Problem 1
Question
Which of the following shows greater reactivity towards \(\mathrm{S}_{\mathrm{N}} 2\) reaction than \(\mathrm{CH}_{3}-\mathrm{CH}_{2} \mathrm{Br}\) with sodium methoxide? (a) \(\mathrm{CH}_{3}-\mathrm{O}-\mathrm{CH}_{2} \mathrm{Br}\) (b) \(\mathrm{CH}_{3} \mathrm{Br}\) (c) \(\mathrm{CH}_{2}=\mathrm{CH}-\mathrm{Br}\) (d) \(\mathrm{CH}_{2}=\mathrm{CH}-\mathrm{CH}_{2} \mathrm{Br}\)
Step-by-Step Solution
Verified Answer
(b) CH_3Br is more reactive towards SN2 than CH_3-CH_2Br.
1Step 1: Understanding SN2 Reaction Mechanism
In an SN2 reaction (nucleophilic substitution bimolecular), a nucleophile attacks the electrophilic carbon atom from the opposite side of the leaving group. This requires a relatively unhindered substrate to facilitate this backside attack.
2Step 2: Analyzing Steric Hindrance
SN2 reactions are quickest with less sterically hindered substrates. Comparing the given molecules, the primary bromide (CH_3Br) will have the least steric hindrance. CH_3-CH_2Br, a primary bromide, is more hindered compared to CH_3Br.
3Step 3: Additional Electronic Effects
Apart from steric factors, the presence of electron-withdrawing or donating groups can affect reactivity. Vinyl and allyl bromides like CH_2=CH-CH_2Br benefit from resonance stabilization in the transition state, potentially enhancing reactivity in certain cases.
4Step 4: Comparing Reactivity Potential
Ultimately, CH_3Br's smaller size and lack of hindrance makes it the most reactive towards SN2 when compared to the more sterically crowded CH_3-CH_2Br.
Key Concepts
Nucleophilic SubstitutionSteric HindranceReactivity in Organic Chemistry
Nucleophilic Substitution
Nucleophilic substitution is a fundamental concept in organic chemistry, particularly observed in the
SN2 reaction mechanism.
In an SN2 reaction, a nucleophile—which is a molecule or ion with a pair of electrons to donate—attacks an electrophilic carbon atom. This carbon is typically bonded to a leaving group, often a halide.
The nucleophile approaches from the opposite side of the leaving group in a process known as a backside attack.
- The reaction proceeds in a single concerted step, without any intermediates.
- It results in the inversion of configuration at the carbon center, often referred to as a "Walden inversion." This can be compared to an umbrella flipping inside out, hence why this reaction is often stereospecific.
Steric Hindrance
Steric hindrance is a crucial factor dictating the speed of SN2 reactions. It refers to the prevention of chemical reactions due to the spatial arrangement of atoms within a molecule.Atoms or groups within a molecule can block the approach of a nucleophile, slowing or inhibiting the reaction.
- In primary substrates, steric hindrance is minimal, allowing smoother and faster nucleophilic attacks.
- In contrast, secondary and tertiary substrates possess more bulky groups, increasing steric hindrance and thus slowing down or even preventing SN2 reactions entirely.
Reactivity in Organic Chemistry
Reactivity in organic chemistry, especially concerning SN2 reactions, is influenced by multiple factors. These include steric hindrance, electronic effects from substituents, and solvent effects.
- Steric Effects: Less bulk around the reactive site increases reactivity as it allows nucleophiles easier access to the electrophilic carbon.
- Electronic Effects: Electron-withdrawing groups can increase the electrophilicity of a carbon atom, making it more susceptible to nucleophilic attack.
- Solvent Effects: Polar aprotic solvents like acetone and DMF enhance SN2 reaction rates by stabilizing ions without solvation of the nucleophile, improving its reactivity.
Other exercises in this chapter
Problem 5
Which of the following statements are correct? (a) The \(\mathrm{S}_{\mathrm{N}} 1\) reaction obeys the Ist order kinetics (b) The solvolysis of ethylbromide in
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Correct statements is (a) Nucleophilicity of \(\mathrm{SH}\) is greater than \(\mathrm{OH}\) in dimethyl formamide (b) Hydrazine is a better nucleophile than \(
View solution Problem 9
Identify the compounds that will undergo substitution reaction faster than \(\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{Br}\) with \(\mathrm{OH}^{-} / \mathrm{DMSO}
View solution