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(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 6 Solutions
EBK ORGANIC AND BIOLOGICAL CHEMISTRY
- Denote the dipole for the indicated bonds in the following molecules. ✓ H3C CH3 B F-CCl3 Br-Cl H3C —Si(CH3)3 CH3 OH HO HO H HO OH vitamin Carrow_forward(a) What is the hybridization of the carbon in the methyl cation (CH3*) and in the methyl anion (CH3)? (b) What is the approximate H-C-H bond angle in the methyl cation and in the methyl anion?arrow_forward10:16 ☑ Vo)) Vo) 4G LTE 76% Complete the following reaction by identifying the principle organic product of the reaction. HO OH ↑ CH2N2 OH ? ○ A. 01 N₂H2C OH ОН B. HO OCH3 OH ○ C. HO OH ŎCH₂N2 ○ D. H3CO OH он Quiz navigation 1 2 3 4 5 11 12 Next page 10 6 7 8 9 10arrow_forward
- Which one of the following statements explain why protecting groups are referred to as “a necessary evil in organic synthesis”? Question 12Select one or more: A. They increase the length and cost of the synthesis B. Every synthesis employs protecting groups C. Protecting group have no role to play in a synthesis D. They minimize the formation of side productsarrow_forwardWhich of the following attributes is a key advantage of the chiral auxiliary approach over the chiral pool approach in asymmetric synthesis? Question 10Select one: A. Chiral auxiliaries are cheaper than chiral pool substrates B. Chiral auxiliary can be recovered and recycled unlike chiral pool substrates. C. The use of chiral auxiliaries provide enantiopure products, while chiral pool reactions are only enantioselective D. The chiral auxiliaries are naturally occurring and do not require synthesisarrow_forwardIn the following molecule, indicate the hybridization and shape of the indicated atoms. CH3 CH3 H3C HO: CI:arrow_forward
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- Draw the major organic product of the Bronsted acid-base reaction. Include all lone pairs and charges as appropriate. Ignore any counterions. :0: NaOH Harrow_forward5. Calculate the total amount of heat transferred as 50 g of wat Specific heat H₂O (g) 2.00 J/g°C -10 °C. Specific heat H₂O (1) Specific heat H₂O (s) 4.18 J/g°C 2.11 J/g°C Heat of vaporization 2260 J/g Heat of fusion 334 J/g Melting point 0°C 6. Calculate the total amount of heat transferred as 25 g of water is heated from 50 °C to 100 °C as a gas. Boiling point 100 °Carrow_forwardCalculate the total amount of heat transferred as 50 g of Water -10°C. Calculate the total amount of heat transferred as 25 g of water is heated from 50°C to 100°C as a gas. \table[[Specific heat H₂O(g), 2.00°C Η 2 g 5. Calculate the total amount of heat transferred as 50 g of wat Specific heat H₂O (g) 2.00 J/g°C -10 °C. 4.18 J/g°C 2.11 J/g°C 2260 J/g 334 J/g Specific heat H₂O (1) Specific heat H₂O (s) Heat of vaporization Heat of fusion Melting point 6. Calculate the total amount of heat transferred as 25 g of water is heated from 50 °C to 100 °C as a gas. Boiling point 100 °C 0°Carrow_forward
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