Concept explainers
(a)
Interpretation:
The given compound is to be identified as
Concept introduction:
Heterocyclic compounds are defined as cyclic compounds which consist of at least one heteroatom, mostly nitrogen, sulfur, or oxygen. Heterocyclic compounds can be aromatic, antiaromatic, or nonaromatic. In some heterocyclic compounds, a noncarbon atom contributes to a p atomic orbital to the aromatic
Huckel’s
1) Aromatic if the number of electrons in that

Answer to Problem 14.20P
The given compound is antiaromatic.
Explanation of Solution
The given compound is:
It is a heterocyclic compound with an oxygen atom. The hybridization of the oxygen atom must be
The total number of electrons in this
The given heterocyclic compound is antiaromatic if the total number of electrons in that
(b)
Interpretation:
The given compound is to be identified as aromatic, antiaromatic, or nonaromatic.
Concept introduction:
Heterocyclic compounds are defined as cyclic compounds which consists of at least one heteroatom mostly nitrogen, sulfur, or oxygen. Heterocyclic compounds can be aromatic, antiaromatic, or nonaromatic. In some heterocyclic compounds, a noncarbon atom contributes a p atomic orbital to the aromatic
Huckel’s rule for aromaticity states that if a species possesses a pi system of molecular orbitals constructed from p-orbitals that are fully conjugated around a ring (
1) Aromatic if the number of electrons in that

Answer to Problem 14.20P
The given compound is aromatic.
Explanation of Solution
The given compound is:
It is a heterocyclic compound with a nitrogen atom. The hybridization of the nitrogen atom must be
Thus, the total number of electrons in this
The given heterocyclic compound is aromatic if the number of electrons in that
(c)
Interpretation:
The given compound is to be identified as aromatic, antiaromatic, or nonaromatic.
Concept introduction:
Heterocyclic compounds are defined as cyclic compounds which consist of at least one heteroatom mostly nitrogen, sulfur, or oxygen. Heterocyclic compounds can be aromatic, antiaromatic, or nonaromatic. In some heterocyclic compounds, a noncarbon atom contributes a p atomic orbital to the aromatic
Huckel’s rule for aromaticity states that if a species possesses a pi system of molecular orbitals constructed from p-orbitals that are fully conjugated around a ring (
1) Aromatic if the number of electrons in that

Answer to Problem 14.20P
The given compound is nonaromatic.
Explanation of Solution
The given compound is:
It is an oxygen containing a heterocycle consisting of a seven-membered ring with three double bonds. Due to geometrical constrains, the lone pairs of electrons on the oxygen atom are not in conjugation with the
If the molecule is assumed to be planar (flat), then too it would contain
The rule for aromaticity or antiaromaticity applies only if the system is planar, cyclic and has overlap of p-orbitals.
(d)
Interpretation:
The given compound is to be identified as aromatic, antiaromatic, or nonaromatic.
Concept introduction:
Heterocyclic compounds are defined as cyclic compounds which consists of at least one heteroatom mostly nitrogen, sulfur, or oxygen. Heterocyclic compounds can be aromatic, antiaromatic, or nonaromatic. In some heterocyclic compounds, a noncarbon atom contributes a p atomic orbital to the aromatic
Huckel’s rule for aromaticity states that if a species possesses a pi system of molecular orbitals constructed from p-orbitals that are fully conjugated around a ring (
1) Aromatic if the number of electrons in that

Answer to Problem 14.20P
The given compound is nonaromatic.
Explanation of Solution
The given compound is:
It is a sulfur containing a heterocycle consisting of a six-membered ring with two double bonds. There are two lone pair of electrons on the sulfur atom, but the sulfur atom is
As the
The rule for aromaticity or antiaromaticity applies only if the system is planar, cyclic and has overlap of p-orbitals.
(e)
Interpretation:
The given compound is to be identified as aromatic, antiaromatic, or nonaromatic.
Concept introduction:
Heterocyclic compounds are defined as cyclic compounds which consists of at least one heteroatom mostly nitrogen, sulfur, or oxygen. Heterocyclic compounds can be aromatic, antiaromatic, or nonaromatic. In some heterocyclic compounds, a noncarbon atom contributes a p atomic orbital to the aromatic
Huckel’s rule for aromaticity states that if a species possesses a pi system of molecular orbitals constructed from p-orbitals that are fully conjugated around a ring (
1) Aromatic if the number of electrons in that

Answer to Problem 14.20P
The given compound is nonaromatic.
Explanation of Solution
The given compound is:
It is a straight chain compound with an oxygen atom and two double bonds. The above compound is not a cyclic compound. The rule for aromaticity or antiaromaticity applies only if the system is planar, cyclic and has overlap of p-orbitals. Thus, the
The rule for aromaticity or antiaromaticity applies only if the system is planar, cyclic and has overlap of p-orbitals.
(f)
Interpretation:
The given compound is to be identified as aromatic, antiaromatic, or nonaromatic.
Concept introduction:
Heterocyclic compounds are defined as cyclic compounds which consists of at least one heteroatom mostly nitrogen, sulfur, or oxygen. Heterocyclic compounds can be aromatic, antiaromatic, or nonaromatic. In some heterocyclic compounds, a noncarbon atom contributes a p atomic orbital to the aromatic
Huckel’s rule for aromaticity states that if a species possesses a pi system of molecular orbitals constructed from p-orbitals that are fully conjugated around a ring (
1) Aromatic if the number of electrons in that

Answer to Problem 14.20P
The given compound is aromatic.
Explanation of Solution
The given compound is:
It is a heterocyclic compound with a nitrogen atom in which one six-membered and one five-membered rings are fused. The hybridization of the nitrogen atom is
Thus, the total number of electrons in this
The given heterocyclic compound is aromatic if the number of electrons in that
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Chapter 14 Solutions
ORG CHEM W/ EBOOK & SW5 + STUDY GUIDE
- Indicate the rate expressions for reactions that have order 0, 1, and 2.arrow_forwardPROBLEMS Q1) Label the following salts as either acidic, basic, or neutral a) Fe(NOx) c) AlBr b) NH.CH COO d) HCOON (1/2 mark each) e) Fes f) NaBr Q2) What is the pH of a 0.0750 M solution of sulphuric acid?arrow_forward8. Draw all the resonance forms for each of the fling molecules or ions, and indicate the major contributor in each case, or if they are equivalent (45) (2) -PH2 سمة مدarrow_forward
- A J то گای ه +0 Also calculate the amount of starting materials chlorobenzaldehyde and p-chloroacetophenone required to prepare 400 mg of the given chalcone product 1, 3-bis(4-chlorophenyl)prop-2-en-1-one molar mass ok 1,3-bis(4-Chlorophenyl) prop-2-en-1-one = 277.1591m01 number of moles= 0.400/277.15 = 0.00144 moles 2 x 0.00 144=0.00288 moves arams of acetophenone = 0.00144 X 120.16 = 0.1739 0.1739x2=0.3469 grams of benzaldehyde = 0.00144X106.12=0.1539 0.1539x2 = 0.3069 Starting materials: 0.3469 Ox acetophenone, 0.3069 of benzaldehyde 3arrow_forward1. Answer the questions about the following reaction: (a) Draw in the arrows that can be used make this reaction occur and draw in the product of substitution in this reaction. Be sure to include any relevant stereochemistry in the product structure. + SK F Br + (b) In which solvent would this reaction proceed the fastest (Circle one) Methanol Acetone (c) Imagine that you are working for a chemical company and it was your job to perform a similar reaction to the one above, with the exception of the S atom in this reaction being replaced by an O atom. During the reaction, you observe the formation of three separate molecules instead of the single molecule obtained above. What is the likeliest other products that are formed? Draw them in the box provided.arrow_forward3. For the reactions below, draw the arrows corresponding to the transformations and draw in the boxes the reactants or products as indicated. Note: Part A should have arrows drawn going from the reactants to the middle structure and the arrows on the middle structure that would yield the final structure. For part B, you will need to draw in the reactant before being able to draw the arrows corresponding to product formation. A. B. Rearrangement ΘΗarrow_forward
- 2. Draw the arrows required to make the following reactions occur. Please ensure your arrows point from exactly where you want to exactly where you want. If it is unclear from where arrows start or where they end, only partial credit will be given. Note: You may need to draw in lone pairs before drawing the arrows. A. B. H-Br 人 C Θ CI H Cl Θ + Br Oarrow_forward4. For the reactions below, draw the expected product. Be sure to indicate relevant stereochemistry or formal charges in the product structure. a) CI, H e b) H lux ligh Br 'Harrow_forwardArrange the solutions in order of increasing acidity. (Note that K (HF) = 6.8 x 10 and K (NH3) = 1.8 × 10-5) Rank solutions from least acidity to greatest acidity. To rank items as equivalent, overlap them. ▸ View Available Hint(s) Least acidity NH&F NaBr NaOH NH,Br NaCIO Reset Greatest acidityarrow_forward
- 1. Consider the following molecular-level diagrams of a titration. O-HA molecule -Aion °° о ° (a) о (b) (c) (d) a. Which diagram best illustrates the microscopic representation for the EQUIVALENCE POINT in a titration of a weak acid (HA) with sodium. hydroxide? (e)arrow_forwardAnswers to the remaining 6 questions will be hand-drawn on paper and submitted as a single file upload below: Review of this week's reaction: H₂NCN (cyanamide) + CH3NHCH2COOH (sarcosine) + NaCl, NH4OH, H₂O ---> H₂NC(=NH)N(CH3)CH2COOH (creatine) Q7. Draw by hand the reaction of creatine synthesis listed above using line structures without showing the Cs and some of the Hs, but include the lone pairs of electrons wherever they apply. (4 pts) Q8. Considering the Zwitterion form of an amino acid, draw the Zwitterion form of Creatine. (2 pts) Q9. Explain with drawing why the C-N bond shown in creatine structure below can or cannot rotate. (3 pts) NH2(C=NH)-N(CH)CH2COOH This bond Q10. Draw two tautomers of creatine using line structures. (Note: this question is valid because problem Q9 is valid). (4 pts) Q11. Mechanism. After seeing and understanding the mechanism of creatine synthesis, students should be ready to understand the first half of one of the Grignard reactions presented in a past…arrow_forwardPropose a synthesis pathway for the following transformations. b) c) d)arrow_forward
- Organic Chemistry: A Guided InquiryChemistryISBN:9780618974122Author:Andrei StraumanisPublisher:Cengage Learning
