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(a)
Interpretation:
The kekule structure should be drawn for the given molecule
Concept introduction:
- Lone-pair electrons are valence electrons that are not used in bonding.
- A proton is positively charged hydrogen ion. A hybrid ion is a negatively charged hydrogen ion.
- When the atom is neutral, C form 2 bonds, N forms 3 bonds O forms 2 bonds and H or a halogen forms 1 bond.
- A carbocation has a positively charge carbon, a carbon has a negatively charge carbon and a radical has an unpaired electron.
kekule structure:
Without lone pairs of electros in Lewis structures is known as kekule structure.
(b)
Interpretation:
The kekule structure should be drawn for the given molecule
Concept introduction:
- Lone-pair electrons are valence electrons that are not used in bonding.
- A proton is positively charged hydrogen ion. A hybrid ion is a negatively charged hydrogen ion.
- Formal charge = the number of valence electron – the number of electrons the atom has to itself (the entire lone pair electron and one – half of the bonding electrons).
- Lewis structure indicates which atom is bonded together and show lone pairs and formal charge.
- When the atom is neutral, C form 2 bonds, N forms 3 bonds O forms 2 bonds and H or a halogen forms 1 bond.
- A carbocation has a positively charge carbon, a carbon has a negatively charge carbon and a radical has an unpaired electron.
kekule structure:
Without lone pairs of electros in Lewis structures is known as kekule structure.
(c)
Interpretation:
The kekule structure should be drawn for the given molecule
Concept introduction:
- Lone-pair electrons are valence electrons that are not used in bonding.
- A proton is positively charged hydrogen ion. A hybrid ion is a negatively charged hydrogen ion.
- Formal charge = the number of valence electron – the number of electrons the atom has to itself (the entire lone pair electron and one – half of the bonding electrons).
- Lewis structure indicates which atom is bonded together and show lone pairs and formal charge.
- When the atom is neutral, C form 2 bonds, N forms 3 bonds O forms 2 bonds and H or a halogen forms 1 bond.
- A carbocation has a positively charge carbon, a carbon has a negatively charge carbon and a radical has an unpaired electron.
kekule structure:
Without lone pairs of electros in Lewis structures is known as kekule structure.
(d)
Interpretation:
The kekule structure should be drawn for the given molecule
Concept introduction:
- Lone-pair electrons are valence electrons that are not used in bonding.
- A proton is positively charged hydrogen ion. A hybrid ion is a negatively charged hydrogen ion.
- Formal charge = the number of valence electron – the number of electrons the atom has to itself (the entire lone pair electron and one – half of the bonding electrons).
- Lewis structure indicates which atom is bonded together and show lone pairs and formal charge.
- When the atom is neutral, C form 2 bonds, N forms 3 bonds O forms 2 bonds and H or a halogen forms 1 bond.
- A carbocation has a positively charge carbon, a carbon has a negatively charge carbon and a radical has an unpaired electron.
kekule structure:
Without lone pairs of electros in Lewis structures is known as kekule structure.
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Chapter 1 Solutions
Organic Chemistry Study Guide and Solutions Manual, Books a la Carte Edition (8th Edition)
- Nonearrow_forwardH2SO4 (cat.), H₂O 100 °C NH₂arrow_forwardX Draw the major products of the elimination reaction below. If elimination would not occur at a significant rate, check the box under the drawing area instead. ది www. Cl + OH Elimination will not occur at a significant rate. Click and drag to start drawing a structure.arrow_forward
- Nonearrow_forward1A H 2A Li Be Use the References to access important values if needed for this question. 8A 3A 4A 5A 6A 7A He B C N O F Ne Na Mg 3B 4B 5B 6B 7B 8B-1B 2B Al Si P 1B 2B Al Si P S Cl Ar K Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn Ga Ge As Se Br Kr Rb Sr Y Zr Nb Mo Tc Ru Rh Pd Ag Cd In Sn Sb Te I Xe * Cs Ba La Hf Ta W Re Os Ir Pt Au Hg Tl Pb Bi Po At Rn Fr Ra Ac Rf Ha ****** Ce Pr Nd Pm Sm Eu Gd Tb Dy Ho Er Tm Yb Lu Th Pa U Np Pu Am Cm Bk Cf Es Fm Md No Lr Analyze the following reaction by looking at the electron configurations given below each box. Put a number and a symbol in each box to show the number and kind of the corresponding atom or ion. Use the smallest integers possible. cation anion + + Shell 1: 2 Shell 2: 8 Shell 3: 1 Shell 1 : 2 Shell 2 : 6 Shell 1 : 2 Shell 2: 8 Shell 1: 2 Shell 2: 8arrow_forwardNonearrow_forward
- IV. Show the detailed synthesis strategy for the following compounds. a. CH3CH2CH2CH2Br CH3CH2CCH2CH2CH3arrow_forwardDo the electrons on the OH participate in resonance with the ring through a p orbital? How many pi electrons are in the ring, 4 (from the two double bonds) or 6 (including the electrons on the O)?arrow_forwardPredict and draw the product of the following organic reaction:arrow_forward
- Nonearrow_forwardRedraw the molecule below as a skeletal ("line") structure. Be sure to use wedge and dash bonds if necessary to accurately represent the direction of the bonds to ring substituents. Cl. Br Click and drag to start drawing a structure. : ☐ ☑ Parrow_forwardK m Choose the best reagents to complete the following reaction. L ZI 0 Problem 4 of 11 A 1. NaOH 2. CH3CH2CH2NH2 1. HCI B OH 2. CH3CH2CH2NH2 DII F1 F2 F3 F4 F5 A F6 C CH3CH2CH2NH2 1. SOCl2 D 2. CH3CH2CH2NH2 1. CH3CH2CH2NH2 E 2. SOCl2 Done PrtScn Home End FA FQ 510 * PgUp M Submit PgDn F11arrow_forward
- World of Chemistry, 3rd editionChemistryISBN:9781133109655Author:Steven S. Zumdahl, Susan L. Zumdahl, Donald J. DeCostePublisher:Brooks / Cole / Cengage LearningIntroductory Chemistry: A FoundationChemistryISBN:9781337399425Author:Steven S. Zumdahl, Donald J. DeCostePublisher:Cengage Learning
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