# Why is the reactivity of primary alkyl halides with nucleophiles (SN2 mechanism) greater than secondary and tertiary alkyl halides?

For the reaction between alkyl halide and a nucleophile, following the SN2 mechanism, the reactivity of alkyl halides is in the order: primary halide > secondary halide > tertiary halide.

If you consider the reactivity of alcohols with halogen acids to form alkyl halides, the reactivity of alcohols is in the order: tertiary alcohol > secondary alcohol > primary alcohol.

Why is there a difference in order of reactivity in both the cases?

The order of reactivity by substitution in these two reactions is difference because they have different mechanisms. The substitution of an alkyl halide by a strong nucleophile in a polar aprotic solvent is an SN2 mechanism. A $\ce{C-Nu}$ bond forms and a $\ce{C-X}$ bond breaks at the same time:

$$\ce{CH3CH2Br + NaOH ->[\text{acetone}] Na}\bigg[\ce{CH3CH2(Br)(OH)}\bigg]^{^\ddagger_{-}}\ce{->CH3CH2OH +NaBr}$$

All else equal, the rates of these reactions are controlled by sterics on the alkyl halide, which control how well the nucleophile can react the $\ce{C-X}\ \sigma^*$ orbital. Thus:

$$\text{Methyl}>1^\circ>2^\circ>>3^\circ$$

The reactions of alcohols with hydrohalic acids follow a different mechanism (mostly). These reactions happen in a protic medium ($\ce{HX}$ is protic) with a poor leaving group ($\ce{OH}$), at least initially. These are conditions that favor the step-wise SN1 mechanism, where $\ce{C-X}$ bond breaking occurs before $\ce{C-Nu}$ bond forming. A carbocation is formed as a key intermediate.

$$\ce{(CH3)3COH + HBr -> [(CH3)3COH2+]Br- -> [(CH3)3C+]Br- + H2O -> (CH3)3CBr + H2O}$$

According to the Hammond postulate, the stability of this carbocation influences the rate of reaction. More stable carbocation intermediates suggest lower energy transition states for the formation of those carbocations, which implies lower activation energies and faster rates. Thus:

$$3^\circ>2^\circ>>1^\circ,\text{Methyl}$$

• @BenNorris.Thank you for the answer. By my other question it follows that tertiary alcohols require no catalyst (more reactive), whereas primary and secondary alcohol (less reactive) require catalyst for the reaction with alkyl halides. Does the use of catalyst really imply that primary and secondary alcohols are less reactive than tertiary and they follow $S_N1$ mechanism? – Immortal Player Nov 10 '13 at 6:54
• The need for a catalyst to make the reaction productive does imply that primary and secondary alcohols are less reactive, as expected based on the $S_N1$ mechanism. The catalysts work by changing the mechanism to be more $S_N2$-like. – Ben Norris Nov 10 '13 at 19:54
• @BenNorris Good answer, but I think you could include steric hindrance as a reason too. Tertiary halides are quite hindered compared to the Primary ones. This also explains why neopentyl halide doesn't undergo Sn2. – Pritt says Reinstate Monica Apr 12 '17 at 2:10