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Organic Chemistry

Topic 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)

SN2 forms and breaks bonds in one concerted step

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

In SN2, a nucleophile donates a pair to the carbon bearing the leaving group while the C-X pair moves to X. These changes occur together through one transition state, with no carbocation intermediate. The nucleophile approaches opposite the leaving group, so bulky groups around the reacting carbon strongly hinder the pathway.

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 elementary SN2 step, rate = k[halogenoalkane][nucleophile]. Methyl and primary substrates are generally more accessible than secondary ones; tertiary substrates are strongly hindered. These are pathway preferences, not a claim that solvent, nucleophile and competing elimination never matter.

Inversion of configuration means the three remaining groups turn through the transition-state geometry as the nucleophile replaces the leaving group from the opposite side. An initially single enantiomer gives the inverted arrangement for this pathway, rather than a racemic mixture. Merely rotating the product drawing on the page is not inversion.

Worked example

Construct the SN2 drawing without adding an intermediate

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

  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.