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Concept explainers
(a)
Interpretation:
“Parent” amine can be regenerated from the given salt or not has to be indicated.
Concept Introduction:
Quaternary ammonium salt is the one that has four carbon atoms attached to the nitrogen atom. This is formed by the reaction of tertiary amine with
Neutralization reaction is the one that takes place between an acid and a base to give salt as product. As
When a strong base is added to the amine salt, the parent amine can be obtained. This is a reverse reaction of the amine salt formation reaction. These reactions can be represented as shown below,
Quaternary ammonium salt does not give the “parent” amine when treated with a strong base as there is no possibility of deprotonation to take place.
(b)
Interpretation:
“Parent” amine can be regenerated from the given salt or not has to be indicated.
Concept Introduction:
Quaternary ammonium salt is the one that has four carbon atoms attached to the nitrogen atom. This is formed by the reaction of tertiary amine with alkyl halide in presence of a strong base.
Neutralization reaction is the one that takes place between an acid and a base to give salt as product. As amines are bases due to the amino group in it, the reaction with inorganic acid or carboxylic acid gives salt as product. The salt formed is an amine salt. Proton is donated from the acid to the nitrogen atom which acts as a proton acceptor. In simple words, it can be said that in an amine‑acid reaction, the acid loses a hydrogen ion and amine gains a hydrogen ion.
When a strong base is added to the amine salt, the parent amine can be obtained. This is a reverse reaction of the amine salt formation reaction. These reactions can be represented as shown below,
Quaternary ammonium salt does not give the “parent” amine when treated with a strong base as there is no possibility of deprotonation to take place.
(c)
Interpretation:
“Parent” amine can be regenerated from the given salt or not has to be indicated.
Concept Introduction:
Quaternary ammonium salt is the one that has four carbon atoms attached to the nitrogen atom. This is formed by the reaction of tertiary amine with alkyl halide in presence of a strong base.
Neutralization reaction is the one that takes place between an acid and a base to give salt as product. As amines are bases due to the amino group in it, the reaction with inorganic acid or carboxylic acid gives salt as product. The salt formed is an amine salt. Proton is donated from the acid to the nitrogen atom which acts as a proton acceptor. In simple words, it can be said that in an amine‑acid reaction, the acid loses a hydrogen ion and amine gains a hydrogen ion.
When a strong base is added to the amine salt, the parent amine can be obtained. This is a reverse reaction of the amine salt formation reaction. These reactions can be represented as shown below,
Quaternary ammonium salt does not give the “parent” amine when treated with a strong base as there is no possibility of deprotonation to take place.
(d)
Interpretation:
“Parent” amine can be regenerated from the given salt or not has to be indicated.
Concept Introduction:
Quaternary ammonium salt is the one that has four carbon atoms attached to the nitrogen atom. This is formed by the reaction of tertiary amine with alkyl halide in presence of a strong base.
Neutralization reaction is the one that takes place between an acid and a base to give salt as product. As amines are bases due to the amino group in it, the reaction with inorganic acid or carboxylic acid gives salt as product. The salt formed is an amine salt. Proton is donated from the acid to the nitrogen atom which acts as a proton acceptor. In simple words, it can be said that in an amine‑acid reaction, the acid loses a hydrogen ion and amine gains a hydrogen ion.
When a strong base is added to the amine salt, the parent amine can be obtained. This is a reverse reaction of the amine salt formation reaction. These reactions can be represented as shown below,
Quaternary ammonium salt does not give the “parent” amine when treated with a strong base as there is no possibility of deprotonation to take place.
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Chapter 17 Solutions
General, Organic, and Biological Chemistry
- 1. For the four structures provided, Please answer the following questions in the table below. a. Please draw π molecular orbital diagram (use the polygon-and-circle method if appropriate) and fill electrons in each molecular orbital b. Please indicate the number of π electrons c. Please indicate if each molecule provided is anti-aromatic, aromatic, or non- aromatic TT MO diagram Number of π e- Aromaticity Evaluation (X choose one) Non-aromatic Aromatic Anti-aromatic || ||| + IVarrow_forward1.3 grams of pottasium iodide is placed in 100 mL of o.11 mol/L lead nitrate solution. At room temperature, lead iodide has a Ksp of 4.4x10^-9. How many moles of precipitate will form?arrow_forwardQ3: Circle the molecules that are optically active: ДДДДarrow_forward
- 6. How many peaks would be observed for each of the circled protons in the compounds below? 8 pts CH3 CH3 ΤΙ A. H3C-C-C-CH3 I (₁₁ +1)= 7 H CI B. H3C-C-CI H (3+1)=4 H LIH)=2 C. (CH3CH2-C-OH H D. CH3arrow_forwardNonearrow_forwardQ1: Draw the most stable and the least stable Newman projections about the C2-C3 bond for each of the following isomers (A-C). Are the barriers to rotation identical for enantiomers A and B? How about the diastereomers (A versus C or B versus C)? H Br H Br (S) CH3 (R) CH3 H3C (S) H3C H Br Br H A C enantiomers H Br H Br (R) CH3 H3C (R) (S) CH3 H3C H Br Br H B D identicalarrow_forward
- 2. Histamine (below structure) is a signal molecule involved in immune response and is a neurotransmitter. Histamine features imidazole ring which is an aromatic heterocycle. Please answer the following questions regarding Histamine. b a HN =N C NH2 a. Determine hybridization of each N atom (s, p, sp, sp², sp³, etc.) in histamine N-a hybridization: N-b hybridization: N-c hybridization: b. Determine what atomic orbitals (s, p, sp, sp², sp³, etc.) of the lone pair of each N atom resided in N-a hybridization: N-b hybridization: N-c hybridization:arrow_forwardNonearrow_forward29. Use frontier orbital analysis (HOMO-LUMO interactions) to decide whether the following dimerization is 1) thermally allowed or forbidden and 2) photochemically allowed or forbidden. +arrow_forward
- 30.0 mL of 0.10 mol/L iron sulfate and 20.0 mL of 0.05 mol/L of silver nitrate solutions are mixed together. Justify if any precipitate would formarrow_forwardDoes the carbonyl group first react with the ethylene glycol, in an intermolecular reaction, or with the end alcohol, in an intramolecular reaction, to form a hemiacetal? Why does it react with the alcohol it does first rather than the other one? Please do not use an AI answer.arrow_forwardThe number of noncyclic isomers that have the composition C4H8Owith the O as part of an OH group, counting a pair of stereoisomers as1, is A. 8; B. 6; C. 9; D. 5; E. None of the other answers is correct.arrow_forward
- General, Organic, and Biological ChemistryChemistryISBN:9781285853918Author:H. Stephen StokerPublisher:Cengage LearningOrganic And Biological ChemistryChemistryISBN:9781305081079Author:STOKER, H. Stephen (howard Stephen)Publisher:Cengage Learning,Introduction to General, Organic and BiochemistryChemistryISBN:9781285869759Author:Frederick A. Bettelheim, William H. Brown, Mary K. Campbell, Shawn O. Farrell, Omar TorresPublisher:Cengage Learning
- Chemistry for Today: General, Organic, and Bioche...ChemistryISBN:9781305960060Author:Spencer L. Seager, Michael R. Slabaugh, Maren S. HansenPublisher:Cengage Learning
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