[email protected]
India's First Online IIT-JEE & NEET Coaching Platform - Trusted Since 2006
+91-87964 74404
Question icon
12 grade chemistry others

Pheol reacts with sodium hydroxide to give sodium phenoxide. The phenoxide ion undergoes electrophilic substitution with carbon dioxide (a weak electrophile) because the phenoxide ion is more reactive than phenol. Salicylic acid is formed as a major product. At a lower temperature, the ortho- isomer is predominant, whereas para- isomer is obtained at higher temperatures. Give the reason behind this:

Profile image of Aniket Singh
1 Year agoGrade
Answers icon

1 Answer

Profile image of Askiitians Tutor Team
ApprovedApproved Tutor Answer1 Year ago

When phenol reacts with sodium hydroxide, it forms sodium phenoxide, which is a more reactive species than phenol itself. This increased reactivity allows the phenoxide ion to undergo electrophilic substitution with carbon dioxide, leading to the formation of salicylic acid. The interesting aspect of this reaction is how temperature influences the distribution of ortho and para isomers in the final product. Let's break this down step by step.

The Role of Temperature in Electrophilic Substitution

In electrophilic aromatic substitution reactions, the position where the electrophile attacks the aromatic ring can lead to different isomers. In the case of salicylic acid formation, the electrophile is carbon dioxide, and the phenoxide ion can react at either the ortho or para positions relative to the hydroxyl group.

Understanding Ortho and Para Positions

  • Ortho Position: This refers to the positions adjacent to the hydroxyl group on the benzene ring.
  • Para Position: This is the position directly opposite the hydroxyl group.

The ortho and para positions have different steric and electronic environments, which affect the stability of the transition states during the reaction. At lower temperatures, the reaction tends to favor the formation of the ortho isomer. This is primarily due to the following reasons:

Lower Temperature Effects

  • Lower Activation Energy: The transition state leading to the ortho product is often lower in energy compared to the para product. This is because the ortho position allows for some stabilization through intramolecular hydrogen bonding between the hydroxyl group and the incoming electrophile.
  • Faster Reaction Rate: At lower temperatures, the reaction kinetics favor the formation of the ortho isomer due to the less sterically hindered nature of the ortho attack, allowing for a quicker reaction.

Higher Temperature Dynamics

As the temperature increases, the situation changes. The reaction becomes more thermodynamically controlled rather than kinetically controlled. At higher temperatures, the energy available allows the system to overcome the activation barriers for both the ortho and para pathways. The para product becomes more favorable for several reasons:

  • Stability of the Para Isomer: The para isomer is generally more stable than the ortho isomer due to less steric hindrance and better distribution of charge within the molecule.
  • Thermodynamic Control: At elevated temperatures, the reaction can proceed to a point where the more stable product (para) is favored, even if it takes longer to form compared to the ortho product.

Summary of the Reaction Pathway

To summarize, the formation of salicylic acid from sodium phenoxide and carbon dioxide is influenced by temperature. At lower temperatures, the ortho isomer predominates due to lower activation energy and faster reaction kinetics. Conversely, at higher temperatures, the more stable para isomer is favored due to thermodynamic control. This interplay between kinetic and thermodynamic factors is a fundamental concept in organic chemistry, illustrating how reaction conditions can significantly influence product distribution.


question mark

is Pradeep Chemistry+Ncert sufficient for all of the theory of Jee Mains Chemistry?


Hi, I’m a self-study JEE 2027 aspirant with a large 11th backlog. Since I don’t think 6–8 hour one-shots are sufficient to cover the full theory + breadth of problem types of a JEE chapter, I’ve shifted from lecture-based preparation to book-based self-study.


I have a problem with using NCERT as my primary chemistry source. In my experience, it is concise and information-heavy, but often lacks detailed explanations, systematic organisation of concepts, and categorisation of problem types. This seems especially problematic for Physical and Organic Chemistry, where I feel that simply reading NCERT may not give enough depth or problem-solving preparation.


So I’m considering this approach:


Pradeep Chemistry → complete the chapter thoroughly from Pradeep → then read NCERT for that chapter → solve JEE Main PYQs/problems.


My main question is:


Is Pradeep + NCERT sufficient as the complete theory source for JEE Main-level Chemistry?


In other words, after thoroughly completing a chapter from Pradeep and then NCERT, can I consider the theory part of that chapter complete for JEE Main, with only PYQs/problem practice remaining?


I’m asking specifically about JEE Main level, not Advanced.


Also, if Pradeep is not sufficient, what specific gap does it leave, and what would you recommend as a better book-based alternative for self-study?

Grade 1212 grade chemistry others

Last Activity: 8 Days ago