Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

In addition to aldehyde group glucose contains (a) one secondary OH and four primary OH groups (b) one primary OH and four secondary OH groups (c) two primary \(\mathrm{OH}\) and three secondary OH groups (d) three primary OH and two secondary OH groups

Short Answer

Expert verified
(b) one primary OH and four secondary OH groups.

Step by step solution

01

Understanding the Structure of Glucose

Glucose is a monosaccharide with the chemical formula \(\mathrm{C}_6\mathrm{H}_{12}\mathrm{O}_6\). It has an aldehyde group at the first carbon, and the rest of the carbons have hydroxyl (OH) groups attached.
02

Identifying Types of OH Groups

Primary OH groups are attached to a carbon that is connected to only one other carbon, whereas secondary OH groups are attached to a carbon connected to two other carbons.
03

Analyzing Carbon Chain of Glucose

Glucose consists of a chain of six carbon atoms. The aldehyde group is on carbon 1. The OH groups on the subsequent carbons are positioned as follows: the OH at carbon 1 is part of the aldehyde group, the OH on carbon 6 is primary because it's at the end of the chain, and the OH groups on carbons 2, 3, 4, and 5 are secondary because each of these carbons is connected to two other carbon atoms.
04

Counting Primary and Secondary OH Groups

From the analysis of the carbon chain, glucose has one primary OH group (on carbon 6) and four secondary OH groups (on carbons 2, 3, 4, and 5).
05

Choosing the Correct Answer

Based on the analysis, glucose has one primary OH group and four secondary OH groups. Therefore, the correct option is (b) one primary OH and four secondary OH.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Monosaccharide
Glucose is part of a fascinating class of sugars known as monosaccharides. These are the simplest forms of carbohydrates. The term 'monosaccharide' itself suggests a "single" sugar unit. Simple sugars like glucose are vital because they serve as the building blocks for more complex carbohydrates.
Monosaccharides have a general formula of \(\mathrm{(CH_2O)}_n\), where \(n\) can be any number from three to seven. In the case of glucose, \(n = 6\), which contributes to its formula \(\mathrm{C_6H_{12}O_6}\). Glucose is particularly crucial because:
  • It is a primary source of energy for cells.
  • It participates in vital biochemical processes, like glycolysis and photosynthesis.
  • It can bond with other monosaccharides to form disaccharides and polysaccharides.

Understanding monosaccharides is essential for unraveling the chemistry of life itself because they are involved in numerous biological processes.
Primary OH Groups
In glucose, knowing the difference between primary and secondary hydroxyl (OH) groups helps in grasping its structure and reactivity. A primary OH group is attached to a carbon atom that is bonded to just one other carbon. This configuration usually appears at the end of a carbon chain.
For glucose, the primary OH group is located at carbon 6. This position makes it distinct from the others and impacts the molecule's properties and reactivity.
Here's why primary OH groups are significant:
  • They often engage in distinct chemical reactions due to being terminal OH groups.
  • They affect the solubility and physical properties of glucose in solution.
  • Some specific reactions target primary OH groups due to their relative accessibility in the molecule.

By identifying the primary OH groups in glucose, students can better understand its chemical behavior and how it interacts with other molecules.
Secondary OH Groups
Secondary OH groups in glucose are attached to a carbon atom bonded to two other carbons. This makes them integral to the internal structure of the glucose molecule, especially concerning its stability and function.
In the glucose molecule, the secondary OH groups are found at carbons 2, 3, 4, and 5. These OH groups influence both the molecule's physical characteristics and its biological work.
Key points about secondary OH groups include:
  • They possess a slightly different reactivity profile compared to primary OH groups.
  • They play a crucial role in glucose's ability to form hydrogen bonds, impacting solubility and binding capacity.
  • These groups allow glucose to partake in various chemical modifications needed for energy metabolism.

Grasping the distribution of secondary OH groups aids in understanding the diverse roles glucose plays in biological systems, especially its solubility and reactivity in aqueous environments.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Which of the following has magnesium? (a) carbonic anhydrase (b) haemocyanin (c) chlorophyll (d) Vitamin \(\mathrm{B}_{12}\)

The reason for double helical structure of DNA is operation of [2003] (a) van der waals forces (b) hydrogen bonding (c) dipole-dipole moment (d) electrostatic attractions

Which of the following statements are correct? (a) All monosaccharides whether aldose or ketose act as reducing sugars. (b) Bromine water can be used to differentiate between aldoses and ketoses. (c) A pair of diastereomeric aldoses which differ only in configuration about \(\mathrm{C}_{2}\) is termed as a pair of anomers. (d) Osazone formation destroys the configuration about \(\mathrm{C}_{2}\) of an aldose but does not affect the configuration of the rest of the molecule.

Which statement is incorrect about peptide bond? (a) \(\mathrm{C}-\mathrm{N}\) bond length in proteins is smaller than usual bond length of \(\mathrm{C}-\mathrm{N}\) bond (b) spectroscopic analysis shows planar structure \(-\) C-NH-bond (c) \(\mathrm{C}-\mathrm{N}\) bond length in proteins is longer than usual bond length of \(\mathrm{C}-\mathrm{N}\) bond (d) none of these

Consider the following statements about amino acids: (1) nitrous acid liberates nitrous oxide from amino acids (2) an important sensitive test for the detection of amino acids is the ninhydrin colour test (3) glycine is the only amino acid which does not possess a chiral centre. (4) all the natural amino acids that occur as constituents of proteins belong to the L-series Which of these statements are correct? (a) 1,2 and 3 (b) 1,2 and 4 (c) 2,3 and 4 (d) 2 and 3 only

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free