in the page, ethyl towards the viewer and H behind.

Read both arrows in the reactant together. The lower drawing shows the resulting butan-2-ol arrangement and Br-. Follow each of CH3, C2H5 and H through the change; inversion is a three-dimensional change at carbon, not a change of optical-rotation sign inferred from the drawing.

For an SN2 step, rate = k[halogenoalkane][nucleophile]. Methyl and primary halogenoalkanes are easier to reach than secondary ones. Tertiary ones are strongly blocked. These are general preferences. The solvent, the nucleophile and competing elimination still matter.

Inversion of configuration means the other three groups flip over, like an umbrella in the wind, as the nucleophile replaces the leaving group from the opposite side. So a single enantiomer gives the inverted form, not a racemic mixture. Turning the drawing of the product round on the page is not inversion.

Worked exampleConstruct the SN2 drawing without adding an intermediate

What must be retained when drawing hydroxide substitution at the chiral carbon of 2-bromobutane?

Try it first. Then show the worked stepsHide the worked steps
  1. Copy the four groups on carbon, including the wedge and hashed bonds. Put the oxygen lone pair on the opposite side from Br and draw both curly arrows shown above in the same step.
  2. If displaying the transition state, show partial O-C and C-Br bonds with dashed lines, bracket the whole arrangement with an overall negative charge and the transition-state symbol. The three retained groups are coplanar at that stage; it is not a stable carbon with five full covalent bonds.
  3. Complete C-O formation and C-Br breaking. Draw the three retained groups turned through the central plane, as in the lower drawing; do not leave their tetrahedral arrangement unchanged while simply relabelling Br as OH.
  4. Check atoms and charge: C4H9Br + OH- gives C4H9OH + Br-. Only the initially selected substrate enantiomer is being followed.
Answer

The SN2 pathway gives the inverted alcohol arrangement. It has no separate carbocation step and does not give equal amounts of both alcohol enantiomers.

Exam-style question

Bromoethane is warmed with NaOH(aq). Name and describe the mechanism, and write the rate equation. [3]

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