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Question

For SN2 reaction, the increasing order of the reactivity of the following alkyl halides is:

(A) CH3CH2CH2CH2Br

(B) CH3CH2CH(Br)CH3

(C) (CH3)3CBr

(D) (CH3)2CHCH2Br

Choose the correct answer from the options given below:

The correct answer is

(C) < (B) < (D) < (A)

Understanding SN2 Reaction Reactivity of Alkyl Halides

SN2 reactions, or bimolecular nucleophilic substitution reactions, are fundamental in organic chemistry. They involve a single step where a nucleophile attacks the substrate and the leaving group departs simultaneously. The rate of an SN2 reaction is heavily influenced by steric hindrance around the carbon atom where the substitution occurs (the electrophilic carbon).

Let's analyze the given alkyl halides and their structures to determine their reactivity towards SN2 reactions.

Analyzing Alkyl Halide Structures and Steric Hindrance

We are given four different alkyl bromides:

  • (A) CH3CH2CH2CH2Br: This is a straight-chain primary alkyl bromide (1°). The carbon attached to the bromine is bonded to only one other carbon and two hydrogen atoms. There is minimal steric hindrance around this carbon.
  • (B) CH3CH2CH(Br)CH3: This is a secondary alkyl bromide (2°). The carbon attached to the bromine is bonded to two other carbon atoms (an ethyl group and a methyl group) and one hydrogen atom. This introduces more steric hindrance compared to a primary carbon.
  • (C) (CH3)3CBr: This is a tertiary alkyl bromide (3°). The carbon attached to the bromine is bonded to three methyl groups. This creates significant steric hindrance around the reaction center, making it very difficult for a nucleophile to approach the carbon from the backside, which is required for an SN2 reaction.
  • (D) (CH3)2CHCH2Br: This is also a primary alkyl bromide (1°). The carbon attached to the bromine is bonded to a hydrogen atom and a (CH3)2CH- group (an isopropyl group). Although the reactive carbon is primary, the branching at the beta-carbon (the carbon adjacent to the one bearing the leaving group) increases steric hindrance near the reaction center compared to a simple straight-chain primary alkyl halide like (A). This is sometimes referred to as the "beta-branching effect".

Comparing SN2 Reactivity Based on Steric Hindrance

The rate of an SN2 reaction decreases as steric hindrance around the electrophilic carbon increases. This is because the nucleophile needs to approach this carbon from the back side of the leaving group. More substituents on the carbon hinder this approach.

Based on the degrees of the alkyl halides and the presence of beta-branching, we can predict the relative SN2 reactivity:

  • Tertiary alkyl halides ((C)) are generally the least reactive in SN2 reactions due to maximum steric hindrance.
  • Secondary alkyl halides ((B)) are less reactive than primary ones but more reactive than tertiary ones.
  • Primary alkyl halides ((A) and (D)) are generally the most reactive. However, branching on the beta-carbon ((D)) increases steric hindrance compared to a straight-chain primary alkyl halide ((A)). Thus, (A) is more reactive than (D) in SN2 reactions.

Therefore, the order of increasing steric hindrance, and thus decreasing SN2 reactivity, is:

Steric Hindrance: (A) < (D) < (B) < (C)

SN2 Reactivity: (C) < (B) < (D) < (A)

Determining the Increasing Order of Reactivity

Arranging the given alkyl halides in order of increasing SN2 reactivity:

  1. Least reactive: The tertiary halide (C) because of significant steric hindrance.
  2. Next least reactive: The secondary halide (B).
  3. More reactive: The primary halide with beta-branching (D), which has some steric hindrance due to the branched group near the reaction center.
  4. Most reactive: The straight-chain primary halide (A), with the least steric hindrance.

So, the increasing order of reactivity for the SN2 reaction is (C) < (B) < (D) < (A).

Alkyl Halide Structure Degree Steric Hindrance around Electrophilic Carbon Relative SN2 Reactivity
(A) CH3CH2CH2CH2Br Primary (1°) Primary Low (straight-chain) Highest
(B) CH3CH2CH(Br)CH3 Secondary (2°) Secondary Moderate Moderate
(C) (CH3)3CBr Tertiary (3°) Tertiary Very High Lowest (practically zero)
(D) (CH3)2CHCH2Br Primary (1°) Primary (with β-branching) Medium-Low (higher than A due to β-branching) High (lower than A)

The order of increasing reactivity is (C) < (B) < (D) < (A).

Revision Table: SN2 Reactivity Factors

Factor Effect on SN2 Rate Explanation
Steric Hindrance on Carbon with Leaving Group Decreases rate Hinders nucleophile attack; order: CH3 > 1° > 2° > 3°
β-Branching Decreases rate Branching near reaction center adds steric bulk.
Strength of Nucleophile Increases rate Stronger nucleophiles react faster.
Leaving Group Ability Increases rate Good leaving groups stabilize negative charge; order: I> > Br> > Cl> > F> (for halides)
Solvent Aprotic polar solvents increase rate Aprotic solvents don't solvate nucleophile strongly, leaving it more reactive.

Additional Information: SN2 Reaction Details

The SN2 reaction is a concerted reaction, meaning bond breaking (C-leaving group) and bond formation (C-nucleophile) happen simultaneously. It proceeds through a transition state where the carbon atom is partially bonded to both the incoming nucleophile and the departing leaving group. This transition state is crowded, making it very sensitive to steric effects.

The stereochemistry of an SN2 reaction involves inversion of configuration at the carbon center. If the starting material is chiral and has a specific configuration (R or S), the product will have the opposite configuration.

For tertiary alkyl halides, SN2 reactions are generally not observed. Instead, they tend to react via SN1 or E1 mechanisms if suitable conditions (e.g., protic solvent, heat) are present.

Primary alkyl halides usually favor SN2 reactions, especially with strong nucleophiles and polar aprotic solvents. Secondary alkyl halides can undergo both SN1 and SN2 reactions, depending on the reaction conditions, particularly the strength of the nucleophile and the nature of the solvent.

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Important Questions from Organic Compounds Containing Nitrogen

  1. The correct increasing order of basic strength of amine is:

    (A) C₆H₅NH₂ < NH₃ < C₆H₅CH₂NH₂ < C₂H₅NH₂ < (C₂H₅)₂NH

    (B) NH₃ < C₆H₅NH₂ < C₆H₅CH₂NH₂ < C₂H₅NH₂ < (C₂H₅)₂NH

    (C) C₆H₅CH₂NH₂ < C₆H₅NH₂ < NH₃ < C₂H₅NH₂ < (C₂H₅)₂NH

    (D) C₂H₅NH₂ < (C₂H₅)₂NH < C₆H₅NH₂ < NH₃

    (E) NH₃ < C₂H₅NH₂ < C₆H₅CH₂NH₂ < (C₂H₅)₂NH < C₆H₅NH₂

    Choose the correct answer from the options given below:

  2. In which of the following molecules carbon atom marked with asterisk (*) is a stereocentre or chiral centre?

  3. Match List-I with List-II:

    List-IList-II
    (A) Urease(I) Maltose
    (B) Maltase(II) Glucose and fructose
    (C) Invertase(III) NH₃ and CO₂
    (D) Diastase(IV) Glucose

    Choose the correct answer from the options given below:

  4. Phenol is manufactured from hydrocarbon, Cumene. Cumene is chemically:

  5. t99.9% with respect to t90% for a first-order reaction is:

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