Organic Chemistry -- Diagnostic Tests

Question

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

(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]

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

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]

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 $OH^{-}$).

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]

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.

flowchart TD
    A[Diag Organic Chemistry] --> B[Key Concepts]
    A --> C[Core Principles]
    A --> D[Practical Applications]
    B --> E[Fundamental definitions]
    C --> F[Design patterns]
    D --> G[Real-world usage]

Carbon’s versatility: Organic chemistry is the study of carbon compounds — like LEGO bricks that can snap together in infinite ways to form chains, rings, and complex structures.

Why it matters: From medicines to plastics, organic chemistry underpins modern life. Understanding functional groups and reactions lets you design and predict molecule behavior.

The key insight: Functional groups determine reactivity — an alcohol reacts differently from an alkene because the functional group changes the electron distribution.

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]

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).

(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.

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]

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.

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]

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.

Confusing SN1 and SN2 mechanisms: SN1 is favoured by tertiary halogenoalkanes (stable carbocation, steric hindrance prevents backside attack). SN2 is favoured by primary halogenoalkanes (no steric hindrance, strong nucleophile needed). Don’t assume all nucleophilic substitutions follow the same mechanism.

Forgetting the requirements for cis-trans isomerism: You need a C=C double bond (or ring) AND each carbon of the double bond must have two different groups attached. But-1-ene doesn’t show cis-trans isomerism because one carbon has two identical H atoms.

Mixing up electrophilic addition with nucleophilic substitution: Alkenes undergo electrophilic addition (electron-rich C=C attacks electrophile). Halogenoalkanes undergo nucleophilic substitution (nucleophile attacks electron-poor carbon). The mechanisms are fundamentally different.

• Atomic Structure: Atomic structure is foundational
• Equilibrium: Equilibrium connects topics
• Organic Chemistry: Organic chemistry is a major area