Concept explainers
(a)
Interpretation:
The complete IUPAC name of the given molecule is to be written.
Concept introduction:
The IUPAC name of a compound consists of three parts, prefix, root and suffix. The root is the longest continuous carbon chain or the largest ring that bears the highest priority functional group. The suffix refers to the highest priority functional group. The functional group name suffix replaces the last ‘e’ in the root alkyl/aryl name. The root chain or ring carbons are numbered so that the highest priority group gets the lowest possible location numbers. The number is inserted between the root and the suffix unless redundant. Any other, low priority
(b)
Interpretation:
The complete IUPAC name of the given molecule is to be written.
Concept introduction:
The IUPAC name of a compound consists of three parts, prefix, root, and suffix. The root is the longest continuous carbon chain or the largest ring that bears the highest priority functional group. The suffix refers to the highest priority functional group. The functional group name suffix replaces the last ‘e’ in the root alkyl/aryl name. The root chain or ring carbons are numbered so that the highest priority group gets the lowest possible location numbers. The number is inserted between the root and the suffix unless redundant. Any other low priority functional groups are listed alphabetically in the prefix, along with their location numbers. If the root is a ring compound, the ring is numbered in the direction that gives the lowest possible numbers to other substituents if present. A prefix di, tri, etc. is used in case two or more instances of the functional group are present.
(c)
Interpretation:
The complete IUPAC name of the given molecule is to be written.
Concept introduction:
The IUPAC name of a compound consists of three parts, prefix, root, and suffix. The root is the longest continuous carbon chain or the largest ring that bears the highest priority functional group. The suffix refers to the highest priority functional group. The functional group name suffix replaces the last ‘e’ in the root alkyl/aryl name. The root chain or ring carbons are numbered so that the highest priority group gets the lowest possible location numbers. The number is inserted between the root and the suffix unless redundant. Any other low priority functional groups are listed alphabetically in the prefix, along with their location numbers. If the root is a ring compound, the ring is numbered in the direction that gives the lowest possible numbers to other substituents if present. A prefix di, tri, etc. is used in case two or more instances of the functional group are present.

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Chapter E Solutions
EBK GET READY FOR ORGANIC CHEMISTRY
- 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
