Concept explainers
Videos
(a)
Interpretation:
The hybridization of each of the numbered carbon, nitrogen and oxygen atoms in tryptophan is to be determined.
Concept introduction:
The atomic orbital is the wave function that is used to find the probability to find an electron around the nucleus of an atom. It is the space around the nucleus of an atom where the electrons are supposed to be found.
Hybridization is the process of intermixing of atomic orbital of slightly different energies to form hybrid orbitals that have similar energy. These orbital have lower energy and more stability than the atomic orbital.
Hybridization of the central atom can be determined from the number of electron groups around the central atom in the Lewis structure of the molecule. Single bond, double bond, triple bond and lone pair all are considered as single electron group.
(b)
Interpretation:
The number of sigma bonds in tryptophan is to be determined.
Concept introduction:
The sigma bond is the covalent bond that is formed by the axial overlap of orbitals. The pi bond is formed by the lateral overlap of orbitals. Both types of bonds can be bonding or anti-bonding interaction. Bonding interaction takes place when there is an overlap of orbitals in the same phase and antibonding interaction occurs when the orbital overlap in opposite phase. Sigma bond is stronger than the pi bond.
All single bonds are sigma bonds. One double bond is made up of one sigma bond and one pi bond. One triple bond is made up of one sigma bond and two pi bonds.
(c)
Interpretation:
The bond angles at points a, b and c are to be determined.
Concept introduction:
The bond angle is the angle that is formed between any two non-central atoms and the central atom. It depends on the hybridization of the central atom and the lone pairs present around the atom.
Hybridization is the process of intermixing of atomic orbital of slightly different energies to form hybrid orbitals that have similar energy. These orbital have lower energy and more stability than the atomic orbital.
Hybridization of the central atom can be determined from the number of electron groups around the central atom in the Lewis structure of the molecule. Single bond, double bond, triple bond and lone pair all are considered as single electron group.
Want to see the full answer?
Check out a sample textbook solution
Chapter 11 Solutions
Chemistry: The Molecular Nature of Matter and Change - Standalone book
- Add conditions above and below the arrow that turn the reactant below into the product below in a single transformation. + More... If you need to write reagents above and below the arrow that have complex hydrocarbon groups in them, there is a set of standard abbreviations you can use. More... T H,N NC Datarrow_forwardIndicate the order of basicity of primary, secondary and tertiary amines.arrow_forward> Classify each of the following molecules as aromatic, antiaromatic, or nonaromatic. Cl Z- N O aromatic O antiaromatic O nonaromatic O aromatic O antiaromatic O nonaromatic O aromatic ○ antiaromatic nonaromaticarrow_forward
- Please help me answer this question. I don't understand how or even if this can happen in a single transformation. Please provide a detailed explanation and a drawing showing how it can happen in a single transformation. Add the necessary reagents and reaction conditions above and below the arrow in this organic reaction. If the products can't be made from the reactant with a single transformation, check the box under the drawing area instead.arrow_forward2) Draw the correct chemical structure (using line-angle drawings / "line structures") from their given IUPAC name: a. (E)-1-chloro-3,4,5-trimethylhex-2-ene b. (Z)-4,5,7-trimethyloct-4-en-2-ol C. (2E,6Z)-4-methylocta-2,6-dienearrow_forwardපිපිම Draw curved arrows to represent the flow of electrons in the reaction on the left Label the reactants on the left as either "Acid" or "Base" (iii) Decide which direction the equilibrium arrows will point in each reaction, based on the given pk, values (a) + H-O H 3-H + (c) H" H + H****H 000 44-00 NH₂ (e) i Дон OH Ө NHarrow_forward
- 3) Label the configuration in each of the following alkenes as E, Z, or N/A (for non-stereogenic centers). 00 E 000 N/A E Br N/A N/A (g) E N/A OH E (b) Oz N/A Br (d) 00 E Z N/A E (f) Oz N/A E (h) Z N/Aarrow_forward6) Fill in the missing Acid, pKa value, or conjugate base in the table below: Acid HCI Approximate pK, -7 Conjugate Base H-C: Hydronium (H₂O') -1.75 H-O-H Carboxylic Acids (RCOOH) Ammonium (NH4) 9.24 Water (H₂O) H-O-H Alcohols (ROH) RO-H Alkynes R--H Amines 25 25 38 HOarrow_forward5) Rank the following sets of compounds in order of decreasing acidity (most acidic to least acidic), and choose the justification(s) for each ranking. (a) OH V SH я вон CH most acidic (lowst pKa) least acidic (highest pKa) Effect(s) Effect(s) Effect(s) inductive effect O inductive effect O inductive effect electronegativity electronegativity O electronegativity resonance polarizability resonance polarizability O resonance O polarizability hybridization Ohybridization O hybridization оarrow_forward
ChemistryChemistryISBN:9781305957404Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCostePublisher:Cengage Learning
ChemistryChemistryISBN:9781259911156Author:Raymond Chang Dr., Jason Overby ProfessorPublisher:McGraw-Hill Education
Principles of Instrumental AnalysisChemistryISBN:9781305577213Author:Douglas A. Skoog, F. James Holler, Stanley R. CrouchPublisher:Cengage Learning
Organic ChemistryChemistryISBN:9780078021558Author:Janice Gorzynski Smith Dr.Publisher:McGraw-Hill Education
Chemistry: Principles and ReactionsChemistryISBN:9781305079373Author:William L. Masterton, Cecile N. HurleyPublisher:Cengage Learning
Elementary Principles of Chemical Processes, Bind...ChemistryISBN:9781118431221Author:Richard M. Felder, Ronald W. Rousseau, Lisa G. BullardPublisher:WILEY