4.2.8-1. False. Orbital hybridization is just a concept, a model to account for many experimental observations, including the tetrahedral geometry of methane.
4.2.8-2. (f).
4.2.8-5. (a) heptane (or n-heptane); (b) 2,3-dimethylbutane; (c) 3,4-dimethylhexane; (d) 3-ethylheptane; (e) 2,2,3,3-tetramethylpentane.
4.2.8-6. (a) the longest carbon chain is missed. The ethyl group in the 2nd position in the name as written should be included in the longest C chain. (b) The branching closest to one of the two ends of the straight chain consisting of four C atoms is at the 2nd, not 3rd atom. Actually, the 2 and 3 positions in butane are equivalent due to symmetry. However, the main chain must always be numbered such that the first branching has the smallest possible number. (c) This name is absurd! A carbon atom in hydrocarbons cannot be pentavalent.
4.2.8-7. Neither. Isomers have the same composition, whereas in pentane there are two more H atoms than in cyclopentane. Homologues are a series of compounds differing from one another by the repeating unit CH2.
4.2.8-8.
(a) 2 C2H6 + 7 O2 = 4 CO2 + 6 H2O
(b) C5H12 + 8 O2 = 5 CO2 + 6 H2O
(a) 2 C2H6 + 7 O2 = 4 CO2 + 6 H2O
(b) C5H12 + 8 O2 = 5 CO2 + 6 H2O
4.2.8-9. The mechanism is exactly as for the chlorination, see 4.2.7. Four organic products (CH3Br, CH2Br2, CHBr3, CBr4) and one inorganic product (HBr). See what a chain reaction is in 4.2.7.
4.3.9-1. (d).
4.3.9-2. False. Ethylene is planar because that is the way it is. The sp2 hybridization model is just a way to rationalize the planar geometry of ethylene.
4.3.9-3. (c).
4.3.9-6. (a) 3,4-diethyl-3-hexene; (b) 2,3-dimethyl-1-butene; (c) cis-3-hexene; (d) trans-3-hexene; (e) 3,3,7-trimethyl-7-ethyl-1-nonene.
4.3.9-7. (a) a terminal alkene (C=C at one of the ends of the chain) cannot be cis or trans. (b) The carbon atoms of the C=C bond should be given the smallest numbers. Correct name: 4,4-dimethyl-2-pentene. (c) This name is absurd! A carbon atom in hydrocarbons cannot be pentavalent.
4.3.9-8. Yes, they are. Isomers have the same composition, but different connectivity (structure). These three hydrocarbons have the same composition, C6H12, but different structures.
4.3.9-10. First, we need to figure out the empirical formulas for these hydrocarbons. If you cannot do it mentally, draw the structures from the names and count the number of C and H atoms in each molecule. Alternatively, count just the number of C atoms and derive the formulas using the general formula for alkenes, CnH2n. The empirical formula for propene is C3H6 and for 2-methyl-1-hexene is C7H14.
(a) 2 C3H6 + 9 O2 = 6 CO2 + 6 H2O
(b) 2 C7H14 + 21 O2 = 14 CO2 + 14 H2O
(a) 2 C3H6 + 9 O2 = 6 CO2 + 6 H2O
(b) 2 C7H14 + 21 O2 = 14 CO2 + 14 H2O
4.3.9-12. The scheme should be similar to the one presented in Figure 4-45, except the phenyl radical should be replaced with an isobutyronitrile radical.
4.3.9-13. (b).
4.4.6-1. (d) and (g).
4.4.6-2. Hint: try to draw a structure of "isobutyne" to find out that it is impossible without making one C atom pentavalent.
4.4.6-3. If necessary, revisit subsection 4.3.3.
4.4.6-5. (c).
4.4.6-7. (a) 4,5-dimethyl-2-heptyne; (b) 3-methyl-4-ethyl-1-hexyne; (c) 2,5-dimethyl-3-hexyne.
4.4.6-9.
4.4.6-10.
(1). CaCO3 = CaO + CO2 (thermal decomposition (calcination) in a kiln at about 1000 oC)
(2). CaO + 3 C = CaC2 + CO (synthesis of calcium carbide from quicklime and coke in an electric arc furnace at about 2,000 oC)
(3). CaC2 + 2 H2O = HC≡CH + Ca(OH)2
(4). HC≡CH + H2O = CH3CHO (in the presence of a catalyst).
(1). CaCO3 = CaO + CO2 (thermal decomposition (calcination) in a kiln at about 1000 oC)
(2). CaO + 3 C = CaC2 + CO (synthesis of calcium carbide from quicklime and coke in an electric arc furnace at about 2,000 oC)
(3). CaC2 + 2 H2O = HC≡CH + Ca(OH)2
(4). HC≡CH + H2O = CH3CHO (in the presence of a catalyst).
4.5.6-1. False.
4.5.6-2. (a), (b), and (e).
4.5.6-3. All carbon-carbon bonds of benzene are equivalent, but not due to the C=C and C-C bond position interchanging.
4.5.6-5. (a) 1,2,4-triethylbenzene; (b) 3,4-dinitrotoluene; (c) absurd.
4.5.6-7. (b).
4.5.6-8. See Figures 4-73 and 4-74.
4.5.6-9. False.
4.5.6-11. False.
4.5.6-12. See 4.5.4.
4.5.6-13. Yes. Look up the definition of homologues above in the Volume or in the Glossary.
4.6.5-4. Test the liquids for miscibility with water. Methanol, ethanol, and acetic acid are miscible with water, whereas alkanes (gasoline) are not. The only water-immiscible liquid of the four is gasoline. Mark the bottle. Place the three bottles containing the as yet unidentified liquids on ice and wait a little. Acetic acid will freeze out to form white crystals. Mark the bottle. To tell methanol from ethanol in the remaining two unmarked bottles, shake samples from each with an approximately equal volume of the already identified gasoline. Ethanol is miscible with gasoline, whereas methanol is not.
4.6.5-7. See 4.6.4. Adding HCl or any other strong acid to soap will diminish or eliminate altogether its cleaning power. Soap is water-soluble sodium or potassium stearate. Stearic acid is a water-insoluble weak acid that is inactive as soap. On addition of a strong acid to soap, stearic acid will be formed and precipitate out due to the reaction: C17H35COO- + H+ = C17H35COOH↓.
4.6.5-8. Hint: See Figure 4-96.
4.6.5-9. (a) sp3; (b) sp3; (d) sp2; (e) sp2.
4.6.5-10. Formyl. See Table 6.
4.6.5-11. In a way, yes. Look at the structural formula of formic acid to see that it contains the C(H)O moiety, which is the aldehyde functional group. However, formic acid is conventionally considered to be a carboxylic acid rather than an aldehyde.
4.6.5-13. (a) methanol, CH3OH; (b) an aqueous solution of formaldehyde, HC(H)O; (c) phenol, C6H5OH; (d) lead (II) acetate, Pb(OOCCH3)2.
4.6.5-14. Being caustic and toxic to tissues, phenol is used to destroy the ingrown part of the toenail. You can learn more here.