CHE-12 TA - AS Level Organic Chemistry Paper 1

​Question:
Structural isomerism and stereoisomerism should be considered when answering this question.

2,5-dibromohexane is heated under reflux with ethanolic KOH.

How many isomeric compounds are formed with molecular formula C₆H₁₀?

A. 3
B. 5
C. 6
D. 7

​Testosterone is an optically active organic molecule.

How many chiral centres are there in one molecule of testosterone?

• ​Steamy fumes with PCl₅ indicate the presence of an –OH group (alcohols and carboxylic acids both react to release HCl).

• 2,4-DNPH gives an orange precipitate with aldehydes/ketones; no visible product ⇒ no carbonyl group present.

• Therefore, we can say with certainty that T is not an aldehyde. We cannot conclude it is specifically an alcohol or a carboxylic acid, and it is not a hydrocarbon (because it contains oxygen).

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​Complete combustion of compound T produces carbon dioxide and water only.
Compound T produces steamy fumes with PCl₅.
Compound T does not give any visible product with 2,4-dinitrophenylhydrazine (2,4-DNPH) reagent.

What can be deduced with certainty from this information?

A) Compound T is a carboxylic acid.
B) Compound T is a hydrocarbon.
C) Compound T is an alcohol.
D) Compound T is not an aldehyde.

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​Explanation:

• In an addition polymer, each repeat unit corresponds to the two carbons that were originally the C=C.

• Repeat unit of J is –CH(C₂H₅)–CH(CH₃)–, so the alkene must have substituents C₂H₅ and CH₃ on adjacent double-bond carbons → CH₃–CH=CH–CH₂CH₃ (pent-2-ene).

• Repeat unit of K is –CH(CH₃)–CH(CH₃)–, meaning both double-bond carbons each had a CH₃ substituent → CH₃–CH=CH–CH₃ (but-2-ene).

​Explanation:

Free-radical substitution occurs when Br₂ + UV light reacts with ethane.

The main product is bromoethane and HBr (hydrogen bromide). ✅

A possible side reaction: two ethyl radicals can combine to form butane. ✅

Hydrogen gas (H₂) is not formed in this mechanism (no pathway produces H· radicals that combine significantly). ❌

So, the possible products are butane (1) and hydrogen bromide (3) only.

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​Explanation:

• Ethanolic NaOH (heat) promotes E2 elimination of HBr from 1-bromopropane, giving propene in high yield.

• Ammonia → nucleophilic substitution to a primary amine (propylamine).

• Aqueous KOH → nucleophilic substitution to a primary alcohol (propan-1-ol).

• KCN → nucleophilic substitution to a nitrile (butanenitrile).
  Elimination is specifically favoured with strong base in ethanol, hence option D.

​Explanation:

Ammonia acts as a nucleophile, attacking the δ⁺ carbon in ethyl iodide and displacing I⁻ in an S_N2 step to give the protonated amine.
A second NH₃ molecule removes the proton, forming ethylamine and NH₄⁺ (which pairs with I⁻ → NH₄I). It’s not addition (no C=C involved) and not a radical process.

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​Explanation:

Sodium borohydride (NaBH₄) is a selective reducing agent for aldehydes and ketones. It reduces the ketone group (C=O) in propanone to a secondary alcohol, giving propan-2-ol. It does not convert the carbonyl to an alkane, nor does it change the carbon skeleton to an aldehyde or primary alcohol.

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​Explanation:

• (CH₃)₃COH is a tertiary alcohol → no H on the C–OH carbon, so it is not oxidised by acidified dichromate.

• All alcohols react with Na(s) to give alkoxides + H₂, so A does react with sodium.

• B is a ketone (butanone): not oxidised by dichromate under these conditions and does not react with Na.

• C (primary) and D (secondary) are oxidised by acidified dichromate.

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Explanation:

The reaction is nucleophilic addition to the carbonyl:

• CN⁻ attacks the electrophilic carbonyl carbon → forms an alkoxide intermediate.

• The alkoxide is protonated by HCN, giving the cyanohydrin (HO–CH₂–CH₂–CN) and regenerating CN⁻ (so CN⁻ acts as a catalyst).

A is wrong because HCN does have a lone pair on N;
B is nonsense;
D is incorrect because it’s addition, not substitution.

​Explanation:

• Alkaline I₂ (iodoform test) is positive for a methyl ketone (–COCH₃) or a secondary alcohol oxidisable to it.

• Tollens’ reagent is positive for an aldehyde (–CHO).

• Option C contains both a methyl ketone (CH₃CO–) and an aldehyde (–CHO), so it gives both tests positive.

• A has only aldehydes (no methyl ketone).

• B has a ketone but not a methyl ketone.

• D has methyl ketones but no aldehyde.

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​Explanation:
Fehling’s reagent is reduced by aldehydes (and by methanoic acid) to give a brick-red precipitate of Cu₂O.

• HCO₂H behaves like an aldehyde under these conditions → positive result.

• CH₃CH₂CHO (propanal) is an aldehyde → positive result.

• Carboxylic acids (A), esters (B), and ketones (C) do not reduce Fehling’s solution.

​Explanation:

• Na(s) deprotonates all acidic –OH groups: a carboxylic acid (–CO₂H) and alcohol –OH groups.

  • For A: two alcohol –OH + one –CO₂H ⇒ total 3 acidic hydrogens ⇒ ion of –3.

• Aqueous NaOH is strong enough to deprotonate carboxylic acids but not ordinary alcohols in water.

  • Hence only the –CO₂H is deprotonated ⇒ ion of –1.

• B has only two acidic –OH groups (–CO₂H and one alcohol) ⇒ would give –2 with Na.

• C has two –CO₂H + one alcohol ⇒ NaOH would give –2 (dicarboxylate), not –1.

• D has only one alcohol –OH ⇒ would give –1 with Na.

Explanation:

• Ammonia acts as a nucleophile and attacks the δ⁺ carbon of ethyl iodide, displacing I⁻ in an S_N2 step to give the protonated amine.

• A second NH₃ removes the proton → ethylamine and NH₄I.

• It’s not addition (no C=C) and not a radical process.

​Explanation:

• The repeat unit alternates –CH₂–CH(Cl)– and –CH₂–CH(OH)–, i.e. an alternating copolymer of chloroethene (vinyl chloride) and ethenol (vinyl alcohol) → statement 2 is true.

• Polymers containing Cl release HCl (and other toxic gases) on burning, making disposal hazardous → statement 1 is true.

• If the polymer came from one monomer, 3-chlorobut-2-ene-1-ol would give –CH(CH₂Cl)–CH(CH₂OH)–, not the shown unit (which has Cl and OH directly attached). Hence 3 is false.

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