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Which is dehydrated to a maximum extent using conc. \(\mathrm{H}_{2} \mathrm{SO}_{4} ?\) (a) O=C1CCCCC1O (b) O=C1CCC(O)CC1 (c) O=C1CCCC(O)C1 (d) CC1C(=O)CCCC1O

Short Answer

Expert verified
Option (d) is dehydrated to the maximum extent due to carbocation stability.

Step by step solution

01

Understand the Reaction

Dehydration using concentrated \(\) is a common way to remove water from alcohols. The goal is to form a double bond by eliminating water \(\( H_2O \)\). The process is favored in molecules that can stabilize the resulting carbocation.
02

Analyze Each Compound for Dehydration

Look at each option to determine how easily it can lose water when treated with concentrated \( \ \).- (a) \smiles{O=C1CCCCC1O}: 5-membered ring with a secondary alcohol.- (b) \smiles{O=C1CCC(O)CC1}: 6-membered ring with a secondary alcohol.- (c) \smiles{O=C1CCCC(O)C1}: similar to (b) with a different position of alcohol.- (d) \smiles{CC1C(=O)CCCC1O}: secondary alcohol on a 6-membered ring with an extra methyl group.
03

Consider Stability of Carbocations

The stability of the carbocation formed during dehydration is crucial. A more substituted carbocation (e.g., tertiary) will be more stable compared to secondary or primary. - Option (d) has an extra methyl group, which enhances carbocation stability through hyperconjugation and inductive effects.
04

Assess Maximum Dehydration

To identify maximum dehydration, consider which compound forms the most stable alkene. Options with greater substitution around the potential alkene will result in more stable, hence more favorable, dehydration. - (d) : The extra methyl group contributes to forming a more substituted, stable double bond upon dehydration.

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Key Concepts

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

Carbocation Stability
Carbocations are key intermediates in many organic reactions, including the dehydration of alcohols. Their stability is vital, as it influences the reaction's rate and outcome. Factors contributing to carbocation stability include:
  • Substitution: Tertiary carbocations are more stable than secondary and primary ones due to hyperconjugation and inductive effects from surrounding alkyl groups.
  • Resonance: Availability of resonance structures can greatly enhance stability.
  • Inductive Effect: Electronegative atoms or groups can stabilize carbocations by electron withdrawal.
In the original exercise, option (d) benefits from having an extra methyl group, which stabilizes the carbocation through hyperconjugation, making it the most favorable for dehydration.
Alkene Formation
The dehydration of alcohols typically aims to form alkenes. This involves the removal of water ( H_2O) from the alcohol. Factors affecting alkene formation include:
  • Position of the Alcohol Group: The location of the OH group impacts which hydrogen atoms can be most easily removed to form a double bond.
  • Regioselectivity: Zaitsev's rule often predicts that the more substituted alkene (more stable one) will be the major product.
  • Stereochemistry: Depending on the starting material, cis or trans isomers may also form, with the trans (or E) form usually being more stable.
In our example, the additional methyl group in option (d) enhances alkene stability, leading to maximum dehydration.
Hyperconjugation
Hyperconjugation is a stabilizing interaction that occurs when electrons in a C-H sigma bond of an adjoining C-H or C-C bond are delocalized into an empty p orbital, or π orbital of an adjacent unsaturated system. This interaction is crucial for:
  • Stabilizing Carbocations: Through hyperconjugation, electrons help to delocalize charge, enhancing stability.
  • Contributing to Alkene Stability: More hyperconjugative structures result in more stable alkenes.
In the context of the problem, the methyl group in option (d) offers additional hyperconjugation, promoting both a stable carbocation intermediate and a stable alkene product upon dehydration.
Chemical Reactions
Chemical reactions involving dehydration often require a reagent like concentrated H_2SO_4, which acts as a strong acid to promote the loss of water. The general mechanism involves:
  • Protonation of the Alcohol: The OH group is converted into a better leaving group, H_2O.
  • Carbocation Formation: Water is removed, resulting in a carbocation.
  • Alkene Formation: A proton is removed from an adjacent carbon, forming a double bond.
For option (d), the presence of the methyl group assists in maintaining the carbocation stability, facilitating the dehydration process.
Organic Chemistry
Organic chemistry is the study of carbon-containing compounds and their reactions. A comprehensive understanding involves:
  • Functional Groups: Recognition of various functional groups like alcohols, ketones, and alkenes.
  • Reaction Mechanisms: A step-by-step approach to how reactions proceed, such as dehydration in this case.
  • Stability Considerations: Various factors like hyperconjugation and substituent effects that influence reaction pathways.
The original exercise highlights these principles, demonstrating how structure affects reactivity, as seen with the preferential dehydration of option (d), the most structurally favorable candidate due to its high carbocation and alkene stability.

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Most popular questions from this chapter

The most stable free radical among the following is (1) \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}_{2}\) (2) \(\mathrm{CH}_{3}-\dot{\mathrm{CH}}-\mathrm{CH}_{3}\) (3) \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CHCH}_{3}\) (4) \(\mathrm{CH}_{3} \mathrm{CH}_{2}\) (a) \(3>1>4>2\) (b) \(1>3>2>4\) (c) \(3>1>2>4\) (d) \(3>2>1>4\)

The increasing order of stability of the following free radicals is (a) \(\left(\mathrm{CH}_{3}\right)_{2} \dot{\mathrm{C}} \mathrm{H}<\left(\mathrm{CH}_{3}\right)_{3} \dot{\mathrm{C}}<\left(\mathrm{C}_{6} \mathrm{H}_{5}\right)_{2} \dot{\mathrm{C}} \mathrm{H}<\left(\mathrm{C}_{6} \mathrm{H}_{3}\right)_{3} \dot{\mathrm{C}}\) (b) \(\left(\mathrm{C}_{6} \mathrm{H}_{3}\right)_{3} \dot{\mathrm{C}}<\left(\mathrm{C}_{6} \mathrm{H}_{5}\right)_{2} \dot{\mathrm{C}} \mathrm{H}<\left(\mathrm{CH}_{3}\right)_{3} \dot{\mathrm{C}}<\left(\mathrm{CH}_{3}\right)_{2} \dot{\mathrm{C}} \mathrm{H}\) (c) \(\left(\mathrm{C}_{6} \mathrm{H}_{5}\right)_{2} \dot{\mathrm{C}} \mathrm{H}<\left(\mathrm{C}_{6} \mathrm{H}_{5}\right)_{3} \dot{\mathrm{C}}<\left(\mathrm{CH}_{3}\right)_{3} \dot{\mathrm{C}}<\left(\mathrm{CH}_{3}\right)_{2} \dot{\mathrm{C}} \mathrm{H}\) (d) \(\left(\mathrm{CH}_{3}\right)_{2} \dot{\mathrm{C}} \mathrm{H}<\left(\mathrm{CH}_{3}\right)_{3} \dot{\mathrm{C}}<\left(\mathrm{C}_{6} \mathrm{H}_{5}\right)_{3} \dot{\mathrm{C}}<\left(\mathrm{C}_{6} \mathrm{H}_{3}\right)_{2} \dot{\mathrm{C}} \mathrm{H}\)

In the presence of peroxide, hydrogen chloride and hydrogen iodide do not undergo antiMarkownikoff's addition to alkenes because (a) both are highly ionic (b) one is oxidizing and the other is reducing (c) one of the steps is endothermic in both the cases (d) all the steps are exothermic in both the cases

Which is correct about SN reaction? (a) Here reaction intermediate is carbocation (b) It is favoured by polar solvents (c) It is favoured by stability of carbocation (d) all of these

Which of the following order/statement (s) is/are correct? (a) In \(S_{N} 1\) reactions, \(\mathrm{AgNO}_{3}\) increases the rates of solvolysis (b) Soft bases have enhanced nucleophilicities towards \(\mathrm{S}_{\mathrm{N}} 2\) reactions. (c) \(\mathrm{PhSO}_{3}^{-}>\mathrm{Cl}_{3} \mathrm{CCOO}^{-}>\mathrm{CH}_{3} \mathrm{COO}^{-}>\mathrm{PhO}^{-}\)(leav- ing group ability) (d) HCOOH may lead a tertiary halide towards \(\mathrm{S}_{\mathrm{N}}^{2} .\)

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