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Jun 4, 2024

Key Concept

Molecular Orbital Theory

Explanation

The molecular orbital shown in the black box is formed by the combination of atomic orbitals. The correct pair of atomic orbitals that would lead to this molecular orbital is represented by answer choice A, which shows the appropriate combination of atomic orbitals.

The complex, [RhCl(H)2(PPh 3)2], is suggested as an intermediate in the catalytic hydrogenation of alkenes. Under the Effective Atomic Number Rule, how many electrons is it considered to have? A. 10 B. 12 C. 14 D. 16 E. 18

The azide ligand, N3 ̄, is a common uninegative, monodentate ligand in transition metal chemistry. One such complex is [W(N3)7] ̄. What is the valence electron count of tungsten (W) in [W(N3) 7] ̄? A. 7 valence electrons B. 3 valence electrons C. 2 valence electrons D. 1 valence electrons E. 0 valence electrons

Key Concept

Valence electron count in transition metal complexes

Explanation

The valence electron count of tungsten (W) in the complex [W(N3)7]⁻ is determined by considering the oxidation state of tungsten and the number of electrons contributed by the azide ligands. Tungsten in this complex is in the +7 oxidation state, and each azide ligand (N₃⁻) is uninegative, contributing one electron. Therefore, tungsten has 7 valence electrons.

Key Concept

Valence electron count in transition metal complexes

Explanation

The valence electron count of tungsten in the complex

[W(N_3)_7]^{-}

is determined by considering the oxidation state of tungsten and the electron contribution from the azide ligands. Tungsten in this complex has an oxidation state of +7, and since each azide ligand is uninegative, the total electron count is 0.

What is the electronic configuration of tungsten (W) in the complex

[W(N_{3})_{7}]

Solution

Determine the oxidation state of tungsten (W): In the complex

[W(N_{3})_{7}]

, each azide ligand (

N_{3}^{-}

) has a charge of -1. Since there are 7 azide ligands, the total charge contributed by the ligands is

7 \times (-1) = -7

. Assuming the complex is neutral, the oxidation state of tungsten must be +7 to balance the -7 charge from the ligands

Identify the electronic configuration of neutral tungsten (W): The atomic number of tungsten (W) is 74. The electronic configuration of a neutral tungsten atom is

[Xe] 4f^{14} 5d^{4} 6s^{2}

Adjust the electronic configuration for the +7 oxidation state: When tungsten is in the +7 oxidation state, it loses 7 electrons. The electrons are removed first from the 6s orbital and then from the 5d orbital. Therefore, the electronic configuration of

W^{7+}

[Xe] 4f^{14} 5d^{0} 6s^{0}

Answer

The electronic configuration of tungsten (W) in the complex

[W(N_{3})_{7}]

[Xe] 4f^{14} 5d^{0} 6s^{0}

Key Concept

The electronic configuration of an element in a complex depends on its oxidation state and the removal of electrons from the outermost orbitals.

Explanation

In the complex

[W(N_{3})_{7}]

, tungsten is in the +7 oxidation state, which means it loses 7 electrons from its neutral state configuration. This results in the electronic configuration

[Xe] 4f^{14} 5d^{0} 6s^{0}

Key Concept

Effective Atomic Number (EAN) Rule

Explanation

The Effective Atomic Number Rule states that a stable complex will have an effective atomic number equal to the nearest noble gas. For the complex

[RhCl(H)_2(PPh_3)_2]

, Rhodium (Rh) has 9 valence electrons, each hydrogen ligand contributes 1 electron, each PPh₃ ligand contributes 2 electrons, and the chloride ligand contributes 2 electrons. Thus, the total electron count is

9 + 1 + 1 + 2 + 2 + 2 = 18

electrons.

Key Concept

Valence Electron Count

Explanation

The valence electron count of tungsten (W) in the complex ion

[\mathrm{W}(\mathrm{N}_3)_7]^{-}

is determined by considering the oxidation state of tungsten and the electron contribution from the azide ligands. Tungsten in this complex is in the +7 oxidation state, and each azide ligand contributes one electron. Therefore, the valence electron count for tungsten is 1.

Key Concept

Optical Isomerism in Complexes

Explanation

Optical isomers are non-superimposable mirror images. The cis-configuration of

[Ni(NH_3)_2(en)_2]^{2+}

allows for such isomerism due to the spatial arrangement of the ligands.

Key Concept

Spin-only magnetic moment

Explanation

For a high spin

Ni^{2+}

compound, the electron configuration is

[Ar] 3d^8

. The number of unpaired electrons is 2, and the spin-only magnetic moment is calculated using the formula

\mu = \sqrt{n(n+2)}

Bohr Magnetons, where

n

is the number of unpaired electrons. For

n=2

\mu = \sqrt{2(2+2)} = \sqrt{8} \approx 2.83

Bohr Magnetons.

Key Concept

Equilibrium Constant Calculation

Explanation

The equilibrium constant for the reaction

[\mathrm{Cu}(\mathrm{H}_{2}\mathrm{O})_{4}]^{2+} + 2\mathrm{NH}_{3} \leftrightarrow [\mathrm{Cu}(\mathrm{H}_{2}\mathrm{O})_{2}(\mathrm{NH}_{3})_{2}]^{2+} + 2\mathrm{H}_{2}\mathrm{O}

is calculated by multiplying the equilibrium constants

K_1

and

K_2

for the stepwise reactions. Thus,

K = K_1 \times K_2 = (1.9 \times 10^4) \times (3.9 \times 10^3) = 7.4 \times 10^7

Key Concept

Base Strength in E2 Reactions

Explanation

In E2 reactions, the base strength is crucial. Stronger bases are more effective in abstracting protons, leading to the formation of the double bond. The ranking from best to worst base for performing an E2 reaction is based on their basicity.

Key Concept

Substitution Reaction Product

Explanation

In a substitution reaction, one functional group in a chemical compound is replaced by another. The correct product is determined by the specific reactants and conditions given in the question.

Key Concept

Substitution Reaction Mechanism

Explanation

In the depicted substitution reaction, the correct product is determined by the movement of electrons as indicated by the curved arrows. Species D represents the correct arrangement of atoms and bonds resulting from this mechanism.

Key Concept

Enol Formation

Explanation

Enol formation occurs when a compound has an alpha hydrogen adjacent to a carbonyl group, allowing for keto-enol tautomerism. Compounds 3, 4, and 5 have this structural feature.

Key Concept

Curly arrows in reaction mechanisms

Explanation

Curly arrows are used to show the movement of electron pairs during chemical reactions. In the decarboxylation of an alpha-keto-acid, the correct set of arrows will depict the loss of

CO_2

and the formation of the enol.

Key Concept

Activating and deactivating groups in electrophilic aromatic substitution

Explanation

The OH group (compound 1) is an activating group that directs electrophilic substitution to the ortho and para positions. The (C=O)NH group (compound 4) is also an activating group that directs substitution to the ortho and para positions. Therefore, compounds 1 and 4 will give the para-substituted product as the major product.

Which compound would give the para-substituted product upon reaction with

Br_2/FeBr_3

in an electrophilic aromatic substitution reaction

Solution

Identify the reaction: The reaction in question is an electrophilic aromatic substitution (EAS) with

Br_2/FeBr_3

. This reaction typically involves the substitution of a hydrogen atom on an aromatic ring with a bromine atom

Determine the directing effects: In EAS reactions, substituents on the aromatic ring can influence the position where the new substituent (bromine in this case) will be added. Substituents can be either ortho/para-directing or meta-directing. Ortho/para-directing groups are typically electron-donating groups (EDGs), while meta-directing groups are usually electron-withdrawing groups (EWGs)

Identify common substituents: Common ortho/para-directing groups include -OH, -OCH₃, -NH₂, -CH₃, etc. Common meta-directing groups include -NO₂, -CN, -COOH, etc

Select the compound: To determine which compound would give the para-substituted product, we need to identify a compound with an ortho/para-directing group. For example, toluene (methylbenzene) has a -CH₃ group, which is an ortho/para-directing group

Answer

Toluene (methylbenzene)

Key Concept

Ortho/para-directing groups in electrophilic aromatic substitution reactions

Explanation

Ortho/para-directing groups, such as -CH₃ in toluene, direct the incoming electrophile (bromine) to the ortho and para positions relative to themselves. Therefore, toluene will give the para-substituted product upon reaction with

Br_2/FeBr_3

Key Concept

Acidity of organic compounds

Explanation

The acidity of organic compounds is influenced by the presence of electron-donating or electron-withdrawing groups. Methoxy groups (OCH3) are electron-donating, which decreases acidity, while the length of the carbon chain can also affect acidity.

Key Concept

Reactivity in Electrophilic Aromatic Substitution

Explanation

Electron-withdrawing groups like

NO_2

decrease reactivity, while electron-donating groups like

OH

and

CH_3

increase reactivity. The order from least to most reactive is NO_2 < Cl < H < CH_3 < OH.

what is the relationship between pKa and pH

Solution

Identify the relationship: The pKa of a substance is a measure of the strength of an acid, specifically the negative logarithm of its acid dissociation constant (

K_a

). The pH of a solution is a measure of its hydrogen ion concentration, specifically the negative logarithm of the hydrogen ion concentration (

[H^+]

)

Mathematical expressions: The relationship between pKa and pH can be described by the Henderson-Hasselbalch equation:

\text{pH} = \text{pKa} + \log \left( \frac{[\text{A}^-]}{[\text{HA}]} \right)

where

[\text{A}^-]

is the concentration of the conjugate base and

[\text{HA}]

is the concentration of the acid

Application: This equation shows that when the concentrations of the acid and its conjugate base are equal, the pH of the solution is equal to the pKa of the acid. This is useful for buffer solutions, where the pH can be controlled by adjusting the ratio of the acid and its conjugate base

Answer

The relationship between pKa and pH is described by the Henderson-Hasselbalch equation, which shows how the pH of a solution is related to the pKa of the acid and the ratio of the concentrations of the conjugate base and the acid.

Key Concept

The Henderson-Hasselbalch equation relates pH, pKa, and the ratio of the concentrations of the conjugate base and the acid.

Explanation

The pH of a solution is determined by the concentration of hydrogen ions, while the pKa is a measure of the acid strength. The Henderson-Hasselbalch equation provides a way to calculate the pH of a solution based on the pKa and the ratio of the concentrations of the acid and its conjugate base.

pKa and acidity

Solution

Define pKa: The pKa of a compound is the negative logarithm of its acid dissociation constant (Ka). It is a measure of the strength of an acid in solution. The formula is

\text{pKa} = -\log(\text{Ka})

Relationship between pKa and acidity: The lower the pKa value, the stronger the acid. This is because a lower pKa indicates a higher Ka, meaning the acid dissociates more completely in solution, releasing more hydrogen ions (H+)

Compare pKa values: To determine the relative acidity of different compounds, compare their pKa values. The compound with the lower pKa is the stronger acid

Answer

The pKa value is inversely related to the strength of an acid; a lower pKa indicates a stronger acid.

Key Concept

pKa and acidity relationship

Explanation

The pKa value helps determine the strength of an acid; a lower pKa means a stronger acid due to higher dissociation in solution.

Key Concept

Electrophilic Aromatic Substitution (EAS)

Explanation

The major product in this reaction is determined by the directing effects of substituents on the aromatic ring. In this case, the substituents direct the incoming electrophile to the position that leads to product C.

The lower pKa, the higher acidity?

Solution

Identify the relationship: The pKa value is the negative logarithm of the acid dissociation constant (

K_a

). This means that

pK_a = -\log(K_a)

Understand the inverse relationship: A lower pKa value indicates a higher

K_a

, which means the acid dissociates more in solution, making it a stronger acid

Conclusion: Therefore, the lower the pKa value, the higher the acidity of the substance

Answer

Yes, the lower the pKa, the higher the acidity.

Key Concept

pKa and acidity relationship

Explanation

The pKa value is inversely related to the strength of an acid. A lower pKa value means a higher acid dissociation constant (

K_a

), indicating a stronger acid.

Key Concept

Infrared Spectroscopy

Explanation

The graph described is an infrared (IR) spectrum, which measures the transmittance of infrared radiation through a sample. The peaks correspond to the vibrational frequencies of different bonds within the molecules.

Sure, let's analyze the given spectrum and identify the functional groups present based on the regions highlighted in the graph.

Solution

Region A (3100 to 2800 cm⁻¹): This region typically corresponds to C-H stretching vibrations. The presence of a cluster of peaks suggests the presence of alkanes (sp³ C-H stretch) or alkenes (sp² C-H stretch)

Region B (2800 to 2500 cm⁻¹): A single broad peak in this region is characteristic of O-H stretching vibrations, indicating the presence of alcohols or carboxylic acids

Region C (2000 to 1900 cm⁻¹): The absence of significant peaks in this region suggests that there are no triple bonds (C≡C or C≡N) present in the molecule

Region D (1800 to 500 cm⁻¹): This region is known as the fingerprint region, which contains a complex pattern of peaks and valleys. It is unique to each molecule and can be used to identify specific functional groups. Peaks in this region can correspond to C=O stretching (around 1700 cm⁻¹), C=C stretching (around 1600 cm⁻¹), and various bending vibrations

Answer

The spectrum suggests the presence of alkanes or alkenes (Region A), alcohols or carboxylic acids (Region B), and various functional groups in the fingerprint region (Region D). There are no triple bonds present (Region C).

Key Concept

Infrared (IR) spectroscopy is used to identify functional groups in a molecule based on the absorption of infrared light at specific wavenumbers.

Explanation

Different functional groups absorb infrared light at characteristic wavenumbers, resulting in peaks in the IR spectrum. By analyzing these peaks, we can determine the functional groups present in the molecule.

Name the four key types of transition represented by arrows A, B (multiple instances), C and D in the diagram, and explain the corresponding absorption or emission

Solution

Identify the transition A: The arrow labeled "A" represents the transition from the ground electronic state

S_0

to the first excited electronic state

S_1

. This is an absorption process where the molecule absorbs a photon and moves to a higher energy state

Identify the transition B: The wavy arrows labeled "B" represent non-radiative transitions between vibrational levels within the same electronic state. These are internal conversions or vibrational relaxations where the molecule loses energy without emitting a photon

Identify the transition C: The straight right arrow labeled "C" represents intersystem crossing from the singlet excited state

S_1

to the triplet state

T_1

. This is a non-radiative process where the molecule changes its spin state

Identify the transition D: The straight down arrow labeled "D" represents the transition from the triplet state

T_1

to the ground state

S_0

. This is a phosphorescence process where the molecule emits a photon and returns to the ground state

Answer

The four key types of transitions are: - A: Absorption (S0 to S1) - B: Internal Conversion/Vibrational Relaxation (within S1 or T1) - C: Intersystem Crossing (S1 to T1) - D: Phosphorescence (T1 to S0)

Key Concept

Types of electronic and vibrational transitions in a Jablonski diagram

Explanation

The Jablonski diagram illustrates various transitions such as absorption, internal conversion, intersystem crossing, and phosphorescence, which are crucial for understanding molecular photophysics.

Summarise three key differences between fluorescence and phosphorescence

Solution

Identify the nature of fluorescence and phosphorescence: Fluorescence and phosphorescence are both types of photoluminescence, where a substance absorbs photons and then re-emits them. However, they differ in their electronic transitions and lifetimes

Fluorescence: In fluorescence, the absorbed energy excites an electron to a higher singlet state (e.g., from

S_0

S_1

). The electron then returns to the ground state (

S_0

) by emitting a photon. This process occurs very quickly, typically within nanoseconds

Phosphorescence: In phosphorescence, the absorbed energy excites an electron to a higher singlet state (e.g., from

S_0

S_1

), but then the electron undergoes intersystem crossing to a triplet state (

T_1

). The electron then returns to the ground state (

S_0

) by emitting a photon. This process is slower, occurring over milliseconds to minutes

Key Differences: - Electronic States: Fluorescence involves singlet-singlet transitions, while phosphorescence involves intersystem crossing to a triplet state. - Lifetimes: Fluorescence has a short lifetime (nanoseconds), whereas phosphorescence has a longer lifetime (milliseconds to minutes). - Energy Emission: Fluorescence typically emits higher energy (shorter wavelength) photons compared to phosphorescence, which emits lower energy (longer wavelength) photons

Answer

Fluorescence and phosphorescence differ in their electronic states, lifetimes, and energy emission.

Key Concept

Fluorescence and phosphorescence are types of photoluminescence with distinct characteristics.

Explanation

Fluorescence involves fast singlet-singlet transitions, while phosphorescence involves slower triplet-singlet transitions, leading to differences in their lifetimes and emitted photon energies.

Using structures, A, B, and C provided above, rationalise any differences in molecular stability with specific reference to how this relates to electron distribution

Solution

Identify the structures: The three compounds A, B, and C are benzene rings with carbon chains attached. The carbon chains have the same number of carbon atoms but differ in the positions of the double bonds

Analyze electron distribution: The stability of these compounds can be rationalized by examining the electron distribution in the carbon chains and how it affects the benzene ring. Double bonds can either be conjugated with the benzene ring or isolated. Conjugation generally increases stability due to delocalization of electrons

Compare stability: - Compound A: If the double bond is conjugated with the benzene ring, it will have increased stability due to electron delocalization. - Compound B: If the double bond is isolated (not conjugated with the benzene ring), it will be less stable compared to A. - Compound C: If the double bond is in a position that allows partial conjugation, it will have intermediate stability between A and B

Answer

Compound A is the most stable due to full conjugation, Compound C has intermediate stability, and Compound B is the least stable due to isolated double bonds.

Key Concept

Conjugation and electron delocalization increase molecular stability.

Explanation

Conjugation allows for the delocalization of electrons, which stabilizes the molecule by lowering its overall energy.

Solution

Identify the structures: Structure A has a benzene ring attached to a carbon chain with one double bond. Structure B has a benzene ring attached to a carbon chain with two conjugated double bonds. Structure C has a benzene ring attached to a carbon chain with three conjugated double bonds

Analyze electron distribution: Conjugation in structures B and C allows for delocalization of π-electrons across the double bonds, increasing stability. Structure A, with only one double bond, has less electron delocalization and thus lower stability

Relate to molecular stability: The stability of these molecules increases with the extent of conjugation. Structure C, with the most conjugated double bonds, is the most stable, followed by structure B, and then structure A. This is due to the increased delocalization of electrons, which lowers the overall energy of the molecule

Answer

Structure C is the most stable due to the highest degree of conjugation, followed by structure B, and then structure A.

Key Concept

Conjugation and electron delocalization increase molecular stability.

Explanation

Conjugation allows for the delocalization of π-electrons, which lowers the energy and increases the stability of the molecule. The more extensive the conjugation, the greater the stability.

Propose and explain what differences or similarities you would observe for the IR spectra of compounds B and C

Solution

Identify the functional groups: Both compounds B and C have benzene rings and carbon chains with multiple double bonds. Compound B has three double bonds separated by single bonds, while compound C has two adjacent double bonds and one separated by a single bond

Analyze the IR spectra: The IR spectra of compounds B and C will show characteristic absorption bands for the functional groups present. The benzene ring will show C-H stretching vibrations around 3100 cm

^{-1}

and C=C stretching vibrations around 1600 cm

^{-1}

. The carbon chains with double bonds will show C=C stretching vibrations around 1650 cm

^{-1}

Compare the spectra: The key difference between the IR spectra of compounds B and C will be due to the different arrangements of double bonds in the carbon chains. Compound B, with isolated double bonds, will show distinct C=C stretching bands. Compound C, with conjugated double bonds, will show a shift in the C=C stretching bands to lower wavenumbers due to conjugation effects

Answer

The IR spectra of compounds B and C will show differences in the C=C stretching bands due to the different arrangements of double bonds in the carbon chains. Compound B will have distinct C=C stretching bands, while compound C will show a shift to lower wavenumbers due to conjugation.

Key Concept

The arrangement of double bonds affects the IR spectra, particularly the C=C stretching bands.

Explanation

Compound B has isolated double bonds, leading to distinct C=C stretching bands, while compound C has conjugated double bonds, causing a shift to lower wavenumbers.

Explain the relative magnitudes of KX for the acids, HF, HCl, HBr, HI.

Solution

Identify the reaction: The given reaction is:

\left[\left(\mathrm{CH}_{3}\right) \mathrm{Hg}\left(\mathrm{H}_{2} \mathrm{O}\right)\right]^{+} + \mathrm{HX} \Leftrightarrow \left(\mathrm{CH}_{3}\right) \mathrm{HgX} + \mathrm{H}_{3} \mathrm{O}^{+}

where

\mathrm{X}^{-}

is varied

Analyze the equilibrium constants ($K_x$): The equilibrium constants for the reaction with different anions are given as:

\begin{array}{ll} \mathrm{X} &amp; K_{x} \\ \hline \mathrm{F} &amp; 4.5 \times 10^{-2} \\ \mathrm{Cl} &amp; 1.8 \times 10^{12} \\ \mathrm{Br} &amp; 4.2 \times 10^{15} \\ \mathrm{I} &amp; 1 \times 10^{18} \\ \mathrm{H}_{2} \mathrm{O} &amp; 5 \times 10^{-7} \end{array}

Determine the relative strengths of the acids: The magnitude of

K_x

indicates the position of equilibrium. A larger

K_x

value means the equilibrium lies more to the right, indicating a stronger acid. Therefore, the relative strengths of the acids are: \mathrm{HI} > \mathrm{HBr} > \mathrm{HCl} > \mathrm{HF}

Answer

The relative magnitudes of

K_x

for the acids are: \mathrm{HI} > \mathrm{HBr} > \mathrm{HCl} > \mathrm{HF}

Key Concept

Equilibrium constant (

K_x

) and acid strength

Explanation

The equilibrium constant

K_x

for the reaction indicates the strength of the acid

\mathrm{HX}

. A larger

K_x

value means the acid is stronger, as the equilibrium lies more to the right, favoring the formation of

\left(\mathrm{CH}_{3}\right) \mathrm{HgX}

and

\mathrm{H}_{3} \mathrm{O}^{+}

Rate H2O as a hard or soft base, relative to the halides.

Solution

Identify the concept: The concept of hard and soft acids and bases (HSAB) is used to predict the stability of compounds and the direction of chemical reactions

Classify H2O: Water (

H_2O

) is generally considered a hard base. This classification is based on its small size, high charge density, and the presence of a highly electronegative oxygen atom

Compare with halides: Halides (such as

F^-

Cl^-

Br^-

, and

I^-

) vary in their hardness. Fluoride (

F^-

) is a hard base, while iodide (

I^-

) is a soft base. Chloride (

Cl^-

) and bromide (

Br^-

) fall in between, with chloride being closer to hard and bromide being closer to soft

Answer

H2O is classified as a hard base relative to halides.

Key Concept

Hard and soft acids and bases (HSAB) theory

Explanation

H2O is a hard base due to its small size, high charge density, and the presence of a highly electronegative oxygen atom. In comparison, halides vary in hardness, with fluoride being hard and iodide being soft.

Solution

Identify the compound and bond distances: The compound in question is

[\mathrm{Mn}(\mathrm{hfacac})_3]

, where

\mathrm{hfacac}^-

stands for hexafluoroacetylacetonate anion (

\mathrm{C}_5\mathrm{HF}_6\mathrm{O}_2^-

). The Mn-O bond distances are given as 2.141, 2.147, 1.906, 1.912, 1.933, and 1.937 Å

Understand the ligand field theory: In an octahedral field, the

d

-orbitals of the central metal ion split into two sets:

t_{2g}

(lower energy) and

e_g

(higher energy). The energy difference between these sets is denoted as

\Delta_0

Analyze the $d$-orbital splitting: For

[\mathrm{Mn}(\mathrm{hfacac})_3]

, the manganese ion is in a +3 oxidation state (

\mathrm{Mn}^{3+}

), which has a

d^4

electronic configuration. The

d

-orbital splitting diagram for

\mathrm{Mn}^{3+}

in an octahedral field will show the

t_{2g}

orbitals (dxy, dxz, dyz) lower in energy and the

e_g

orbitals (dz^2, dx^2-y^2) higher in energy

Explain the bond distances: The variation in Mn-O bond distances can be attributed to the Jahn-Teller effect, which occurs in octahedral complexes with an uneven distribution of electrons in the

e_g

orbitals. For

\mathrm{Mn}^{3+}

(

d^4

), the Jahn-Teller distortion leads to elongation or compression of bonds along one axis, resulting in different bond lengths

Answer

The Mn-O bond distances in

[\mathrm{Mn}(\mathrm{hfacac})_3]

vary due to the Jahn-Teller effect, which causes distortion in the octahedral geometry of the

\mathrm{Mn}^{3+}

ion.

Key Concept

Jahn-Teller effect in octahedral complexes

Explanation

The Jahn-Teller effect causes distortion in octahedral complexes with uneven electron distribution in the

e_g

orbitals, leading to variations in bond lengths.

Solution

Identify the mechanism in profile B: Profile B shows a reaction with one transition state labeled Q. This suggests a single-step reaction mechanism

Determine the type of mechanism: Since there is only one transition state, it is likely an elementary reaction, which could be a simple bimolecular reaction (e.g.,

S_N2

E2

mechanism)

Answer

The mechanism corresponding to profile B is likely an elementary reaction, such as

S_N2

E2

Key Concept

Elementary reaction mechanisms involve a single transition state.

Explanation

Profile B shows a single transition state, indicating a one-step reaction mechanism, typical of elementary reactions like

S_N2

E2

Solution

Analyze profile D: Profile D shows two reactions on the same axes, leading to products Y and Z

Determine the favored product under high temperatures and long reaction times: Under these conditions, the product with the lower energy (more stable) will be favored

Identify the more stable product: Compare the energy levels of products Y and Z. The product with the lower energy level is more stable

Answer

Product Z would be favored under conditions of high temperatures and long reaction times.

Key Concept

The more stable product is favored under high temperatures and long reaction times.

Explanation

In profile D, the product with the lower energy level (Z) is more stable and thus favored under these conditions.

In profile D, which product, “Y” or “Z”, would be favoured under conditions of high temperatures and long reaction times?

Solution

Identify the reaction pathways: In profile D, there are two reaction pathways, each with its own transition state and product. The energy of the reactants is marked as

X

, the energy of the products for the first pathway is marked as

Z

, and the energy of the products for the second pathway is marked as

Y

Determine the energy differences: The product

Y

has a lower energy than the product

Z

, indicating that

Y

is the thermodynamically more stable product

Consider the conditions: Under conditions of high temperatures and long reaction times, the reaction is more likely to reach thermodynamic equilibrium, favoring the formation of the more stable product

Conclusion: Therefore, under conditions of high temperatures and long reaction times, product

Y

would be favored

Answer

Product

Y

would be favored under conditions of high temperatures and long reaction times.

Key Concept

Thermodynamic stability

Explanation

Under high temperatures and long reaction times, reactions tend to favor the formation of the thermodynamically more stable product, which in this case is

Y

iii) What type of mechanism corresponds to profile C? iv) In profile A, is transition state N best considered early or late? In profile D, which product is favoured under conditions of kinetic control? vi) In profile B, what term is given to Q?

Solution

Identify the mechanism in profile C: Profile C shows a reaction with two transition states, indicating a two-step mechanism. The first transition state is higher in energy than the second, suggesting the formation of an intermediate

Determine the nature of transition state N in profile A: In profile A, the transition state N is best considered early because it is closer in energy to the reactants (M) than to the products (O). This suggests that the transition state resembles the reactants more than the products

Identify the favored product under kinetic control in profile D: In profile D, the product formed via pathway Z is favored under kinetic control because it has a lower activation energy compared to pathway Y

Define the term given to Q in profile B: In profile B, Q represents the activation energy of the reaction, which is the energy barrier that must be overcome for the reaction to proceed

Answer

1. The mechanism corresponding to profile C is a two-step mechanism with an intermediate. 2. In profile A, transition state N is best considered early. 3. In profile D, the product formed via pathway Z is favored under conditions of kinetic control. 4. In profile B, the term given to Q is the activation energy.

Key Concept

Reaction mechanisms and transition states

Explanation

Profile C indicates a two-step mechanism with an intermediate. Profile A's transition state N is early, suggesting it resembles the reactants. In profile D, the product with the lower activation energy (pathway Z) is favored under kinetic control. In profile B, Q represents the activation energy.

What type of mechanism corresponds to profile C?

Solution

Identify the reaction profile: Profile C shows a reaction coordinate with multiple energy changes, indicating a multi-step mechanism. The curve starts at a structure, rises to a peak (T), falls to a valley (U), rises again to a smaller peak (V), and finally falls to a different structure

Determine the mechanism: The presence of multiple peaks and valleys suggests that the reaction involves intermediates and transition states. The first peak (T) represents the transition state of the first step, the valley (U) represents an intermediate, and the second peak (V) represents the transition state of the second step

Classify the mechanism: This profile is characteristic of a stepwise mechanism, where the reaction proceeds through distinct intermediates and transition states. An example of such a mechanism is the SN1 reaction in organic chemistry, which involves the formation of a carbocation intermediate

Answer

The mechanism corresponding to profile C is a stepwise mechanism involving intermediates and transition states.

Key Concept

Stepwise mechanism

Explanation

Profile C shows multiple energy peaks and valleys, indicating the presence of intermediates and transition states, characteristic of a stepwise mechanism.

In profile B, what term is given to Q?

Solution

Identify the reaction coordinate diagram: In profile B, the reaction coordinate diagram shows the energy changes during the conversion of tert-butyl bromide to tert-butyl methyl ether. The peak labeled Q represents the highest energy point along the reaction path

Determine the term for Q: The peak Q in a reaction coordinate diagram is typically referred to as the "transition state." This is the point at which the reactants are in a high-energy, unstable configuration before converting into products

Answer

Transition state

Key Concept

Transition state

Explanation

The transition state is the highest energy point along the reaction path, representing the configuration of atoms at the maximum energy barrier that must be overcome for the reaction to proceed.

Alkyl halides can undergo substitution reactions (via S N1 or S N2 mechanisms) and/or elimination reactions (via E1 or E2 mechanisms). Illustrate your understanding of these reactions in the following examples (Reactions 1 & 2 on separate pages below) by: i. Indicating the reaction pathway(s) favoured by the particular type of substrate, nucleophile, leaving group and solvent employed. ii. Predicting the major reaction mechanism that will occur under these conditions. iii. Indicating which of the structures provided is likely to be the major product(s) of the reaction. If you consider two products equally likely to form you should give both structures as the major product

Solution

Identify the reaction pathway(s): The substrate is an alkyl halide with iodine as the leaving group, and the nucleophile is methanol (

CH_3OH

). Methanol is a weak nucleophile and a polar protic solvent, which typically favors

S_N1

and

E1

mechanisms

Predict the major reaction mechanism: Given the conditions (weak nucleophile and polar protic solvent), the

S_N1

mechanism is more likely to occur. The

E1

mechanism is also possible but less favored compared to

S_N1

Indicate the major product(s): For the

S_N1

mechanism, the major product will be the one where the nucleophile (

CH_3OH

) replaces the iodine atom. This leads to the formation of Product J. However, due to the possibility of carbocation rearrangement, Product K could also form. For the

E1

mechanism, the major products would be the alkenes, which are Product L and Product M

Answer

The major products are likely to be Product J and Product K for the

S_N1

mechanism, and Product L and Product M for the

E1

mechanism.

Key Concept

The type of mechanism (substitution or elimination) depends on the nature of the substrate, nucleophile, leaving group, and solvent.

Explanation

In this reaction, the weak nucleophile and polar protic solvent favor the

S_N1

mechanism, leading to substitution products. The

E1

mechanism is also possible, leading to elimination products.

Solution

Identify the reaction pathway(s): Given the substrate is an alkyl halide with iodine as the leaving group, and the nucleophile is potassium tert-butoxide (tBuOK) in dimethylformamide (DMF), we need to consider the possible reaction pathways. tBuOK is a strong, bulky base, which favors elimination reactions (E2 mechanism) over substitution reactions (S

_N

1 or S

_N

Predict the major reaction mechanism: Under these conditions, the E2 mechanism is favored. This is because tBuOK is a strong base and DMF is a polar aprotic solvent, which does not stabilize carbocations well, making S

_N

1 and E1 mechanisms less likely. The steric hindrance of tBuOK also makes S

_N

2 less favorable

Determine the major product(s): In an E2 elimination reaction, the major product is typically the more substituted alkene, following Zaitsev's rule. Therefore, we need to identify the most substituted alkene among Products E, F, G, and H

Answer

The major product is likely to be the most substituted alkene, which can be determined by examining the structures of Products E, F, G, and H.

Key Concept

E2 elimination reactions are favored by strong, bulky bases and polar aprotic solvents, leading to the formation of the most substituted alkene.

Explanation

Potassium tert-butoxide (tBuOK) is a strong, bulky base that favors E2 elimination reactions, especially in a polar aprotic solvent like DMF. The major product is the most substituted alkene, following Zaitsev's rule.

Explain Sn1 Sn2

Solution

Identify the Mechanism: The SN1 and SN2 mechanisms are two types of nucleophilic substitution reactions that occur in alkyl halides

SN1 Mechanism: The SN1 mechanism stands for unimolecular nucleophilic substitution. It involves two steps: - Step 1: Formation of a carbocation intermediate by the departure of the leaving group. - Step 2: Nucleophile attacks the carbocation to form the final product. The rate-determining step is the formation of the carbocation, and the rate of reaction depends only on the concentration of the substrate

SN2 Mechanism: The SN2 mechanism stands for bimolecular nucleophilic substitution. It involves a single, concerted step: - The nucleophile attacks the substrate from the opposite side of the leaving group, leading to a backside attack and inversion of configuration. The rate of reaction depends on the concentration of both the substrate and the nucleophile

Factors Affecting SN1 and SN2: - SN1: Favored by tertiary alkyl halides, polar protic solvents, and weak nucleophiles. - SN2: Favored by primary alkyl halides, polar aprotic solvents, and strong nucleophiles

Examples: - SN1:

(CH_3)_3CBr + H_2O \rightarrow (CH_3)_3COH + HBr

- SN2:

CH_3Br + OH^- \rightarrow CH_3OH + Br^-

Answer

SN1 and SN2 mechanisms are two types of nucleophilic substitution reactions in alkyl halides, with SN1 involving a two-step process and SN2 involving a single-step process.

Key Concept

SN1 and SN2 mechanisms

Explanation

SN1 involves a two-step process with a carbocation intermediate, while SN2 involves a single-step, concerted process with backside attack and inversion of configuration.

Solution

Identify the reaction: The reaction involves a six-membered ring with a carbonyl group reacting with a nitrogen compound in the presence of a PhSO3H catalyst to form a six-membered ring with a nitrogen atom and water as a byproduct

Analyze intermediates: The intermediates C through G need to be placed in the correct order of formation

Determine the order: Based on the given structures and the reaction mechanism, the intermediates can be ordered as follows: - Intermediate C: The initial attack of the nitrogen compound on the carbonyl group forms a protonated intermediate. - Intermediate D: Proton transfer leads to the formation of a protonated hydroxyl group. - Intermediate F: Dehydration occurs, leading to the formation of a positively charged nitrogen intermediate. - Intermediate G: Further rearrangement and stabilization of the intermediate. - Intermediate E: Final product formation with the nitrogen atom bonded to two methyl groups and the release of water

Answer

C → D → F → G → E

Key Concept

Reaction intermediates in a chemical reaction

Explanation

The intermediates in a chemical reaction represent transient species that form during the conversion of reactants to products. Their correct order of formation is crucial for understanding the reaction mechanism.

Place the intermediates C – G in correct order of formation during this reaction.

Solution

Identify the initial reactants and catalyst: The reaction starts with a six-membered cyclohexane ring with a double bond to oxygen on carbon 1 and a nitrogen atom bonded to two methyl groups. The catalyst used is PhSO3H

Determine the first intermediate (C): The first intermediate, C, is formed by the addition of a hydroxyl group and a dimethylamine group to the ring. This suggests a nucleophilic addition reaction

Formation of intermediate D: Intermediate D is formed next, characterized by a double-bonded oxygen with a positive charge on the oxygen and a hydrogen atom bonded to the ring. This indicates a protonation step

Formation of intermediate E: Intermediate E is formed by the conversion of the double-bonded oxygen to a double-bonded nitrogen with a positive charge on the nitrogen and two methyl groups bonded to the nitrogen. This suggests an imine formation

Formation of intermediate F: Intermediate F is formed by the addition of a hydroxyl group and a dimethylamine group with a positive charge on the nitrogen to the ring. This indicates another nucleophilic addition

Formation of intermediate G: Intermediate G is formed by the addition of a hydroxyl group and a dimethylamine group with a positive charge on the oxygen to the ring. This suggests a final protonation step

Answer

C → D → E → F → G

Key Concept

Sequence of formation of reaction intermediates

Explanation

The intermediates are formed in a specific order based on the type of chemical reactions occurring, such as nucleophilic addition, protonation, and imine formation.

Provide the name of the functional group in product A

Solution

Identify the reaction: The reaction involves cyclohexanone and ammonia in the presence of a phosphoric acid catalyst to form cyclohexanamine and water

Determine the product: The product formed is cyclohexanamine, which is an amine

Identify the functional group: The functional group in cyclohexanamine is the amine group, characterized by the presence of a nitrogen atom bonded to one or more alkyl or aryl groups

Answer

The functional group in product A is an amine.

Key Concept

Amine functional group

Explanation

An amine is a functional group characterized by a nitrogen atom bonded to one or more alkyl or aryl groups. In cyclohexanamine, the nitrogen is bonded to a cyclohexyl group.

Solution

Step 1: Sulfonation of Phenol: To convert phenol (

C_6H_5OH

) to the intermediate with a sulfonic acid group (

-SO_3H

) attached to the benzene ring, we use concentrated sulfuric acid (

H_2SO_4

). The reaction is as follows:

C_6H_5OH + H_2SO_4 \rightarrow C_6H_4(OH)(SO_3H) + H_2O

Step 2: Substitution of Sulfonic Acid Group with Bromine: The intermediate product from Step 1 undergoes a substitution reaction where a bromine atom replaces the sulfonic acid group. This can be achieved using bromine (

Br_2

) in the presence of a catalyst such as iron(III) bromide (

FeBr_3

). The reaction is:

C_6H_4(OH)(SO_3H) + Br_2 \rightarrow C_6H_4(OH)(Br) + H_2SO_4

Step 3: Bromination of Phenol: The product from Step 2 undergoes another substitution reaction, this time replacing the hydrogen atom on the benzene ring with a bromine atom, resulting in the formation of 4-bromophenol. This can be done using bromine (

Br_2

) in the presence of a catalyst such as iron(III) bromide (

FeBr_3

). The reaction is:

C_6H_4(OH)(Br) + Br_2 \rightarrow C_6H_3(OH)(Br_2) + HBr

Answer

The reagents required for each of the three steps are: 1. Concentrated sulfuric acid (

H_2SO_4

) 2. Bromine (

Br_2

) and iron(III) bromide (

FeBr_3

) 3. Bromine (

Br_2

) and iron(III) bromide (

FeBr_3

)

Key Concept

The key concept here is the sequence of chemical reactions involving sulfonation and bromination of phenol.

Explanation

The sulfonation reaction introduces a sulfonic acid group to the benzene ring, which is then replaced by a bromine atom in the subsequent bromination steps.

Explain why conversion of phenol to 2-bromophenol cannot be readily achieved in a single step.

Solution

Identify the reaction: The conversion of phenol to 2-bromophenol involves the introduction of a bromine atom at the ortho position relative to the hydroxyl group on the benzene ring

Consider the reactivity of phenol: Phenol is highly reactive due to the electron-donating effect of the hydroxyl group, which activates the ortho and para positions for electrophilic substitution reactions

Electrophilic substitution: Direct bromination of phenol typically leads to substitution at both the ortho and para positions, resulting in a mixture of 2-bromophenol and 4-bromophenol, and often even 2,4,6-tribromophenol

Regioselectivity challenge: Achieving selective bromination at the ortho position (to form 2-bromophenol) without affecting the para position is difficult in a single step due to the high reactivity and lack of control over the regioselectivity of the reaction

Alternative methods: To achieve selective ortho bromination, a multi-step process involving protection and deprotection of the hydroxyl group or the use of directing groups may be required

Answer

The conversion of phenol to 2-bromophenol cannot be readily achieved in a single step due to the high reactivity of phenol and the difficulty in controlling regioselectivity during electrophilic substitution reactions.

Key Concept

Regioselectivity in electrophilic substitution reactions

Explanation

Phenol's hydroxyl group activates both ortho and para positions, making it challenging to selectively brominate only the ortho position without affecting the para position.

Solution

Identify the intermediate A: Intermediate A is the structure labeled at the bottom center of the image. It is crucial to understand the spatial arrangement of atoms around the Cβ-Cα bond, where Cα is bound to Ni and Cβ is the next carbon atom

Analyze the dihedral angle: The dihedral angle refers to the angle between two planes, in this case, the plane containing the Cα-Ni bond and the plane containing the Cβ-Cα bond. This angle will influence the steric interactions and the stability of the intermediate

Predict the product: To determine whether the cis (Z) or trans (E) product is kinetically favored, consider the steric hindrance and electronic effects. The trans (E) product is generally more stable due to less steric hindrance between substituents on the double bond

Justify the prediction: Given the view along the Cβ-Cα bond, the trans (E) product will be kinetically favored because the larger groups will be positioned further apart, reducing steric clashes and leading to a more stable transition state

Answer

The trans (E) product will be kinetically favored.

Key Concept

Steric hindrance and stability of the transition state determine the favored product.

Explanation

The trans (E) product is favored due to reduced steric hindrance, leading to a more stable transition state.

Solution

Identify the reaction: The reaction in question is an electrophilic aromatic substitution (EAS) where nitrobenzene reacts with the nitronium ion (

\mathrm{NO}_{2}^{+}

) to form a meta-substituted product

Draw the initial attack: The nitronium ion (

\mathrm{NO}_{2}^{+}

) acts as the electrophile and attacks the benzene ring of nitrobenzene at the meta position. This forms a sigma complex (arenium ion)

Show resonance structures: Draw all possible resonance structures of the sigma complex to illustrate the delocalization of the positive charge

Reform the aromatic system: The final step involves the loss of a proton (deprotonation) to restore the aromaticity of the benzene ring, resulting in the meta-substituted nitrobenzene product

Answer

The curly arrow mechanism for the reaction of nitrobenzene with the nitronium ion (

\mathrm{NO}_{2}^{+}

) to form the meta-substituted product involves the following steps: 1. The nitronium ion attacks the benzene ring at the meta position. 2. Formation of the sigma complex with resonance structures. 3. Deprotonation to restore aromaticity.

Key Concept

Electrophilic Aromatic Substitution (EAS)

Explanation

In EAS, an electrophile replaces a hydrogen atom on an aromatic ring. The nitronium ion (

\mathrm{NO}_{2}^{+}

) is the electrophile in this reaction, and the meta position is favored due to the electron-withdrawing nature of the nitro group already present on the benzene ring.

The para-substituted product.

Solution

Identify the reactants and products: The reactant is nitrobenzene, which is a benzene ring with a nitro group (

\mathrm{NO}_2

) attached. The electrophile in this reaction is the nitronium ion (

\mathrm{NO}_2^+

). The product is the para-substituted nitrobenzene, where a second nitro group is added to the para position relative to the first nitro group

Draw the initial attack: The nitronium ion (

\mathrm{NO}_2^+

) attacks the benzene ring at the para position relative to the existing nitro group. This forms a sigma complex (arenium ion) intermediate. Use curly arrows to show the movement of electrons from the benzene ring to the nitronium ion

Show resonance structures: Draw all possible resonance structures of the sigma complex intermediate. This involves delocalizing the positive charge around the ring

Reform the aromatic system: The sigma complex loses a proton to reform the aromatic system, resulting in the para-substituted product. Use curly arrows to show the movement of electrons that restore the aromaticity of the benzene ring

Answer

The para-substituted product is formed by the electrophilic aromatic substitution of nitrobenzene with the nitronium ion (

\mathrm{NO}_2^+

), resulting in the addition of a second nitro group at the para position.

Key Concept

Electrophilic Aromatic Substitution

Explanation

In electrophilic aromatic substitution, an electrophile (in this case, the nitronium ion) attacks the benzene ring, forming a sigma complex intermediate. The intermediate then loses a proton to restore the aromaticity, resulting in the substitution product.

Solution

Identify the reaction: The reaction involves benzoyl nitrite (C6H5CO2N) and benzaldehyde (C6H5CHO) in the presence of NaOH, H2O, ethanol, and heat to form trans-stilbene (C6H5CH=CHC6H5) with a nitro (NO2) group attached

Initiation of the reaction: The NaOH in the reaction mixture deprotonates the benzaldehyde, forming a benzaldehyde anion. This anion is a strong nucleophile

Nucleophilic attack: The benzaldehyde anion attacks the carbonyl carbon of benzoyl nitrite, leading to the formation of a tetrahedral intermediate

Formation of intermediate: The intermediate collapses, resulting in the elimination of a nitrite ion (NO2-) and the formation of a double bond between the two benzene rings, leading to the formation of trans-stilbene

Final product: The final product is trans-stilbene with a nitro group attached to one of the benzene rings

Answer

The detailed mechanism involves the deprotonation of benzaldehyde, nucleophilic attack on benzoyl nitrite, formation of a tetrahedral intermediate, and elimination of nitrite ion to form trans-stilbene.

Key Concept

Electrophilic aromatic substitution and nucleophilic attack mechanisms

Explanation

The reaction proceeds through a series of steps involving nucleophilic attack and elimination, leading to the formation of the final product.