Problem 38
Question
Statement -1: Phenol is more reactive than benzene towards electrophilic substitution reactions. Statement-2:In the case of phenol, the intermediate carbocation is more resonance stabilized. (a) If both Statement-1 and Statement-2 are correct, and Statement \(-2\) is the correct explanation of the Statement-2. (b) If both Statement-1 and Statement-2 are correct, but Statement \(-2\) is not the correct explanation of the Statement-1. (c) If Statement-1 is correct but Statement- 2 is incorrect. (d) If Statement-1 is incorrect but Statement-2 is correct. \(\theta\) 10 Subjeet Problems
Step-by-Step Solution
Verified Answer
(a) Both statements are correct, and Statement-2 explains Statement-1.
1Step 1: Understand Statement-1
Statement-1 states that phenol is more reactive than benzene towards electrophilic substitution reactions. This is true because the electron-donating -OH group in phenol increases the electron density in the benzene ring, making it more reactive to electrophiles.
2Step 2: Understand Statement-2
Statement-2 claims that the intermediate carbocation in phenol is more resonance stabilized. This is also true because the -OH group in phenol provides additional resonance structures by delocalizing the positive charge from the carbocation formed during the reaction.
3Step 3: Check if Statement-2 Explains Statement-1
Statement-2 directly explains why Statement-1 is true: the increased resonance stabilization in phenol due to the -OH group makes the electrophilic substitution more favorable compared to benzene, which lacks such stabilization.
Key Concepts
Phenol ReactivityResonance StabilizationCarbocation Intermediate
Phenol Reactivity
Phenol is significantly more reactive than benzene when it comes to electrophilic substitution reactions. This difference in reactivity is primarily due to the presence of the hydroxyl group (-OH) attached to the benzene ring. The -OH group is a powerful electron-donating group and enhances the electron density on the aromatic ring.
As a result, phenol becomes an attractive target for electrophiles, which are species that seek out areas of high electron density to react with.
As a result, phenol becomes an attractive target for electrophiles, which are species that seek out areas of high electron density to react with.
- The electron-donating nature of the -OH group increases the likelihood of reactions occurring.
- This makes the chemical environment of phenol more favorable for electrophilic substitution.
Resonance Stabilization
Resonance stabilization is a key concept when delving into why phenol is more reactive than benzene. In phenol, the carbocation intermediate formed during electrophilic substitution can be stabilized by resonance. This is possible because the positive charge on the carbocation can be delocalized over the aromatic ring.
The -OH group contributes to this stabilization by forming multiple resonance structures.
The -OH group contributes to this stabilization by forming multiple resonance structures.
- This delocalization allows the positive charge to be spread out over the structure, reducing energy and increasing stability.
- The more resonance structures you can draw, the greater the stabilization of the intermediate.
Carbocation Intermediate
A carbocation intermediate is an essential player in phenol’s electrophilic substitution reaction pathway. When phenol undergoes such reactions, an intermediate with a positively charged carbon atom, known as a carbocation intermediate, is briefly formed.
This intermediate is quite reactive and usually unstable, but in phenol, it benefits from additional stabilization.
This intermediate is quite reactive and usually unstable, but in phenol, it benefits from additional stabilization.
- The -OH group on phenol provides a means to delocalize the positive charge through resonance.
- This results in several resonance structures that distribute the positive charge across the molecule.
Other exercises in this chapter
Problem 37
A ketone 'A' which undergoes haloform reaction gives compound B on reduction. \(\mathrm{B}\) on heating with sulphuric acid gives compound \(\mathrm{C}\), which
View solution Problem 38
A hydrocarbon A (molecular formula \(\mathrm{C}_{5} \mathrm{H}_{10}\) ) yields 2-methylbutane on catalytic hydrogenation. A adds \(\mathrm{HBr}\) (in accordance
View solution Problem 40
Give a chemical test/suggest a reagent to distinguish between methanol and ethanol.
View solution Problem 41
An alcohol \(A\), when heated with conc. \(\mathrm{H}_{2} \mathrm{SO}_{4}\) gives an alkene \(B\). When \(B\) is bubbled through bromine water and the product o
View solution