DSE Chemistry Diagnostic: Organic Chemistry

Unit Test 1: Functional Group Identification from Tests

Question

An unknown organic compound $X$ has the molecular formula $C_{3}H_{6}O_{2}$. The following test results were obtained:

Test	Observation
Add solid $Na_{2}CO_{3}$	Effervescence observed
Add Fehling's solution and heat	No change
Add acidified $K_{2}Cr_{2}O_{7}$ and warm	Orange to green
Sweet-smelling compound obtained on heating with ethanol and $H_{2}SO_{4}$

(a) Identify the functional group(s) present in $X$. [3 marks]

(b) Deduce the structure of $X$ and name it. [2 marks]

(c) Write an equation for the reaction of $X$ with sodium carbonate. [2 marks]

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Worked Solution

(a) - Effervescence with $Na_{2}CO_{3}$: $X$ is acidic, producing $CO_{2}$. This indicates a carboxylic acid ($-COOH$) group.

• No reaction with Fehling's solution: $X$ does not contain an aldehyde ($-CHO$) group. (Fehling's solution is reduced by aldehydes, giving a brick-red precipitate.)

• Orange to green with $K_{2}Cr_{2}O_{7}$: $X$ is oxidised, but since we already know it has a $-COOH$ group (which cannot be further oxidised by this reagent), the colour change is likely due to an alcohol group also being present. However, the molecular formula is $C_{3}H_{6}O_{2}$, which accounts for exactly one $-COOH$ group ($C_{2}H_{3}O_{2}$) plus one additional carbon ($CH_{3}$). The only structure is $CH_{3}CH_{2}COOH$ (propanoic acid), which has no alcohol group.

Note: $CH_{3}CH_{2}COOH$ would NOT give orange-to-green with $K_{2}Cr_{2}O_{7}$ since carboxylic acids are not oxidised by this reagent. However, the $Na_{2}CO_{3}$ result is definitive for carboxylic acid. No other $C_{3}H_{6}O_{2}$ structure (e.g., hydroxypropanal) would react with $Na_{2}CO_{3}$. The orange-to-green result may be due to a slight impurity, or the question is designed to test whether students recognise that carboxylic acids are the endpoint of oxidation. Given the $Na_{2}CO_{3}$ result, $X$ must be propanoic acid.

(b) Structure: $CH_{3}CH_{2}COOH$

Name: propanoic acid

(c) $2CH_{3}CH_{2}COOH + Na_{2}CO_{3} \rightarrow 2CH_{3}CH_{2}COONa + H_{2}O + CO_{2}$

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Unit Test 2: Nucleophilic Substitution Mechanism

Question

Consider the reaction of 2-bromo-2-methylpropane with aqueous sodium hydroxide:

$(CH_{3})_{3}CBr + NaOH(aq) \rightarrow (CH_{3})_{3}COH + NaBr$

(a) State the type of mechanism for this reaction and explain why this mechanism is favoured for tertiary halogenoalkanes. [3 marks]

(b) Outline the mechanism, showing the movement of electrons using curly arrows. [3 marks]

(c) Explain why this reaction is first order with respect to the halogenoalkane and zero order with respect to NaOH. [2 marks]

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Worked Solution

(a) The mechanism is $S_{N}1$ (unimolecular nucleophilic substitution).

Tertiary halogenoalkanes favour $S_{N}1$ because:

1. The three alkyl groups on the carbon bearing the halogen stabilise the carbocation intermediate through electron-donating inductive effects and hyperconjugation.
2. Steric hindrance from the three bulky alkyl groups makes it difficult for the nucleophile ($OH^{-}$) to approach the carbon in a single step ($S_{N}2$).
3. The formation of a stable tertiary carbocation lowers the activation energy for the rate-determining step.

(b) Step 1 (slow, rate-determining): Heterolytic fission of the $C-Br$ bond.

The $C-Br$ bond breaks, with both electrons going to the bromine atom. A tertiary carbocation intermediate is formed.

$(CH_{3})_{3}C-Br \rightarrow (CH_{3})_{3}C^{+} + Br^{-}$

(Curly arrow from the $C-Br$ bond to the $Br$ atom.)

Step 2 (fast): Nucleophilic attack.

The hydroxide ion ($OH^{-}$) attacks the carbocation, forming the alcohol.

$(CH_{3})_{3}C^{+} + OH^{-} \rightarrow (CH_{3})_{3}COH$

(Curly arrow from the lone pair on $O$ of $OH^{-}$ to the carbocation carbon.)

(c) The rate-determining step (step 1) involves only the halogenoalkane molecule. The $OH^{-}$ is not involved until the fast second step. Therefore:

$\text{Rate} = k[(CH_{3})_{3}CBr]$

The reaction is first order with respect to the halogenoalkane and zero order with respect to NaOH (or $OH^{-}$).

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Unit Test 3: Structural Isomerism vs Stereoisomerism

Question

The molecular formula $C_{4}H_{8}$ can give rise to multiple isomers.

(a) Draw all structural isomers of $C_{4}H_{8}$ (including cyclic isomers). Name each. [4 marks]

(b) Which of these isomers exhibit cis-trans (geometric) isomerism? Explain the requirement for cis-trans isomerism. [3 marks]

(c) Does but-1-ene exhibit cis-trans isomerism? Explain. [2 marks]

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Worked Solution

(a) Structural isomers of $C_{4}H_{8}$:

1. But-1-ene: $CH_{2}=CH-CH_{2}-CH_{3}$
2. cis-But-2-ene: $CH_{3}-CH=CH-CH_{3}$ (cis)
3. trans-But-2-ene: $CH_{3}-CH=CH-CH_{3}$ (trans)
4. 2-Methylpropene: $CH_{2}=C(CH_{3})_{2}$
5. Cyclobutane: (square ring of 4 CH$_{2}$ groups)
6. Methylcyclopropane: (3-membered ring with $CH_{3}$ substituent)

Note: cis- and trans-but-2-ene are stereoisomers of the same structural isomer (but-2-ene). As structural isomers (different connectivity): but-1-ene, but-2-ene, 2-methylpropene, cyclobutane, methylcyclopropane = 5 structural isomers.

(b) But-2-ene exhibits cis-trans isomerism.

Requirements for cis-trans isomerism:

1. There must be a $C=C$ double bond (or a ring structure that restricts rotation).
2. Each carbon of the double bond must be bonded to two different groups.

In but-2-ene ($CH_{3}-CH=CH-CH_{3}$), each carbon of the $C=C$ is bonded to a $CH_{3}$ group and an $H$ atom -- two different groups on each carbon. This allows cis (same side) and trans (opposite sides) arrangements.

2-Methylpropene does NOT exhibit cis-trans isomerism because one carbon of the $C=C$ is bonded to two identical $CH_{3}$ groups.

(c) But-1-ene ($CH_{2}=CH-CH_{2}-CH_{3}$) does not exhibit cis-trans isomerism.

Reason: One carbon of the $C=C$ double bond (the $CH_{2}=$ end) is bonded to two hydrogen atoms (identical groups). Since each carbon of the double bond must have two different groups for cis-trans isomerism to occur, but-1-ene fails this requirement.

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Integration Test 1: Reaction Scheme Completion

Question

Complete the following reaction scheme by identifying compounds $A$ to $E$:

$CH_{3}CH_{2}OH \xrightarrow[\text{excess}]{K_{2}Cr_{2}O_{7}/H^{+}} A \xrightarrow[\text{heat}]{\text{alkaline } I_{2}} B + CHI_{3} \downarrow$

$B \xrightarrow{LiAlH_{4}} C \xrightarrow{HBr} D \xrightarrow{NaOH(aq)} C$

$A \xrightarrow{CH_{3}CH_{2}OH / H^{+}} E$

(a) Identify compounds $A$ to $E$ and name each. [5 marks]

(b) State the type of reaction occurring in each step. [4 marks]

(c) What observation confirms the formation of compound $B$ in the iodoform reaction? [1 mark]

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Worked Solution

(a) $A$: $CH_{3}CH_{2}OH$ is a primary alcohol. Oxidation with excess acidified $K_{2}Cr_{2}O_{7}$ gives the carboxylic acid: $CH_{3}COOH$ (ethanoic acid).

$B$: The iodoform reaction works for methyl ketones ($CH_{3}CO-$) and ethanol ($CH_{3}CH_{2}OH$). Ethanoic acid ($CH_{3}COOH$) does not give the iodoform test. The scheme therefore requires $A$ to be the aldehyde $CH_{3}CHO$ (ethanal), formed by controlled oxidation (interpreting the question as intending the aldehyde stage). Ethanal undergoes the iodoform reaction:

$CH_{3}CHO + 3I_{2} + 4NaOH \rightarrow CHI_{3} + HCOONa + 3NaI + 3H_{2}O$

So $B =$ HCOOH (methanoic acid).

Compound	Identity	Name
$A$	$CH_{3}CHO$	Ethanal
$B$	$HCOOH$	Methanoic acid
$C$	$CH_{3}OH$	Methanol
$D$	$CH_{3}Br$	Bromomethane
$E$	$CH_{3}CH(OCH_{2}CH_{3})_{2}$	1,1-Diethoxyethane (acetal)

(b) Step 1: $CH_{3}CH_{2}OH \rightarrow A$: Oxidation (controlled)

Step 2: $A \rightarrow B$: Iodoform reaction (haloform reaction)

Step 3: $B \rightarrow C$: Reduction ($LiAlH_{4}$ reduces carboxylic acid to alcohol)

Step 4: $C \rightarrow D$: Nucleophilic substitution ($-OH$ replaced by $-Br$)

Step 5: $D \rightarrow C$: Nucleophilic substitution ($-Br$ replaced by $-OH$)

Step 6: $A \rightarrow E$: Acetal formation (nucleophilic addition)

(c) The formation of a bright yellow precipitate of $CHI_{3}$ (iodoform) confirms the iodoform reaction.

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Integration Test 2: Mechanism + Product Stereochemistry

Question

Consider the addition of $HBr$ to but-2-ene.

(a) Write the mechanism for the electrophilic addition of $HBr$ to trans-but-2-ene. Show curly arrows and intermediates. [4 marks]

(b) Does the product show optical activity? Explain. [2 marks]

(c) If but-2-ene is reacted with bromine ($Br_{2}$) in an inert solvent, state the stereochemistry of the product and explain. [3 marks]

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Worked Solution

(a) Step 1 (electrophilic attack): The electron-rich $C=C$ double bond attacks the electrophilic $H$ atom of $HBr$.

The $H-Br$ bond breaks heterolytically, with the electron pair going to $Br$.

A carbocation intermediate is formed: $CH_{3}-\overset{+}{C}H-CH_{2}CH_{3}$ (secondary carbocation).

(Curly arrow from the $C=C$ pi bond to the $H$ of $HBr$; curly arrow from the $H-Br$ bond to $Br$.)

Step 2 (nucleophilic attack): The bromide ion ($Br^{-}$) attacks the carbocation from either side.

$CH_{3}-\overset{+}{C}H-CH_{2}CH_{3} + Br^{-} \rightarrow CH_{3}-CHBr-CH_{2}CH_{3}$

Product: 2-bromobutane.

(Curly arrow from lone pair on $Br^{-}$ to the carbocation carbon.)

(b) The product (2-bromobutane) has a chiral centre at $C-2$ (carbon bonded to four different groups: $CH_{3}$-, $H$, $Br$, $CH_{2}CH_{3}$). However, the reaction produces a racemic mixture (equal amounts of both enantiomers) because the $Br^{-}$ can attack the planar carbocation intermediate with equal probability from either face. Since equal amounts of both enantiomers are produced, the mixture is optically inactive (no net rotation of plane-polarised light).

(c) The addition of $Br_{2}$ to but-2-ene proceeds via a bromonium ion intermediate, which is a three-membered ring involving the two carbons and one bromine atom.

The bromonium ion is attacked by $Br^{-}$ from the opposite side (anti-addition), giving:

• From trans-but-2-ene: the product is a meso compound (2,3-dibromobutane) because the molecule has a plane of symmetry. The product is optically inactive.

• From cis-but-2-ene: the product is a racemic mixture of $(2R,3R)$- and $(2S,3S)$-2,3-dibromobutane. The two enantiomers are produced in equal amounts, so the mixture is optically inactive.

In both cases, the addition is stereospecific anti-addition.

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Integration Test 3: Multi-Step Organic Synthesis

Question

Starting from propene ($CH_{2}=CHCH_{3}$), propose a synthesis of 1,2-dibromopropane ($CH_{2}Br-CHBr-CH_{3}$).

(a) Give the reagent(s) and conditions for each step. [3 marks]

(b) Write balanced equations for each step. [3 marks]

(c) Explain why direct addition of $HBr$ to propene followed by reaction with $Br_{2}$ would NOT give the desired product. [2 marks]

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Worked Solution

(a) Step 1: Addition of $Br_{2}$ to propene to give 1,2-dibromopropane.

Reagent: Bromine ($Br_{2}$) in an inert organic solvent (e.g., $CCl_{4}$ or cyclohexane)

Conditions: Room temperature, in the dark (or with light excluded)

$CH_{2}=CHCH_{3} + Br_{2} \rightarrow CH_{2}Br-CHBr-CH_{3}$

This is a one-step synthesis: direct addition of $Br_{2}$ across the double bond gives exactly the desired 1,2-dibromopropane.

(b) $CH_{2}=CHCH_{3} + Br_{2} \rightarrow CH_{2}BrCHBrCH_{3}$

(1,2-dibromopropane)

(c) If $HBr$ is added first:

$CH_{2}=CHCH_{3} + HBr \rightarrow CH_{3}-CHBr-CH_{3}$

(2-bromopropane -- Markovnikov addition places $Br$ on the more substituted carbon)

Then adding $Br_{2}$ to 2-bromopropane would give:

$CH_{3}-CHBr-CH_{3} + Br_{2} \rightarrow CH_{3}-CBr_{2}-CH_{3} + HBr$

This is a substitution reaction (not addition, since there is no $C=C$), giving 2,2-dibromopropane, NOT 1,2-dibromopropane.

Alternatively, if the addition of $HBr$ gave 1-bromopropane (anti-Markovnikov, requiring peroxide), then $Br_{2}$ would substitute at the terminal position, still not giving the desired 1,2-dibromo product.

The key point: once the $C=C$ is consumed in step 1, the subsequent $Br_{2}$ reaction is substitution (not addition), placing $Br$ at a different position. The direct addition of $Br_{2}$ to the alkene is the only efficient route.