Question

Which of the following alcohols reacts fastest with concentrated hydrochloric acid in the presence of anhydrous zinc chloride? (A) Butan-2-ol (B) Butan-1-ol (C) 2-Methylpropan-1-ol (D) 2-Methylpropan-2-ol

Short answer

The alcohol that reacts fastest with concentrated hydrochloric acid in the presence of anhydrous zinc chloride is (D) 2-Methylpropan-2-ol, as it forms the most stable tertiary carbocation during the rate-determining step of the nucleophilic substitution reaction.

Step-by-step solution

Identify the carbocations formed from the given alcohols

The first step is to identify the carbocations formed when each of the given alcohols reacts with concentrated hydrochloric acid. In each case, the carbocation will form at the carbon atom that was bound to the hydroxyl (OH) group. (A) Butan-2-ol: \(CH_{3}CH(OH)CH_{2}CH_{3}\) will form the carbocation \(CH_{3}CH^{+}CH_{2}CH_{3}\) (B) Butan-1-ol: \(CH_{3}CH_{2}CH_{2}CH_{2}OH\) will form the carbocation \(CH_{3}CH_{2}CH_{2}CH_{2}^{+}\) (C) 2-Methylpropan-1-ol: \(CH_3CH(CH_3)CH_2OH\) will form the carbocation \(CH_3CH(CH_3)CH_2^{+}\) (D) 2-Methylpropan-2-ol: \(CH_{3}C(OH)(CH_{3})CH_{3}\) will form the carbocation \(CH_{3}C^{+}(CH_{3})CH_{3}\)

Compare the stability of carbocations

Now, we must compare the stability of the carbocations formed in each case. Carbocations are more stable when there are more alkyl groups (methyl or ethyl) attached to the positively charged carbon atom. This is due to the inductive effect, where electron-donating alkyl groups help to stabilize the positive charge on the carbocation. In general, the stability of carbocations follows the order: tertiary > secondary > primary > methyl. (A) Butan-2-ol: The carbocation is secondary, with two alkyl groups attached to the positively charged carbon. (B) Butan-1-ol: The carbocation is primary, with only one alkyl group attached to the positively charged carbon. (C) 2-Methylpropan-1-ol: The carbocation is primary, with only one alkyl group attached to the positively charged carbon. (D) 2-Methylpropan-2-ol: The carbocation is tertiary, with three alkyl groups attached to the positively charged carbon.

Determine the alcohol with the fastest reaction

Based on the carbocation stability order and the stability of the carbocations formed from each of the given alcohols, we can now determine which alcohol reacts fastest with concentrated hydrochloric acid. Since a more stable carbocation leads to a faster reaction, 2-Methylpropan-2-ol (D) forms the most stable tertiary carbocation and will, therefore, react fastest with concentrated hydrochloric acid in the presence of anhydrous zinc chloride. Answer: (D) 2-Methylpropan-2-ol.

Key concepts

Carbocation Stability

Understanding carbocation stability is crucial for grasping why certain reactions occur faster than others in organic chemistry. A carbocation is an ion with a positively charged carbon atom that has only six electrons in its outer shell, making it electron-deficient and highly reactive.

The stability of these carbocations is greatly influenced by the number of alkyl groups attached to the positively charged carbon. Alkyl groups are electron-donating, which helps to disperse the positive charge over a larger area, thereby stabilizing the carbocation. The general order of carbocation stability is tertiary (three alkyl groups) > secondary (two alkyl groups) > primary (one alkyl group) > methyl (no alkyl groups).

In the exercise provided, when comparing butan-2-ol (secondary) and 2-methylpropan-2-ol (tertiary), for instance, the latter is more stable due to the presence of three alkyl groups that help stabilize the positive charge. This increased stability translates to a faster reaction with hydrochloric acid when catalyzed by anhydrous zinc chloride.

Inductive Effect

The inductive effect plays a fundamental role in the stability of carbocations. It refers to the electrostatic effect that atoms or groups of atoms have on the distribution of charges within a molecule. Atoms or groups that can donate electron density through sigma bonds exert a positive inductive effect, which is observed in carbocations.

Alkyl groups, due to their electron-donating nature, exert a positive inductive effect and help to delocalize the charge in carbocations, resulting in enhanced stability. The extent of the inductive effect depends on the number of alkyl groups and their proximity to the positively charged carbon.

For example, the inductive effect is stronger in the tertiary carbocation derived from 2-methylpropan-2-ol because it has more electron-donating alkyl groups close to the carbocation center than the carbocations formed from butan-1-ol or butan-2-ol, which have fewer alkyl groups to donate electron density.

Hydrochloric Acid Interaction with Alcohols

The interaction of hydrochloric acid (HCl) with alcohols forms an alkyl halide via a substitution reaction. This interaction typically requires a catalyst, such as anhydrous zinc chloride (ZnCl2), which activates the alcohol, making the hydroxyl group a better leaving group.

The reaction proceeds with the formation of a carbocation intermediary when the hydroxyl group leaves, forming water. The positively charged carbon is then attacked by the chloride ion from HCl, forming the corresponding alkyl halide.

The rate at which this reaction occurs is heavily dependent on the formation of the carbocation, as seen in the exercise provided. The tertiary alcohol forms the most stable carbocation and thus reacts the fastest with HCl. It is important for students to recognize that the reaction rate is not just about the presence of HCl, but also about how favorably the carbocation intermediate can form and be stabilized through factors such as the inductive effect.