Concept explainers
(a)
Interpretation:
Lewis structure for the given molecule is to be completed.
Concept introduction:
Lewis structures involve only valence electrons. When drawing a Lewis structure, the first step is to calculate the total number of valence electrons. For a complete Lewis structure of a molecule, the atoms must complete their normal valency by bond formation and lone pairs of electrons. Maximum number of covalent bonds formed by any neutral atom with maximum number of lone pairs is
Atom | Number of bond | Number of lone pairs |
C | 4 | 0 |
H | 1 | 0 |
O | 2 | 2 |
N | 1 | 1 |
F | 1 | 3 |

Answer to Problem 1.46P
The complete Lewis structure for the given molecule is
Explanation of Solution
The given structure is
Total valence electron count for the given molecule is
The other oxygen atom has formed only one bond with carbon. This is converted to a double bond and two lone pairs are placed on the oxygen atom so that its octet is complete. A double bond is placed between C and N atom to complete the octet of carbon and a lone pair is placed in nitrogen to complete its octet.
A triple bond is placed between the other C and N to complete the octet of carbon and a lone pair is placed in the nitrogen to complete its octet.
This structure now accounts for all 54 electrons and the octet of each atom, except hydrogen, is complete. The duet for all hydrogens is complete.
The Lewis structure for the given molecule is completed from total valence electron count.
(b)
Interpretation:
Lewis structure for the given molecule is to be completed.
Concept introduction:
Lewis structures involve only valence electrons. When drawing a Lewis structure, the first step is to calculate the total number of valence electrons. For a complete Lewis structure of a molecule, every carbon atom must form four covalent bonds whereas the hydrogen atom forms one bond.

Answer to Problem 1.46P
The complete Lewis structure for the given molecule is
Explanation of Solution
The given structure is
Total valence electron count for the given molecule must be
This structure now accounts for all 38 electrons and the octet of each atom, except hydrogen, is complete. The duet for all hydrogen atoms is complete.
The Lewis structure for the given molecule is completed from total valence electron count.
(c)
Interpretation:
Lewis structure for the given molecule is to be completed.
Concept introduction:
Lewis structures involve only valence electrons. When drawing a Lewis structure, the first step is to calculate the total number of valence electrons. For a complete Lewis structure of a molecule, the atoms must complete their normal valency by bond formation and lone pairs of electrons. Maximum numbers of covalent bonds formed by any neutral atom with maximum number of lone pair are
Atom | Number of bond | Number of lone pairs |
C | 4 | 0 |
H | 1 | 0 |
O | 2 | 2 |
N | 1 | 1 |

Answer to Problem 1.46P
The complete Lewis structure for the given molecule is
Explanation of Solution
The given structure is
Total valence electron count for the given molecule is
The oxygen atom has formed only one bond with nitrogen. This is converted to a double bond and two lone pairs are placed on the oxygen atom so that its octet is complete. Another lone pair is placed on the nitrogen atom so that its octet is complete.
A double bond is placed between the C atoms attached to one hydrogen each. This completes the octet of both carbon atoms
This structure now accounts for all the 28 electrons, and the octet of each atom, except hydrogen, is complete. The duet for all hydrogen atoms is complete.
The Lewis structure for the given molecule is completed from total valence electron count.
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Chapter 1 Solutions
EBK ORGANIC CHEMISTRY: PRINCIPLES AND M
- ✓ Check the box under each molecule that has a total of five ẞ hydrogens. If none of the molecules fit this description, check the box underneath the table. tab OH CI 0 Br xx Br None of these molecules have a total of five ẞ hydrogens. esc Explanation Check caps lock shift 1 fn control 02 F2 W Q A N #3 S 80 F3 E $ t 01 205 % 5 F5 & 7 © 2025 McGraw Hill LLC. All Rights Reserved. Terms of Use | Privacy Center | Accessibility FT * 8 R T Y U כ F6 9 FIG F11 F D G H J K L C X V B < N M H option command P H + F12 commandarrow_forwardDraw the major product of this reaction. Ignore inorganic byproducts and the carboxylic acid side product. O 1. CHзMgBr (excess) 2. H₂O ✓ W X 人arrow_forwardIf cyclopentyl acetaldehyde reacts with NaOH, state the product (formula).arrow_forward
- Draw the major product of this reaction. Ignore inorganic byproducts. N S S HgCl2, H2SO4 く 8 W X Parrow_forwardtab esc く Drawing the After running various experiments, you determine that the mechanism for the following reaction occurs in a step-wise fashion. Br + OH + Using this information, draw the correct mechanism in the space below. 1 Explanation Check F2 F1 @2 Q W A os lock control option T S # 3 80 F3 Br $ 4 0105 % OH2 + Br Add/Remove step X C F5 F6 6 R E T Y 29 & 7 F D G H Click and drag to start drawing a structure. © 2025 McGraw Hill LLC. All Rights Reserved. Terms of Use | Privacy Ce A F7 DII F8 C Ո 8 * 9 4 F10 F C J K L C V Z X B N M H command P ge Coarrow_forwardIndicate compound A that must react with ethylbenzene to obtain 4-ethylbenzene-1-sulfonic acid. 3-bromo-4-ethylbenzene-1-sulfonic acid.arrow_forward
- Part 1 of 2 Draw the structure of A, the minor E1 product of the reaction. esc I Skip Part Check H₂O, D 2 A + Click and drag to start drawing a structure. -0- F1 F2 1 2 # 3 Q A 80 F3 W E S D F4 $ 4 % 5 F5 ㅇ F6 R T Y F G X 5 & 7 + Save 2025 McGraw Hill LLC. All Rights Reserved. DII F7 F8 H * C 80 J Z X C V B N 4 F9 6arrow_forwardFile Preview The following is a total synthesis of the pheromone of the western pine beetle. Such syntheses are interesting both because of the organic chemistry, and because of the possibility of using species specific insecticides, rather than broad band insecticides. Provide the reagents for each step. There is some chemistry from our most recent chapter in this synthesis, but other steps are review from earlier chapters. (8 points) COOEt COOEt A C COOEt COOEt COOH B OH OTS CN D E See the last homework set F for assistance on this one. H+, H₂O G OH OH The last step is just nucleophilic addition reactions, taking the ketone to an acetal, intramolecularly. But it is hard to visualize the three dimensional shape as it occurs. Frontalin, pheromone of the western pine beetlearrow_forwardFor the reaction below: 1. Draw all reasonable elimination products to the right of the arrow. 2. In the box below the reaction, redraw any product you expect to be a major product. C Major Product: Check + ◎ + X ง © Cl I F2 80 F3 I σ F4 I F5 NaOH Click and drawing F6 A 2025 McGraw Hill LLC. All Rights E F7 F8 $ # % & 2 3 4 5 6 7 8 Q W E R T Y U A S D F G H Jarrow_forward
- Can I please get help with this graph. If you can show exactly where it needs to pass through.arrow_forwardN Draw the major product of this reaction. Ignore inorganic byproducts. D 1. H₂O, pyridine 2. neutralizing work-up V P W X DE CO e C Larrow_forwardDraw the major product of this reaction. Ignore inorganic byproducts. N O' 1. H2O, pyridine 2. neutralizing work-up く 8 W aarrow_forward
- Introduction to General, Organic and BiochemistryChemistryISBN:9781285869759Author:Frederick A. Bettelheim, William H. Brown, Mary K. Campbell, Shawn O. Farrell, Omar TorresPublisher:Cengage LearningOrganic Chemistry: A Guided InquiryChemistryISBN:9780618974122Author:Andrei StraumanisPublisher:Cengage Learning

