Concept explainers
Videos
(a)
Interpretation:
The hybrid orbitals used by the central atom nitrogen and the types of bonds that are formed in
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.
Sigma bond is formed by the end to end overlapping of atomic or hybrid orbitals. 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.
(b)
Interpretation:
The hybrid orbitals used by the central atom carbon and the types of bonds that are formed in
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.
Sigma bond is formed by the end to end overlapping of atomic or hybrid orbitals. 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 hybrid orbitals used by the central atom carbon and the types of bonds that are formed in
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.
Sigma bond is formed by the end to end overlapping of atomic or hybrid orbitals. 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.
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Chapter 11 Solutions
CHEMISTRY(LOOSELEAF) W/CONNECT+EBOOK
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- Add substituents to draw the conformer below (sighting down the indicated bond), then rotate the back carbon to provide the conformation that will be capable of an E2 elimination. R/S stereochemistry is graded. + I I H CH3 Ph Досн Br OCH 3 Drawing Q H Atoms, Bonds and Rings Charges Tap a node to see suggestions. H H H H H Undo Reset Remove Done Rotatearrow_forward20.17 Predict the structure of the major product formed by 1,2-addition of HBr to 3-methylenecyclohexene. 3-Methylenecyclohexene 20.18 Predict the major product formed by 1,4-addition of HBr to 3-methylenecyclohexene.arrow_forward+ Draw a vicinal alkyl bromide that would produce the following alkene in an E2 elimination. Use a dash or wedge bond to indicate stereochemistry on asymmetric centers, where applicable. Ignore any inorganic byproducts. Br Drawing Strong Base H Q Atoms, Bonds Charges and Rings Draw or tap a new bond to see suggestions. Remove Done 語 Reset Undo + Drag To Panarrow_forward
- Draw a vicinal alkyl bromide that would produce the following alkene in an E2 elimination. Use a dash or wedge bond to indicate stereochemistry on asymmetric centers, where applicable. Ignore any inorganic byproducts. + Drawing Į Strong Base H Br Q Atoms, Bonds and Rings Charges Draw or tap a new bond to see suggestions. Undo Reset 謂 Remove Done Drag To Pan +arrow_forwardDraw the product of the E2 reaction shown below. Include the correct stereochemistry. Ignore any inorganic byproducts. + Br CH3 Q Strong Base Drawing Atoms, Bonds and Rings Charges Undo Reset H "Br H N Br. Remove Done .N. Drag To Panarrow_forwardCurved arrows are used to illustrate the flow of electrons. Use the reaction conditions provided and follow the curved arrows to draw the product of this elementary step in an elimination mechanism. Include all lone pairs and charges as appropriate. Ignore stereochemistry. Ignore byproducts. + Br: .. 8 0.01 M NaOH heat Drawing Q Atoms, Bonds and Rings Charges and Lone Pairs Draw or tap a new bond to see suggestions. Undo Reset Remove Done + Drag To Panarrow_forward
- + Draw the product of the E2 reaction shown below. Include the correct stereochemistry. Ignore any inorganic byproducts. Ph CH2CH3 H H3C H Br DBN [૪] Drawing Atoms, Bonds and Rings H | OH Charges ―00 H. C | Undo Reset Br I Remove Done Drag To Pan +arrow_forwardReaction A Now the production A Œ In the product of reaction i 12 Dear the product of actionarrow_forwardMacmillan Learnin When an unknown amine reacts with an unknown acid chloride, an amide with a molecular mass of 163 g/mol (M* = 163 m/z) is formed. In the infrared spectrum, important absorptions appear at 1661, 750 and 690 cm-1. The 13C NMR and DEPT spectra are provided. Draw the structure of the product as the resonance contributor lacking any formal charges. 13C NMR DEPT 90 200 160 120 80 40 0 200 160 120 80 DEPT 135 200 160 120 80 40 0 Draw the unknown amide. 40 40 0arrow_forward
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