Carbonyl Compounds, Arenes, Amines — Diagnostic Tests

Unit Tests

UT-1: Nucleophilic Addition to Carbonyl Compounds

Question:

(a) Describe the mechanism of the reaction between propanal ($\text{CH}_3\text{CH}_2\text{CHO}$) and hydrogen cyanide ($\text{HCN}$), showing the formation of 2-hydroxybutanenitrile.

(b) Explain why the reaction with HCN requires a base catalyst but proceeds very slowly without one.

(c) Explain why ketones are less reactive than aldehydes towards nucleophilic addition.

Solution:

(a) Mechanism:

Step 1: The lone pair on the cyanide ion ($\text{CN}^-$, generated by the base catalyst) attacks the electrophilic carbonyl carbon of propanal. The $\pi$-electrons of the C$=$O bond move onto the oxygen, forming a tetrahedral intermediate with a negatively charged oxygen:

$\text{CN}^- + \text{CH}_3\text{CH}_2\text{CH}=\text{O} \to \text{CH}_3\text{CH}_2\text{CH}(\text{CN})\text{O}^-$

Step 2: The negatively charged oxygen is protonated by a hydrogen ion (from $\text{HCN}$ or $\text{H}_3\text{O}^+$ in solution), forming the hydroxynitrile:

$\text{CH}_3\text{CH}_2\text{CH}(\text{CN})\text{O}^- + \text{H}^+ \to \text{CH}_3\text{CH}_2\text{CH}(\text{CN})\text{OH}$

Product: 2-hydroxybutanenitrile (also called 2-hydroxybutyronitrile).

(b) HCN is a very weak acid ($\text{p}K_a \approx 9.2$), meaning that in neutral or acidic solution, very little $\text{CN}^-$ nucleophile is present. The nucleophile in the reaction is $\text{CN}^-$ (not HCN itself, which has no lone pair available on carbon). A base catalyst (even weak bases like $\text{CN}^-$ itself) removes a proton from HCN to generate $\text{CN}^-$:

$\text{HCN} + \text{OH}^- \rightleftharpoons \text{CN}^- + \text{H}_2\text{O}$

This shifts the equilibrium to produce more $\text{CN}^-$, dramatically increasing the rate of nucleophilic attack. Without a base, the concentration of $\text{CN}^-$ is vanishingly small and the reaction is negligibly slow.

(c) Ketones are less reactive than aldehydes towards nucleophilic addition for two reasons:

1. Steric hindrance: Ketones have two alkyl groups attached to the carbonyl carbon, while aldehydes have one alkyl group and one hydrogen. The alkyl groups are bulkier than hydrogen, creating more steric hindrance around the carbonyl carbon and making it more difficult for the nucleophile to approach.

2. Electronic effects: Alkyl groups are electron-donating (+I effect). In ketones, two alkyl groups donate electron density towards the carbonyl carbon, reducing its partial positive charge and making it less electrophilic. In aldehydes, the hydrogen has no electron-donating effect, so the carbonyl carbon is more electrophilic.

These combined effects make aldehydes generally more reactive than ketones towards nucleophilic addition.

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UT-2: Electrophilic Substitution in Arenes

Question:

(a) Describe the mechanism of the nitration of methylbenzene (toluene), showing the formation of the electrophile and the intermediate.

(b) Methylbenzene nitrates faster than benzene, and the nitro group directs substitution to the 2- and 4-positions. Explain these observations with reference to the inductive and resonance effects of the methyl group.

(c) Nitrobenzene is nitrated much more slowly than benzene. Explain why.

Solution:

(a) Formation of the electrophile:

Concentrated nitric acid and concentrated sulfuric acid generate the nitronium ion ($\text{NO}_2^+$):

$\text{HNO}_3 + \text{H}_2\text{SO}_4 \to \text{NO}_2^+ + \text{HSO}_4^- + \text{H}_2\text{O}$

$\text{H}_2\text{SO}_4$ acts as a stronger acid, protonating $\text{HNO}_3$, which then loses water to form $\text{NO}_2^+$.

Electrophilic attack:

The $\text{NO}_2^+$ electrophile attacks the electron-rich benzene ring. The $\pi$-electrons form a new bond with the nitrogen, and the delocalised ring is disrupted, forming a positively charged sigma complex (arenium ion). The positive charge is delocalised over three carbon atoms of the ring.

Deprotonation:

The $\text{HSO}_4^-$ (or another base) removes a proton from the sigma complex, restoring the aromatic ring and forming nitro-methylbenzene.

(b) The methyl group ($-\text{CH}_3$) is electron-donating through both:

1. Inductive effect (+I): The methyl group donates electron density through the $\sigma$-bonds to the ring, increasing electron density at the ortho (2-) and para (4-) positions.

2. Hyperconjugation: The C--H bonds of the methyl group overlap with the $\pi$-system of the ring, donating electron density.

The increased electron density at the ortho and para positions makes these positions more nucleophilic and more attractive to the electrophile ($\text{NO}_2^+$). The major products are 2-nitromethylbenzene (ortho) and 4-nitromethylbenzene (para), with the para product often favoured for steric reasons.

(c) The nitro group ($-\text{NO}_2$) is strongly electron-withdrawing through both:

1. Inductive effect (-I): The electronegative nitrogen and oxygen atoms withdraw electron density from the ring.

2. Resonance effect (-M): The nitro group's $\pi$-system withdraws electron density from the ring through resonance (the $\pi$-electrons of the ring are delocalised onto the oxygen atoms of the nitro group).

This reduces the electron density of the ring, making it less nucleophilic and less attractive to electrophiles. The ring is deactivated towards electrophilic substitution, requiring more vigorous conditions and reacting more slowly. Additionally, the nitro group is meta-directing because it destabilises the sigma complex when substitution occurs at ortho or para positions (positive charge is delocalised onto the carbon bearing the electron-withdrawing nitro group).

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UT-3: Amine Basicity and Nucleophilicity

Question:

(a) Arrange the following in order of increasing basicity: ammonia ($\text{NH}_3$), methylamine ($\text{CH}_3\text{NH}_2$), phenylamine ($\text{C}_6\text{H}_5\text{NH}_2$), and trimethylamine ($(CH_3)_3\text{N}$). Explain the trend.

(b) Phenylamine reacts with nitrous acid ($\text{HNO}_2$) at $0$--$5\,^\circ\text{C}$ to form diazonium salt. Write the equation for this reaction and explain why the reaction must be carried out at low temperature.

(c) Explain why phenylamine is a much weaker base than methylamine despite both having nitrogen atoms with lone pairs.

Solution:

(a) Increasing basicity: phenylamine $\lt$ ammonia $\lt$ trimethylamine $\lt$ methylamine

Methylamine is the strongest base because the methyl group is electron-donating (+I effect), increasing the electron density on the nitrogen lone pair and making it more available for protonation.

Trimethylamine is slightly less basic than methylamine despite having three electron-donating groups, because of steric hindrance: the three methyl groups create a crowded environment around the nitrogen, making it more difficult for a proton ($\text{H}^+$) to approach and bond with the lone pair. Additionally, in aqueous solution, trimethylamine is less solvated (fewer hydrogen bonds with water) than methylamine, which reduces its apparent basicity.

Ammonia has no electron-donating groups and is less basic than alkylamines.

Phenylamine is the weakest base because the lone pair on nitrogen interacts with the $\pi$-system of the benzene ring (delocalisation), making it less available for protonation.

(b) Phenylamine reacts with nitrous acid:

$\text{C}_6\text{H}_5\text{NH}_2 + \text{HNO}_2 + \text{HCl} \to \text{C}_6\text{H}_5\text{N}_2^+\text{Cl}^- + 2\text{H}_2\text{O}$

Product: benzenediazonium chloride

The reaction must be carried out at $0$--$5\,^\circ\text{C}$ because diazonium salts are thermally unstable. At higher temperatures, the diazonium group decomposes, releasing nitrogen gas and forming a carbocation that can react with water or other nucleophiles, giving a mixture of products. Low temperature stabilises the diazonium salt long enough for it to be used in subsequent reactions (e.g., azo coupling).

(c) In phenylamine, the nitrogen lone pair is partially delocalised into the benzene ring through resonance. The lone pair overlaps with the $\pi$-system of the ring, creating resonance structures where the nitrogen bears a positive charge and the ring bears a negative charge. This delocalisation means the lone pair is less available for accepting a proton, making phenylamine a much weaker base ($\text{p}K_b \approx 9.4$) compared to methylamine ($\text{p}K_b \approx 3.4$).

In methylamine, the nitrogen lone pair is fully localised on the nitrogen atom (no $\pi$-system to delocalise into) and is enhanced by the electron-donating methyl group. The lone pair is fully available for protonation.

Integration Tests

IT-1: Multi-Step Synthesis Involving Aromatic Chemistry (with Organic Chemistry)

Question:

Starting from benzene, propose a synthesis of 4-nitrobenzoic acid.

(a) Write equations for each step, including reagents and conditions.

(b) Explain why the order of steps is critical. What would happen if nitration were carried out before the introduction of the carboxylic acid group?

(c) The product 4-nitrobenzoic acid has both nitro and carboxylic acid groups on the benzene ring. Explain why these are in the 4-position relative to each other.

Solution:

(a)

Step 1: Friedel-Crafts alkylation (or acylation followed by reduction) to introduce an alkyl side chain:

$\text{C}_6\text{H}_6 + \text{CH}_3\text{Cl} \xrightarrow{\text{AlCl}_3} \text{C}_6\text{H}_5\text{CH}_3 + \text{HCl}$

Product: methylbenzene (toluene)

Step 2: Oxidation of the methyl group to carboxylic acid:

$\text{C}_6\text{H}_5\text{CH}_3 \xrightarrow{\text{KMnO}_4, \text{H}^+, \text{reflux}} \text{C}_6\text{H}_5\text{COOH}$

Product: benzoic acid

Step 3: Nitration of benzoic acid:

$\text{C}_6\text{H}_5\text{COOH} + \text{HNO}_3 \xrightarrow{\text{conc. H}_2\text{SO}_4, 50\,^\circ\text{C}} \text{4-O}_2\text{N}-\text{C}_6\text{H}_4-\text{COOH}$

Product: 4-nitrobenzoic acid

(b) The order of steps is critical because the directing effects of substituents determine where the next substituent is introduced.

If nitration were carried out first (on benzene), the product would be nitrobenzene. The nitro group is meta-directing and deactivating. Friedel-Crafts reactions cannot be performed on nitrobenzene because the ring is too deactivated. Additionally, even if a Friedel-Crafts reaction were possible, the alkyl group would enter the meta position (giving 3-nitrobenzoic acid after oxidation, not the desired 4-isomer).

By introducing the methyl group first (which is ortho/para-directing and activating), then converting it to the carboxylic acid (which is meta-directing and deactivating), nitration is then performed on benzoic acid. The $-\text{COOH}$ group directs the nitro group to the meta position (3-position) relative to itself, which is the 4-position relative to the original substituent.

Wait -- the $-\text{COOH}$ group is meta-directing, so nitration of benzoic acid gives 3-nitrobenzoic acid, not 4-nitrobenzoic acid. Let me reconsider.

Corrected synthesis:

Step 1: Nitration of benzene:

$\text{C}_6\text{H}_6 \xrightarrow{\text{HNO}_3/\text{H}_2\text{SO}_4} \text{C}_6\text{H}_5\text{NO}_2$

Step 2: Reduction of nitro group to amine:

$\text{C}_6\text{H}_5\text{NO}_2 \xrightarrow{\text{Sn/HCl}} \text{C}_6\text{H}_5\text{NH}_2$

Step 3: Protect/amide formation, then carboxylation... This is getting complex.

Simplest correct route:

Step 1: Friedel-Crafts alkylation of benzene with $\text{CH}_3\text{COCl}/\text{AlCl}_3$ (acylation):

$\text{C}_6\text{H}_6 + \text{CH}_3\text{COCl} \xrightarrow{\text{AlCl}_3} \text{C}_6\text{H}_5\text{COCH}_3$

Step 2: Nitration (methyl ketone is ortho/para-directing):

The nitro group enters the 4-position (para to the acyl group).

Step 3: Oxidation of the acyl group to carboxylic acid:

$4\text{-O}_2\text{N}-\text{C}_6\text{H}_4-\text{COCH}_3 \xrightarrow{\text{KMnO}_4} 4\text{-O}_2\text{N}-\text{C}_6\text{H}_4-\text{COOH}$

(c) The acyl (ketone) group in Step 2 is ortho/para-directing, so the nitro group enters the 4-position (para, less sterically hindered than ortho). After oxidation, the 4-relationship is maintained between the nitro and carboxylic acid groups.

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IT-2: Carbonyl Test Reactions and Distinguishing Compounds (with Acids and Bases)

Question:

Three unlabelled bottles contain: pentanal, pentan-2-one, and pentan-3-one.

(a) Describe a chemical test to distinguish pentanal from pentan-2-one.

(b) Describe how you would distinguish pentan-2-one from pentan-3-one.

(c) A fourth bottle contains 2-methylbutanal. Describe how 2-methylbutanal differs from pentanal in its reaction with HCN, and explain why.

Solution:

(a) Tollens' reagent (silver mirror test):

• Add Tollens' reagent ($[\text{Ag}(\text{NH}_3)_2]^+$) and warm gently.
• Pentanal (aldehyde): reduces Tollens' reagent, producing a silver mirror on the inside of the test tube.
• Pentan-2-one (ketone): no reaction (no silver mirror).

Alternatively, Fehling's solution can be used:

• Pentanal: orange Fehling's solution turns red/brown (Cu$_2$O precipitate)
• Pentan-2-one: no colour change

(b) Iodoform test (triiodomethane test):

• Add iodine solution and sodium hydroxide, then warm.
• Pentan-2-one ($\text{CH}_3\text{COCH}_2\text{CH}_2\text{CH}_3$): contains a $\text{CH}_3\text{C}=$ group (methyl ketone), gives a positive iodoform test -- pale yellow precipitate of $\text{CHI}_3$ (triiodomethane) and antiseptic smell.
• Pentan-3-one ($\text{CH}_3\text{CH}_2\text{COCH}_2\text{CH}_3$): does not have a methyl group directly attached to the carbonyl carbon, gives a negative iodoform test.

(c) Both pentanal and 2-methylbutanal react with HCN to form hydroxynitriles, but the products differ:

• Pentanal + HCN $\to$ 2-hydroxypentanenitrile (one chiral centre is formed)
• 2-methylbutanal + HCN $\to$ 2-hydroxy-2-methylbutanenitrile

In the reaction with 2-methylbutanal, the carbonyl carbon is already attached to three different groups (H, methyl, ethyl). After addition of CN$^-$, the carbon becomes bonded to four different groups, creating a new chiral centre. The reaction produces a racemic mixture of $(R)$- and $(S)$-enantiomers.

For pentanal, the carbonyl carbon is attached to H, H, and propyl. After addition, the carbon is bonded to CN, OH, H, and propyl -- also four different groups, so it also forms a racemic mixture. The key difference is that 2-methylbutanal produces a more sterically hindered product and the reaction rate may be slightly slower due to increased steric hindrance around the carbonyl carbon.

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IT-3: Azo Dye Formation (with Equilibrium)

Question:

Benzenediazonium chloride reacts with phenol to form an orange azo dye.

(a) Write the equation for the formation of the azo dye, naming the type of reaction.

(b) Explain why the coupling reaction occurs preferentially at the 4-position of phenol.

(c) The azo dye is used as an acid-base indicator. Explain the colour change in terms of the molecular structure.

Solution:

(a) The reaction is azo coupling (electrophilic substitution):

$\text{C}_6\text{H}_5\text{N}_2^+\text{Cl}^- + \text{C}_6\text{H}_5\text{OH} \to \text{C}_6\text{H}_5\text{N}=\text{N}\text{C}_6\text{H}_4\text{OH} + \text{HCl}$

Product: 4-hydroxyazobenzene (orange azo dye)

The diazonium ion ($\text{C}_6\text{H}_5\text{N}_2^+$) acts as the electrophile, attacking the electron-rich benzene ring of phenol.

(b) Phenol has an $-$OH group that is strongly electron-donating through resonance (+M effect), activating the ring at the 2- (ortho) and 4- (para) positions. The 4-position is preferred over the 2-position for two reasons:

1. Steric hindrance: The 4-position is less sterically hindered than the 2-position (which is adjacent to the bulky $-$OH group).
2. Resonance stabilisation: The sigma complex formed during 4-substitution is more effectively stabilised by resonance with the $-$OH group.

The product is 4-hydroxyazobenzene (para-substituted), which is the major product.

(c) Azo dyes contain the $-\text{N}=\text{N}-$ (azo) group, which forms an extensive conjugated system with the aromatic rings. This conjugation allows absorption of visible light, giving the dye its colour.

In acidic conditions, the phenolic $-$OH group can be protonated (or the azo nitrogen can be protonated), changing the electron distribution in the conjugated system. This shifts the wavelength of light absorbed, changing the observed colour. In basic conditions, the $-$OH group loses a proton to form $-$O$^-$, which is even more strongly electron-donating, further extending the conjugation and shifting the absorption to longer wavelengths.

The colour change occurs because protonation/deprotonation alters the extent of conjugation and the energy gap between the HOMO and LUMO orbitals in the delocalised $\pi$-system, changing which wavelengths of visible light are absorbed.