4.7.6-1. In the presence of water, CO₂ forms (reversibly) carbonic acid, H₂CO₃. Being stronger an acid than phenol, H₂CO₃ displaces phenol from its salt:

4.7.6-2. For oxidation of methanol to formaldehyde, see Figure 4-102. Phenol cannot be oxidized to an aldehyde.

4.7.6-3. This is a somewhat tricky one. One of the two carbon atoms of the ethylene molecule gets to bind to a hydrogen atom of water in the reaction, so it is certainly reduced. The other carbon atom binds to the O atom of the OH moiety of the water molecule, so it is oxidized. That is why the addition of water to alkenes is not considered to be an oxidation-reduction reaction. To confirm this conclusion, determine the oxidations states of the carbon atoms of ethylene and ethanol, the product of the H₂O addition to CH₂=CH₂.

4.7.6-4. Mass percentage concentration is the gram amount of a solute in 100 g of solution. Therefore, to solve the problem, we need to know the total mass of the solution after the reaction and the amount of sodium ethoxide produced in the reaction. The equation is: 2 C₂H₅OH + 2 Na = 2 C₂H₅ONa + H₂. The hydrogen gas produced bubbles off. For the reaction, we have 1 g of Na and 100 mL of ethanol. The amount of ethanol in grams is 100 mL x 0.79 g/mL = 79 g. The total mass of the solution after the reaction is therefore 79 g (ethanol) plus 1 g (sodium) minus the mass of the H₂ gas that was formed and left the reaction mixture. This follows from the law of conservation of matter.

We need to determine which of the two reactants, ethanol or sodium, is the limiting reagent. The atomic weight of sodium is 23 and the molecular weight of ethanol is 46, twice as much. Therefore, 2 g of ethanol are needed for the reaction with 1 g of sodium metal. We have 79 g of ethanol, vastly more than 2 g that is needed for the reaction. Hence, sodium is our limiting reagent.

It is clear from the equation that 2 moles of Na (2 x 23 g = 46 g) react with ethanol to give 2 moles of sodium ethoxide (2 x 68 g = 136 g) and 1 mole of H₂ (2 g). Solving the proportion for x, the amount of sodium ethoxide formed from 1 g of Na,

46 g Na --- gives 136 g C₂H₅ONa
1 g Na --- gives x g C₂H₅ONa

we get: x = (136 x 1)/46 = 2.96 g. This is the amount of sodium ethoxide produced.

Solving the proportion for y, the amount of H₂ formed from 1 g of Na,

46 g Na --- gives 2 g H₂
1 g Na --- gives y H₂

we get: y = (2 x 1)/46 = 0.04 g. This is the amount of H₂ produced.

The total weight of the solution after the reaction has gone to completion and the hydrogen gas produced has escaped, is 79 g (ethanol) + 1 g (Na) - 0.04 g (H₂ formed and lost) = 79.96 g.

Knowing the weight of the solution (79.96 g) and the amount of sodium ethoxide in it (2.96 g), we determine the percentage concentration:

79.96 g of solution ------ contains 2.96 g of sodium ethoxide
100 g of such solution ------ would contain z g of sodium ethoxide;

z = (100 x 2.96)/79.96 = 3.7 g. Thus the concentration of C₂H₅ONa in the solution produced is 3.7%.

4.7.6-5. (b).

4.7.6-6. Light-initiated free-radical bromination of methane will produce bromomethane (among other brominated methanes). Phosphoric acid-catalyzed addition of water to ethylene will give ethanol. Reacting the ethanol with Na will give sodium ethoxide. The Williamson ether synthesis using sodium ethoxide and bromomethane will give methyl ethyl ether. Write down and balance chemical equations for all of these transformations.

4.7.6-7. (d).

4.7.6-8. (a).

4.7.6-9. (b) and (c).

4.7.6-12. (c). Note that ether does produce explosive hydroperoxides by slow reaction with oxygen of the air. The industry can avoid this problem, though, by storing ether under an inert atmosphere such as nitrogen, or by using freshly produced ether. It is the exceptionally high flammability of ether that makes it industrially unfeasible as a solvent.

4.7.6-13. n = 25. First, we calculate the molecular weight of the repeat unit of PET (Figure 4-122). It is 192. By dividing the molecular weight of the polymer by the molecular weight of the repeat unit we find the degree of polymerization n = 4,800/192 = 25.

4.7.6-15. They are, because they have the same composition, C₃H₈O, but different structures.

4.7.6-16. Now that you know what triglycerides are, try to find an answer to this question using the Internet.

4.7.6-17. Only (a) and (d) can (and do) form hydroperoxides because their molecules have C-H bonds next to the oxygen atom. It is one of those bonds that a molecule of O₂ inserts into to produce a hydroperoxide, see Figure 4-128.

4.7.6-18. First off, I would not touch the bottles. I would restrict or, better, block access to the storeroom and immediately report the dangerous finding to Departmental Head and a Safety Officer. I would insist that a special service dealing with explosives be contacted. Such issues should be dealt with by professional explosives experts. I am not one. I would never -

- touch, let alone open those bottles;
- put the supernatant down the drain;
- add anything to those crystals, not even water;
- leave such bottles "to stand around for several weeks awaiting disposal";
- place those bottles in a box along with other chemicals, especially "sodium scraps and unlabeled chemicals";
- drive such a 'gift box', especially a long distance;
- throw stones at those bottles. I'd rather use a slingshot or, better, a pump shotgun from a safer distance.

4.7.6-20. (a) and (b).

4.7.6-21. Yes. Water, methanol, and ethanol are all weak acids yet are sufficiently acidic to react with sodium metal. Being much more acidic than water, methanol, and ethanol, phenol does react with Na to give sodium phenoxide and H₂.

4.7.6-22. (d).

4.7.6-24. Distill off the water co-product as the reaction occurs.

4.7.6-25. CH₃I + C₂H₅OK = CH₃OC₂H₅ + KI

4.8.3-3. Because these reactions employ alkaline conditions, under which fructose isomerizes to glucose (and another aldose, mannose).

4.8.3-4. The OH groups of glucose form hydrogen bonds with molecules of water, which greatly enhances the affinity of glucose for water and its solubility.

4.8.3-5. Because the linear and cyclic forms of glucose are in fast equilibrium in solution. This equilibrium is shifted to the linear aldehyde form as it undergoes the oxidation. In this way, the equilibrium is completely driven away from the cyclic forms in the reaction.

4.8.3-8. Look at Figure 4-132 and 4-133 to see how the cyclization occurs. In the reverse process, the proton of the OH group on the carbon in the first position jumps back to the O atom of the C(5)-O-C(1) moiety with the simultaneous cleavage of the C(1)-O bond and restoration of the C(1)=O double bond. Such a process is impossible for sucrose because the C atoms in the reactive positions no longer bear the OH groups but are rather involved in the C-O-C glycosidic bond.

4.8.3-9. As sucrose does not have an aldehyde group, it does not react with Tollens' reagent, nor with Cu(OH)₂. Hydrolysis of sucrose, which is catalyzed by acids, gives rise to fructose and glucose, which is an aldehyde and is therefore easily oxidized by Tollens' reagent or Cu(OH)₂.

4.8.3-12. (b) and (d).

4.8.3-15. (c).

4.8.3-16. Probably not. In all likelihood, the $100 bill was genuine. During the laundry in hot water, a small amount of pure cotton cellulose that the bill was made of must have been converted to starch, which made the counterfeit bill detection test positive.

4.9.6-2. Yes.

4.9.6-3. (CH₃)₂NH > CH₃NH₂ > (CH₃)₃N > NH₃ > C₆H₅NH₂

4.9.6-4. Propylamine is C₃H₇NH₂. For convenience, we can use the empirical formula C₃H₉N.

4 C₃H₉N + 21 O₂ = 12 CO₂ + 18 H₂O + 2 N₂

4.9.6-5. Make nitrobenzene from benzene and nitric acid in the presence of sulfuric acid. Make H₂S from Na₂S and H₂SO₄. Do the Zinin reduction of nitrobenzene with H₂S.

4.9.6-6. Hint: Draw the structures and look for carbon atoms bearing four different substituents.

4.9.6-8. Hint: Figure 4-154.

4.9.6-9. Hint: Figure 4-153.

4.9.6-11. See Figure 4-161 and accompanying text.