Carbocations MCQ Quiz - Objective Question with Answer for Carbocations - Download Free PDF

Concept:-

Stability of Carbocation:

• The stability of a carbocation depends on several factors like hyperconjugation, resonance, and inductive effect. 
• Out of these factors, the resonance effect is the most important factor for the stability of a carbocation.
• The observed order of stability for carbocations is as follows:

tertiary>secondary>primary>methyl.

Explanation:-

• The carbocations I and III are bridgehead carbocations. A carbocation is sp² hybridized. Hence it has a planar geometry.

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• The C atom at the bridgehead position of a  bridgehead carbocation is not planar. This is because the C atom experiences a sufficient angle strain at the bridgehead position due to sp² hybridization. Hence carbocations I and III are unstable carbocations.
• Among the bridgehead carbocations, the carbocation with more number of C atoms in the ring will have less steric strains than the other carbocations and hence will be stable to some extent.\
• Thus, Carbocation II with a higher number of C atoms in the ring will be more stable than Carbocation I.
• Now, carbocation II is a tertiary carbocation. The tertiary Carbocation III is the most stable carbocation among all the three carbocations as it is a tertiary carbocation, which is stabilized by both the inductive and hyperconjugation effect.
• Thus the stability order will be II>III>I

Conclusion:-

• Hence, the correct order of stability of the following carbocation is II>III>I.

CONCEPT:

Carbocation Stability

• A carbocation is a positively charged carbon atom with three bonds instead of four.
• The stability of a carbocation is influenced by several factors including hyperconjugation, resonance, and inductive effects.
• In general, the order of stability for carbocations is as follows:
  • Tertiary (3°) > Secondary (2°) > Primary (1°) > Methyl (CH₃⁺)

EXPLANATION:

Therefore, the most stable carbocation is option 1 (a).

CONCEPT:

Dehydration of Alcohols to Form Alkenes

• In the presence of acid (H⁺) and heat (Δ), tertiary alcohols undergo dehydration to form alkenes through an E1 elimination mechanism.
• The reaction proceeds by first protonating the hydroxyl group, making it a good leaving group (water).
• The formation of a carbocation intermediate occurs after the departure of water.
• Carbocation rearrangement can occur if a more stable carbocation can be formed, leading to the most substituted (and stable) alkene as the major product.

EXPLANATION:

• The given alcohol is a tertiary alcohol, so it will undergo an E1 elimination.
• Step 1: Protonation of the hydroxyl group (–OH) leads to the formation of a good leaving group (H₂O), which then leaves, generating a carbocation.
• Step 2: The resulting carbocation is less stable and require further rearrangement.
• Step 3: A hydrogen atom is removed from the adjacent carbon, leading to the formation of a double bond. The product formed will be the most substituted alkene (Zaitsev's rule), which is the most stable form.
• Reaction:

CONCLUSION:

The correct option is: Option 3

CONCEPT:

Dehydration of Alcohol and Ring Expansion

• When alcohol reacts with sulfuric acid, it typically undergoes dehydration, forming a carbocation intermediate.
• In strained rings like cyclopropyl, the carbocation undergoes rearrangement to relieve ring strain, resulting in a more stable carbocation.
• The final product is determined by the stability of the rearranged carbocation, often leading to ring expansion and formation of carbonyl compounds like ketones.

MECHANISM:

• Step 1: Formation of Cyclopropylmethyl Carbocation
  • 
• Step 3: Ring Expansion to Cyclobutyl Carbocation
  • The unstable carbocation undergoes ring expansion, where the strained cyclopropyl ring opens to form a more stable four-membered cyclobutyl carbocation.
  • 
• Step 4: Formation of Cyclobutanone
  • The rearranged cyclobutyl carbocation loses a proton, resulting in the formation of cyclobutanone .
  • 

The correct product is Cyclobutanone (Option 3)

CONCEPT:

Carbocation Stability Factors

• Electron-donating groups (EDGs) like -OCH₃, -OH, and -NH₂ stabilize carbocations by donating electrons through resonance and inductive effects.
• More resonance structures lead to better stabilization of the positive charge.
• Inductive effects (+I effect) also help to stabilize carbocations by dispersing the positive charge.

EXPLANATION:

• (P) is a benzyl carbocation without any electron-donating or withdrawing groups. It has moderate stability due to resonance but lacks additional stabilizing groups.
• (Q) has a -OCH₃ group, which is an electron-donating group via resonance (+R effect). However, the inductive effect of oxygen makes it less stabilizing than groups like -OH and -NH₂.
• (R) has a hydroxyl group -OH, which is an electron-donating group through resonance, stabilizing the carbocation better than -OCH₃.
• (S) has an amine group -NH₂, which is the strongest electron-donating group through resonance (+R effect), providing the highest stability to the carbocation.

CONCLUSION:

The correct answer is: Option 4 - S > R > Q > P

Concept:

Acid Strength and Aromatacity:

• A substance that gives H⁺ ion in the solution and forms salts by combining with some metals is called acid. According to Arrhenius, acid is a substance that releases an H+ ion in the solution and a base is a substance that accepts an H+ ion from the solution.
• The pKa value indicates the strength of an acid in the solution. The lower the value of pKa, the more the acid's strength will be.
• Aromatic compounds follow Huckel's rule, according to which a cyclic, planar, and conjugated species having (4n+2)pi electrons (n = 0,1,3...) is aromatic.
• Aromatic compounds are very stable.
• Whereas, the anti-aromatic compounds follow the 4npi electrons rule. According to this a cyclic, planar, and conjugated species having (4n+2)pi electrons (n = 0,1,3...) is aromatic.
• Anti-aromatic compounds are very unstable.

Explanation:

• The conjugate base formed by compound A after deprotonation is an anti-aromatic compound with 4npi electrons for n=1, which is unstable. Thus, compound A has lower acid strength and a high value of pKa.
• The conjugate base formed by compound C after deprotonation is an anti-aromatic compound with 8npi electrons for n=2, which is unstable. Thus, compound B also has a lower acid strength and a high pKa value.
• The conjugate base formed by compound B after deprotonation is an aromatic compound with 6npi electrons for n=1, which is very stable. Thus, compound B has the highest acid strength and lowest value of pKa.

Conclusion:

• Thus, the correct order of the pKa of the following compounds is A > C > B.

Concept:

• The dienone phenol rearrangement is the rearrangement of a dienone into a phenol. The most common example is the conversion of 4,4 disubstitutedcyclohexadienone into 3,4 disubstituted phenol.
• The reaction involves the migration of groups from the 2nd or 4th position depending on the stability of the carbocation formed.
• A bicyclic dienone is converted to tetrahydronapthol system.
• The first step of the reaction is the protonation of the most basic atom of the system, the carbonyl oxygen atom.
• The intermediate carbocation then undergoes an alkyl shift to produce a more stable carbocation.
• The driving force is the aromatization of the ring. The general reaction is:

• The general mechanism is shown below:

Explanation:

• The molecule undergoes the reaction in the following manner.

​

​

Hence, the product formed is 

• Solvolysis is a type of substitution or elimination reaction in which the solvent acts as a nucleophile.
• A nucleophile is anything that can act as an electron-pair donor in a chemical reaction and form new bonds with electrophiles, which are electron-pair acceptors.
•  

   it is least stabilized.

So, The order of solvolysis is T > R > Q > S > P  .

Concept:

Acid Strength and Aromatacity:

• A substance that gives H⁺ ion in the solution and forms salts by combining with some metals is called acid. According to Arrhenius, acid is a substance that releases an H+ ion in the solution and a base is a substance that accepts an H+ ion from the solution.
• The pKa value indicates the strength of an acid in the solution. The lower the value of pKa, the more the acid's strength will be.
• Aromatic compounds follow Huckel's rule, according to which a cyclic, planar, and conjugated species having (4n+2)pi electrons (n = 0,1,3...) is aromatic.
• Aromatic compounds are very stable.
• Whereas, the anti-aromatic compounds follow the 4npi electrons rule. According to this a cyclic, planar, and conjugated species having (4n+2)pi electrons (n = 0,1,3...) is aromatic.
• Anti-aromatic compounds are very unstable.

Explanation:

• The conjugate base formed by compound A after deprotonation is an anti-aromatic compound with 4npi electrons for n=1, which is unstable. Thus, compound A has lower acid strength and a high value of pKa.
• The conjugate base formed by compound C after deprotonation is an anti-aromatic compound with 8npi electrons for n=2, which is unstable. Thus, compound B also has a lower acid strength and a high pKa value.
• The conjugate base formed by compound B after deprotonation is an aromatic compound with 6npi electrons for n=1, which is very stable. Thus, compound B has the highest acid strength and lowest value of pKa.

Conclusion:

• Thus, the correct order of the pKa of the following compounds is A > C > B.

Concept:-

Stability of Carbocation:

• The stability of a carbocation depends on several factors like hyperconjugation, resonance, and inductive effect. 
• Out of these factors, the resonance effect is the most important factor for the stability of a carbocation.
• The observed order of stability for carbocations is as follows:

tertiary>secondary>primary>methyl.

Explanation:-

• The carbocations I and III are bridgehead carbocations. A carbocation is sp² hybridized. Hence it has a planar geometry.

​

• The C atom at the bridgehead position of a  bridgehead carbocation is not planar. This is because the C atom experiences a sufficient angle strain at the bridgehead position due to sp² hybridization. Hence carbocations I and III are unstable carbocations.
• Among the bridgehead carbocations, the carbocation with more number of C atoms in the ring will have less steric strains than the other carbocations and hence will be stable to some extent.\
• Thus, Carbocation II with a higher number of C atoms in the ring will be more stable than Carbocation I.
• Now, carbocation II is a tertiary carbocation. The tertiary Carbocation III is the most stable carbocation among all the three carbocations as it is a tertiary carbocation, which is stabilized by both the inductive and hyperconjugation effect.
• Thus the stability order will be II>III>I

Conclusion:-

• Hence, the correct order of stability of the following carbocation is II>III>I.

Concept:

• The dienone phenol rearrangement is the rearrangement of a dienone into a phenol. The most common example is the conversion of 4,4 disubstitutedcyclohexadienone into 3,4 disubstituted phenol.
• The reaction involves the migration of groups from the 2nd or 4th position depending on the stability of the carbocation formed.
• A bicyclic dienone is converted to tetrahydronapthol system.
• The first step of the reaction is the protonation of the most basic atom of the system, the carbonyl oxygen atom.
• The intermediate carbocation then undergoes an alkyl shift to produce a more stable carbocation.
• The driving force is the aromatization of the ring. The general reaction is:

• The general mechanism is shown below:

Explanation:

• The molecule undergoes the reaction in the following manner.

​

​

Hence, the product formed is 

Concept:-

Stability of Carbocation:

• The stability of a carbocation depends on several factors like hyperconjugation, resonance, and inductive effect. 
• Out of these factors, the resonance effect is the most important factor for the stability of a carbocation.
• The observed order of stability for carbocations is as follows:

tertiary>secondary>primary>methyl.

Explanation:-

• The carbocations I and III are bridgehead carbocations. A carbocation is sp² hybridized. Hence it has a planar geometry.

​

• The C atom at the bridgehead position of a  bridgehead carbocation is not planar. This is because the C atom experiences a sufficient angle strain at the bridgehead position due to sp² hybridization. Hence carbocations I and III are unstable carbocations.
• Among the bridgehead carbocations, the carbocation with more number of C atoms in the ring will have less steric strains than the other carbocations and hence will be stable to some extent.\
• Thus, Carbocation II with a higher number of C atoms in the ring will be more stable than Carbocation I.
• Now, carbocation II is a tertiary carbocation. The tertiary Carbocation III is the most stable carbocation among all the three carbocations as it is a tertiary carbocation, which is stabilized by both the inductive and hyperconjugation effect.
• Thus the stability order will be II>III>I

Conclusion:-

• Hence, the correct order of stability of the following carbocation is II>III>I.

CONCEPT:

Carbocation Stability

• A carbocation is a positively charged carbon atom with three bonds instead of four.
• The stability of a carbocation is influenced by several factors including hyperconjugation, resonance, and inductive effects.
• In general, the order of stability for carbocations is as follows:
  • Tertiary (3°) > Secondary (2°) > Primary (1°) > Methyl (CH₃⁺)

EXPLANATION:

Therefore, the most stable carbocation is option 1 (a).

CONCEPT:

Dehydration of Alcohols to Form Alkenes

• In the presence of acid (H⁺) and heat (Δ), tertiary alcohols undergo dehydration to form alkenes through an E1 elimination mechanism.
• The reaction proceeds by first protonating the hydroxyl group, making it a good leaving group (water).
• The formation of a carbocation intermediate occurs after the departure of water.
• Carbocation rearrangement can occur if a more stable carbocation can be formed, leading to the most substituted (and stable) alkene as the major product.

EXPLANATION:

• The given alcohol is a tertiary alcohol, so it will undergo an E1 elimination.
• Step 1: Protonation of the hydroxyl group (–OH) leads to the formation of a good leaving group (H₂O), which then leaves, generating a carbocation.
• Step 2: The resulting carbocation is less stable and require further rearrangement.
• Step 3: A hydrogen atom is removed from the adjacent carbon, leading to the formation of a double bond. The product formed will be the most substituted alkene (Zaitsev's rule), which is the most stable form.
• Reaction:

CONCLUSION:

The correct option is: Option 3

CONCEPT:

Dehydration of Alcohol and Ring Expansion

• When alcohol reacts with sulfuric acid, it typically undergoes dehydration, forming a carbocation intermediate.
• In strained rings like cyclopropyl, the carbocation undergoes rearrangement to relieve ring strain, resulting in a more stable carbocation.
• The final product is determined by the stability of the rearranged carbocation, often leading to ring expansion and formation of carbonyl compounds like ketones.

MECHANISM:

• Step 1: Formation of Cyclopropylmethyl Carbocation
  • 
• Step 3: Ring Expansion to Cyclobutyl Carbocation
  • The unstable carbocation undergoes ring expansion, where the strained cyclopropyl ring opens to form a more stable four-membered cyclobutyl carbocation.
  • 
• Step 4: Formation of Cyclobutanone
  • The rearranged cyclobutyl carbocation loses a proton, resulting in the formation of cyclobutanone .
  • 

The correct product is Cyclobutanone (Option 3)