(A) Salicylic acid $(C_6H_4(OH)COOH)$ undergoes decarboxylation when heated with sodalime $(NaOH + CaO)$.First,the carboxylic acid group is removed as

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

(A) Salicylic acid $(C_6H_4(OH)COOH)$ undergoes decarboxylation when heated with sodalime $(NaOH + CaO)$.First,the carboxylic acid group is removed as $Na_2CO_3$,and the phenolic group remains attached to the benzene ring.The reaction is: $C_6H_4(OH)COOH + 2NaOH \xrightarrow{\Delta, CaO} C_6H_5OH + Na_2CO_3 + H_2O$.The product formed is phenol.

Class 12 Chemistry Alcohols, Phenols and Ethers Properties of Phenols

Difficult

What is the product formed when salicylic acid is heated with sodalime?

A
Phenol
B
Benzyl alcohol
C
Benzene
D
Benzoic acid

Explore More

Browse All

Question Bank

All Subjects

Class 12 Questions

Class 12

Chemistry Questions

Chapter

Alcohols, Phenols and Ethers

Subtopic

Properties of Phenols

The major product of the following reaction is
$2-(2-hydroxyethyl)phenol$ $\xrightarrow[2. \, CH_3I \, (1. \, eq.)]{1. \, K_2CO_3}$

DifficultJEE MAIN 2017

View Solution

Assertion : Resorcinol turns $FeCl_3$ solution purple.
Reason : Resorcinol has a phenolic group.

MediumAIIMS 2000

View Solution

The correct order of boiling point for the given compounds is:

Difficult

View Solution

Which group is present in phenol? What is the effect of this group on the ring and its reactions?

Medium

View Solution

The $IUPAC$ name of picric acid is

Medium

View Solution

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial

For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free

For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo

What is the product formed when salicylic acid is heated with sodalime?

Similar Questions

The major product of the following reaction is $2-(2-hydroxyethyl)phenol$ $\xrightarrow[2. \, CH_3I \, (1. \, eq.)]{1. \, K_2CO_3}$

Assertion : Resorcinol turns $FeCl_3$ solution purple.Reason : Resorcinol has a phenolic group.

The correct order of boiling point for the given compounds is:

Which group is present in phenol? What is the effect of this group on the ring and its reactions?

The $IUPAC$ name of picric acid is

Vedclass Test Series

Exam Paper Generator

Online Exam Module

The major product of the following reaction is
$2-(2-hydroxyethyl)phenol$ $\xrightarrow[2. \, CH_3I \, (1. \, eq.)]{1. \, K_2CO_3}$

Assertion : Resorcinol turns $FeCl_3$ solution purple.
Reason : Resorcinol has a phenolic group.