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Haloalkanes and Haloarenes - Mechanism of substitution reactions (SN1 and SN2)

Grade 12ICSEChemistry

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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SN1S_N1 Mechanism (Unimolecular Nucleophilic Substitution): This occurs in two steps. The first step involves the slow ionization of the haloalkane to form a carbocation intermediate (R+R^+). The second step is the fast attack by the nucleophile (Nu−Nu^-).

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SN2S_N2 Mechanism (Bimolecular Nucleophilic Substitution): This is a concerted, single-step process where the nucleophile attacks the carbon atom from the side opposite to the leaving group (backside attack), leading to a pentacoordinate transition state.

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Stereochemistry of SN1S_N1: Since the carbocation is planar (sp2sp^2 hybridized), the nucleophile can attack from either side, resulting in racemization (a 50:50 mixture of enantiomers).

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Stereochemistry of SN2S_N2: The backside attack results in a complete inversion of configuration, famously known as Walden Inversion.

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Reactivity Trends: For SN1S_N1, the order is 3∘>2∘>1∘>CH3X3^\circ > 2^\circ > 1^\circ > CH_3X (based on carbocation stability). For SN2S_N2, the order is CH3X>1∘>2∘>3∘CH_3X > 1^\circ > 2^\circ > 3^\circ (based on steric hindrance).

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Solvent Effects: Polar protic solvents (e.g., H2OH_2O, C2H5OHC_2H_5OH) favor SN1S_N1 by stabilizing the carbocation and the leaving group. Polar aprotic solvents (e.g., DMSODMSO, Acetone) favor SN2S_N2 as they do not solvate the nucleophile strongly.

📐Formulae

RateSN1=k[R−X]Rate_{S_N1} = k[R-X]

RateSN2=k[R−X][Nu−]Rate_{S_N2} = k[R-X][Nu^-]

R−X→slowR++X−(Step 1:SN1)R-X \xrightarrow{slow} R^+ + X^- \quad (Step \, 1: S_N1)

R++Nu−→fastR−Nu(Step 2:SN1)R^+ + Nu^- \xrightarrow{fast} R-Nu \quad (Step \, 2: S_N1)

💡Examples

Problem 1:

Which of the following will react faster in an SN2S_N2 reaction with OH−OH^-: CH3BrCH_3Br or (CH3)3CBr(CH_3)_3CBr?

Solution:

CH3BrCH_3Br will react faster.

Explanation:

SN2S_N2 reactions are sensitive to steric hindrance. CH3BrCH_3Br is a primary halide with minimum steric crowding, allowing the OH−OH^- nucleophile to attack the carbon easily. (CH3)3CBr(CH_3)_3CBr is a tertiary halide where bulky methyl groups block the approach of the nucleophile.

Problem 2:

Arrange the following in increasing order of reactivity towards SN1S_N1 reaction: CH3CH2BrCH_3CH_2Br, (CH3)2CHBr(CH_3)_2CHBr, (CH3)3CBr(CH_3)_3CBr.

Solution:

CH3CH2Br<(CH3)2CHBr<(CH3)3CBrCH_3CH_2Br < (CH_3)_2CHBr < (CH_3)_3CBr

Explanation:

SN1S_N1 reactivity depends on the stability of the carbocation formed in the rate-determining step. The stability order of carbocations is: Tertiary(3∘)>Secondary(2∘)>Primary(1∘)Tertiary (3^\circ) > Secondary (2^\circ) > Primary (1^\circ). Thus, (CH3)3CBr(CH_3)_3CBr forms the most stable carbocation and reacts fastest.

Problem 3:

Explain why haloarenes (like C6H5ClC_6H_5Cl) are less reactive towards nucleophilic substitution than haloalkanes.

Solution:

Haloarenes are less reactive due to resonance effect and sp2sp^2 hybridization.

Explanation:

In haloarenes, the lone pair on the halogen is in resonance with the benzene ring, giving the C−XC-X bond a partial double bond character which is harder to break. Also, the sp2sp^2 hybridized carbon of the ring is more electronegative than the sp3sp^3 carbon of haloalkanes, making the C−XC-X bond shorter and stronger.