For the given compounds, the order from smallest heat of combustion to largest heat of combustion is to be determined. Concept introduction: A cyclohexane is lower in energy (more stable) when the bulkier substituent occupies an equatorial position. The equatorial conformer is more stable than the axial conformer because there is more room for the substituent in the equatorial position. The increase in energy that occurs from electron repulsion between groups of atoms or atoms that are not directly bonded together but occupy the same space is nothing but steric strain. The higher the ring strain lower is the heat of combustion and vice versa.

Question

Want to see the full answer?

Answer 1

Question

Chapter 4, Problem 4.61P

Interpretation Introduction

Interpretation:

For the given compounds, the order from smallest heat of combustion to largest heat of combustion is to be determined.

Concept introduction:

A cyclohexane is lower in energy (more stable) when the bulkier substituent occupies an equatorial position. The equatorial conformer is more stable than the axial conformer because there is more room for the substituent in the equatorial position.

The increase in energy that occurs from electron repulsion between groups of atoms or atoms that are not directly bonded together but occupy the same space is nothing but steric strain. The higher the ring strain lower is the heat of combustion and vice versa.

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Students have asked these similar questions

I

Draw the anti-Markovnikov product of the hydration of this alkene. this problem. Note for advanced students: draw only one product, and don't worry about showing any stereochemistry. Drawing dash and wedge bonds has been disabled for esc esc ☐ Explanation Check F1 1 2 F2 # 3 F3 + $ 14 × 1. BH THE BH3 2. H O NaOH '2 2' Click and drag to start drawing a structure. F4 Q W E R A S D % 905 LL F5 F6 F7 © 2025 McGraw Hill LLC. All Rights Reserved. Terms of Use | Privacy Center | Accessibility < & 6 7 27 8 T Y U G H I F8 F9 F10 F11 F12 9 0 J K L P + // command option Z X C V B N M H H rol option command

AG/F-2° V 3. Before proceeding with this problem you may want to glance at p. 466 of your textbook where various oxo-phosphorus derivatives and their oxidation states are summarized. Shown below are Latimer diagrams for phosphorus at pH values at 0 and 14: -0.93 +0.38 -0.50 -0.51 -0.06 H3PO4 →H4P206 →H3PO3 →→H3PO₂ → P → PH3 Acidic solution Basic solution -0.28 -0.50 3--1.12 -1.57 -2.05 -0.89 PO HPO H₂PO₂ →P → PH3 -1.73 a) Under acidic conditions, H3PO4 can be reduced into H3PO3 directly (-0.28V), or via the formation and reduction of H4P206 (-0.93/+0.38V). Calculate the values of AG's for both processes; comment. (3 points) 0.5 PH P 0.0 -0.5 -1.0- -1.5- -2.0 H.PO, -2.3+ -3 -2 -1 1 2 3 2 H,PO, b) Frost diagram for phosphorus under acidic conditions is shown. Identify possible disproportionation and comproportionation processes; write out chemical equations describing them. (2 points) H,PO 4 S Oxidation stale, N