Problem 73
Question
\(\mathrm{R}-\mathrm{OH}+\mathrm{HX} \longrightarrow \mathrm{R}-\mathrm{X}+\mathrm{H}_{2} \mathrm{O}\) In the above reaction, the reactivity of different alcohols is (a) tertiary \(>\) secondary \(>\) primary (b) tertiary \(>\) secondary \(<\) primary (c) tertiary \(<\) secondary \(>\) primary (d) secondary \(<\) primary < tertiary
Step-by-Step Solution
Verified Answer
Option (a) tertiary > secondary > primary.
1Step 1: Understanding the Reaction
The given reaction involves the conversion of an alcohol (R-OH) into an alkyl halide (R-X) by reacting it with a hydrogen halide (HX). Water (H2O) is also produced as a byproduct.
2Step 2: Considering Carbocation Stability
In the mechanism of this reaction, the alcohol is protonated and forms a water molecule and a carbocation (R+). The stability of this carbocation plays a crucial role in determining the reaction rate.
3Step 3: Analyzing Carbocation Stability Order
Carbocations are more stable when they are more substituted. Thus, a tertiary carbocation is more stable than a secondary, which is more stable than a primary. This order is: tertiary > secondary > primary.
4Step 4: Matching Options to Stability Order
According to the stability of the carbocations and their reactivity, tertiary alcohols should react faster than secondary, and secondary faster than primary, to form alkyl halides.
Key Concepts
Carbocation StabilityAlkyl Halide FormationReaction Mechanism
Carbocation Stability
Carbocation stability is a critical factor in chemical reactions involving alcohols and hydrogen halides. When an alcohol reacts with a hydrogen halide, the oxygen atom in the alcohol gets protonated, forming a positively charged water molecule, which eventually leaves. This departure creates a carbocation (R+), a molecule with a positively charged carbon atom.
Carbocations differ in stability depending on their substitution level:
Carbocations differ in stability depending on their substitution level:
- Tertiary carbocations: These are the most stable because the central carbon is bonded to three alkyl groups. The electron-donating nature of these groups helps stabilize the positive charge.
- Secondary carbocations: These have two alkyl groups and are more stable than primary carbocations but less stable than tertiary.
- Primary carbocations: The least stable form, where the carbocation is only bonded to one alkyl group.
Alkyl Halide Formation
During the transformation from alcohols to alkyl halides, the stability of the intermediate carbocation significantly impacts how rapidly the reaction occurs. An alcohol reacts with a hydrogen halide (HX) to generate an alkyl halide and water.
Steps involved in the reaction include:
Steps involved in the reaction include:
- Protonation of Alcohol: The process begins with the alcohol group (R-OH) accepting a proton from the hydrogen halide. This protonation makes the hydroxyl group a better leaving group by converting it into water.
- Carbocation Formation: Once water leaves, a carbocation is formed. The stability of this carbocation is a decisive factor for the reaction rate.
- Nucleophilic Attack: Finally, the halide ion (X-) attacks the carbocation, forming the alkyl halide (R-X).
Reaction Mechanism
Understanding the reaction mechanism is fundamental to mastering the conversion of alcohol into alkyl halides. This mechanism typically proceeds through an SN1 (substitution nucleophilic unimolecular) process for tertiary alcohols and perhaps SN2 (bimolecular nucleophilic substitution) for primary alcohols due to structural differences.
- SN1 Mechanism:
- The reaction begins with the protonation of the alcohol, turning it into a good leaving group.
- Subsequently, the molecule dissociates to form a stable carbocation. This is a unimolecular step, hence the term SN1.
- The halide ion then attacks the carbocation, completing the substitution.
- SN2 Mechanism:
- This is characterized by a concerted, single-step process where the nucleophile attacks the substrate as the leaving group is leaving.
- Primary substrates usually undergo this due to an unstable primary carbocation that can't sufficiently stabilize a positive charge in a multistep process.
Other exercises in this chapter
Problem 71
The end product [B] in the following sequence of reactions is \(\mathrm{CH}_{3} \mathrm{Cl} \stackrel{\mathrm{KCN}}{\longrightarrow}[\mathrm{A}] \stackrel{\math
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\(\mathrm{CH}_{3}-\mathrm{CH}_{2}-\mathrm{CH}_{2}-\mathrm{Cl} \stackrel{\text { alc. } \mathrm{KOH}}{\longrightarrow}(\mathrm{X}) \stackrel{\mathrm{HBr}}{\longr
View solution Problem 74
Under which of the following conditions would toluene \(\mathrm{C}_{6} \mathrm{H}_{5}-\mathrm{CH}_{3}\), be converted into bromomethyl benzene, \(\mathrm{C}_{6}
View solution Problem 75
The number of isomeric alkyl halides possible for \(\mathrm{C}_{5} \mathrm{H}_{11} \mathrm{Cl}\) is (a) 5 (b) 6 (c) 4 (d) 8
View solution