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Which of the following is the correct order of decreasing \(\mathrm{SN}^{2}\) reactivity? (a) \(\mathrm{RCH}_{2} \mathrm{X}>\mathrm{R}_{2} \mathrm{CHX}>\mathrm{R}_{3} \mathrm{CX}\) (b) \(\mathrm{R}_{3} \mathrm{CX}>\mathrm{R}_{2} \mathrm{CHX}>\mathrm{RCH}_{2} \mathrm{X}\) (c) \(\mathrm{R}_{2} \mathrm{CHX}>\mathrm{R}_{3} \mathrm{CX}>\mathrm{RCH}_{2} \mathrm{X}\) (d) \(\mathrm{RCH}_{2} \mathrm{X}>\mathrm{R}_{3} \mathrm{CX}>\mathrm{R}_{2} \mathrm{CHX}\)

Short Answer

Expert verified
The correct order of decreasing SN2 reactivity is option (a): \(\mathrm{RCH}_{2} \mathrm{X}>\mathrm{R}_{2} \mathrm{CHX}>\mathrm{R}_{3} \mathrm{CX}\.\)

Step by step solution

01

Understand SN2 Reactivity

The reactivity in an SN2 mechanism is inversely related to steric hindrance. This means that the less crowded the carbon center where the reaction occurs, the more reactive it will be. Typically, primary alkyl halides react faster than secondary, which react faster than tertiary due to increasing steric hindrance.
02

Analyzing the Options

Let's analyze each option based on what we just learned about steric hindrance and SN2 reactivity: - (a) suggests primary > secondary > tertiary. - (b) suggests tertiary > secondary > primary. - (c) suggests secondary > tertiary > primary. - (d) suggests primary > tertiary > secondary.
03

Identify the Correct Order

Based on the theory of SN2 reactivity, primary alkyl halides ({RCH2X}) are the most reactive, followed by secondary ({R2CHX}), and tertiary ({R3CX}) are the least reactive. Therefore, the correct order is primary (least steric hindrance) > secondary > tertiary.
04

Choose the Correct Option

Now, compare the theoretical order to the given options. Option (a) matches the expected order of primary > secondary > tertiary. Thus, option (a) is the correct answer for decreasing SN2 reactivity.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Steric Hindrance
Steric hindrance is a crucial concept in understanding the reactivity of chemical compounds, especially within organic chemistry. Imagine trying to squeeze through a crowded room; the more people there, the harder it is to move. Similarly, in a chemical structure, the more atoms or groups near a reactive center, the harder it is for a reaction to occur.

In an SN2 reaction, a nucleophile attacks a substrate, displacing a leaving group in a one-step process. If there are fewer atoms around the reactive center (usually a carbon atom), the nucleophile can approach more easily and react faster. That's why compounds with low steric hindrance, like primary alkyl halides, react more readily in SN2 reactions.
  • Less crowded = more reactive in SN2
  • More crowded = less reactive in SN2
Understanding steric hindrance helps us predict which molecules will react most quickly or slowly, making it easier to control chemical reactions effectively.
Primary Alkyl Halide
Primary alkyl halides have the structure RCH2X, where the halogen (X) is bonded to a carbon that is attached to only one other carbon. This configuration leads to minimal steric hindrance, allowing nucleophiles easy access to the carbon center. Because of this lack of crowding, primary alkyl halides are generally the most reactive in SN2 reactions.

  • Structure: RCH2X
  • Little crowding around carbon makes Nucleophilic Attack easy
  • Most reactive in SN2 reactions
Consequently, given a choice among primary, secondary, and tertiary, primary alkyl halides will consistently react fastest in SN2 pathways. This makes them ideal candidates for many organic synthesis procedures where quick and efficient reactions are desirable.
Secondary Alkyl Halide
Secondary alkyl halides are positioned between primary and tertiary in terms of reactivity in SN2 reactions. Their general formula is R2CHX, where the halogen is attached to a carbon linked to two other carbons. This introduces more steric hindrance than in primary alkyl halides, but still less than tertiary ones.

  • Formula: R2CHX
  • More steric hindrance than primary; less than tertiary
  • Moderate reactivity in SN2 reactions
This level of steric hindrance means secondary alkyl halides react slower than primary ones but are still capable of undergoing SN2 reactions. They provide a balance, offering reactivity while maintaining some additional stability that may be useful in more complex synthetic applications.
Tertiary Alkyl Halide
Tertiary alkyl halides, with the structure R3CX, are characterized by a carbon atom bonded to three other carbon atoms. This structure makes them the most crowded and thus the least reactive in SN2 reactions due to the high steric hindrance.

  • Structure: R3CX
  • Maximum steric hindrance
  • Least reactive in SN2 reactions
Because the nucleophile faces significant difficulty accessing the crowded carbon center, SN2 reactions with tertiary alkyl halides are usually ineffective. Instead, tertiary alkyl halides are more likely to react via SN1 mechanisms, where the rate-determining step involves the loss of the leaving group, independent of the nucleophile's initial attack.

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Most popular questions from this chapter

Arrange the following compounds in the order of increasing tendency to undergo electrophillic substitution: (1) O=[N+]([O-])c1ccccc1 (2) c1ccccc1(3) Oc1ccccc1 (4) Cc1ccccc1 (5) (a) \(5<1<2<4<3\) (b) \(3>2>1>4>5\) (c) \(1>2>5>3>4\) (d) \(5<1<4<2<3\)

The relative order of reactivity of \(\mathrm{F}^{-}, \mathrm{Cl}^{-}, \mathrm{Br}\) and \(\mathrm{I}\) is/ are (a) \(\mathrm{F}^{-}<\mathrm{Cl}^{-}<\mathrm{Br}<\mathrm{I}^{-}\)in polar solvent (b) \(\mathrm{F}^{-}>\mathrm{Cl}^{->} \mathrm{Br}>\mathrm{I}^{-}\)in non polar solvent (c) \(\mathrm{F}^{-}<\mathrm{Cl}^{-}<\mathrm{Br}<\mathrm{I}^{-}\)in non polar solvent (d) \(\mathrm{F}^{-}>\mathrm{Cl}^{-}>\mathrm{Br}>\mathrm{I}^{-}\)in polar solvent

Tertiary alkyl halides are practically inert to substitution by \(\mathrm{SN}^{2}\) mechanism because of (a) insolubility (b) instability (c) inductive effect (d) steric hindrance

Match the following: List I List II (Reagents) (Nature) 1\. \((\mathrm{CN})\) (i) Neutral nucleophile 2\. (ONO) (ii) electrophile 3\. \(\mathrm{NH}_{3}\) (iii) Ambident nucleophile 4\. \(\mathrm{AlCl}_{3}\) (iv) both El and Nu The correct matching is: 1 \(2 \quad 3\) 4 (a) (i) (iii) (ii) (iv) (b) (iii) (iii) (i) (ii) (c) (iv) (iii) (iii) (i) (d) (ii) (iv) (i) (iii)

The reaction \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{C}-\mathrm{Br} \stackrel{\mathrm{H}_{2} \mathrm{O}}{\longrightarrow}\left(\mathrm{CH}_{3}\right)_{3} \mathrm{C}-\mathrm{OH}\) is (a) addition reaction (b) substitution reaction (c) free radical reaction (d) elimination reaction

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