Appendix 3. Answers to problems

TO THE STUDENT

As it was mentioned earlier, problem solving is, probably, the best key to studying chemistry. This work is very often like solving a puzzle; therefore never look at the answer until you succeed (or failed) in your solution. The best way to use this Appendix is to compare your solutions and those given below, or to find explanations for the unsolved problems only after all attempts to solve them have failed.

Appendix 3 presents solutions to all the problems (in-text and additional problems). In many cases not only the answer is given, but also the explanation of the reasoning, even if it has not been questioned. Some problems have more than one solution, but only one is generally given. Thus, your answer may differ from the one presented here. The main thing is that your solution must be correct.

To Chapter 1

1.2. It is similar to methane tetrahedral structure, except for appreciable deviation from the value of 109.5°.

1.3. The CH₃ carbon is sp³-hybridized and forms the H-C-H bond angles near 109°. Double-bond carbons are sp²-hybridized and form the H-C-H and HC-C angles near 120°.

1.4. Both oxygen and nitrogen are sp³-hybridized and have tetrahedral geometry. Three-dimensional structures are similar to those in Figures 1.10, c and 1.11, b, respectively.

1.5. CH₃CH₂CH₂CH₂Br, CH₃CH₂CHBrCH₃, (CH₃)₂CHCH₂Br, and (CH₃)₃CBr.

1.6. (a) CH₃NH₂; (b) CH₃CH=CH₂; (c) CH₃CH₂CH₂Cl; (d) (CH₃)₃CH; (e) CH₃CH(OH)CH₃. Two isomers (in total) are possible for compounds from (b) to (d), and three isomers - for the compound (e).

1.8. Sodium and oxygen form the ionic bond, CH₃CH₂O^- Na⁺; other bonds in the molecule are covalent.

1.9. All bonds are covalent, except for CaCl₂.