Chapter 9: Hydrocarbons Application Questions | chemistry Class 11 CBSE

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Chapter 9: Hydrocarbons Application Questions | chemistry Class 11 CBSE | ByteNotes.in

Application Question

Medium difficulty • Concept in a practical situation

Medium

Question 1

Applied Concept

A laboratory technician is given an unknown liquid and told it is either an alkane or an alkene. Describe the chemical tests the technician can perform to identify the compound, and explain the expected observations for each test.

Bromine water test: When bromine solution in CCl4 is added to the sample, an alkene immediately decolourises the reddish-orange colour by adding Br2 across the double bond (CH2=CH2 + Br2 → CH2Br-CH2Br); an alkane does not decolourise it under normal conditions (no UV light).
Baeyer's reagent test: Cold dilute KMnO4 (purple) is added; an alkene decolourises it by forming a vicinal glycol (oxidation at the double bond), while an alkane does not cause decolourisation under these mild conditions.
If neither test causes a change, the compound is an alkane; if both tests give positive decolourisation, the compound is an alkene.
Confirmation: The alkene can further be identified by its characteristic ozonolysis products (treating with O3 followed by Zn-H2O gives aldehydes/ketones that can be identified), while the alkane can be identified by combustion behaviour and IR spectroscopy showing absence of C=C stretch.

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Application Question

Hard difficulty • Concept in a practical situation

Hard

Question 2

Applied Concept

In an industrial plant, propene reacts with HBr to produce a bromopropane. Two different products are obtained depending on reaction conditions. Identify the products formed under (i) normal ionic conditions and (ii) in the presence of benzoyl peroxide, and explain the reason for the difference.

Under normal ionic conditions (absence of peroxide), HBr adds to propene following Markovnikov's rule. H+ acts as the electrophile and attaches to the terminal CH2 (more H atoms), forming the more stable secondary carbocation (CH3-CH+-CH3). Br- then attacks to give 2-bromopropane (CH3-CHBr-CH3) as the major product.
In the presence of benzoyl peroxide, the Kharash (peroxide) effect operates via a free radical mechanism. Benzoyl peroxide generates Br• radicals; the Br• adds to the terminal CH2 of propene (forming the more stable secondary carbon radical CH3-•CH-CH2Br), while H attaches to the internal CH, giving 1-bromopropane (CH3-CH2-CH2Br) as the major product.
The reason for the reversal is the difference in intermediate stability: secondary carbocation is more stable than primary (Markovnikov, ionic path); secondary free radical is more stable than primary (peroxide, free radical path) — in both cases the more stable intermediate forms preferentially, but the orientation of Br addition differs.
This reaction (peroxide effect) is specific to HBr; HCl does not show it (H-Cl bond too strong to be cleaved by radicals), and HI does not show it (I radicals are stable and recombine rather than adding).

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Application Question

Medium difficulty • Concept in a practical situation

Medium

Question 3

Applied Concept

With reference to the boiling point data in Table 9.2, explain why n-pentane (b.p. 309.1 K), 2-methylbutane (b.p. 300.9 K), and 2,2-dimethylpropane (b.p. 282.5 K) have different boiling points despite having the same molecular formula C5H12.

All three compounds are structural isomers with the same molecular formula (C5H12) and hence the same molecular mass. The difference in boiling points is solely due to the difference in molecular shape resulting from varying degrees of chain branching.
n-Pentane has a continuous chain of five carbons, giving a relatively elongated shape with maximum surface area for intermolecular contact. This allows the strongest van der Waals dispersion forces among the three isomers, requiring the most energy to vaporise (highest b.p., 309.1 K).
2-Methylbutane has one branch, making it more compact, reducing the surface area compared to n-pentane. The weaker intermolecular forces result in a lower boiling point (300.9 K).
2,2-Dimethylpropane (neopentane) has maximum branching and the most spherical shape, minimising surface area and intermolecular contact. It has the weakest van der Waals forces among the three isomers and the lowest boiling point (282.5 K).

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Application Question

Medium difficulty • Concept in a practical situation

Medium

Question 4

Applied Concept

A chemist wants to synthesise n-butane using the Wurtz reaction. Write the equation for the reaction and then explain why this method is unsuitable for synthesising pentane (C5H12).

To synthesise n-butane (C4H10) by Wurtz reaction: 2 C2H5Br + 2Na (dry ether) → C2H5-C2H5 + 2NaBr. Using two moles of ethyl bromide and sodium metal, n-butane is obtained cleanly because both alkyl groups are identical.
The Wurtz reaction works best for even-carbon alkanes because it involves coupling of two identical alkyl halides; when two identical halides are used, only one alkane product is obtained.
To synthesise pentane (C5H12 = n-pentane) one would need to couple a methyl halide (CH3X) with a butyl halide (C4H9X). However, when two different alkyl halides are used, three different alkane products can form: CH3-CH3 (ethane from CH3X + CH3X), C4H9-C4H9 (octane from C4H9X + C4H9X), and CH3-C4H9 (pentane, the desired product).
The formation of a mixture of three alkanes makes isolation of pure pentane inefficient and impractical. Therefore, Wurtz reaction is not preferred for synthesising alkanes with an odd number of carbon atoms.

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Application Question

Hard difficulty • Concept in a practical situation

Hard

Question 5

Applied Concept

A student performs ozonolysis on an unknown alkene 'X' and obtains ethanal (CH3CHO) and pentan-3-one (CH3-CO-CH2CH3) as products. Deduce the structure and IUPAC name of alkene 'X'.

In ozonolysis, the C=C double bond is cleaved and each carbon of the double bond becomes the carbonyl carbon of the product. The two carbonyl carbons in the products were originally joined by the double bond in the alkene.
Ethanal is CH3-CHO: the carbon bearing the H in the CHO was part of the double bond. The rest of the molecule is CH3- attached to the double bond carbon.
Pentan-3-one is CH3CH2-CO-CH2CH3: the carbonyl carbon (C3) was part of the double bond; the groups attached are -CH2CH3 on each side (actually CH3CO- means C2 is the carbonyl so the groups are -CH2CH3 and -CH3; re-examining: pentan-3-one is CH3-CH2-CO-CH2-CH3, so the double bond carbon had groups -CH2-CH3 and -CH2-CH3... wait: pentan-3-one = CH3CH2COCH2CH3, so the C=O carbon (C3) had -CH2CH3 on both sides and becomes the ketone; but since it is a ketone, this C was not terminal).
Combining: one end of the double bond gave ethanal (CH3-CH=), the other gave pentan-3-one (=C(CH2CH3)2 — i.e., =C(C2H5)2). Therefore, alkene X is: CH3-CH=C(C2H5)2, i.e., 3-ethylpent-2-ene. IUPAC name: 3-ethylpent-2-ene.

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Application Question

Medium difficulty • Concept in a practical situation

Medium

Question 6

Applied Concept

In a petrochemical refinery, dodecane (C12H26) from kerosene oil is subjected to catalytic cracking. Write the reaction for one possible cracking and explain the industrial importance of this process.

Dodecane undergoes pyrolysis (cracking) at 973 K in presence of Pt/Pd/Ni catalyst: C12H26 → C7H16 + C5H10 + other products. Heptane and pentene are among the key products.
Industrial importance: Cracking converts heavier, less valuable petroleum fractions into lighter, more valuable products. Petrol fractions (C5-C9 alkanes) obtained by cracking are directly used as automobile fuel, and the alkene co-products (ethene, propene) serve as essential monomers for plastics and chemical synthesis.
Since crude petroleum naturally contains a large proportion of heavier hydrocarbons (kerosene, gas oil), cracking allows refineries to maximise the yield of high-demand light fractions (petrol/LPG) from heavy crude, making it economically crucial for the petrochemical industry.
The mechanism is believed to be free radical in nature, similar to the halogenation of alkanes, where initial C-C bond homolysis generates radicals that propagate the chain to give diverse smaller products.

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Application Question

Medium difficulty • Concept in a practical situation

Medium

Question 7

Applied Concept

With reference to the concept of resonance in benzene, explain why all C-C bonds in benzene are equal in length and why benzene does not readily undergo addition reactions.

X-ray diffraction data shows that all six C-C bond lengths in benzene are identical at 139 pm, which is intermediate between a C-C single bond (154 pm) and a C=C double bond (133 pm). This uniformity cannot be explained by either Kekulé structure alone (which would predict alternating short and long bonds).
Benzene is a resonance hybrid of the two Kekulé structures (and other minor contributors). In the hybrid, no individual C-C bond is purely single or double; instead, the six pi electrons are fully delocalised over all six C-C bonds equally. Each C-C bond has the same fractional double bond character (~1.5 bond order), explaining the uniform bond length.
Due to delocalisation, the six pi electrons are more strongly attracted to all six carbon nuclei than localised pi electrons would be to just two carbons. This extra stabilisation (resonance/delocalisation energy) makes benzene much more stable than hypothetical cyclohexatriene.
Benzene resists addition reactions because addition would break the delocalised pi system, destroying aromaticity and losing the resonance stabilisation energy. The energy cost is prohibitive under normal conditions. Instead, electrophilic substitution preserves the aromatic ring, which is strongly thermodynamically favoured.

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Application Question

Medium difficulty • Concept in a practical situation

Medium

Question 8

Applied Concept

A student needs to distinguish between but-1-yne and but-2-yne. Suggest a simple chemical test based on the acidic character of alkynes, and write the relevant equations.

But-1-yne (HC≡C-CH2-CH3) is a terminal alkyne — it has a hydrogen atom directly attached to the triply bonded carbon. But-2-yne (CH3-C≡C-CH3) is an internal alkyne with no such H atom.
Test: Add sodium metal (Na) or sodamide (NaNH2) solution to each sample. But-1-yne (terminal alkyne) will react and liberate hydrogen gas: HC≡C-CH2-CH3 + Na → NaC≡C-CH2-CH3 + 1/2 H2 (effervescence); with NaNH2: HC≡C-CH2-CH3 + NaNH2 → NaC≡C-CH2-CH3 + NH3.
But-2-yne (internal alkyne) will NOT react with sodium metal or sodamide under these conditions because neither triply bonded carbon carries a hydrogen atom. No gas is evolved.
Conclusion: Effervescence of H2 gas with Na (or liberation of NH3 with NaNH2) identifies but-1-yne. No reaction confirms but-2-yne. This test exploits the acidic character unique to terminal alkynes due to the high electronegativity of sp hybridised carbons.

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Application Question

Medium difficulty • Concept in a practical situation

Medium

Question 9

Applied Concept

A polymer chemist uses ethene and propene to manufacture polythene and polypropylene respectively. Write the polymerisation reactions for both and briefly discuss one environmental concern associated with their widespread use.

Polymerisation of ethene: n(CH2=CH2) under high temperature, high pressure, and in presence of a catalyst → -(CH2-CH2)n- (polythene). Polythene is used for plastic bags, squeeze bottles, refrigerator dishes, toys, and pipes.
Polymerisation of propene: n(CH3-CH=CH2) under similar conditions → -(CH-CH2)n- with CH3 side chains (polypropylene). Polypropylene is used for milk crates, plastic buckets, and moulded articles.
Environmental concern: Both polythene and polypropylene are non-biodegradable polymers. They persist in the environment for centuries, accumulating in landfills, oceans, and soil. Plastic waste harms wildlife (animals ingest plastic fragments) and pollutes water bodies.
The chapter explicitly notes: 'excessive use of polythene and polypropylene is a matter of great concern for all of us,' highlighting the need for reduced use, improved recycling, and development of biodegradable alternatives.

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Application Question

Hard difficulty • Concept in a practical situation

Hard

Question 10

Applied Concept

Explain why toluene (methylbenzene) undergoes nitration more easily than benzene, while nitrobenzene undergoes nitration much more slowly and gives predominantly the meta-nitro product.

The methyl group (-CH3) in toluene is an ortho/para activating group. It donates electron density to the benzene ring by hyperconjugation and inductive effect (+I), increasing the pi electron density especially at the ortho and para positions, making the ring more electron-rich and therefore more reactive towards electrophilic attack (nitronium ion, NO2+) than benzene.
The nitro group (-NO2) in nitrobenzene is a strongly electron-withdrawing group with both -I and -M (mesomeric) effects. Its resonance structures show that it withdraws electron density from the ring especially from ortho and para positions, making them electron-poor.
The meta position in nitrobenzene is comparatively less electron-deficient, so the electrophile (NO2+) attacks there preferentially, giving predominantly the meta-dinitrobenzene product.
Overall reactivity order for nitration: toluene > benzene > nitrobenzene. Toluene's activating methyl group lowers the energy of activation; nitrobenzene's deactivating nitro group raises it significantly, making nitration of nitrobenzene require harsher conditions.

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Chapter 9: Hydrocarbons | Application Questions

A laboratory technician is given an unknown liquid and told it is either an alkane or an alkene. Describe the chemical tests the technician can perform to identify the compound, and explain the expected observations for each test.

In an industrial plant, propene reacts with HBr to produce a bromopropane. Two different products are obtained depending on reaction conditions. Identify the products formed under (i) normal ionic conditions and (ii) in the presence of benzoyl peroxide, and explain the reason for the difference.

With reference to the boiling point data in Table 9.2, explain why n-pentane (b.p. 309.1 K), 2-methylbutane (b.p. 300.9 K), and 2,2-dimethylpropane (b.p. 282.5 K) have different boiling points despite having the same molecular formula C5H12.

A chemist wants to synthesise n-butane using the Wurtz reaction. Write the equation for the reaction and then explain why this method is unsuitable for synthesising pentane (C5H12).

A student performs ozonolysis on an unknown alkene 'X' and obtains ethanal (CH3CHO) and pentan-3-one (CH3-CO-CH2CH3) as products. Deduce the structure and IUPAC name of alkene 'X'.

In a petrochemical refinery, dodecane (C12H26) from kerosene oil is subjected to catalytic cracking. Write the reaction for one possible cracking and explain the industrial importance of this process.

With reference to the concept of resonance in benzene, explain why all C-C bonds in benzene are equal in length and why benzene does not readily undergo addition reactions.

A student needs to distinguish between but-1-yne and but-2-yne. Suggest a simple chemical test based on the acidic character of alkynes, and write the relevant equations.

A polymer chemist uses ethene and propene to manufacture polythene and polypropylene respectively. Write the polymerisation reactions for both and briefly discuss one environmental concern associated with their widespread use.

Explain why toluene (methylbenzene) undergoes nitration more easily than benzene, while nitrobenzene undergoes nitration much more slowly and gives predominantly the meta-nitro product.

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