Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 3: Problem 19

The hydrolysis of 1,3 -bisphosphoglycerate is favorable, due in part to the increased resonance stabilization of the products of the reaction. Draw resonance structures for the reactant and the products of this reaction to establish that this statement is true..

Short Answer

Expert verified
Upon drawing the structures, you will see that the products of the hydrolysis have more resonance structures than the reactant, which indicates higher resonance stabilization, making the reaction favorable.

Step by step solution

01

Draw the Reactant

First, draw the structure of the 1,3-bisphosphoglycerate. It includes an atom of Carbon in the center surrounded by two atoms of Oxygen at both ends and a Hydrogen atom on the side. On two sides, there are also Phosphate groups attached.
02

Draw Products of the Hydrolysis

Now draw the products of this reaction. Upon hydrolysis, 1,3-bisphosphoglycerate is broken down into three-phosphoglycerate and inorganic phosphate. Three-phosphoglycerate consists of a central carbon atom surrounded by an OH group, a phosphate group, a hydrogen atom, and an oxygen atom. While the inorganic phosphate consists of a phosphorus atom in the center with four atoms of oxygen surrounding it.
03

Draw Resonance Structures

Finally, for resonance structures, they are shown for those molecules or ions that can't be represented by a single Lewis structure and therefore have multiple structures. Draw these multiple structures for the reactant and the product molecules. Remember to represent the delocalized electrons that are involved in resonance.

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!

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

Draw all possible resonance structures for creatine phosphate and discuss their possible effects on resonance stabilization of the molecule.

Calculate the free energy of hydrolysis of ATP in a rat liver cell in which the ATP, ADP, and \(P\), concentrations are \(3.4,1.3,\) and \(4.8 \mathrm{m} M\) respectively.

The standard-state free energy of hydrolysis for acetyl phosphate is \\[ \begin{aligned} \Delta G^{\circ}=-42.3 \mathrm{kJ} / \mathrm{mol} \\ \text { Actyl-P }+\mathrm{H}_{2} \mathrm{O} \longrightarrow \text { acctate }+\mathrm{P}_{\mathrm{i}} \end{aligned} \\] Calculate the free energy change for acetyl phosphate hydrolysis in a solution of \(2 \mathrm{m} M\) acctate, \(2 \mathrm{m} M\) phosphate, and \(3 \mathrm{n} M\) acetyl phosphate.

You are studying the various components of the venom of a poisonous lizard. One of the venom components is a protein that appears to be temperature sensitive. When heated, it denatures and is no longer toxic. The process can be described by the following simple equation: \\[ \mathbf{T}(\text { toxic }) \rightleftharpoons \mathrm{N} \text { (nontoxic) } \\] There is only enough protein from this venom to carry out two equilibrium measurements. At \(298 \mathrm{K}\), you find that \(98 \%\) of the protein is in its toxic form. However, when you raise the temperature to \(320 \mathrm{K},\) you find that only \(10 \%\) of the protein is in its toxic form. a. Calculate the equilibrium constants for the T to N conversion at these two temperatures. b. Use the data to determine the \(\Delta H^{\circ}, \Delta S^{\circ},\) and \(\Delta G^{\circ}\) for this process.

An enzymatic hydrolysis of fructose- \(1-P\) \\[\text { Fructose- } 1-\mathrm{P}+\mathrm{H}_{2} \mathrm{O} \rightleftharpoons \text { fructose }+\mathrm{P}_{1}\\] was allowed to proceed to equilibrium at \(25^{\circ} \mathrm{C}\). The original concentration of fructose-1-P was \(0.2 \mathrm{M}\), but when the system had reached equilibrium the concentration of fructose-1-P was only \(6.52 \times 10^{-5} \mathrm{M}\). Calculate the equilibrium constant for this reaction and the free energy of hydrolysis of fructose- \(1-P\).
See all solutions

Recommended explanations on Chemistry Textbooks

Chemical Reactions

Read Explanation

Physical Chemistry

Read Explanation

Ionic and Molecular Compounds

Read Explanation

Organic Chemistry

Read Explanation

Kinetics

Read Explanation

Nuclear Chemistry

Read Explanation
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