The SN1 Mechanism and Comparing SN1 vs SN2 Reactivity

In the previous topic, we explored the $S_N2$ mechanism, where the nucleophile actively pushes the leaving group out in one smooth, concerted step. But what happens when the carbon bearing the halogen is too crowded for a backside attack? Nature has a second route: the substrate simply falls apart on its own first, and the nucleophile steps in afterward. This is the $S_N1$ mechanism, and it flips the reactivity rules of $S_N2$ completely on their head.

Breaking Apart First, Reacting Second: The $S_N1$ Pathway

$S_N1$ stands for Substitution, Nucleophilic, Unimolecular. The “unimolecular” part is the key difference from $S_N2$. In the rate-determining (slowest) step, only one molecule participates: the alkyl halide itself. The nucleophile does not appear until the second, faster step.

Consider what happens when tert-butyl bromide reacts with hydroxide ion:

$(CH_3)_3CBr + \;^-OH \longrightarrow (CH_3)_3COH + Br^-$

$\text{2-Bromo-2-methylpropane} \qquad \text{2-Methylpropan-2-ol}$

Experiments show that the rate of this reaction depends only on the concentration of tert-butyl bromide. Doubling the hydroxide concentration has zero effect on how fast the reaction goes. This is first-order kinetics, meaning only one reactant controls the rate.

The two steps in detail

Step I (slow, rate-determining): The polarised $C-Br$ bond breaks on its own. The bromine departs with the bonding electrons, leaving behind a positively charged carbon, a carbocation (a carbon atom carrying a formal positive charge). This step is slow because breaking a bond without any help from an incoming group requires significant energy.

Step II (fast): The carbocation, now desperately electron-poor, is immediately attacked by the nucleophile ($OH^-$). The hydroxide donates its electron pair to the positively charged carbon, forming the new $C-O$ bond and producing tert-butyl alcohol.

Why the rate depends only on the substrate

Since Step I is the slowest step, it acts as the bottleneck. Everything else must wait for the carbocation to form. The nucleophile only enters in Step II, which is fast. So no matter how much hydroxide you add, the reaction cannot go any faster than the rate at which the $C-Br$ bond breaks. This is why the reaction follows first-order kinetics: rate depends on the alkyl halide concentration alone.

The Role of the Solvent: Why Polar Protic Solvents Matter

$S_N1$ reactions are carried out in polar protic solvents such as water, alcohols, and acetic acid. These solvents have hydrogen atoms bonded to electronegative atoms ($O-H$, $N-H$), which means those hydrogens carry a partial positive charge ($\delta+$).

Here is why this matters. Step I requires the $C-X$ bond to break, and that costs energy. Where does the energy come from? The departing halide ion ($X^-$) is immediately surrounded by the partially positive hydrogens of the solvent molecules, a process called solvation (the surrounding and stabilising of an ion by solvent molecules). This solvation releases energy that helps pay the cost of breaking the $C-X$ bond. Without a protic solvent providing this stabilisation, the first step would be far too energetically unfavourable to occur at any useful rate.

Step I is also reversible: the carbocation and halide can recombine. However, because the solvent quickly solvates the halide, pulling it away from the carbocation, the forward direction is favoured.

Carbocation Stability: The Factor That Controls SN1 Speed

Since the rate-determining step is the formation of a carbocation, anything that makes the carbocation more stable will make it form more easily and speed up the overall reaction.

Recall the stability order of carbocations:

$3° \text{ (tertiary)} > 2° \text{ (secondary)} > 1° \text{ (primary)} > \text{methyl}$

Tertiary carbocations have three alkyl groups pushing electron density toward the positive centre through hyperconjugation (delocalisation of electrons from adjacent $C-H$ bonds into the empty $p$-orbital) and the inductive effect (electron donation through sigma bonds). This extra electron density partially offsets the positive charge, lowering the energy of the carbocation and making it easier to form.

For $S_N1$, therefore, the reactivity order is:

$\text{Tertiary halide} > \text{Secondary halide} > \text{Primary halide} > \text{Methyl halide}$

This is exactly the opposite of the $S_N2$ order. In $S_N2$, bulky groups block the nucleophile; in $S_N1$, those same groups stabilise the carbocation.

Side by Side: SN1 vs SN2 Reactivity Orders

Putting the two mechanisms together gives a clear, mirror-image picture:

Substrate class	$S_N1$ reactivity	$S_N2$ reactivity
Tertiary ($R_3CX$)	Fastest	Essentially zero
Secondary ($R_2CHX$)	Moderate	Slow
Primary ($RCH_2X$)	Very slow	Fast
Methyl ($CH_3X$)	Essentially zero	Fastest

The governing factors are different for each mechanism:

• $S_N1$ is controlled by carbocation stability. More substitution means a more stable carbocation and a faster reaction.
• $S_N2$ is controlled by steric accessibility. Fewer groups around the reacting carbon means the nucleophile can reach the back side more easily.

Allylic and Benzylic Halides: Fast in SN1 Thanks to Resonance

Allylic halides (halogen on a carbon next to a $C=C$ double bond) and benzylic halides (halogen on a carbon directly attached to a benzene ring) show high $S_N1$ reactivity, even when the halogen-bearing carbon is only primary. The reason is resonance stabilisation of their carbocations.

Allylic carbocation

When an allylic halide ionises, the positive charge does not stay on one carbon. The empty $p$-orbital overlaps with the $\pi$ electrons of the adjacent double bond, spreading the charge over two carbon atoms:

This delocalisation makes the allylic carbocation far more stable than a comparable isolated primary carbocation.

Benzylic carbocation

For benzylic halides, the carbocation formed after ionisation has its positive charge spread not just over two atoms, but across the entire benzene ring. The empty $p$-orbital on the benzylic carbon overlaps with the ring’s $\pi$ system, and resonance structures show the positive charge appearing on the benzylic carbon itself plus the ortho and para positions of the ring:

With so many resonance contributors sharing the positive charge, benzylic carbocations are very stable, and benzylic halides undergo $S_N1$ readily.

The Leaving-Group Trend: Same for Both Mechanisms

Regardless of whether the reaction follows $S_N1$ or $S_N2$, the ease with which the halogen departs follows the same order for a given alkyl group:

$R-I > R-Br > R-Cl >> R-F$

Iodide is the best leaving group because it is the largest halide ion, spreads its negative charge over a big electron cloud, and is very comfortable departing with the bonding electrons. Fluoride is the worst leaving group because it is small, holds its electrons tightly, and forms a very strong $C-F$ bond that is difficult to break.

This trend holds in both mechanisms because, regardless of the pathway, the halogen must ultimately leave. The only question is whether it leaves before (in $S_N1$) or simultaneously with (in $S_N2$) the nucleophile’s arrival.

Solved Example 6.6: Comparing SN2 Rates Within Pairs

Problem: In the following pairs, which compound undergoes $S_N2$ faster?

(a) Benzyl chloride ($C_6H_5CH_2Cl$) vs chlorobenzene ($C_6H_5Cl$)

(b) 1-Iodopropane ($CH_3CH_2CH_2I$) vs 1-chloropropane ($CH_3CH_2CH_2Cl$)

Solution:

(a) Benzyl chloride wins. In benzyl chloride, the chlorine sits on an $sp^3$ carbon (the $CH_2$ group attached to the ring). This is a primary halide, and the nucleophile can approach the back side without difficulty.

In chlorobenzene, the chlorine is bonded directly to an $sp^2$ ring carbon. The $C-Cl$ bond here is shorter and stronger (partial double-bond character from resonance between the lone pairs on chlorine and the ring), and the ring blocks the nucleophile’s approach. Chlorobenzene is essentially unreactive in $S_N2$.

(b) 1-Iodopropane wins. Both are primary halides with identical carbon skeletons, so the steric environment is the same. The difference lies in the leaving group: iodide is much larger and more polarisable than chloride, making it a better leaving group. It departs faster when the nucleophile pushes in.

Solved Example 6.7: Ordering Reactivity of Isomeric Halides

Problem: Predict the order of reactivity in both $S_N1$ and $S_N2$ for:

(i) The four isomeric bromobutanes

(ii) $C_6H_5CH_2Br$, $C_6H_5CH(C_6H_5)Br$, $C_6H_5CH(CH_3)Br$, $C_6H_5C(CH_3)(C_6H_5)Br$

Solution:

Part (i): The four isomeric bromobutanes

The four isomers and their classifications:

Compound	Formula	Type
1-Bromobutane	$CH_3CH_2CH_2CH_2Br$	Primary
1-Bromo-2-methylpropane	$(CH_3)_2CHCH_2Br$	Primary
2-Bromobutane	$CH_3CH_2CH(Br)CH_3$	Secondary
2-Bromo-2-methylpropane	$(CH_3)_3CBr$	Tertiary

$S_N1$ order (increasing reactivity):

$CH_3CH_2CH_2CH_2Br < (CH_3)_2CHCH_2Br < CH_3CH_2CH(Br)CH_3 < (CH_3)_3CBr$

The tertiary bromide is fastest because its carbocation ($3°$) is the most stable. The secondary bromide comes next. Between the two primary bromides, the carbocation from $(CH_3)_2CHCH_2Br$ gets extra stabilisation from the isopropyl group ($(CH_3)_2CH-$), which donates more electron density through the inductive effect than a straight-chain butyl group. So $(CH_3)_2CHCH_2Br$ is slightly more reactive than $CH_3CH_2CH_2CH_2Br$ in $S_N1$.

$S_N2$ order (increasing reactivity):

$(CH_3)_3CBr < CH_3CH_2CH(Br)CH_3 < (CH_3)_2CHCH_2Br < CH_3CH_2CH_2CH_2Br$

The order reverses because $S_N2$ is governed by steric hindrance. The tertiary bromide is most crowded, so the nucleophile cannot get through. The two primary bromides are the most accessible, with the straight-chain primary being slightly less hindered than the branched primary.

Part (ii): Benzylic bromides with varying substituents

Compound	Type of carbon
$C_6H_5CH_2Br$	Primary benzylic
$C_6H_5CH(CH_3)Br$	Secondary benzylic
$C_6H_5CH(C_6H_5)Br$	Secondary benzylic
$C_6H_5C(CH_3)(C_6H_5)Br$	Tertiary benzylic

$S_N1$ order (increasing reactivity):

$C_6H_5CH_2Br < C_6H_5CH(CH_3)Br < C_6H_5CH(C_6H_5)Br < C_6H_5C(CH_3)(C_6H_5)Br$

The tertiary benzylic bromide is the fastest because its carbocation is stabilised by two phenyl groups (resonance) and one methyl group (inductive effect). Among the two secondary bromides, $C_6H_5CH(C_6H_5)Br$ forms a carbocation stabilised by two phenyl groups through resonance, which is more effective than stabilisation by one phenyl plus one methyl (as in $C_6H_5CH(CH_3)Br$). The primary benzylic bromide is slowest because its carbocation has the least stabilisation.

$S_N2$ order (increasing reactivity):

$C_6H_5C(CH_3)(C_6H_5)Br < C_6H_5CH(C_6H_5)Br < C_6H_5CH(CH_3)Br < C_6H_5CH_2Br$

$S_N2$ is controlled by steric crowding. A phenyl group takes up much more space than a methyl group, so $C_6H_5CH(C_6H_5)Br$ is more sterically blocked than $C_6H_5CH(CH_3)Br$, even though both are secondary. The primary benzylic bromide ($C_6H_5CH_2Br$) has the least crowding and reacts fastest by $S_N2$.