Understanding Advanced Chemistry Through Problem Solving

The Learner's ApproachVolume 2

Kim Seng Chan, Jeanne Tan

• Publisher: WSPC
• Year: 2014
• ISBN:
  9789814596510
• Topic: Ciencias físicas
• Subtopic: Química inorgánica

Written for students taking either the University of Cambridge Advanced Level examinations or the International Baccalaureate examinations, this guidebook covers essential topics and concepts under both stipulated chemistry syllabi. The book is written in such a way as to guide the reader through the understanding and applications of essential chemical concepts using the problem solving approach. The authors have also retained the popular discourse feature from their previous two books — Understanding Advanced Physical Inorganic Chemistry and Understanding Advanced Organic and Analytical Chemistry — to help the learners better understand and see for themselves, how the concepts should be applied during solving problems. Based on the Socratic Method, questions are implanted throughout the book to help facilitate the reader's development in forming logical conclusions of concepts and the way they are being applied to explain the problems. In addition, the authors have also included important summaries and concept maps to help the learners to recall, remember, reinforce and apply the fundamental chemical concepts in a simple way.

Topics are explored through an explanatory and inquiry-based approach. They are interrelated and easy to understand, with succinct explanations/examples being included, especially on areas that students frequently find difficult. Topics address the whys and hows behind key concepts to be mastered, so that the concepts are made understandable and intuitive for students. The focus is on conceptual learning so as to equip students with knowledge for critical learning and problem solving.

Existing A-level or IB guidebooks generally introduce concepts in a matter-of-fact manner. This book adds a unique pedagogical edge which few can rival. Through their many years of teaching experiences, the authors have acquired a sound awareness of common students' misconceptions which are relayed through the questions and thus help to reinforce concepts learnt. This book is essential and useful to help the students to be adequately prepared for their high stake examinations.

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Contents:

• Structure and Bonding
• Isomerism in Organic Compounds
• Organic Reactions and Mechanisms
• Alkanes
• Alkenes
• Arenes
• Halogen Derivatives
• Alcohols and Phenol
• Carbonyl Compounds
• Carboxylic Acids and Their Derivatives
• Amines
• Amino Acids
• Polymers
• Summary of Important Organic Reactions

Readership: JC students and teaching professionals in Chemistry.

CHAPTER 1

STRUCTURE AND BONDING

1. (a) (i) The enthalpies of the formation of ethane, ethene, ethyne, and benzene are –84.7, +52.3, +227, and +82.9 kJ mol^–1, respectively. With the given bond energies of H–H and C–H as +436 and +412 kJ mol^–1, respectively, calculate the values for the carbon-carbon bond energies in the four hydrocarbons.

Explanation:

For ethane:

By Hess’ Law,

For ethene:

By Hess’ Law,

For ethyne:

By Hess’ Law,

For benzene:

By Hess’ Law,

Q	Why did you use the term ‘carbon–carbon’ when you calculated the bond energy for the carbon–carbon bond in benzene?

A: This is because the carbon–carbon bond in benzene is neither a C–C single nor a C=C double bond. Hence, it is inappropriate to use the term ‘C–C.’

(ii) Account for the differences in the carbon–carbon bond length in the compounds ethane, ethene, ethyne, and benzene.

Explanation:

The carbon–carbon bond energies increase in the order: ethane < benzene < ethene < ethyne. Hence, the bond length would also decrease in this similar trend, with the C–C bond in ethane the longest. As the number of bonds between the two carbon atoms increases, the bond becomes stronger and shorter because there is stronger attractive force acting on the increasing number of shared electrons exerted by the two nuclei.

For benzene, the bond energy value indicates that the carbon–carbon bond is intermediate between a C–C single and a C=C double bond. This is because the pi electrons between any two carbon atoms in benzene can actually delocalize throughout all the six carbon atoms. As a result, there is only on the average one pi electron between two carbon atoms. This accounts for the weaker carbon–carbon bond in benzene as compared to ethene, but the bond is stronger than that in ethane.

Do you know?

— As the number of shared electrons increases, the attractive force by the nuclei on the shared electrons increases but at the same time, the inter-electronic repulsion between the shared electrons also increases. This increase in inter-electronic repulsion would act against the attractive force. This is why it is very difficult to form four bonds between two atoms as the inter-electronic repulsion would be too strong to be contained.

— BE(C=C) – BE(C–C) = 601.7 – 350.7 = +251.0 kJ mol^–1, which is equivalent to the bond energy to overcome a pi bond between two carbon atoms. Did you notice that the strength of the pi bond is weaker than that of the sigma bond? This is because the formation of a sigma bond results in the accumulation of electron density within the inter-nuclei region, which buffers the inter-nuclei repulsion much better than the accumulation of electron density out of the inter-nuclei region for the pi bond. The sigma bond is able to accumulate electron density within the inter-nuclei region due to the head-on overlap of the atomic orbitals. While for pi bond, it is the side-on overlap!

(iii) Draw the dot-and-cross diagrams of ethane, ethane, and ethyne.

Explanation:

(b) Explain why CH₂=C=CH₂ is not a flat molecule and the carbon skeleton of the following molecule is not planar

Explanation:

The CH₂=C=CH₂ is not a flat molecule because the two C–H bonds on C₁ are lying on a plane which is perpendicular to the plane that contains the two C–H bonds on C₃.

Let us label the following molecule A:

The 16 carbon atoms of molecule A cannot be lying on the same plane because there would be too much steric effect between the four –CH₃ groups. So instead, the two benzene rings are in fact perpendicular to each other, causing the two –CH₃ groups on each benzene ring to be perpendicular to each other as shown below:

Q	Why are the two C–H bonds on C1 lying on a plane that is perpendicular to the plane that contains the two C–H bonds on C3 for CH2=C=CH2?

A: C₁ is an sp² hybridized atom, which means it uses an s orbital and two p orbitals to form three sp² hybrid orbitals. Let’s say that it uses both the p_x and p_y orbitals for hybridization. It would then be left with the p_z orbital to form a pi bond with C₂. Hence, the three sp² hybrid orbitals on C₁ must be sitting on the x–y plane.

Now C₂ is an sp hybridized atom, which means it uses an s orbital and p orbitals. Let’s say that it uses the p_x orbital. But do not forget that it has already used up the p_z orbital to form a pi bond with C₁. Hence, C₂ can only use its p_y orbital to form a pi bond with C₃.

As for C₃, it is also an sp² hybridized atom, which means it uses an s orbital an...

Table of contents

• Cover
• HalfTitle Page
• Title Page
• Copyright Page
• Preface
• Acknowledgements
• Contents
• Chapter 1 Structure and Bonding
• Chapter 2 Isomerism in Organic Compounds
• Chapter 3 Organic Reactions and Mechanisms
• Chapter 4 Alkanes
• Chapter 5 Alkenes
• Chapter 6 Arenes
• Chapter 7 Halogen Derivatives
• Chapter 8 Alcohols and Phenol
• Chapter 9 Carbonyl Compounds
• Chapter 10 Carboxylic Acids and Their Derivatives
• Chapter 11 Amines
• Chapter 12 Amino Acids
• Chapter 13 Polymers
• Chapter 14 Summary of Important Organic Reactions
• Index