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Concept explainers
(a)
Interpretation:
Lone pair has to be identified, lone pairs are delocalized or not have to be indicated and also geometry has to be predicted.
Concept Introduction:
The unshared electrons of an atom that does not involve in bonding are called lone pair.
Resonance is an electron displacement effect for stabilizing a molecule through delocalization of bonding electrons in the pi orbital.
Delocalized electrons stabilize a compound. The extra stability gains from having delocalized electrons are called resonance stabilization or resonance energy.
The steric number is the combination of both number of
The geometry of the central atom will be determined by counting the steric number followed by the hybridization state of that central atom and finally electronic arrangement of atoms in space.
If the steric number is 4, the central atom is sp3 hybridized and the electronic arrangement of atoms in space (i.e. geometry) will be tetrahedral.
If the steric number is 3, the central atom is sp2 hybridized and the electronic arrangement of atoms in space (i.e. geometry) will be trigonal planar.
If the steric number is 2, the central atom is sp hybridized and the electronic arrangement of atoms in space (i.e. geometry) will be linear.
trigonal pyramidal(if one lone pair of electron is present).
bent geometry (if two lone pair of electrons are present).
(b)
Interpretation:
Lone pair has to be identified, lone pairs are delocalized or not have to be indicated and also geometry has to be predicted.
Concept Introduction:
The unshared electrons of an atom that does not involve in bonding are called lone pair.
Resonance is an electron displacement effect for stabilizing a molecule through delocalization of bonding electrons in the pi orbital.
Delocalized electrons stabilize a compound. The extra stability gains from having delocalized electrons are called resonance stabilization or resonance energy.
The steric number is the combination of both number of
The geometry of the central atom will be determined by counting the steric number followed by the hybridization state of that central atom and finally electronic arrangement of atoms in space.
If the steric number is 4, the central atom is sp3 hybridized and the electronic arrangement of atoms in space (i.e. geometry) will be tetrahedral.
If the steric number is 3, the central atom is sp2 hybridized and the electronic arrangement of atoms in space (i.e. geometry) will be trigonal planar.
If the steric number is 2, the central atom is sp hybridized and the electronic arrangement of atoms in space (i.e. geometry) will be linear.
trigonal pyramidal(if one lone pair of electron is present).
bent geometry (if two lone pair of electrons are present).
(c)
Interpretation:
Lone pair has to be identified, lone pairs are delocalized or not have to be indicated and also geometry has to be predicted.
Concept Introduction:
The unshared electrons of an atom that does not involve in bonding are called lone pair.
Resonance is an electron displacement effect for stabilizing a molecule through delocalization of bonding electrons in the pi orbital.
Delocalized electrons stabilize a compound. The extra stability gains from having delocalized electrons are called resonance stabilization or resonance energy.
The steric number is the combination of both number of
The geometry of the central atom will be determined by counting the steric number followed by the hybridization state of that central atom and finally electronic arrangement of atoms in space.
If the steric number is 4, the central atom is sp3 hybridized and the electronic arrangement of atoms in space (i.e. geometry) will be tetrahedral.
If the steric number is 3, the central atom is sp2 hybridized and the electronic arrangement of atoms in space (i.e. geometry) will be trigonal planar.
If the steric number is 2, the central atom is sp hybridized and the electronic arrangement of atoms in space (i.e. geometry) will be linear.
trigonal pyramidal(if one lone pair of electron is present).
bent geometry (if two lone pair of electrons are present).
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Chapter 2 Solutions
ORGANIC CHEMISTRY-STD.WILEY PLUS CARD
- Diels Alder Cycloaddition: Focus on regiochemistry (problems E-F) –> match + of thedienophile and - of the diene while also considering stereochemistry (endo).arrow_forwardHELP! URGENT! PLEASE RESOND ASAP!arrow_forwardQuestion 4 Determine the rate order and rate constant for sucrose hydrolysis. Time (hours) [C6H12O6] 0 0.501 0.500 0.451 1.00 0.404 1.50 0.363 3.00 0.267 First-order, k = 0.210 hour 1 First-order, k = 0.0912 hour 1 O Second-order, k = 0.590 M1 hour 1 O Zero-order, k = 0.0770 M/hour O Zero-order, k = 0.4896 M/hour O Second-order, k = 1.93 M-1-hour 1 10 ptsarrow_forward
- Determine the rate order and rate constant for sucrose hydrolysis. Time (hours) [C6H12O6] 0 0.501 0.500 0.451 1.00 0.404 1.50 0.363 3.00 0.267arrow_forwardDraw the products of the reaction shown below. Use wedge and dash bonds to indicate stereochemistry. Ignore inorganic byproducts. OSO4 (cat) (CH3)3COOH Select to Draw ઘarrow_forwardCalculate the reaction rate for selenious acid, H2SeO3, if 0.1150 M I-1 decreases to 0.0770 M in 12.0 minutes. H2SeO3(aq) + 6I-1(aq) + 4H+1(aq) ⟶ Se(s) + 2I3-1(aq) + 3H2O(l)arrow_forward
- Problem 5-31 Which of the following objects are chiral? (a) A basketball (d) A golf club (b) A fork (c) A wine glass (e) A spiral staircase (f) A snowflake Problem 5-32 Which of the following compounds are chiral? Draw them, and label the chirality centers. (a) 2,4-Dimethylheptane (b) 5-Ethyl-3,3-dimethylheptane (c) cis-1,4-Dichlorocyclohexane Problem 5-33 Draw chiral molecules that meet the following descriptions: (a) A chloroalkane, C5H11Cl (c) An alkene, C6H12 (b) An alcohol, C6H140 (d) An alkane, C8H18 Problem 5-36 Erythronolide B is the biological precursor of erythromycin, a broad-spectrum antibiotic. How H3C CH3 many chirality centers does erythronolide B have? OH Identify them. H3C -CH3 OH Erythronolide B H3C. H3C. OH OH CH3arrow_forwardPLEASE HELP! URGENT! PLEASE RESPOND!arrow_forward2. Propose a mechanism for this reaction. ہلی سے ملی N H (excess)arrow_forward
- Steps and explanationn please.arrow_forwardProblem 5-48 Assign R or S configurations to the chirality centers in ascorbic acid (vitamin C). OH H OH HO CH2OH Ascorbic acid O H Problem 5-49 Assign R or S stereochemistry to the chirality centers in the following Newman projections: H Cl H CH3 H3C. OH H3C (a) H H H3C (b) CH3 H Problem 5-52 Draw the meso form of each of the following molecules, and indicate the plane of symmetry in each: OH OH (a) CH3CHCH2CH2CHCH3 CH3 H3C. -OH (c) H3C CH3 (b) Problem 5-66 Assign R or S configurations to the chiral centers in cephalexin, trade-named Keflex, the most widely prescribed antibiotic in the United States. H2N H IHH S Cephalexin N. CH3 CO₂Harrow_forwardSteps and explanationn please.arrow_forward
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