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(a)
Interpretation:
Depending on the number and type of hydrogen nonhydrogen atoms in the given compound and its IHD, the number of hydrogen atoms in each compound is to be determined.
Concept introduction:
The index of hydrogen deficiency of a molecule is the extent to which the molecule is unsaturated. It is half the number of hydrogen atoms missing from the molecule as compared to a completely saturated molecule. The contribution of each double bond in a molecule to the molecule’s index of hydrogen deficiency is 1. The contribution of each triple bond in a molecule to the molecule’s index of hydrogen deficiency is 2. The contribution of each ring in a molecule to the molecule’s index of hydrogen deficiency is 1. If a molecule is saturated and has no rings, double bonds, and triple bonds, its index of hydrogen deficiency is 0.
(b)
Interpretation:
Depending on the number and type of hydrogen nonhydrogen atoms in the given compound and its IHD, the number of hydrogen atoms in each compound is to be determined.
Concept introduction:
The index of hydrogen deficiency of a molecule is the extent to which the molecule is unsaturated. It is half the number of hydrogen atoms missing from the molecule as compared to a completely saturated molecule. The contribution of each double bond in a molecule to the molecule’s index of hydrogen deficiency is 1. The contribution of each triple bond in a molecule to the molecule’s index of hydrogen deficiency is 2. The contribution of each ring in a molecule to the molecule’s index of hydrogen deficiency is 1. If a molecule is saturated and has no rings, double bonds, and triple bonds, its index of hydrogen deficiency is 0.
(c)
Interpretation:
Depending on the number and type of hydrogen nonhydrogen atoms in the given compound and its IHD, the number of hydrogen atoms in each compound is to be determined.
Concept introduction:
The index of hydrogen deficiency of a molecule is the extent to which the molecule is unsaturated. It is half the number of hydrogen atoms missing from the molecule as compared to a completely saturated molecule. The contribution of each double bond in a molecule to the molecule’s index of hydrogen deficiency is 1. The contribution of each triple bond in a molecule to the molecule’s index of hydrogen deficiency is 2. The contribution of each ring in a molecule to the molecule’s index of hydrogen deficiency is 1. If a molecule is saturated and has no rings, double bonds, and triple bonds, its index of hydrogen deficiency is 0.
(d)
Interpretation:
Depending on the number and type of hydrogen nonhydrogen atoms in the given compound and its IHD, the number of hydrogen atoms in each compound is to be determined.
Concept introduction:
The index of hydrogen deficiency of a molecule is the extent to which the molecule is unsaturated. It is half the number of hydrogen atoms missing from the molecule as compared to a completely saturated molecule. The contribution of each double bond in a molecule to the molecule’s index of hydrogen deficiency is 1. The contribution of each triple bond in a molecule to the molecule’s index of hydrogen deficiency is 2. The contribution of each ring in a molecule to the molecule’s index of hydrogen deficiency is 1. If a molecule is saturated and has no rings, double bonds, and triple bonds, its index of hydrogen deficiency is 0.
(e)
Interpretation:
Depending on the number and type of hydrogen nonhydrogen atoms in the given compound and its IHD, the number of hydrogen atoms in each compound is to be determined.
Concept introduction:
The index of hydrogen deficiency of a molecule is the extent to which the molecule is unsaturated. It is half the number of hydrogen atoms missing from the molecule as compared to a completely saturated molecule. The contribution of each double bond in a molecule to the molecule’s index of hydrogen deficiency is 1. The contribution of each triple bond in a molecule to the molecule’s index of hydrogen deficiency is 2. The contribution of each ring in a molecule to the molecule’s index of hydrogen deficiency is 1. If a molecule is saturated and has no rings, double bonds, and triple bonds, its index of hydrogen deficiency is 0.
(f)
Interpretation:
Depending on the number and type of hydrogen nonhydrogen atoms in the given compound and its IHD, the number of hydrogen atoms in each compound is to be determined.
Concept introduction:
The index of hydrogen deficiency of a molecule is the extent to which the molecule is unsaturated. It is half the number of hydrogen atoms missing from the molecule as compared to a completely saturated molecule. The contribution of each double bond in a molecule to the molecule’s index of hydrogen deficiency is 1. The contribution of each triple bond in a molecule to the molecule’s index of hydrogen deficiency is 2. The contribution of each ring in a molecule to the molecule’s index of hydrogen deficiency is 1. If a molecule is saturated and has no rings, double bonds, and triple bonds, its index of hydrogen deficiency is 0.
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Chapter 4 Solutions
Organic Chemistry: Principles And Mechanisms (second Edition)
- A solution containing 100.0 mL of 0.155 M EDTA buffered to pH 10.00 was titrated with 100.0 mL of 0.0152 M Hg(ClO4)2 in a cell: calomel electrode (saturated)//titration solution/Hg(l) Given the formation constant of Hg(EDTA)2-, logKf= 21.5, and alphaY4-=0.30, find out the cell voltage E. Hg2+(aq) + 2e- = Hg(l) E0= 0.852 V E' (calomel electrode, saturated KCl) = 0.241 Varrow_forwardFrom the following reduction potentials I2 (s) + 2e- = 2I- (aq) E0= 0.535 V I2 (aq) + 2e- = 2I- (aq) E0= 0.620 V I3- (aq) + 2e- = 3I- (aq) E0= 0.535 V a) Calculate the equilibrium constant for I2 (aq) + I- (aq) = I3- (aq). b) Calculate the equilibrium constant for I2 (s) + I- (aq) = I3- (aq). c) Calculate the solubility of I2 (s) in water.arrow_forward2. (3 pts) Consider the unit cell for the spinel compound, CrFe204. How many total particles are in the unit cell? Also, show how the number of particles and their positions are consistent with the CrFe204 stoichiometry - this may or may not be reflected by the particle colors in the diagram. (HINT: In the diagram, the blue particle is in an interior position while each red particle is either in a corner or face position.)arrow_forward
- From the following potentials, calculate the activity of Cl- in saturated KCl. E0 (calomel electrode)= 0.268 V E (calomel electrode, saturated KCl)= 0.241 Varrow_forwardCalculate the voltage of each of the following cells. a) Fe(s)/Fe2+ (1.55 x 10-2 M)//Cu2+ (6.55 x 10-3 M)/Cu(s) b) Pt, H2 (0.255 bar)/HCl (4.55 x 10-4 M), AgCl (sat'd)/Ag Fe2+ +2e- = Fe E0= -0.44 V Cu2+ + 2e- = Cu E0= 0.337 V Ag+ + e- = Ag E0= 0.799 V AgCl(s) + e- = Ag(s) + Cl- E0= 0.222 V 2H+ + 2e- = H2 E0= 0.000 Varrow_forwardA solution contains 0.097 M Ce3+, 1.55x10-3 M Ce4+, 1.55x10-3 M Mn2+, 0.097 M MnO4-, and 1.00 M HClO4 (F= 9.649 x 104 C/mol). a) Write a balanced net reaction that can occur between species in this solution. b) Calculate deltaG0 and K for the reaction. c) Calculate E and deltaG for the conditions given. Ce4+ + e- = Ce3+ E0= 1.70 V MnO4- + 8H+ + 5e- = Mn2+ + 4H2O E0= 1.507 Varrow_forward
- 1. Provide a step-by-step mechanism for formation of ALL STEREOISOMERS in the following reaction. Na HCO3 (Sodium bicarbonate, baking soda) is not soluble in CH2Cl2. The powder is a weak base used to neutralize strong acid (pKa < 0) produced by the reaction. Redraw the product to show the configuration(s) that form at C-2 and C-4. Br2 OH CH2Cl2 Na* HCO3 Br HO OH + Na Br +arrow_forward2. Specify the solvent and reagent(s) required to carry out each of the following FGI. If two reagent sets must be used for the FGI, specify the solvent and reagent(s) for each reagent set. If a reaction cannot be carried out with reagents (sets) class, write NP (not possible) in the solvent box for reagent set #1. Use the letter abbreviation for each solvent; use a number abbreviation for reagent(s). Solvents: CH2Cl2 (A); H₂O (B); Reagents: HBr (1); R₂BH (6); H2SO4 (2); CH3OH (C); Br₂ (3); CH3CO₂H (D) NaHCO3 (4); Hg(OAc)2 (5); H₂O2/HO (7); NaBH4 (8) Reagent Set #1 Reagent Set #2 FGI + enant OH Solvent Reagent(s) Solvent Reagent(s)arrow_forwardGermanium (Ge) is a semiconductor with a bandgap of 2.2 eV. How could you dope Ge to make it a p-type semiconductor with a larger bandgap? Group of answer choices It is impossible to dope Ge and have this result in a larger bandgap. Dope the Ge with silicon (Si) Dope the Ge with gallium (Ga) Dope the Ge with phosphorus (P)arrow_forward
- Which of the following semiconductors would you choose to have photons with the longest possible wavelengths be able to promote electrons to the semiconductor's conduction band? Group of answer choices Si Ge InSb CdSarrow_forwardWhich of the following metals is the only one with all of its bands completely full? Group of answer choices K Na Ca Alarrow_forward2. Specify the solvent and reagent(s) required to carry out each of the following FGI. If two reagent sets must be used for the FGI, specify the solvent and reagent(s) for each reagent set. If a reaction cannot be carried out with reagents (sets) class, write NP (not possible) in the solvent box for reagent set #1. Use the letter abbreviation for each solvent; use a number abbreviation for reagent(s). Solvents: CH2Cl2 (A); Reagents: H₂O (B); CH3CO₂H (D) NaHCO3 (4); Hg(OAc)2 (5); HBr (1); R₂BH (6); H2SO4 (2); CH3OH (C); Br₂ (3); H₂O₂ / HO- (7); NaBH4 (8) Reagent Set #1 Reagent Set #2 FGI OH - α-α Br + enant Solvent Reagent(s) Solvent Reagent(s)arrow_forward
Organic Chemistry: A Guided InquiryChemistryISBN:9780618974122Author:Andrei StraumanisPublisher:Cengage Learning
Chemistry: The Molecular ScienceChemistryISBN:9781285199047Author:John W. Moore, Conrad L. StanitskiPublisher:Cengage Learning