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

The alcohol which does not give a stable compound on dehydration is (a) methyl alcohol (b) ethyl alcohol (c) \(\mathrm{n}\)-butyl alcohol (d) n-propyl alcohol

Short Answer

Expert verified
(a) Methyl alcohol does not give a stable compound on dehydration.

Step by step solution

01

Understanding Dehydration Reaction

Dehydration of alcohols typically involves the removal of a water molecule, leading to the formation of an alkene. This process is influenced by the stability of the resulting carbocation intermediate. Tertiary carbocations are more stable than secondary, which are more stable than primary carbocations.
02

Analyzing Methyl Alcohol

Methyl alcohol ( ext{CH}_3 ext{OH}) undergoes dehydration but does not lead to a stable carbocation, as it would require the formation of a primary carbocation, which is unstable.
03

Analyzing Ethyl Alcohol

Ethyl alcohol ( ext{C}_2 ext{H}_5 ext{OH}) can be dehydrated to form ethene by forming a primary carbocation, which is not very stable but commonly occurs in reactions.
04

Analyzing n-Butyl Alcohol

n-Butyl alcohol ( ext{C}_4 ext{H}_9 ext{OH}) can undergo dehydration to form 1-butene via a primary carbocation, which is relatively unstable but can still occur.
05

Analyzing n-Propyl Alcohol

n-Propyl alcohol ( ext{C}_3 ext{H}_7 ext{OH}) dehydrates to form propene through a primary carbocation. While not as stable as secondary or tertiary carbocations, it can nonetheless occur.
06

Identifying the Least Stable Option

Among the given options, methyl alcohol results in the least stable intermediate upon dehydration, as its primary carbocation is the most unstable, leading to no stable compound.

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 positively charged ions, typically involving a carbon atom that has only six electrons, making it quite reactive. The stability of these ions is crucial in many organic reactions, including the dehydration of alcohols. An easy way to remember the stability of carbocations is by considering their degree of substitution:
  • Tertiary carbocation: Three alkyl groups attached. Most stable due to the inductive effect and hyperconjugation.
  • Secondary carbocation: Two alkyl groups attached. Moderately stable.
  • Primary carbocation: One alkyl group attached. Least stable and less commonly observed.
The reasons behind this ranking involve both inductive effects, where electron-donating alkyl groups stabilize the positive charge, and hyperconjugation, which allows delocalization of charge through \(\sigma\) bonds. Because of this, reactions that form more stable carbocations are typically faster and more favored.
Primary Carbocation
A primary carbocation is a specific type of carbocation where the carbon atom bearing the positive charge is bonded to only one other carbon. This limited substitution makes primary carbocations the least stable among the different types.When discussing reactions like alcohol dehydration, primary carbocations are often intermediates. However, because they are not very stable, these reactions can sometimes need extra energy or a catalyst to proceed. In synthetic chemistry, chemists often try to avoid reactions that will form primary carbocations or instead find ways to stabilize them with neighboring groups or rearrangement.Knowing that primary carbocations are unstable illuminates why reactions like the dehydration of methyl alcohol (\( ext{CH}_3 ext{OH} \)) are problematic. The resulting primary carbocation is so unstable that the reaction does not proceed effectively under normal conditions.
Alcohol Dehydration Mechanism
The mechanism of alcohol dehydration involves multiple steps where the hydroxyl group (\( - ext{OH} \)) is removed to eventually form a double bond between carbon atoms (an alkene). This process generally requires acidic conditions, often using sulfuric acid (\( ext{H}_2 ext{SO}_4 \)) as a catalyst. The steps involved in the dehydration mechanism typically include:
  • Protonation: The alcohol oxygen attracts a proton, making the water molecule a better leaving group.
  • Carbocation Formation: The protonated alcohol loses water to form a carbocation. This is the step where the stability of the carbocation plays a critical role.
  • Alkene Formation: Finally, a base removes a proton adjacent to the carbocation, leading to the formation of an alkene.
The entire process is an example of an E1 mechanism, dependent on the formation and stability of the carbocation. Because it requires carbocation stability, the dehydration of primary alcohols (like n-butyl and n-propyl alcohol) can be less efficient than that of secondary or tertiary alcohols.

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

An organic compound 'A' reacts with methyl magnesium iodide, to form an addition product which on hydrolysis forms the compound 'B'. Compound 'B' gives blue colour salt in Victor Meyer's test. The compounds 'A' and 'B' are respectively (a) acetaldehyde and isopropyl alcohol (b) acetone and isopropyl alcohol (c) acetaldehyde and tertiary butyl alcohol (d) acetaldehyde and ethyl alcohol

The reaction of an aromatic aryl chloride and phenol in the presence of the base \(\mathrm{NaOH}\) or pyridine, is called (a) Kolbe's reaction (b) Perkin reaction (c) Sandmeyer's reaction (d) Schotten-Baumann reaction

Three moles of ethanol react with one mole of phosphorus tribromide to form 3 moles of bromoethane and one mole of \(\mathrm{X}\). Which of the following is \(X ?\) (a) \(\mathrm{H}_{3} \mathrm{PO}_{3}\) (b) \(\mathrm{H}_{3} \mathrm{PO}_{2}\) (c) \(\mathrm{H}_{3} \mathrm{PO}_{4}\) (d) \(\mathrm{HPO}_{3}\)

A compound is soluble in concentrated \(\mathrm{H}_{2} \mathrm{SO}_{4}\). It does not decolourize bromine in carbon tetrachloride but is oxidized by chromic anhydride in aqueous sulphuric acid within two seconds, turning the orange solution to blue, then green and finally opaque. The original compound is (a) an alkane (b) an ether (c) a tertiary alcohol (d) a primary alcohol

Vinyl carbinol is (a) \(\mathrm{HO}-\mathrm{CH}_{2}-\mathrm{CH}=\mathrm{CH}_{2}\) (b) \(\mathrm{CH}_{3}-\mathrm{C}\left(\mathrm{CH}_{2} \mathrm{OH}\right)=\mathrm{CH}_{2}\) (c) \(\mathrm{CH}_{3}-\mathrm{CH}=\mathrm{CH}-\mathrm{OH}\) (d) \(\mathrm{CH}_{3} \mathrm{C}(\mathrm{OH})=\mathrm{CH}_{2}\)

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