Q3(a). E2 of 2-fluoropentane: Hofmann orientation

Introduction. Methoxide is a strong base and 2-fluoropentane is a secondary halide, so the main reaction is E2 elimination. Two kinds of β-hydrogen can be removed: from C-1 (CH₃) or from C-3 (CH₂). Because F⁻ is a very poor leaving group, the less substituted alkene is the major product, the reverse of the Zaitsev rule.

Scheme: Elimination: pent-1-ene 70 %, pent-2-ene 30 % (E > Z); a little 2-methoxypentane (SN2) also forms.

1. Products. Major: pent-1-ene (about 70 % of the alkene). Minor: pent-2-ene (about 30 %), as a mixture with E more than Z (the E isomer has no methyl–ethyl cis strain). A small amount of 2-methoxypentane forms by SN2 substitution.
2. Leaving group. The C–F bond is the strongest carbon–halogen bond and F⁻ is the most basic halide ion, i.e. the poorest leaving group. In the E2 transition state the base has already pulled off much of the proton while the C–F bond is still largely intact.
3. Carbanion-like transition state. Negative charge therefore builds up on the β-carbon (an E1cb-like or variable E2 transition state). Its stability follows carbanion stability: primary > secondary, because alkyl groups destabilise a negative charge.
4. Orientation. Removing H from C-1 puts the developing charge on a primary carbon; removing H from C-3 puts it on a secondary carbon. The C-1 pathway has the lower energy transition state, so pent-1-ene (the Hofmann product) is formed faster. There are also three H atoms on C-1 against two on C-3.
5. Comparison. With Cl, Br or I the halide leaves early, the transition state is alkene-like and the more substituted (Zaitsev) alkene is major: for 2-halohexanes the less substituted alkene is only 19 % (I), 28 % (Br), 33 % (Cl).

Conclusion. Major: pent-1-ene; minor: (E)- and (Z)-pent-2-ene (with a little 2-methoxypentane). Fluoride leaves late, the E2 transition state is carbanion-like, and the more stable primary carbanion-like transition state gives the less substituted alkene.

Application. The same carbanion-like E2 explains Hofmann elimination of quaternary ammonium hydroxides, used to locate nitrogen in alkaloid structures and to make terminal alkenes.