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JEE Main 202227 JulyMorning Shift
Exam
JEE Main
Year
2022
Date
27 July
Shift
Morning Shift
Subject
Chemistry
Chapter
Biomolecules

JEE Main 2022 · 27 July · Morning Shift · Chemistry · Biomolecules

A sugar ' X ' dehydrates very slowly under acidic condition to give...

A sugar 'XX' dehydrates very slowly under acidic condition to give furfural which on further reaction with resorcinol gives the coloured product after sometime. Sugar 'XX' isA sugar ' X ' dehydrates very slowly under acidic condition to give furfural which on further reaction with resorcinol

Options

  1. AAldopentose
  2. BAldotetrose
  3. COxalic acid
  4. DKetotetrose

Correct answer

A. Aldopentose

Step-by-step solution

Seliwanoff’s test is a chemical test which distinguishes between aldose and ketose sugars. If the sugar contains a ketone group, it is a ketose. If a sugar contains an aldehyde group, it is an aldose. This test relies on the principle that, when heated, ketoses are more rapidly dehydrated than aldoses. It is named after Theodor Seliwanoff, the chemist that devised the test. When added to a solution containing ketoses, a red color is formed rapidly indicating a positive test. When added to a solution containing aldoses, a slower forming light pink is observed instead. The acidic dehydration of the different aldopentoses gives furfural. This furfural combines with resorcinol to give cherry red product.

A sugar ' X ' dehydrates very slowly under acidic condition to give furfural which on further reaction with resorcinol

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