Organic Chemistry Introduction — Diagnostic Tests

Unit Tests

UT-1: IUPAC Nomenclature with Complex Substituents

Question:

Give the IUPAC name for each of the following compounds:

(a) $\text{CH}_3\text{CH}(\text{CH}_3)\text{CH}_2\text{CH}(\text{Cl})\text{CH}_2\text{CH}_3$

(b) $\text{CH}_3\text{CH}=\text{C}(\text{CH}_3)\text{CH}_2\text{CH}_3$

(c) $\text{HOCH}_2\text{CH}_2\text{CH}(\text{OH})\text{CH}_3$

Solution:

(a) 3-chloro-2-methylhexane

Working: The longest carbon chain is 6 carbons (hexane). Number from the end that gives the lowest locants to substituents. Numbering from the left: Cl at C-3, methyl at C-2. Alphabetical order: chloro before methyl.

(b) 2-methylpent-2-ene

Working: The longest chain containing the double bond is 5 carbons (pent-2-ene). The methyl group is at C-2 (same carbon as the double bond). Alphabetical: methyl is the only substituent.

(c) pentane-1,3-diol

Working: The longest chain is 5 carbons (pentane). The two $-$OH groups are at C-1 and C-3. The suffix "-diol" is used with locants separated by commas.

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UT-2: Stereoisomerism: E/Z and Optical Activity

Question:

(a) For the compound $\text{CH}_3\text{CH}=\text{C}(\text{Cl})\text{CH}_2\text{CH}_3$, identify whether E/Z isomerism is possible and assign the configurations.

(b) Explain why the compound $\text{CH}_3\text{CH}(\text{OH})\text{COOH}$ (lactic acid) exhibits optical isomerism but $\text{CH}_3\text{COOH}$ (ethanoic acid) does not.

(c) A sample of 2-chlorobutane has an observed optical rotation of $+13.5^\circ$. If the pure $(R)$-enantiomer has a specific rotation of $+23.1^\circ$, calculate the enantiomeric excess and the ratio of $(R)$- to $(S)$-2-chlorobutane in the mixture.

Solution:

(a) E/Z isomerism is possible because:

• There is a C$=$C double bond (restricted rotation)
• Each carbon of the double bond has two different groups attached

Assigning priorities (Cahn-Ingold-Prelog rules) for each carbon of the double bond:

Left C of double bond: $-\text{CH}_3$ (priority 2) vs $-\text{H}$ (priority 1) Right C of double bond: $-\text{Cl}$ (priority 1, highest atomic number) vs $-\text{CH}_2\text{CH}_3$ (priority 2)

• Z isomer: Higher priority groups on the same side: both on top or both on bottom
• E isomer: Higher priority groups on opposite sides

(b) Lactic acid ($\text{CH}_3\text{CH}(\text{OH})\text{COOH}$) has a chiral centre at the second carbon. This carbon is bonded to four different groups: $-\text{H}$, $-\text{OH}$, $-\text{CH}_3$, and $-\text{COOH}$. A chiral centre (asymmetric carbon) gives rise to two non-superimposable mirror images (enantiomers).

Ethanoic acid ($\text{CH}_3\text{COOH}$) has no chiral centre. The carbonyl carbon is bonded to $-\text{CH}_3$, $=\text{O}$, and $-\text{OH}$ (three different groups, but with the double bond, it is not $sp^3$ hybridised and cannot be a chiral centre). The methyl carbon is bonded to three hydrogens and one carbon (not four different groups).

(c)

$\text{Enantiomeric excess (ee)} = \frac◆LB◆\text{observed rotation}◆RB◆◆LB◆\text{rotation of pure enantiomer}◆RB◆ \times 100 = \frac{13.5}{23.1} \times 100 = 58.4\%$

This means the mixture contains $58.4\%$ excess of the $(R)$-enantiomer.

Ratio: $[(R) - (S)]/[(R) + (S)] = 0.584$

$(R) + (S) = 100\%$ $(R) - (S) = 58.4\%$

$(R) = 79.2\%, \quad (S) = 20.8\%$

Ratio $(R):(S) = 79.2 : 20.8 \approx 3.81 : 1$

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UT-3: Functional Group Identification and Classification

Question:

(a) Classify each of the following reactions by type (addition, elimination, substitution, oxidation, reduction, hydrolysis, or polymerisation):

(i) $\text{C}_2\text{H}_4 + \text{HBr} \to \text{CH}_3\text{CH}_2\text{Br}$

(ii) $\text{CH}_3\text{CH}_2\text{Br} + \text{NaOH}(aq) \to \text{CH}_3\text{CH}_2\text{OH} + \text{NaBr}$

(iii) $\text{CH}_3\text{CH}_2\text{OH} \xrightarrow{[\text{O}]} \text{CH}_3\text{CHO}$

(b) State the reagents and conditions needed to convert: (i) propene to propan-1-ol (ii) propan-1-ol to propanal (iii) propanal to propanoic acid

(c) A compound has the molecular formula $\text{C}_4\text{H}_8\text{O}$. It reacts with 2,4-DNP to form an orange precipitate but does not react with Fehling's solution. It does not decolourise bromine water. Identify the compound and explain your reasoning.

Solution:

(a) (i) Electrophilic addition -- an alkene reacts with HBr, with the $\pi$-bond breaking and new $\sigma$-bonds forming. (ii) Nucleophilic substitution -- the $-$OH nucleophile replaces the $-$Br leaving group. (iii) Oxidation -- the alcohol is oxidised (loss of hydrogen/ gain of oxygen).

(b) (i) Propene to propan-1-ol: React propene with steam ($\text{H}_2\text{O}$) and a phosphoric acid ($\text{H}_3\text{PO}_4$) catalyst at $300\,^\circ\text{C}$ and $60\,\text{atm}$. This gives predominantly propan-2-ol via Markovnikov addition. For propan-1-ol specifically, use hydroboration-oxidation (not on A-Level syllabus in detail). Alternatively, react propene with $\text{HBr}$, then $\text{NaOH}(aq)$ to get propan-2-ol (Markovnikov product).

(ii) Propan-1-ol to propanal: Oxidise with acidified potassium dichromate(VI) ($\text{K}_2\text{Cr}_2\text{O}_7/\text{H}^+$) under distillation conditions to prevent over-oxidation to the carboxylic acid.

(iii) Propanal to propanoic acid: Oxidise with acidified potassium dichromate(VI) under reflux conditions.

(c) $\text{C}_4\text{H}_8\text{O}$ with the degree of unsaturation: $\text{C}_n\text{H}_{2n}\text{O}$ suggests one double bond equivalent.

• Reacts with 2,4-DNP: Confirms the presence of a carbonyl group (aldehyde or ketone)
• Does not react with Fehling's solution: The carbonyl is a ketone (not an aldehyde -- aldehydes reduce Fehling's)
• Does not decolourise bromine water: The C$=$O double bond does not react with bromine water (unlike C$=$C), confirming the unsaturation is a carbonyl, not an alkene

Possible ketones with formula $\text{C}_4\text{H}_8\text{O}$:

• Butanone ($\text{CH}_3\text{COCH}_2\text{CH}_3$)
• 2-methylpropanone (acetone, $\text{CH}_3\text{COCH}(\text{CH}_3)_2$) -- actually this is the same as propanone which is $\text{C}_3\text{H}_6\text{O}$

The only ketone with $\text{C}_4\text{H}_8\text{O}$ is butanone ($\text{CH}_3\text{COCH}_2\text{CH}_3$).

Integration Tests

IT-1: Isomer Counting and Structure-Property Relationships (with Bonding)

Question:

(a) Draw all the structural isomers of $\text{C}_4\text{H}_9\text{Br}$ and classify each as primary, secondary, or tertiary.

(b) Arrange the isomers from part (a) in order of increasing boiling point and explain the trend.

(c) Which isomer would react fastest with aqueous NaOH? Explain using mechanisms.

Solution:

(a) There are four structural isomers of $\text{C}_4\text{H}_9\text{Br}$:

1. 1-bromobutane ($\text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{Br}$) -- primary
2. 2-bromobutane ($\text{CH}_3\text{CH}(\text{Br})\text{CH}_2\text{CH}_3$) -- secondary
3. 1-bromo-2-methylpropane ($\text{CH}_3\text{CH}(\text{CH}_3)\text{CH}_2\text{Br}$) -- primary
4. 2-bromo-2-methylpropane ($\text{CH}_3\text{C}(\text{Br})(\text{CH}_3)_2$) -- tertiary

(b) Increasing boiling point: 2-bromo-2-methylpropane $\lt$ 1-bromo-2-methylpropane $\lt$ 2-bromobutane $\lt$ 1-bromobutane

Boiling point depends on van der Waals forces, which increase with surface area. More branched isomers have a more compact shape with less surface area for intermolecular interactions:

• 2-bromo-2-methylpropane (most branched) has the lowest boiling point
• 1-bromobutane (straight chain) has the largest surface area and highest boiling point

(c) 2-bromo-2-methylpropane (tertiary) reacts fastest with aqueous NaOH. It undergoes SN1 (unimolecular nucleophilic substitution), where the rate-determining step is the formation of a tertiary carbocation. Tertiary carbocations are stabilised by the electron-donating inductive effect of three methyl groups, making them relatively easy to form. The rate equation is $\text{rate} = k[\text{RBr}]$, independent of $[\text{OH}^-]$.

1-bromobutane (primary) would be slowest, undergoing SN2 where both the halogenoalkane and nucleophile are involved in the rate-determining step.

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IT-2: Reaction Pathways and Functional Group Interconversions (with Organic Mechanisms)

Question:

Starting from ethene ($\text{CH}_2=\text{CH}_2$), propose a reaction pathway to produce ethanoic acid ($\text{CH}_3\text{COOH}$). Include reagents, conditions, and intermediate compounds.

(a) Write equations for each step.

(b) State the type of reaction occurring at each step.

(c) Explain why direct oxidation of ethene to ethanoic acid is not straightforward.

Solution:

Step 1: Ethene to ethanol

$\text{CH}_2=\text{CH}_2 + \text{H}_2\text{O} \xrightarrow{\text{H}_3\text{PO}_4, 300\,^\circ\text{C}, 60\,\text{atm}} \text{CH}_3\text{CH}_2\text{OH}$

Reaction type: Electrophilic addition (hydration)

Note: This gives ethanol. Markovnikov addition is not an issue here since both carbons of ethene are equivalent.

Step 2: Ethanol to ethanal

$\text{CH}_3\text{CH}_2\text{OH} \xrightarrow{\text{K}_2\text{Cr}_2\text{O}_7/\text{H}^+, \text{distillation}} \text{CH}_3\text{CHO} + \text{H}_2\text{O}$

Reaction type: Oxidation

Distillation prevents further oxidation to the carboxylic acid by removing the volatile aldehyde as it forms.

Step 3: Ethanal to ethanoic acid

$\text{CH}_3\text{CHO} \xrightarrow{\text{K}_2\text{Cr}_2\text{O}_7/\text{H}^+, \text{reflux}} \text{CH}_3\text{COOH}$

Reaction type: Oxidation

Reflux allows the aldehyde to remain in contact with the oxidising agent long enough for complete oxidation to the carboxylic acid.

(b)

1. Electrophilic addition -- $\text{H}_2\text{O}$ adds across the C$=$C double bond
2. Oxidation -- primary alcohol oxidised to aldehyde
3. Oxidation -- aldehyde oxidised to carboxylic acid

(c) Direct oxidation of ethene to ethanoic acid is not straightforward because:

• Ethene contains only C--C and C--H bonds, which are not easily oxidised by standard laboratory oxidising agents like acidified dichromate.
• The C$=$C bond reacts with oxidising agents by cleavage (e.g., with hot concentrated $\text{KMnO}_4$, ethene is cleaved to $\text{CO}_2$ and $\text{H}_2\text{O}$, not selectively to ethanoic acid).
• Industrial oxidation of ethene to ethanoic acid uses a two-step process: ethene is first converted to ethanol (or ethanal via the Wacker process), which is then oxidised.
• Selective partial oxidation of alkenes is difficult because the C$=$C bond is highly reactive and tends to undergo complete cleavage.

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IT-3: Chirality and Reaction Mechanisms (with Organic Chemistry)

Question:

(R)-2-chlorobutane undergoes nucleophilic substitution with NaOH in aqueous ethanol.

(a) If the reaction proceeds via an SN2 mechanism, predict the configuration of the product and explain your reasoning.

(b) If the reaction proceeds via an SN1 mechanism, predict the optical activity of the product mixture and explain your reasoning.

(c) Explain why the product from the SN2 reaction has the opposite configuration to the starting material (Walden inversion).

Solution:

(a) In an SN2 mechanism, the nucleophile ($\text{OH}^-$) attacks the carbon from the back side (opposite the leaving group). This causes a single concerted step where the C--Cl bond breaks as the C--OH bond forms. The three other substituents are "flipped" like an umbrella inverting, giving the product (S)-butan-2-ol.

The stereochemistry is inverted: $(R)$ reactant gives $(S)$ product. This is called Walden inversion.

(b) In an SN1 mechanism, the C--Cl bond breaks first to form a planar carbocation intermediate. The nucleophile can then attack from either side of the planar intermediate with equal probability:

• Attack from one side gives the $(R)$-alcohol
• Attack from the other side gives the $(S)$-alcohol

This produces a racemic mixture (50:50 $(R)$ and $(S)$), which is optically inactive because the rotations of the two enantiomers cancel out.

(c) In the SN2 transition state, the nucleophile and leaving group are collinear (180$^\circ$) with the central carbon. As the nucleophile approaches from one side and the leaving group departs from the other, the three remaining substituents flip through a trigonal planar arrangement. This backside attack results in inversion of the stereochemistry at the chiral centre. The configuration changes because the spatial arrangement of substituents around the chiral carbon is reversed -- what was the front becomes the back, and vice versa.

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Additional Practice Problems

UT-4: Nomenclature Challenge

Question: Name the following compounds using IUPAC nomenclature:

(a) $\mathrm{CH}_3\mathrm{CH}(\mathrm{CH}_3)\mathrm{CH}_2\mathrm{CH}(\mathrm{Cl})\mathrm{CH}_2\mathrm{CH}_3$

(b) $\mathrm{BrCH}_2\mathrm{CH}_2\mathrm{CH}(\mathrm{OH})\mathrm{CH}_2\mathrm{CH}_3$

(c) $\mathrm{HOOC}\mathrm{CH}_2\mathrm{CH}(\mathrm{NH}_2)\mathrm{COOH}$

(d) $\mathrm{CH}_3\mathrm{COCH}_2\mathrm{CH}_2\mathrm{CHO}$

Solution:

(a) Find the longest carbon chain: 6 carbons (hexane). Number to give the lowest locants: substituents at C-2 (methyl) and C-4 (chloro).

Name: 4-chloro-2-methylhexane (1 mark).

(b) Longest chain: 5 carbons (pentane). Substituents at C-1 (bromo) and C-3 (hydroxy).

Name: 3-bromopentan-1-ol (1 mark).

(c) The longest chain containing the carboxylic acid group has 3 carbons. Number from the carboxylic acid end. Substituent at C-2 (amino).

Name: 2-aminopropanedioic acid (also known as aspartic acid) (1 mark).

(d) The highest priority functional group is the aldehyde (suffix: -al). The ketone is named as a substituent (prefix: oxo-). Longest chain: 5 carbons.

Name: 4-oxopentanal (1 mark).

UT-5: Functional Group Identification

Question: A compound $\mathrm{C}_4\mathrm{H}_8\mathrm{O}$ gives the following test results:

• Reacts with 2,4-DNP to give an orange precipitate
• Gives a silver mirror with Tollens' reagent
• Does not decolourise bromine water

Identify the compound and explain each test result.

Solution:

The positive 2,4-DNP test indicates a carbonyl group (aldehyde or ketone) (1 mark).

The positive Tollens' test (silver mirror) indicates an aldehyde (not a ketone) (1 mark).

The failure to decolourise bromine water confirms there is no C=C double bond (1 mark).

The molecular formula $\mathrm{C}_4\mathrm{H}_8\mathrm{O}$ with an aldehyde group: possible structures are butanal ($\mathrm{CH}_3\mathrm{CH}_2\mathrm{CH}_2\mathrm{CHO}$) and 2-methylpropanal ($\mathrm{CH}_3\mathrm{CH}(\mathrm{CH}_3)\mathrm{CHO}$). Both are consistent with the test results.

IT-4: Multi-Step Synthesis with Stereochemistry

Question: Starting from $(R)$-butan-2-ol, propose reagents and conditions for a two-step synthesis of $(S)$-butan-2-amine, explaining the stereochemical outcome.

Solution:

Step 1: Convert $(R)$-butan-2-ol to a leaving group. Use $\mathrm{SOCl}_2$ (thionyl chloride) with pyridine to give $(S)$-2-chlorobutane via SN2 with Walden inversion (1 mark).

Step 2: Nucleophilic substitution with ammonia. Excess $\mathrm{NH}_3$ in ethanol gives $(R)$-butan-2-amine via SN2 with a second Walden inversion (1 mark).

Overall: $(R) \xrightarrow{\text{SN2}} (S) \xrightarrow{\text{SN2}} (R)$

This gives the wrong enantiomer. To obtain $(S)$-butan-2-amine from $(R)$-butan-2-ol, we need an odd number of inversions.

Correct route: Use an SN1 reaction for step 1 (to give a racemic mixture) or use a two-step sequence with one SN2 step:

Step 1: Oxidise $(R)$-butan-2-ol to butan-2-one (using acidified $\mathrm{K}_2\mathrm{Cr}_2\mathrm{O}_7$). This destroys the chiral centre (1 mark).

Step 2: Reductive amination: butan-2-one + $\mathrm{NH}_3$ + $\mathrm{NaBH}_3\mathrm{CN}$ gives racemic butan-2-amine (1 mark).

This produces a racemic mixture, not a single enantiomer. To obtain a single enantiomer, a chiral resolving agent or enzymatic resolution would be needed. This illustrates the challenge of stereospecific synthesis with amines.