Question Bank

Explain why benzene undergoes substitution reactions.

Delocalisation of the 6 π electrons around the ring confers extra stability through its resonance structure. Hence, electrophilic addition would destroy the stable delocalised π system, and requires a large input of energy which makes it highly unfavourable. Hence, it undergoes electrophilic substitution instead which preserves it aromaticity after reaction.

Describe and explain, with the aid of suitable equations, the ROLE of NO₂ in the oxidation of atmospheric sulfur dioxide.

SO₂ + NO₂ —> SO₃ + NO

NO + ½ O₂ —> NO₂

NO₂ acts as a homogenous catalyst as it is in the same phase as the products & reactants, and catalyses the oxidation of SO₂ to form SO₃

Explain why the pH of the electrolyte in the hydrogen-oxygen fuel cell remains constant, with reference to the half-equations at each half cell.

Anode: H₂ + 2OH^- —> 2H₂O + 2e^-

Cathode: O₂ + 2H₂O + 4e^- → 4OH^-

Since the amount of electrons (4 mol) transferred must be the same for both halfcells, there will be no net change in the amount of OH^-. Hence, there is no net change to the concentration of OH^-, causing the pH to remain constant.

OR

2H₂ + O₂ → 2H₂O

OH^- not present in the final overall equation, hence there is no net change in the concentration of OH^-, so the pH remains constant

Describe sp³ hybridisation, sp² hybridisation and sp hybridisation. Explain the shapes of the bond angles in, ethane, ethene, benzene and ethyne in relation to ర and π C-C bonds.

sp³ (ethane) : 1 2s, 3 2p orbitals on each C mix to give 4 equivalent sp³ hybrid orbitals, arranged tetrahedrally at 109.5 to one another. Each sp³ orbital forms a ర bond (head-on overlap), with 1 C-C ర bond, 3 C-H ర bonds per carbon.

sp² (ethene) : 1 2s, 2 2p orbitals mix to give 3 sp² hybrid orbitals lying in a plane at 120 to each other, with 1 2p orbital remaining unhybridised, perpendicular to the plane. the sp² orbitals form ర bonds by head-on overlap, while the unhybridised 2p orbital on the 2 C overlap side-on form one π bond, with π-electron cloud lying above & below the molecular plane. Ethene is trigonal planar ABOUT each C, with bond angle 120

sp² (benzene) : Each C is sp² unhybridised, with 6 C-C ర and 6 C-H ర bonds formed by head-on overlap of sp² orbitals. The remaining 6 unhybridised 2p orbitals, being parallel and adjacent, overlap side-on equally with each other around the ring to form a delocalised π-electron cloud above & below the ring. Hence, benzene is a planar regular hexagon, with a C-C-C bond angle of 120, with all C-C bond lengths equal.

sp (ethyne): 1 2s & 1 2p orbital on each C mix to form 2 sp hybrid orbitals arranged linearly at 180, 2 2p orbitals remain unhybridised and are mutually perpendicular to each other and to the sp orbitals. The sp orbitals form ర bonds by head-on overlap, so 1 C-C ర bond, and 1 C-H ర per carbon. 2 unhyrbidised 2p orbitals overlap side-on with the corresponding parallel 2p orbitals on the adjacent carbon, forming 2 π bonds merging into a cylindirical π-electron cloud surrounding the C-C ర bond. Linear molecular, 180

Explain why this indicator is chosen.

The pH range of the indicator concides with the sharp pH change at equivalence point.

Explain why __ can resist changes in pH when NaOH is added.

1. (equation of buffer)

2. As there is a large reservior of conjugate base & acid,

3. the addition of OH^- leads to a small change in the ratio of [acid] : [conjugate base]

4. Since [H⁺] = K_a x [acid]/[conjugate base], where K_a is constant, this results in a very small pH change

Explain what is meant by a chiral centre.

A chiral centre is a carbon atom that is bonded to 4 different atoms or groups.

Explain how CFCs destroy the ozone layer.

1. C-F and C-Cl bonds are strong, so CFCs are very unreactive & long-lived, allowing them time to diffuse into the stratosphere without decomposing

2. Strong UV radiation is able to homolytically cleave the C-Cl bond to generate reactive Cl radicals…

Explain ‘amines are Lewis bases', with an equation.

A Lewis base is a species that can donate a lone pair of electrons to form a dative bond with a Lewis acid.

Amines have a lone pair of electrons on the N atom, which can be donated to form a new dative bond CH₃NH₂ + H⁺ → CH₃ NH₃⁺

Describe the relative bascities of phenylamine, ethylamine and ammonia in aqueous medium, in terms of structure.

Increasing basicity: phenylamine > ammonia > ethylamine

1. Ethylamine is more basic as the ethyl group is electron donating, so this increases electron density on the N atom, making its lone pair more available for protonation, and hence giving a greater extent of ionisation than ammonia

2. Phenylamine is less basic than ammonia because lone pair of electrons on N delocalises intot he benzene ring, as the p-orbital on N overlaps with the ring’s π electron-cloud, decreasing electron density on N, making lone pair less available to accept a proton, smaller extent of ionisation

Explain why amides are neutral rather than basic.

The lone pair of electrons on N delocalises into the adjacent C=O bond,, reducing the electron density on N, making the lone pair unavailable for protonation by na acid, hence they are neutral

For each SN1 and SN2 mechanism, state & explain whether a single enantiomer or racemic mixture is formed.

Sn1: racemic mixture is formed, as the carbonation intermediate formed is trigonal planar about the electron deficient C, hence the nucloephile is able to attack from the top and bottom plane with equal probability, resulting in the forming on equal amounts of enantiomers, resulting in forming a racemic mixture

SN2: A single enantiomer is formed, because the nucleophile attacks from the side directly opposite to the leaving halide, since the halide can still block approach from the front, hence this will invert the spatial arrangement of the other three groups at the chiral carbon, so there is only 1 stereochemical outcome possible.

Compare reactivities of chloroethane, chlorobenzene, and ethanoyl chloride with water & aqueuous medium.

Ethanoyl chloride reacts vigorously with cold water alone, as the acyl carbon is highly electron deficient due to being polarized by 2 electronegative Cl and O atoms, making it strongly susceptible to nucleophile attack, and the planar sp² carbon is also less sterically hindered to nucleophilic attack

Chloroethane: does not react with cold water, hydrolysed upon heating with aqueous NaOH, with its C-Cl bond polarized by only 1 electronegative Cl atom, and the tetrahedral C is more sterically hindered, so more harsh conditions are required

Chlorobenzene: does not react with anything, as the lone pair on Cl delocalises into the benzene ring, giving the C-Cl bond partial double-bond character, and hence it is stronger and harder to break. Also, the ring sterically blocks any backside attack on the C-CL carbon, and electron-rich π electron-cloud of the ring repels any approaching nucleophiles.

Compare acidities of phenol, water and ethanol.

Phenol: strongest acid, as the phenoxide ion is the most stable, since the negative charge on O can be delocalized into the benzene ring, dispersing the negative charge to a greater extent than in the hydroxide ion (OH^-), hence they are less likely to re-accept a proton and dissociates to a greater extent than water

Ethanol: weakest acid, ethyl group is electron donating, so it will intensify the negative charge on O, destabilizing it relative to the hydroxide ion, resulting in the alkoxide ion readily accepting a proton to reform ethanol, hence it dissociates to a much smaller extent

Explain why alkenes, aldehydes & ketones undergo addition.

They all contain a π bond. Since the π bond is weaker than a sigma bond, this can break to allow for the formation of 2 stronger new sigma bonds in its place, hence it is energetically favourable so all 3 classes will undergo addition reactions.

Both N and P are in Group 15 of the Periodic Table. Explain why P can form H₃PO₄ and H₃PO₃ while N cannot.

N is in Period 2 while P is in Period 3. Hence, P has energetically accessible 3d orbitals to accomodate beyond 8 electrons in its valence shell, allowing expansion of octet. However, for N, the next available subshell is 3s, which is not energetically accessible.

The pKa₁ of H₃PO₄ is greater than H₃PO₃. Based on the structures of their conjugate bases, suggest whether the -OH groups are electron-donating or electron-withdrawing. Explain

Since H₃PO₃ has a lower pKa₁, it ionises to a greater extent

H₂PO₄^- is hence less stable than H₂PO₃^-, as it has 2 -OH groups compared to 1 -OH group, suggesting that the -OH group intensifies the negative charge on the O^- in the conjugate base, and is hence electron-donating

State and explain 2 differences in physical properties of s block elements and transition elements.

1. TE has greater melting point than s block elements, due to all of their 3d and 4s electrons being involved in metallic bonding, hence more energy is required to overcome the stronger metallic bonding between the TE cation and sea of delocalized electrons

2. TE has greater density, as it has a larger A_r but a smaller atomic radius, hence it has more mass per unit volume

Given that the rate equation is rate = k[HCN][alkene][L], and a large excess of HCN and alkene is used, explain why the rate law can be simplified to rate = k’[L]

Since alkene and HCN are present in large excess, any changes in concentration will be negligible, hence they can be subsumed into k’

When the electrolyte solution becomes very viscous, a lower rate of reaction occurs. Explain.

Electrical conductivity declines due to the electron carriers being less mobile

Arrange benzene, nitrobenzene and phenylamine in order of ease of bromination. Explain with reference to their structures.

Nitrobenzene < benzene < phenylamine

The –NO2 group in nitrobenzene is a strong electron-withdrawing group by both inductive and resonance effects, which decreases the electron density in the ring. It deactivates the benzene ring. The benzene ring becomes the least susceptible to electrophilic substitution, making bromination the most difficult.

Benzene has no activating or deactivating substituents, so it undergoes bromination under normal conditions.

The –NH2 group in phenylamine is electron-donating by resonance, which increases electron density in the ring. This activates the ring. The benzene ring becomes the most susceptible to electrophilic substitution, making bromination the easiest.

Explain the non-ideal behaviour of a gas at moderately high pressure.

At moderately high pressure, gas molecules come closer together and the intermolecular attractive forces between gas molecules become significant. This causes the gas to occupy a volume smaller than that of an ideal gas, and less pressure is exerted on the walls of the container.

When both ∆H and ∆S is negative, explain the effect of temperature on the spontaneity of the reaction.

∆G = ∆H - T∆S

Since ∆H and ∆S are both negeative, ∆G is more negative at lower temperature as the negative ∆H term outweighs the positive -T∆S term, hence it is spontaneous at lower temperatures

Explain why nitrogen oxides are not formed at lower temperatures / outside of catalytic converters.

1. At lower temperatures, particles have insufficient energy to overcome high activation energy needed to break strong N triple bond and O=O to form NO_x

2. The N triple bond is very strong, so forming nitrogen oxides from N₂ requires a lot of energy to break the bond, which is not favourable at low temperature

Predict the effect of adding a small amount of inert gas into the reaction vessel containing H₂, N₂ and NH₃, under constant temperature & pressure, on the POE of the Haber Process equation.

When small amount of inert gas is added into the reaction vessel under constant

temperature and pressure, total volume of the gaseous system is increased (as the

system must expand to keep its total pressure constant). Concentrations (or partial

pressures) of the reactants and products are decreased. The system counteract the

change shifting the position of equilibrium so as to re-establish the equilibrium,

hence, the equilibrium position will shift to the left, i.e. the side involving greater

number of moles of gas.

Assuming that the rate equation = k[H₂O][alkene], what is the effect on the graph of volume of CO₂ (product) against time when the concenration of alkene is doubled.

1. Gradient will be steeper, and rate will increase by 2x due to the 2x of concentration of alkene, hence there is an proportionate 2x increase in rate of the reaction

2. t_1/2 is half of the original value as t_1/2 = ln 2 / k’, where k’ = k[alkene]

Explain the variation in the boiling points of group 15 hydrides from nirogen to antiomy. Sketch a graph to show this, and explain it in terms of structure and bonding.

1. Group 15 hydrides have a simple molecular lattice structure

2. Down the group, the size of the electron cloud of the molecule increases, hence the electron cloud is more polarisable

3. Allows for stronger dispersion forces between molecules down the group

4. Hence, the energy required to overcome the stronger dispersion forces increases down the group, so the boiling point increases

5. However, more energy is needed to overcome the strong hydrogen bonding in NH₃ than to overcome the weak dispersion forces in PH₃, AsH₃ and SbH₃

6. Hence, the boiling point of NH₃ is much higher than the other group 15 hydrides.

Explain why [Cu(H₂O)₆]²⁺ and [Mn(H₂O)₆]³⁺ have different colours.

1. Electronic configuration of Cu²⁺ and Mn³⁺ is different, so

2. extent of repulsion between ligands & electrons around the metal centre is different

3. The energy gap, ΔE, of a different magnitude

4. Light absorbed by manganese complex different from light absorbed by copper complex, hence complement colour observed is different