Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

From amongst the following alcohols the one that would react fastest with conc. HCl and anhydrous \(\mathrm{ZnCl}_{2}\), is (a) 2-Butanol (b) 2 -Methylpropan \(-2\)-ol (c) 2-Methylpropanol (d) 1 -Butanol

Short Answer

Expert verified
2-Methylpropan-2-ol reacts fastest due to forming a stable tertiary carbocation.

Step by step solution

01

Understanding the Concept

To determine which alcohol reacts fastest with conc. HCl and anhydrous \(\text{ZnCl}_2\), we need to remember that this is a Lucas test. Primary alcohols react slowly, secondary somewhat faster, and tertiary alcohols react immediately. The rate of reaction depends on the carbocation stability formed during the substitution reaction.
02

Analyzing Each Option

Let's examine each alcohol: - 2-Butanol is a secondary alcohol. - 2-Methylpropan-2-ol is a tertiary alcohol. - 2-Methylpropanol is a primary alcohol. - 1-Butanol is a primary alcohol.
03

Determining Carbocation Stability

The tertiary carbocation formed from 2-methylpropan-2-ol is more stable than secondary and primary carbocations. Carbocation stability follows the order: tertiary > secondary > primary.
04

Conclusion

Since 2-methylpropan-2-ol forms the most stable carbocation, it will react the fastest with conc. HCl and anhydrous \(\text{ZnCl}_2\). Tertiary alcohols react immediately compared to secondary and primary ones.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Carbocation Stability
Carbocations are intermediates in many organic reactions, acting as key players in determining the reaction's course and rate. They are positively charged carbon atoms, making them electron-deficient. This electron deficiency makes carbocations highly reactive, as they seek to stabilize their charge by gaining electrons.

The stability of these intermediates is crucial. It is primarily influenced by the nature and number of alkyl groups around the positively charged carbon. For instance, carbocation stability follows a clear hierarchy:
  • Tertiary carbocations: These are the most stable due to the presence of three alkyl groups. Alkyl groups donate electrons via inductive and hyperconjugative effects, thereby stabilizing the positive charge.

  • Secondary carbocations: These have two alkyl groups and show moderate stability.

  • Primary carbocations: With only one alkyl group, they are less stable and form less readily than tertiary and secondary carbocations.
Understanding carbocation stability is crucial in predicting the behavior of alcohols in reactions like the Lucas test. Since the stability of the carbocation directly affects the reaction rate, tertiary alcohols react fastest as they form the most stable carbocation.
Tertiary Alcohol Reaction
Tertiary alcohols like 2-methylpropan-2-ol are well known for their rapid reaction in the presence of reagents such as concentrated HCl and anhydrous \(\text{ZnCl}_2\). This particular setup is known as the Lucas test, which is employed to classify alcohols based on their reaction rate.

The chemistry behind this rapid reaction involves the formation of a highly stable tertiary carbocation when the tertiary alcohol undergoes a substitution reaction with HCl. This carbocation is exceptionally stable due to the electron-donating nature of the surrounding alkyl groups, which neutralize the positive charge more effectively.

Here's a simplified breakdown:
  • Upon contact with HCl and anhydrous \(\text{ZnCl}_2\), the \(\text{OH}\) group is replaced by a chloride ion.

  • This leads to the generation of a tertiary carbocation as an intermediate.

  • The reaction’s completion is marked by the formation of chlorides and commonly results in turbidity or a cloudy solution, indicating a successful reaction.
Tertiary alcohols' rapid reaction is due to the immediate formation of a stable carbocation, making them distinguishable from primary and secondary alcohols in such tests.
Primary vs Secondary vs Tertiary Alcohols
Understanding the differences between primary, secondary, and tertiary alcohols is crucial in predicting their behavior in chemical reactions.
  • Primary alcohols have the \(-\text{OH}\) group attached to a carbon atom that is bonded to only one other carbon atom. They react slowly with HCl in the presence of \(\text{ZnCl}_2\) due to the formation of relatively less stable primary carbocations.

  • Secondary alcohols feature an \(-\text{OH}\) group attached to a carbon atom connected to two other carbon atoms. They react at a moderate speed as the stability of the secondary carbocation is greater than that of primary carbocations but less than tertiary ones.

  • Tertiary alcohols, such as 2-methylpropan-2-ol, show rapid reaction rates, attributed to the attachment of the \(-\text{OH}\) group to a carbon atom with three carbon-carbon bonds. The tertiary carbocation formed during reaction is highly stable, leading to a quick and efficient reaction.
The differences in bonding and resultant stability of the carbocations formed during reactions allow for the effective classification of alcohols, helping chemists predict and understand their behavior under different conditions.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Under different conditions, nitration of phenol yields 1\. o-nitrophenol 2\. p-nitrophenol 3\. \(2,4,6\)-trinitrophenol The correct sequence of decreasing order of acidic nature of these phenols is (a) \(3,2,1\) (b) \(1,2,3\) (c) \(2,1,3\) (d) \(3,1,2\)

Which one of the following undergoes reaction with \(50 \%\) sodium hydroxide solution to give the corresponding alcohol and acid? (a) phenol (b) benzaldehyde (c) butanal (d) benzoic acid

During dehydration of alcohols to alkenes, by heating with concentrated \(\mathrm{H}_{2} \mathrm{SO}_{4}\) the initiation step is (a) formation of an ester (b) elimination of water (c) protonation of alcohol molecule (d) formation of a carbocation

Luca's test of alcohols involves following reaction: \(\mathrm{R}-\mathrm{OH}+\mathrm{HCl} \frac{\text { anhydrous }}{\mathrm{ZnCl}_{2}} \underset{\text { white turbidity }}{\mathrm{RCl}+\mathrm{H}_{2} \mathrm{O}}\) Select the correct statement(s) for the Luca's test. (a) Lesser the acidic character of alcohol, greater is its reactivity towards Luca's reagent. (b) ROH behaves as a base. (c) \(\mathrm{CH}_{3} \mathrm{OH}\) gives Luca's test most quickly. (d) Reactivity of \(1^{\circ}, 2^{\circ}, 3^{\circ}\) alcohol, lies in the following sequence (for Luca's reaction \(3^{\circ}>2^{\circ}>1^{\circ}\) ).

Ortho \(-\) Nitrophenol is less soluble in water than pand \(\mathrm{m}-\) Nitrophenols because (a) Melting point of o- Nitrophenol is lower than those of \(\mathrm{m}\) - and \(\mathrm{p}\) - isomers (b) o- Nitrophenol is more volatile in steam than those of \(\mathrm{m}\) - and \(\mathrm{p}\) - isomers (c) o- Nitrophenol shows Intramolecular H-bonding (d) o- Nitrophenol shows Intermolecular H-bonding

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free