Chapter 7: Problem 82

The reaction of glucose with methanol and a small amount of \(\mathrm{HCl}\) to form methyl glucoside shows that (a) it is a pentahydroxy compound. (b) it is a polyhydroxy aldehyde. (c) it contains six carbon atoms in a straight chain. (d) it exists as cyclic hemiacetal.

Short Answer

Expert verified

Answer: Glucose is a polyhydroxy aldehyde (statement b) and exists as a cyclic hemiacetal (statement d).

Step by step solution

Statement (a): It is a pentahydroxy compound.

When glucose reacts with methanol and HCl, it forms methyl glucoside. This statement suggests that there are five hydroxy groups (OH) present in the glucose molecule. But, glucose is a hexose sugar, meaning it has six hydroxy (OH) groups. So, statement (a) is incorrect.

Statement (b): It is a polyhydroxy aldehyde.

Glucose is an aldose sugar, meaning it has an aldehyde group at one end. Furthermore, glucose has multiple hydroxyl (OH) groups. Therefore, glucose is a polyhydroxy aldehyde. Thus, statement (b) is correct.

Statement (c): It contains six carbon atoms in a straight chain.

Glucose contains six carbon atoms, and it is a linear molecule in its open-chain form. However, glucose predominantly exists as cyclic structures. Therefore, statement (c) can be considered true in reference to the open-chain form but is not fully accurate.

Statement (d): It exists as cyclic hemiacetal.

In its cyclic structure, glucose forms a hemiacetal by the reaction between the aldehyde group at C1 and the hydroxyl group at C5. Therefore statement (d) is correct. In conclusion, both statements (b) and (d) are correct. The reaction of glucose with methanol and HCl forming methyl glucoside supports that glucose is a polyhydroxy aldehyde and exists as a cyclic hemiacetal.

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.

Start your free trial

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

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

The reaction of glucose with methanol and a small amount of \(\mathrm{HCl}\) to form methyl glucoside shows that (a) it is a pentahydroxy compound. (b) it is a polyhydroxy aldehyde. (c) it contains six carbon atoms in a straight chain. (d) it exists as cyclic hemiacetal.

Short Answer

Step by step solution

Statement (a): It is a pentahydroxy compound.

Statement (b): It is a polyhydroxy aldehyde.

Statement (c): It contains six carbon atoms in a straight chain.

Statement (d): It exists as cyclic hemiacetal.

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Chemistry Textbooks

Chemistry Branches

Making Measurements

Chemical Analysis

Ionic and Molecular Compounds

Nuclear Chemistry

Chemical Reactions

Study anywhere. Anytime. Across all devices.