Q.A hydrocarbon 'A', ($C_4H_8$) on reaction with HCl gives a compound 'B', ($C_4H_9Cl$), which on reaction with 1 mol of $NH_3$ gives compound 'C', ($C_4H_{11}N$). On reacting with $NaNO_2$ and HCl followed by treatment with water, compound 'C' yields an optically active alcohol, 'D'. Ozonolysis of 'A' gives 2 mols of acetaldehyde. Identify compounds 'A' to 'D'. Explain the reactions involved.

Concept: Structural Isomerism – The molecular formula $C_4H_8$ corresponds to two alkene isomers (but-1-ene and but-2-ene) and a cycloalkane. The ozonolysis clue uniquely identifies the alkene.

Step 1 – Identify A from ozonolysis

Ozonolysis of A gives 2 moles of acetaldehyde ($CH_3CHO$). This means the double bond is symmetrically placed between two identical halves: $CH_3CH=CHCH_3$. Hence A is but-2-ene ($CH_3CH=CHCH_3$).

Step 2 – Addition of HCl to A

But-2-ene adds HCl following Markovnikov’s rule. The product B is 2-chlorobutane ($CH_3CHClCH_2CH_3$).

Step 3 – Reaction of B with NH$_3$

Alkyl halide B undergoes nucleophilic substitution with 1 mol of NH$_3$ to give a primary amine. C is 2-aminobutane ($CH_3CH(NH_2)CH_2CH_3$). …

Concept and Intuition

This is a classic organic chemistry deduction problem from the hydrocarbons and amines chapter. The clues are layered: start from the end (ozonolysis) to fix the alkene structure, then trace the functional group transformations forward.

Why this approach works: Ozonolysis is the most definitive clue — it cleaves alkenes at the double bond and tells you exactly which carbonyl fragments are produced. Two moles of acetaldehyde means the alkene must be symmetrical and give two identical three-carbon fragments after cleavage. That immediately narrows down A.

The rest is a chain of standard reactions: electrophilic addition of HCl (Markovnikov), nucleophilic substitution with ammonia (SN2), and diazotisation followed by hydrolysis (a way to replace –NH₂ with –OH, retaining stereochemistry if the amine is chiral).

Step-by-step reasoning

1. Determine the structure of hydrocarbon A (C₄H₈) from ozonolysis

Ozonolysis of A gives 2 moles of acetaldehyde (CH₃CHO). Acetaldehyde has the formula C₂H₄O. Two moles of it account for all 4 carbons of A.

The ozonolysis reaction cleaves the C=C bond and adds oxygen atoms. For a symmetrical alkene, the double bond is in the middle:

So A is but-2-ene (CH₃–CH=CH–CH₃). It can exist as cis and trans isomers, but both give the same ozonolysis products. The question does not specify stereochemistry of A, so either form is acceptable.

2. Reaction of A with HCl to give B (C₄H₉Cl)

But-2-ene is a symmetrical alkene — both sides of the double bond are identical (CH₃–CH=), so HCl addition gives the same product regardless of which carbon receives the Cl:

B is 2-chlorobutane (a secondary alkyl halide). Note: 1-chlorobutane is not formed because the double bond is internal.

3. Reaction of B with 1 mol of NH₃ to give C (C₄H₁₁N)

This is a nucleophilic substitution (SN2) where NH₃ replaces Cl. Since B is a secondary halide, SN2 is possible but slow; however, with excess NH₃ (or here, 1 mol under pressure), the reaction proceeds:

C is butan-2-amine (a primary amine). The carbon bearing the –NH₂ group is chiral (it has four different substituents: H, CH₃, C₂H₅, NH₂). Because this carbon carries four different groups, butan-2-amine is a chiral molecule — and that chirality is exactly what carries through to the alcohol in the next step. (Strictly, a laboratory preparation from racemic B would give racemic C; the question's phrase "optically active alcohol" is the NCERT Exemplar's way of flagging that D possesses a chiral carbon. What the deduction needs is simply that C — and hence D — is chiral.)

4. Reaction of C with NaNO₂ + HCl, then water, to give D (optically active alcohol) …

Here is the clear solution method for this problem.

Method Name: Retro-synthetic Analysis with Reaction Pathway Mapping

This method works backwards from the final product and known reactions to deduce the structure of the starting hydrocarbon.

Step 1: Identify Compound 'A' from Ozonolysis Data

• The Clue: Ozonolysis of 'A' ($C_4H_8$) gives 2 moles of acetaldehyde ($CH_3CHO$).
• The Concept: Ozonolysis cleaves the double bond. If both fragments are the same (acetaldehyde), the double bond must be symmetrically placed in the middle of a 4-carbon chain.
• The Structure: The only alkene with formula $C_4H_8$ that yields two $CH_3CHO$ molecules is But-2-ene.
  • $CH_3-CH=CH-CH_3 \xrightarrow{O_3} 2 CH_3CHO$

Therefore, A is But-2-ene ($CH_3CH=CHCH_3$).

Step 2: Identify Compound 'B' from the Addition Reaction

• The Reaction: A ($C_4H_8$) + HCl → B ($C_4H_9Cl$).
• The Concept: This is an electrophilic addition of HX to an alkene. For an unsymmetrical alkene, Markovnikov's rule applies. However, But-2-ene is symmetrical. Adding H to either carbon of the double bond gives the same product.
• The Structure: The product is 2-Chlorobutane.
  • $CH_3-CH=CH-CH_3 + HCl \rightarrow CH_3-CHCl-CH_2-CH_3$

Therefore, B is 2-Chlorobutane ($CH_3CHClCH_2CH_3$).

Step 3: Identify Compound 'C' from the Amination Reaction

• The Reaction: B ($C_4H_9Cl$) + $NH_3$ (1 mol) → C ($C_4H_{11}N$).
• The Concept: This is a nucleophilic substitution ($S_N2$). The halogen (Cl) is replaced by an amino group ($-NH_2$). The product is a primary amine.
• The Structure: The product is Butan-2-amine.
  • $CH_3-CHCl-CH_2-CH_3 + NH_3 \rightarrow CH_3-CH(NH_2)-CH_2-CH_3 + HCl$

Therefore, C is Butan-2-amine ($CH_3CH(NH_2)CH_2CH_3$).

Step 4: Identify Compound 'D' from the Diazotization Reaction

• The Reaction: C ($C_4H_{11}N$) + $NaNO_2$ + HCl → (intermediate) + $H_2O$ → D (Optically active alcohol, $C_4H_{10}O$). …

Common Mistakes & How to Avoid Them

1. Mistake: Ignoring the "2 moles of acetaldehyde" clue

Students often jump to draw all possible $C_4H_8$ isomers without first using the ozonolysis result.

Why it matters:

Ozonolysis of alkene 'A' gives 2 moles of acetaldehyde ($CH_3CHO$). This means the double bond is exactly in the middle, splitting the molecule into two identical 2-carbon fragments.

Correct structure:

Only but-2-ene ($CH_3CH=CHCH_3$) gives 2 moles of acetaldehyde upon ozonolysis.

How to avoid:

Always start with the ozonolysis product — it directly reveals the position of the double bond.

2. Mistake: Forgetting Markovnikov's rule in HCl addition

When 'A' reacts with HCl to give 'B' ($C_4H_9Cl$), students often place the chlorine on the wrong carbon.

Why it matters:

But-2-ene is symmetrical, so Markovnikov addition gives the same product from either end — 2-chlorobutane.

Correct structure of B:

$CH_3CHClCH_2CH_3$ (2-chlorobutane)

How to avoid:

For symmetrical alkenes, Markovnikov's rule still applies but both ends are equivalent — so only one product forms.

3. Mistake: Misinterpreting the reaction with $NH_3$

Students sometimes think $NH_3$ replaces Cl directly, but forget that only 1 mole of $NH_3$ is used.

Why it matters:

1 mole of $NH_3$ with an alkyl halide gives a primary amine via nucleophilic substitution ($S_N2$).

Correct structure of C:

$CH_3CH(NH_2)CH_2CH_3$ (butan-2-amine)

How to avoid:

Remember: $R-X + NH_3 \rightarrow R-NH_2 + HX$ (with excess $NH_3$, further substitution can occur, but here only 1 mole is used).

4. Mistake: Confusing the diazotization reaction

Students often think $NaNO_2 + HCl$ with an amine gives a nitro compound or something else.

Why it matters:

With a primary aliphatic amine, $NaNO_2/HCl$ forms a diazonium salt, which upon treatment with water gives an alcohol.

Correct reaction:

$R-NH_2 \xrightarrow{NaNO_2/HCl} R-N_2^+Cl^- \xrightarrow{H_2O} R-OH + N_2 + HCl$

How to avoid:

Memorise the diazotization sequence:

• Primary amine → diazonium salt (unstable for aliphatic)
• Water → alcohol (with retention of configuration if $S_N1$)

5. Mistake: Missing the "optically active" condition …