Which of the following is the correct order of decreasing SN2 reactivity? (X = a halogen)
(a) RCH2X

Q: Why doesn't the nucleophile attack from the front?

The front side is occupied by the leaving group (the halogen), which is electron-rich and repels the incoming electron-rich nucleophile. Also, the antibonding orbital ($\sigma^*$) of the C-X bond is located on the backside, which the nucleophile must overlap with to break the bond.

Q: Can a tertiary alkyl halide ever undergo an $S_N2$ reaction?

Practically, no. The steric hindrance from three bulky alkyl groups is so severe that the activation energy for the backside attack is too high. Tertiary alkyl halides prefer the $S_N1$ pathway if a stable carbocation can be formed.

Q: What if the alkyl groups are very small, like methyl groups?

A methyl halide ($CH_3X$) has the least steric hindrance of all (only three small hydrogen atoms) and undergoes $S_N2$ reactions even faster than primary alkyl halides!

Q: Does the type of halogen affect the $S_N2$ reactivity order?

The type of halogen (leaving group) affects the overall rate of the reaction (e.g., I- is a better leaving group than Cl-), but it does not change the relative order of reactivity based on steric hindrance ($1^\circ > 2^\circ > 3^\circ$).

Q: Why is $S_N2$ called a 'bimolecular' reaction?

Because the rate-determining step (which is the only step in this concerted mechanism) involves the collision of two molecules: the nucleophile and the alkyl halide.

Question

Which of the following is the correct order of decreasing $S_N2$ reactivity? (X = a halogen)

Options:

(A) $RCH_2X > R_2CX > R_3CX$

(B) $RCH_2X > R_2CHX > R_3CX$

(C) $R_3CX > R_2CHX > RCH_2X$

(D) $R_2CHX > R_3CX > RCH_2X$

Sigma Percentile · Accepted Answer

### The $S_N2$ Battlefield

Imagine a microscopic battlefield where a nucleophile is trying to attack a carbon atom. This is the essence of the **$S_N2$ reaction**, which stands for Substitution Nucleophilic Bimolecular.

In this concerted, single-step process, the nucleophile doesn't just attack from anywhere. It must approach the electrophilic carbon from the **backside**, exactly opposite to the leaving group (the halogen).

Why the backside? Because the front is guarded by the electron-rich leaving group, which repels the incoming nucleophile. The nucleophile must precisely target the empty antibonding orbital located at the rear.

### The Role of Steric Hindrance

Since the nucleophile must physically reach the carbon atom, the size of the groups attached to that carbon becomes the deciding factor. This physical blocking is known as **steric hindrance**.

Think of bulky alkyl groups ($R$) as bouncers at a club. The more bouncers there are, the harder it is for the nucleophile to get in.

Therefore, the rate of an $S_N2$ reaction is heavily dependent on how "open" or "crowded" the backside of the carbon atom is.

### Analyzing the Contenders

Let's evaluate our three contenders: primary ($1^\circ$), secondary ($2^\circ$), and tertiary ($3^\circ$) alkyl halides.

**Primary Alkyl Halides ($RCH_2X$):** Here, the carbon is attached to only one bulky $R$ group and two small hydrogen atoms. The backside is wide open! The nucleophile can attack with minimal resistance, making this the **fastest** to react.

**Secondary Alkyl Halides ($R_2CHX$):** Now we have two bulky $R$ groups. The path is partially blocked. The nucleophile has to squeeze through, which slows down the reaction significantly. It reacts, but at a **moderate** pace.

**Tertiary Alkyl Halides ($R_3CX$):** This is the ultimate fortress. Three bulky $R$ groups completely surround the carbon atom. The backside is totally blocked. The nucleophile simply cannot reach the target, making the $S_N2$ reaction practically **impossible**.

### The Final Verdict

Putting it all together, the reactivity perfectly inversely correlates with steric bulk. Less bulk means a faster reaction.

The decreasing order of $S_N2$ reactivity is:
**$1^\circ > 2^\circ > 3^\circ$**

Translating this to our chemical formulas, we get:
**$RCH_2X > R_2CHX > R_3CX$**

This perfectly matches option (b). Always remember: in the world of $S_N2$, smaller is faster!

Animated Solution for Chemistry - Organic Chemistry: Which of the following is the correct order of decreasing $S_{N} 2$ reactivity? (X = a halogen)

Select Answer:

Visualized Solution

Order of $S_{N} 2$ Reactivity

Mechanism of $S_{N} 2$

Steric Hindrance

Primary Alkyl Halide ( $1^{\circ}$ )

Secondary Alkyl Halide ( $2^{\circ}$ )

Tertiary Alkyl Halide ( $3^{\circ}$ )

Final Reactivity Order

The Sigma Insight: Haloalkanes & Haloarenes

The $S_{N} 2$ Battlefield

The Role of Steric Hindrance

Analyzing the Contenders

The Final Verdict

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Animated Solution for Chemistry - Organic Chemistry: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Which of the following is the correct order of decreasing SN​2 reactivity? (X = a halogen)