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

Which of the following compounds will undergo hydrolysis most readily by \(\mathrm{S}_{\mathrm{N}} 1\) mechanism?

Short Answer

Expert verified
Answer: ____________ To determine the compound that will most readily undergo hydrolysis by an SN1 mechanism, consider the following factors: 1. Substrate structure: Tertiary carbocations (3°) are more reactive than secondary (2°) and primary (1°) carbocations. 2. Leaving group: Good leaving groups, such as halogens or sulfonate ions, increase the likelihood of an SN1 reaction. 3. Solvent: Polar protic solvents, like water, alcohols, and carboxylic acids, are more suitable for SN1 reactions. Compare the given compounds based on these factors, and identify the one with the most desirable combination for favorable SN1 reactivity.

Step by step solution

01

Identifying the substrate structure

The substrate structure affects the reactivity towards SN1 reactions. Tertiary carbocations (3°) are the most stable and will readily undergo SN1 reactions, followed by secondary (2°) and primary (1°) carbocations. We should examine the given compounds for their structural classification as primary, secondary, or tertiary substrates.
02

Assessing the leaving group

The likelihood of a molecule undergoing an SN1 reaction can also be influenced by the leaving group. A good leaving group will make it more likely for a compound to undergo an SN1 reaction. Generally, good leaving groups are those that can stabilize the negative charge after leaving. Halogens like Cl-, Br-, and I- are typically good leaving groups, as are sulfonate ions like \(\mathrm{CH}_{3}\mathrm{SO}_{3}^{-}\). Compare the leaving groups of the given compounds to help determine which compound will most readily undergo hydrolysis by an SN1 mechanism.
03

Examining the solvent

The solvent used in the reaction can also influence the probability of an SN1 reaction occurring. Generally, polar protic solvents are more suitable for SN1 reactions because they can stabilize intermediate carbocations and transition states. Common polar protic solvents include water (H2O), alcohols (ROH), and carboxylic acids (R-COOH). If the given compounds include information about the solvent used, take this into consideration when determining which compound will most readily undergo hydrolysis by an SN1 mechanism.
04

Comparing the compounds

After identifying the substrate structure, leaving group, and solvent used for each of the given compounds, compare them to determine which compound best meets the criteria for favorable SN1 reactivity. The compound with the most desirable combination of these factors will likely be the one that undergoes hydrolysis by an SN1 mechanism the most readily.
05

Concluding the most readily hydrolysed compound

Based on the analysis of the substrate structure, leaving group, and solvent, identify which compound among the given list will most readily undergo hydrolysis by an SN1 mechanism.

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.

Tertiary Carbocations
In SN1 reactions, the formation of carbocations is a crucial step. Tertiary carbocations are the most stable among the carbocation types. This is because they benefit from the presence of three alkyl groups attached to the positively charged carbon. These groups help stabilize the carbocation by dispersing the positive charge through hyperconjugation and inductive effects.
Moreover, tertiary carbocations have a lower activation energy for formation, allowing the reaction to proceed more rapidly compared to primary or secondary carbocations.
In summary, when evaluating compounds for SN1 reactions, look for tertiary carbocations, as they greatly enhance the likelihood of the reaction occurring.
Leaving Groups
A key factor in determining whether a compound will readily undergo an SN1 reaction is the quality of the leaving group. Good leaving groups can leave the substrate, taking their electron pair, and thus facilitate the formation of a carbocation.
  • Stability is crucial: A leaving group must be able to stabilize itself after departing. This usually means the ability to hold a negative charge effectively.
  • Common examples of effective leaving groups include halogens like Cl-, Br-, and I-, due to their ability to delocalize charge. Another effective type includes sulfonate ions such as \(\mathrm{CH}_{3}\mathrm{SO}_{3}^{-}\).
When comparing compounds in a reaction, one with a superior leaving group is more apt to undergo the SN1 mechanism.
Polar Protic Solvents
Polar protic solvents play an indispensable role in SN1 reactions by stabilizing transition states and intermediates such as carbocations. These solvents possess hydrogen atoms capable of forming hydrogen bonds, which helps surround and stabilize carbocations through solvation.
Water (\(\mathrm{H}_2\mathrm{O}\)), alcohols (\(\mathrm{ROH}\)), and carboxylic acids are typical examples of effective polar protic solvents.
Moreover, such solvents can also engage the leaving group, enhancing their departure from the molecule. This creates a favorable environment for SN1 reactions to occur by lowering the energy barrier for the reaction.
Carbocation Stability
Carbocation stability is a cornerstone in understanding SN1 reactions. The ability of a carbocation to remain stable once formed is what drives the reaction forward.
Stability in carbocations comes from several factors:
  • Hyperconjugation: Involves the delocalization of electrons in sigma bonds (like C-H bonds) adjacent to the positively charged carbon.
  • Inductive effects: Electron-donating groups help to spread out the positive charge over a larger area.
These stabilizing effects are particularly strong in tertiary carbocations due to more substituents participating in these interactions, compared to primary or secondary carbocations. In any SN1 reaction, examining the potential carbocation's stability is essential to predict the success and rate of the reaction.

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

In the preparation of \(\mathrm{CH}_{3}-\mathrm{O}-\mathrm{C}\left(\mathrm{CH}_{3}\right)_{3}\), two methods, namely (i) and (ii) are suggested. Which method is preferable and why? (i) \(\mathrm{CH}_{3} \mathrm{ONa}+\left(\mathrm{CH}_{3}\right)_{3} \mathrm{C}-\mathrm{Br} \rightarrow \mathrm{CH}_{3}-\mathrm{O}-\mathrm{C}\left(\mathrm{CH}_{3}\right)_{3}\) (ii) \(\mathrm{CH}_{3} \mathrm{Br}+\left(\mathrm{CH}_{3}\right)_{3} \mathrm{C}-\mathrm{ONa} \rightarrow \mathrm{CH}_{3}-\mathrm{O}-\mathrm{C}\left(\mathrm{CH}_{3}\right)_{3}\)

The products formed in the above reaction are (a) \(\mathrm{CH}_{3}-\mathrm{CHO}, \mathrm{HCHO}, \mathrm{HCOOH}\) (b) \(\mathrm{CH}_{3}-\mathrm{CHO}, \mathrm{HCOOH}\) (c) \(\mathrm{CH}_{3}-\mathrm{CHO}, \mathrm{OHC}-\mathrm{CHO}\) (d) \(\mathrm{CH}_{3}-\mathrm{COOH}, \mathrm{HCOOH}\)

Preparation and reactions of alcohols Alcohols can by prepared from alkyl halides, epoxides, alkenes, etc. When treated with acid, \(\mathrm{C}-\mathrm{OH}\) bond is cleaved heterolytically to form a carbocation which could then undergo substitution, elimination or rearrangement depending on the conditions and stability of the carbocation. When \(\mathrm{BrCH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{OH}\) is refluxed with metallic sodium in dry ethers the major product obtained is (a) \(\mathrm{BrCH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{ONa}\) (b) \(\mathrm{CH}_{2}=\mathrm{CH}-\mathrm{CH}=\mathrm{CH}_{2}\)

When the alkyl iodide \(\mathrm{CH}_{3}-\mathrm{CH}_{2} \mathrm{I}\) is slowly decomposed by light, the organic product formed is (a) \(\mathrm{CH}_{3}-\mathrm{CH}_{2}-\mathrm{CH}_{2}-\mathrm{CH}_{3}\) (b) \(\mathrm{CH}_{3}-\mathrm{CH}_{3}\) (c) \(\mathrm{CH}_{3}-\mathrm{CHO}\) (d) \(\mathrm{CH}_{3}-\mathrm{CH}_{2} \mathrm{OH}\)

Preparation and reactions of alcohols Alcohols can by prepared from alkyl halides, epoxides, alkenes, etc. When treated with acid, \(\mathrm{C}-\mathrm{OH}\) bond is cleaved heterolytically to form a carbocation which could then undergo substitution, elimination or rearrangement depending on the conditions and stability of the carbocation. When 1,2 -dimethyl cyclohexene is treated with cold dilute alkaline \(\mathrm{KMnO}_{4}\), the product obtained is

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