(a)
Interpretation:
The product of stereoisomers that formed in the given reaction should be given.
Concept introduction:
Isomer:
The compound is having same molecular formula but arrangement of the atom is different is called isomer.
In the nucleophilic substitution reaction, the
In
The rate determination step is formation of carbocation.
The stability order of carbocation is,
Tertiary > Secondary > Primary
Therefore, tertiary alcohols undergo substitution very fast than the secondary alcohols because tertiary carbocation is more stable than the secondary carbocation than the primary carbocation. Primary alcohol is less stable therefore it won’t undergoes
In the nucleophilic substitution reaction, the rate of reaction depends on reactant as well as nucleophile, which are involved in reaction is called bimolecular nucleophilic substitution reaction.
In
Reactant and nucleophile are present at the rate determination step.
The order of species involving in
Tertiary < Secondary < Primary
(b)
Interpretation:
The product of stereoisomers that formed in the given reaction should be given.
Concept introduction:
Isomer:
The compound is having same molecular formula but arrangement of the atom is different is called isomer.
In the nucleophilic substitution reaction, the rate of reaction depends only on one reactant, which is involved in reaction is called unimolecular nucleophilic substitution reaction.
In
The rate determination step is formation of carbocation.
The stability order of carbocation is,
Tertiary > Secondary > Primary
Therefore, tertiary alcohols undergo substitution very fast than the secondary alcohols because tertiary carbocation is more stable than the secondary carbocation than the primary carbocation. Primary alcohol is less stable therefore it won’t undergoes
In the nucleophilic substitution reaction, the rate of reaction depends on reactant as well as nucleophile, which are involved in reaction is called bimolecular nucleophilic substitution reaction.
In
Reactant and nucleophile are present at the rate determination step.
The order of species involving in
Tertiary < Secondary < Primary
(c)
Interpretation:
The product of stereoisomers that formed in the given reaction should be given.
Concept introduction:
Isomer:
The compound is having same molecular formula but arrangement of the atom is different is called isomer.
In the nucleophilic substitution reaction, the rate of reaction depends only on one reactant, which is involved in reaction is called unimolecular nucleophilic substitution reaction.
In
The rate determination step is formation of carbocation.
The stability order of carbocation is,
Tertiary > Secondary > Primary
Therefore, tertiary alcohols undergo substitution very fast than the secondary alcohols because tertiary carbocation is more stable than the secondary carbocation than the primary carbocation. Primary alcohol is less stable therefore it won’t undergoes
In the nucleophilic substitution reaction, the rate of reaction depends on reactant as well as nucleophile, which are involved in reaction is called bimolecular nucleophilic substitution reaction.
In
Reactant and nucleophile are present at the rate determination step.
The order of species involving in
Tertiary < Secondary < Primary
(d)
Interpretation:
The product of stereoisomers that formed in the given reaction should be given.
Concept introduction:
Isomer:
The compound is having same molecular formula but arrangement of the atom is different is called isomer.
In the nucleophilic substitution reaction, the rate of reaction depends only on one reactant, which is involved in reaction is called unimolecular nucleophilic substitution reaction.
In
The rate determination step is formation of carbocation.
The stability order of carbocation is,
Tertiary > Secondary > Primary
Therefore, tertiary alcohols undergo substitution very fast than the secondary alcohols because tertiary carbocation is more stable than the secondary carbocation than the primary carbocation. Primary alcohol is less stable therefore it won’t undergoes
In the nucleophilic substitution reaction, the rate of reaction depends on reactant as well as nucleophile, which are involved in reaction is called bimolecular nucleophilic substitution reaction.
In
Reactant and nucleophile are present at the rate determination step.
The order of species involving in
Tertiary < Secondary < Primary
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Chapter 11 Solutions
Organic Chemistry
- What are the major products of the following reaction? Draw all the major products. If there are no major products, then there is no reaction that will take place. Use wedge and dash bonds when necessary.arrow_forwardZeolites. State their composition and structure. Give an example.arrow_forwardDon't used hand raiting and show all reactionsarrow_forward
- Don't used hand raiting and don't used Ai solutionarrow_forwardIX) By writing the appropriate electron configurations and orbital box diagrams briefly EXPLAIN in your own words each one of the following questions: a) The bond length of the Br2 molecule is 2.28 Å, while the bond length of the compound KBr is 3.34 Å. The radius of K✶ is 1.52 Å. Determine the atomic radius in Å of the bromine atom and of the bromide ion. Br = Br b) Explain why there is a large difference in the atomic sizes or radius of the two (Br and Br). Tarrow_forwardWhen 15.00 mL of 3.00 M NaOH was mixed in a calorimeter with 12.80 mL of 3.00 M HCl, both initially at room temperature (22.00 C), the temperature increased to 29.30 C. The resultant salt solution had a mass of 27.80 g and a specific heat capacity of 3.74 J/Kg. What is heat capacity of the calorimeter (in J/C)? Note: The molar enthalpy of neutralization per mole of HCl is -55.84 kJ/mol.arrow_forward
- When 15.00 mL of 3.00 M NaOH was mixed in a calorimeter with 12.80 mL of 3.00 M HCl, both initially at room temperature (22.00 C), the temperature increased to 29.30 C. The resultant salt solution had a mass of 27.80 g and a specific heat capacity of 3.74 J/Kg. What is heat capacity of the calorimeter (in J/C)? Note: The molar enthalpy of neutralization per mole of HCl is -55.84 kJ/mol. Which experimental number must be initialled by the Lab TA for the first run of Part 1 of the experiment? a) the heat capacity of the calorimeter b) Mass of sample c) Ti d) The molarity of the HCl e) Tfarrow_forwardPredict products for the Following organic rxn/s by writing the structurels of the correct products. Write above the line provided" your answer D2 ①CH3(CH2) 5 CH3 + D₂ (adequate)" + 2 mited) 19 Spark Spark por every item. 4 CH 3 11 3 CH 3 (CH2) 4 C-H + CH3OH CH2 CH3 + CH3 CH2OH 0 CH3 fou + KMnDy→ C43 + 2 KMn Dy→→ C-OH ") 0 C-OH 1110 (4.) 9+3 =C CH3 + HNO 3 0 + Heat> + CH3 C-OH + Heat CH2CH3 - 3 2 + D Heat H 3 CH 3 CH₂ CH₂ C = CH + 2 H₂ → 2 2arrow_forwardWhen 15.00 mL of 3.00 M NaOH was mixed in a calorimeter with 12.80 mL of 3.00 M HCl, both initially at room temperature (22.00 C), the temperature increased to 29.30 C. The resultant salt solution had a mass of 27.80 g and a specific heat capacity of 3.74 J/Kg. What is heat capacity of the calorimeter (in J/C)? Note: The molar enthalpy of neutralization per mole of HCl is -55.84 kJ/mol.arrow_forward
- Q6: Using acetic acid as the acid, write the balanced chemical equation for the protonation of the two bases shown (on the -NH2). Include curved arrows to show the mechanism. O₂N- O₂N. -NH2 -NH2 a) Which of the two Bronsted bases above is the stronger base? Why? b) Identify the conjugate acids and conjugate bases for the reactants. c) Identify the Lewis acids and bases in the reactions.arrow_forwardQ5: For the two reactions below: a) Use curved electron-pushing arrows to show the mechanism for the reaction in the forward direction. Redraw the compounds to explicitly illustrate all bonds that are broken and all bonds that are formed. b) Label Bronsted acids and bases in the left side of the reactions. c) For reaction A, which anionic species is the weakest base? Which neutral compound is the stronger acid? Is the forward or reverse reaction favored? d) Label Lewis acids and bases, nucleophiles and electrophiles in the left side of the reactions. A. 용 CH3OH я хон CH3O OH B. HBr CH3ONa NaBr CH3OHarrow_forwardpotential energy Br b) Translate the Newman projection below to its wedge-and-dash drawing. F H. OH CH3 CI c) Isopentane (2-methylbutane) is a compound containing a branched carbon chain. Draw a Newman projection of six conformations about the C2-C3 bond of isopentane. On the curve of potential energy versus angle of internal rotation for isopentane, label each energy maximum and minimum with one of the conformations. 0° 。 F A B D C angle of internal rotation E F 360° (=0°) JDownlarrow_forward
- Organic ChemistryChemistryISBN:9781305580350Author:William H. Brown, Brent L. Iverson, Eric Anslyn, Christopher S. FootePublisher:Cengage Learning