Class 11 Chemistry (Chemistry Part-II) Chapter 9 Hydrocarbons

Key Concept: Reaction Mechanism

b) Sodium propynide

[Solution Description] But-1-yne ($CH_3-C \equiv C-H$) can react with sodium to form monosodium derivative through the removal of a proton from the terminal carbon, leading to $CH_3-C \equiv C^-Na^+$. This compound is known as sodium propynide.

Key Concept: Resonance Stability

d) Due to resonance stability

[Solution Description] Benzene prefers substitution reactions because they allow the aromatic system to maintain its stable resonance energy. Addition reactions would disrupt this stability by breaking the conjugated pi-electron system.

Key Concept: Real-world Application, Reaction Optimization

c) Pressure

[Solution Description] Optimizing the dehydration of ethanol to ethylene requires focusing on factors that favor elimination over competing reactions such as oxidation or polymerization. These factors include maintaining high temperature to drive the reaction forward and ensuring low concentration of water which otherwise could reverse the reaction through hydration. While acid concentration influences the rate, pressure does not significantly alter the equilibrium towards alkene formation, thus making it the least critical factor for optimizing yield.

Key Concept: Reaction Type

b) Electrophilic addition

[Solution Description]

The addition of HBr to alkenes involves the attack of the proton (H+) on the alkene to form a carbocation intermediate, followed by the nucleophilic attack by Br-. This is characterized as electrophilic addition because the electron-rich double bond attacks the electrophilic proton.

Key Concept: Product Prediction, Comparative Analysis

b) Para-haloacetophenone

[Solution Description] Initially, benzene reacts with a halogen ($X_2$) in the presence of a Lewis acid leading to chlorination or bromination, producing chlorobenzene or bromobenzene. Subsequently, in Friedel-Crafts acylation using an acyl chloride and AlC$l_3$, the acylium ion attacks the ring. The halogen substituent, being ortho/para directing, facilitates acyl attachment at those positions. Thus, the major product would be a halo-ketone derivative, likely para-substituted due to steric reasons.

Key Concept: Free Radicals

a) $CH_3\cdot$

[Solution Description] During the propagation step of methane chlorination, a chlorine radical reacts with methane to produce a methyl radical ($CH_3\cdot$). This radical then participates further in the propagation steps to continue the reaction process.

Key Concept: Hückel’s Rule

b) 10 π electrons

[Solution Description] For a compound to be aromatic, it must follow Hückel’s rule which states that the compound should have $(4n + 2)$ π electrons where $n$ is an integer. Let’s consider each option:

a) 8 π electrons: If $4n + 2 = 8$, solving gives $4n = 6$, so $n = 1.5$. Not an integer, hence not aromatic.

b) 10 π electrons: If $4n + 2 = 10$, solving gives $4n = 8$, so $n = 2$. An integer, hence aromatic.

c) 12 π electrons: If $4n + 2 = 12$, solving gives $4n = 10$, so $n = 2.5$. Not an integer, hence not aromatic.

d) 14 π electrons: If $4n + 2 = 14$, solving gives $4n = 12$, so $n = 3$. An integer, however, only even numbered $n$ satisfies common ring sizes for stable aromatic systems.

Thus, the correct answer is b) as it has 10 π electrons.

Key Concept: Reaction Efficiency, Complex Reaction Pathway

b) Reduction using zinc and dilute hydrochloric acid

[Solution Description] To convert an alkyl halide like 1-chlorohexane into hexane, reduction can be performed. This involves using zinc and dilute hydrochloric acid.

The reaction proceeds as follows:

$$\text{C}_6\text{H}_{13}\text{Cl} + \text{Zn} + 2\text{HCl} \rightarrow \text{C}_6\text{H}_{14} + \text{ZnCl}_2$$

This reaction efficiently converts the chloroalkane to the corresponding alkane directly, without requiring a multi-step synthesis, making it optimal for resource management.

Key Concept: Position Isomerism Understanding

a) CH$_3$-CH=CH-CH$_2$-CH$_3$ and CH$_3$-CH$_2$-CH=CH-CH$_3$

[Solution Description] Position isomers differ in the position of a functional group or substituent on the carbon chain. For these compounds to be position isomers, they must have the same molecular formula and same carbon skeleton but different positions for the double bond. The pair A) CH$_3$-CH=CH-CH$_2$-CH$_3$ and CH$_3$-CH$_2$-CH=CH-CH$_3$ satisfies this condition as they both have the molecular formula C$_5$H$_{10}$, with the double bond located at different positions.

Key Concept: Real-world Application, Conformational Dynamics

b) 68 kJ

[Solution Description] To solve this problem, we first need to calculate the energy required to overcome the torsional strain for one mole. The given energy barrier for isobutane is 13.6 kJ/mol. Since there are 5 moles, the total energy required is calculated by multiplying the energy per mole with the number of moles:

$$\text{Total Energy} = 13.6 \, \text{kJ/mol} \times 5\, \text{mol} = 68 \, \text{kJ}$$

Therefore, the total energy required is 68 kJ.

Key Concept: Structural Analysis, Real World Application

a) 2,2,4-trimethylhexane

[Solution Description] Begin by identifying the longest continuous carbon chain in the structure. Counting through the main backbone, we find a chain of six carbons, making this a hexane derivative.

The main chain: CH$_3$-C-C-C-CH$_2$-CH$_3$ with substituents observed at $C_2$ and $C_4$:

– There are two methyl groups attached to the second carbon (2,2).

– Another methyl group at the fourth carbon.

Since there are three methyl groups overall, the naming follows:

“2,2,4-trimethylhexane” correctly represents the compound’s structure.

Key Concept: Reaction Optimization, Reaction Efficiency

d) $Cu$, 523K

[Solution Description] The reaction $2CH_4 + O_2 \xrightarrow{Cu/523K/100atm} 2CH_3OH$ suggests that a $Cu$ catalyst at 523K and 100 atm optimizes both yield and selectivity for methanol production. This is because these conditions provide an appropriate environment for the catalytic action lowering activation energy, crucial for efficient conversion while minimizing side reactions.

Key Concept: Electrophilic Reagents

b) Electrophilic

[Solution Description] Alkenes have loosely held pi electrons available for reactions, making them susceptible to electrophiles. Electrophiles are reagents that seek out and attack areas of high electron density because they are electron deficient themselves. As such, electrophilic reagents typically react with alkenes.

Key Concept: Alkyl Group Recognition

b) Propyl

[Solution Description] An alkyl group is formed by removing one hydrogen atom from an alkane. For propane (C$_3$H$_8$), removing a hydrogen atom forms the propyl group with the formula C$_3$H$_7$. This can be either n-propyl or isopropyl depending on which hydrogen atom is removed. However, the straightforward removal from the end gives n-propyl group.

Key Concept: Fuel Types

c) LPG

[Solution Description] LPG (Liquefied Petroleum Gas) is primarily composed of hydrocarbons, such as propane and butane. It serves as an energy source consisting mainly of these hydrocarbon gases.

Key Concept: Reaction Identification

b) Electrophilic addition

[Solution Description] The addition of cold concentrated sulphuric acid to an alkene involves the breaking of the double bond and the formation of new bonds with the sulphuric acid, classifying it as electrophilic addition.

Key Concept: Proton Transfer Mechanism

b) Donates a proton to generate $NO_2^+$

[Solution Description] Sulfuric acid ($H_2SO_4$) acts as a proton donor in the formation of the nitronium ion ($NO_2^+$). The reaction proceeds as follows:

$$HNO_3 + H_2SO_4 \rightarrow NO_2^+ + H_2O + HSO_4^-$$

Here, sulfuric acid donates a proton to nitric acid, facilitating the generation of the electrophile $NO_2^+$.

Key Concept: Stability and Reactivity

b) Weaker pi bonds

[Solution Description] Alkenes contain a weak pi bond due to poor sideways overlapping, making them more reactive compared to alkanes which have stronger single sigma bonds.

Key Concept: Synthesis Analysis, Structural Transformation

c) Tertiary

[Solution Description] Butane has the following structural formula:

$$\text{CH}_3-\text{CH}_2-\text{CH}_2-\text{CH}_3$$

If a methyl group is added to one of the secondary carbons, for instance, the second carbon, the resulting compound would have the following structure:

$$\text{CH}_3-\text{C}(\text{CH}_3)(\text{H})-\text{CH}_2-\text{CH}_3$$

This results in the formation of 2-methylbutane, where the second carbon now makes three bonds with other carbon atoms, classifying it as a tertiary carbon.

Key Concept: Fuel Types

d) LPG

[Solution Description] LPG is considered a clean fuel for domestic use because it produces fewer pollutants compared to solid fuels like coal and wood.

Key Concept: Real-world Application, Role of Free Radicals

c) Adding inhibitors

[Solution Description] Here, the focus is on controlling radical polymerization to enhance yield. Industrially, controlling radical concentration helps balance between growth and termination phases. By adding inhibitors, such as chemical agents that slow down radical termination, or removing excess radicals from the system, one can limit unnecessary termination steps and reduce byproduct formation.

Thus, adding inhibitors is a practical solution to control the termination phase.

Key Concept: Energy Calculations, Torsional Strain Analysis

b) The eclipsed conformation becomes predominant.

[Solution Description] If the energy barrier is reduced from 12.5 kJ mol$^{-1}$ to 6.25 kJ mol$^{-1}$, the ease of rotation increases, leading to a higher proportion of molecules existing in the less stable eclipsed conformation at any given moment due to lower hindrance to rotation.

Key Concept: Advanced Conformation Analysis, Stereochemistry

b) Ethyl group at 180°, Hydrogen atoms at 60° and 300°

[Solution Description]

To minimize torsional strain in staggered conformations, bulky groups should be positioned as far apart as possible. In the case of 2-methylbutane, where there is a methyl group on the front carbon at 60°, placing the larger ethyl group opposite to it on the rear carbon will minimize strain. This occurs when the ethyl group on the rear carbon is located at 180° relative to the front carbon’s methyl group.

Key Concept: Comparative Mechanism, Reaction Pathway Analysis

d) The presence of SO$_3$H promotes ortho/para substitution in halogenation more efficiently.

[Solution Description] The sulphonation process generates a deactivating group (-SO$_3$H), which is a meta-director due to its electron-withdrawing nature. In contrast, halogens are ortho/para directors despite being deactivating. Thus, an initial sulphonation step will direct subsequent sulphonations to meta positions relative to -SO$_3$H. The inductive effect of the sulphonic group withdraws electrons, stabilizing the intermediate cation at meta position stronger than ortho/para.

Key Concept: Detailed Structure

d) At 120°

[Solution Description]

In a Sawhorse projection, each carbon atom has three bonds that are inclined at 120° to each other. This is due to the tetrahedral geometry around the carbon atoms, where bond angles are approximately 109.5° in space but appear as 120° in the planar Sawhorse representation.

Key Concept: Conformation Isolation, Projection Analysis

b) Low energy barriers allow easy interconversion

[Solution Description] Ethane has low energy barriers for rotation around the C-C sigma bond, approximately $12.5 \, \text{kJ/mol}$, which allows rapid interconversion between various conformations like staggered and eclipsed forms. This makes it difficult to isolate specific conformers. Moreover, visualizing these interconversions via Sawhorse or Newman projections shows that both involve continuous transition through all possible conformations rather than discrete stable structures, highlighting the dynamic nature of such molecular systems.

Key Concept: Complex Isomerism, Structural Analysis

d) 2-methylheptane

[Solution Description] The molecular formula $C_7H_{16}$ represents heptane and its isomers. Possible chain isomers include linear heptane and various branched structures. To identify the incorrect structure, we must analyze each option to ensure it follows the general alkane rule of $C_nH_{2n+2}$ where $n = 7$. Also, verify that each option is a plausible rearrangement of the carbon skeleton. Common isomers are n-heptane, 2-methylhexane, 3-methylhexane, and others with more branching. Determine which does not match these criteria.

Key Concept: Reaction Specificity

c) Concentrated sulphuric acid

[Solution Description] In the nitration of benzene, concentrated sulfuric acid acts as a catalyst to generate the nitronium ion ($NO_2^+$), which is the active electrophile that attacks the benzene ring.

Key Concept: Reaction Pathway Analysis, Advanced Mechanism

c) Alkyl free radical

[Solution Description] Pyrolysis involves a free radical mechanism where the initial step involves homolytic cleavage that produces free radicals. The most plausible initial intermediate for dodecane is the formation of alkyl radicals.

Key Concept: Stereochemistry, Reaction Optimization

b) Cis-butene; low pressure optimizes selectivity

[Solution Description] Partial hydrogenation with Lindlar’s catalyst is known for converting alkynes to cis-alkenes, as it selectively stops the reaction at the alkene stage. Under low-pressure conditions, the reaction kinetics are controlled, minimizing the energy input and preventing further hydrogenation to alkanes.

Key Concept: Comparative Analysis, Process Optimization

b) Catalysts are optimized for greater selectivity towards aromatic compounds

[Solution Description] Aromatization involves dehydrogenation and cyclization, whereas other reforming processes like cracking primarily involve breaking carbon chains. The catalysts used in aromatization, such as vanadium or chromium oxide supported on alumina, are specifically efficient at facilitating these transformations. In terms of product yield, aromatization is more selective for forming benzene and homologues, unlike cracking which produces a wider range of smaller hydrocarbons.

Key Concept: Stereochemistry Application, Complex Property Analysis

b) Trans-but-2-ene due to efficient packing

[Solution Description] The trans-isomer of but-2-ene typically exhibits a higher melting point than the cis-isomer because it packs more efficiently in the crystal lattice owing to its linearity, resulting in stronger van der Waals forces.

Key Concept: Polymer Structure Analysis, Environmental Impact Analysis

b) Methyl side-groups providing steric hindrance in polypropene

[Solution Description]

Polythene consists of a simple hydrocarbon chain whereas polypropene contains methyl groups attached to every other carbon. These methyl groups increase the sterical hindrance preventing effective attack by microbes or environmental factors, thus influencing degradation rates.

Key Concept: Structural Formula Construction

d) $CH_2=C(CH_3)-CH_2-CH_3$

[Solution Description]

The IUPAC name 3-Methylbut-1-ene indicates a five-carbon chain with a methyl group on the third carbon and a double bond starting from the first carbon. Therefore, the structure will be:

– $CH_2=C(CH_3)-CH_2-CH_3$

The correct structural formula according to this description is option D.

Key Concept: Aromaticity Rule

d) Hückel’s Rule

[Solution Description]

Aromaticity in benzene is determined by Hückel’s Rule, which states that for a molecule to be aromatic, it must have $(4n + 2)\ \pi$ electrons, where $n$ is an integer. Benzene satisfies this with six $\pi$ electrons.

Key Concept: Comparative Analysis

d) Polyacetylene

[Solution Description] Polyacetylene is known for its conductive properties. Unlike other commonly used plastics like polythene and polypropene, polyacetylene has a conjugated system which allows it to conduct electricity, making it suitable for use as electrodes in batteries. Other polymers do not have this property.

Key Concept: Physical Properties

b) Hexane

[Solution Description] Alkanes with carbon numbers from 1 to 4 are gases at room temperature, while those having carbon numbers from 5 to 17 are liquids. Hence, hexane ($C_6H_{14}$) is a liquid at room temperature.

Key Concept: Product Prediction

a) Hexane

[Solution Description] Hex-3-yne ($CH_3-CH_2-C \equiv C-CH_2-CH_3$) reacts with excess dihydrogen in the presence of a nickel catalyst. This leads to full saturation of the triple bond through a stepwise reduction process: $R-C \equiv C-R’ \rightarrow R-CH=CH-R’ \rightarrow R-CH_2-CH_2-R’$. Thus, the complete hydrogenation yields hexane ($CH_3-CH_2-CH_2-CH_2-CH_2-CH_3$).

Key Concept: Dipole Moment Analysis

b) Cis-but-2-ene = 0.33 Debye; Trans-but-2-ene = 0 Debye

[Solution Description] Dipole moment measures the separation of charge in a molecule. In cis-but-2-ene, identical groups are on the same side of the C=C bond, creating an overall dipole moment because of the asymmetrical distribution of electron density. On the other hand, trans-but-2-ene has symmetrical distribution about the double bond, leading to cancelation of dipoles. Thus, cis-but-2-ene has a dipole moment of 0.33 Debye, while trans-but-2-ene has zero.

Key Concept: Real-world Application, Environmental Strategy

a) Use of catalytic converters

[Solution Description] The combustion of octane, like other alkanes, results in carbon dioxide and water. To mitigate the environmental impact, especially reducing greenhouse gas emissions:

$$C_8H_{18} + \frac{25}{2}O_2 \rightarrow 8CO_2 + 9H_2O$$

Adopting strategies such as using catalytic converters can significantly decrease harmful emissions by converting exhaust gases into less harmful substances before they are released into the atmosphere. Strategies must focus on both controlling emissions and increasing combustion efficiency to lower CO$_2$ production.

Key Concept: Anti-Markovnikov Product

a) 1-bromohexane

[Solution Description] In the presence of peroxides, the anti-Markovnikov addition occurs due to the free radical mechanism. In hex-1-ene, the bromine atom will add to the less substituted end of the double bond, producing 1-bromohexane as the major product.

Key Concept: Electrophile Formation

b) Acylation

[Solution Description] Anhydrous AlCl$_3$ is a common Lewis acid used in generating electrophiles such as Cl⊕, R⊕, and RC⊕O. Among the given options, acylation involves the use of anhydrous AlCl$_3$ to create an acylium ion (RC⊕O) as the electrophile.

Key Concept: Real-World Solubility, Solubility and Molecular Interactions

d) Acetone

[Solution Description] Benzoic acid consists of a polar carboxylic acid group and a non-polar benzene ring. For solubility, the principle “like dissolves like” applies. The solvent should ideally dissolve both polar and non-polar regions. Therefore, a semi-polar solvent that can interact via hydrogen bonding as well as van der Waals forces would be suitable. Acetone has such properties; it can participate in hydrogen bonding due to its carbonyl group, while also having a non-polar region.

Key Concept: Conceptual Application, Detailed Mechanism

c) Octane

[Solution Description]

Electrolysis of lithium pentanoate ($\text{C}_4\text{H}_9\text{COO}^-\text{Li}^+$) follows Kolbe’s method, generating free radicals that lead to the formation of the alkane:

$$\text{2C}_4\text{H}_9\text{COO}^- \rightarrow 2\text{C}_4\text{H}_9\cdot + 2\text{CO}_2 + 2e^-$$

The resulting butyl radicals ($\text{C}_4\text{H}_9\cdot$) join to form octane:

$$\text{C}_8\text{H}_{18}$$

Therefore, octane is the primary hydrocarbon product.

Key Concept: Isomerism Type

b) Position isomerism

[Solution Description]

But-1-yne and But-2-yne differ in the position of the triple bond along the carbon chain. This change in position constitutes position isomerism as both compounds have the same molecular formula but different structural formulas due to the location of the functional group (the triple bond).

Therefore, these compounds exhibit position isomerism.

Key Concept: Mechanism Details

a) It changes from $sp^2$ to $sp^3$

[Solution Description] During the nitration of benzene, the formation of the $\sigma$-complex involves a change in hybridization from $sp^2$ to $sp^3$ for the carbon atom that forms a bond with the nitronium ion. This allows the addition of the electrophile to one of the carbon atoms in the benzene ring, temporarily forming a tetrahedral geometry at the site of attack.

Key Concept: Source Identification

d) Natural Gas

[Solution Description] CNG (Compressed Natural Gas) is primarily sourced from natural gas fields and consists mostly of methane ($CH_4$), which is extracted during the processing of natural gas.

Key Concept: Torsional Strain

b) Repulsive forces between adjacent hydrogen atoms

[Solution Description] In Newman projections, torsional strain occurs when groups on adjacent carbon atoms repel each other due to their proximity in the eclipsed conformation. This repulsion increases the potential energy of the molecule, making it less stable compared to staggered conformation where such group interactions are minimized.

Key Concept: Directive Influence Analysis, Electrophilic Substitution Mechanism

d) Both ortho and para-nitroanisoles

[Solution Description] Anisole has a methoxy (-OCH$_3$) group that strongly activates the benzene ring toward electrophilic substitution and directs the incoming electrophile to the ortho and para positions via resonance donation. During nitration, the nitronium ion ($NO_2^+$) acts as the electrophile. The resonance effect of the -OCH$_3$ group makes the ortho and para positions highly reactive toward the electrophile. Therefore, the major products are ortho-nitroanisole and para-nitroanisole, with para being typically more stable and predominant due to steric factors.

Key Concept: Reaction Mechanisms, Synthesis Pathways

c) 2,2-Dichlorobutane

[Solution Description] During the reaction of an alkyne like 1-butyne ($CH_3C\equiv CCH_3$) with excess HCl, the addition of HCl occurs in two steps due to the presence of a triple bond. In the first step, one molecule of HCl adds across the triple bond resulting in a chloroalkene (2-chlorobutene). As more HCl is available, it further adds to convert the double bond into a single bond, forming a dichloroalkane. The expected final product will be 2,2-dichlorobutane as follows:

$$CH_3C\equiv CCH_3 + 2HCl \rightarrow CH_3CH(Cl)CH_2CH_2(Cl)$$

Thus, the major product is 2,2-Dichlorobutane.

Key Concept: Synthesis Challenge, Reaction Condition Analysis

c) Ethylene $\xrightarrow{O_3}$ Ozonide $\xrightarrow{Zn-H_2O}$$ Ethanal $\xrightarrow{K_2Cr_2O_7}$ Ethanoic acid

[Solution Description] Converting ethylene to ethanoic acid requires a systematic approach involving several steps:

1. Ethylene reacts with ozone to form ozonides, which are then reduced with $Zn-H_2O$ to produce ethanal.

2. Further oxidation of ethanal with $K_2Cr_2O_7$ or $KMnO_4$ under acidic conditions converts it to ethanoic acid.

Thus, understanding the sequential reactions helps achieve the desired compound efficiently.

Key Concept: Industrial Applications, Advanced Reaction Pathways

b) Steam reforming for hydrogen production

[Solution Description] Partial oxidation of methane involves converting it into syngas (a mixture of $CO$ and $H_2$). This process is pivotal in steam reforming for hydrogen production:

$$CH_4 + \frac{1}{2}O_2 \rightarrow CO + 2H_2$$

This method is widely used in producing hydrogen for various industrial applications.

Key Concept: Role of Catalysts

c) It generates the carbocation by accepting halide ions.

[Solution Description] Anhydrous AlC$l_3$ acts as a Lewis acid catalyst. In the Friedel-Crafts alkylation process, it helps to generate a carbocation by abstracting a halide ion from the alkyl halide. This carbocation then acts as the electrophile, which attacks the benzene ring to form a σ-complex or arenium ion.

Key Concept: Electrophilic Substitution

c) Hydrogenation

[Solution Description] Electrophilic substitution reactions involve the replacement of a hydrogen atom in benzene by an electrophile. Common reactions include nitration, halogenation, sulphonation, and Friedel-Crafts alkylation/acylation. Hydrogenation is not an electrophilic substitution because it involves adding hydrogen to benzene under different conditions.

Key Concept: Basic Naming

c) 4-tert-Butylnonane

[Solution Description] The given structure can be named by identifying the longest carbon chain and then naming substituents. The longest chain contains 9 carbons, making it nonane. The substituent $-C(CH_3$)$_2$ appears at position 4, which is a tert-butyl group.

Thus, the IUPAC name is 4-tert-Butylnonane.

Key Concept: Directive Influence

d) -NO$_2$

[Solution Description] In electrophilic substitution reactions, substituents on the benzene ring can influence the position at which new groups attach. Meta directors typically have an electron-withdrawing nature, such as -NO$_2$, -CN, and -COOH. These groups pull electron density away from the ortho and para positions, directing incoming electrophiles to the meta position. Among the options provided, the nitro group (-NO$_2$) is a classic example of a meta-director due to its strong electron-withdrawing properties.

Key Concept: Boiling Point Trends

c) 2,2-Dimethylpropane

[Solution Description] Boiling points in alkanes are affected by chain length and branching. Generally, longer chains have higher boiling points due to increased surface area and stronger van der Waals forces. Branching reduces these interactions, lowering boiling points. Here, 2,2-dimethylpropane, being highly branched, will have the lowest boiling point compared to other less-branched options with similar molar mass.

Key Concept: Isomeric Structures, Boiling Point Anomalies

c) 2,2-dimethylpropane

[Solution Description] To determine which isomer has the lowest boiling point, consider the branching in each molecule. More branched isomers tend to have lower boiling points due to decreased surface area and weaker van der Waals forces. The isomers in consideration are:

– n-pentane: no branching, higher surface area.

– 2-methylbutane (isopentane): one branch, reduced surface area.

– 2,2-dimethylpropane (neopentane): two branches, much reduced surface area.

Among these, 2,2-dimethylpropane, being highly branched, will have the lowest boiling point as it has the least surface contact for intermolecular interactions.

Key Concept: Source Identification

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

[Solution Description] Benzene is indeed extracted on a commercial scale from coal tar because it contains various aromatic compounds such as benzene, toluene, etc. The assertion correctly states that benzene is commercially isolated from coal tar, and the reason provides an accurate explanation of why this is possible due to the presence of aromatic compounds in coal tar.

Key Concept: Resonance Structures, Orbital Hybridization

c) Delocalization of $\pi$ electrons results in identical bond lengths that provide high stability.

[Solution Description] Benzene’s stability comes from the delocalization of $\pi$ electrons across the ring structure, leading to equivalent resonance structures. The $sp^2$ hybridization of carbon atoms ensures a planar geometry, allowing optimal overlap of p orbitals and formation of a continuous $\pi$ electron cloud.

Key Concept: Mechanism Details, Reaction Pathway Analysis

b) Addition of bromine to cis-2-butene

[Solution Description] To determine which reaction results in a meso compound, we need to analyze the stereochemistry of the products formed in each case. A meso compound is an achiral compound that has chiral centers and an internal plane of symmetry. Consider the addition of bromine to alkenes:

– For 2-butene, the chlorination would lead to the anti-addition, resulting in a racemic mixture or a meso compound depending on the starting material being cis or trans.

– In the specific case of cis-2-butene, both additions occur from opposite sides leading to identical substituents on opposite sides forming a meso compound.

Thus, adding bromine to cis-2-butene results in a meso compound.

Key Concept: Structure Identification

b) 3-Methylpentane: CH$_3$-CH$_2$-CH(CH$_3$)-CH$_2$-CH$_3$

[Solution Description]

Chain isomers have the same molecular formula but different carbon chain structures. The molecular formula for 2-Methylpentane is C$_6$H$_{14}$. To find its chain isomer, we need to rearrange the carbon atoms without altering the total number of carbons or hydrogens.

The molecular structure of 2-Methylpentane: CH$_3$-CH(CH$_3$)-CH$_2$-CH$_2$-CH$_3$

Possible chain isomers:

a) Hexane: CH$_3$-CH$_2$-CH$_2$-CH$_2$-CH$_2$-CH$_3$

b) 3-Methylpentane: CH$_3$-CH$_2$-CH(CH$_3$)-CH$_2$-CH$_3$

c) 2,3-Dimethylbutane: CH$_3$-C(CH$_3$)(CH$_3$)-CH$_2$-CH$_3$

Among these options, 3-Methylpentane (Option b) is a plausible chain isomer as it shares the same molecular formula, C$_6$H$_{14}$, but has a different branching pattern.

Key Concept: Radical Chain Reaction

c) CH$_3$ radical reacts with Cl$_2$

[Solution Description] The propagation step of a radical chain reaction involves the sequence where radicals react with stable molecules to form new radicals, perpetuating the reaction. For methane chlorination:

Step 1: $\text{Cl} \cdot + \text{CH}_4 \rightarrow \text{CH}_3 \cdot + \text{HCl}$

Step 2: $\text{CH}_3 \cdot + \text{Cl}_2 \rightarrow \text{CH}_3\text{Cl} + \text{Cl} \cdot$

In Step 2, the methyl radical reacts with molecular chlorine to form chloromethane and regenerate the chlorine radical.

Key Concept: Comparison with Unsymmetrical Alkenes, Reaction Pathway

b) Symmetrical alkenes lead to a single type of carbocation; unsymmetrical produce multiple possibilities

[Solution Description] In the case of ethylene, a symmetrical alkene, the carbocation formed after protonation is identical on both sides and thus equally stabilized. For an unsymmetrical alkene like propene, the more substituted secondary carbocation is preferred, making the decision pathway important for regioselectivity.

Key Concept: Reaction Mechanism, Advanced Reasoning

c) It serves as an electron donor, enabling complete deprotonation.

[Solution Description] When ethyne first reacts with sodium, monosodium ethynide forms with the release of hydrogen gas. The second sodium atom further donates an electron, facilitating the removal of the remaining hydrogen as gas and forming a di-anionic species $Na^+C^-Na^+ \equiv C^-Na^+$, leading to increased ionic character and stability through additional electron donation.

Key Concept: Reaction Application, Complex Reaction Analysis

c) Lower reaction temperature drastically

[Solution Description] Methane production in Wurtz reactions using methyl chloride poses challenges since the typical route leads to ethane:

$$ 2\,CH_3Cl + 2\,Na \rightarrow C_2H_6 + 2\,NaCl $$

For selective methane synthesis, strong control over rate and medium conditions is crucial. One approach:

– **Alter solvent property:** Use solvents that favor radical stabilization rather than propagation to larger alkanes.

– **Catalyst introduction:** Introduce catalysts that break down intermediate radicals back to methane instead of allowing coupling.

Another key aspect lies in employing inhibitors that halt dimerization – possibly yielding some fraction of methane.

Key Concept: Product Prediction

b) $C_2H_5CH=CH_2$

[Solution Description] The reaction involves the removal of two bromine atoms from adjacent carbons by zinc, forming $ZnBr_2$. This leaves a double bond between the two carbons, yielding an alkene. Thus, the product formed is $C_2H_5CH=CH_2$.

Key Concept: Isomerism

c) Position and chain isomerism

[Solution Description] Alkynes primarily exhibit position and chain isomerism. Position isomerism occurs due to different positions of the triple bond along the carbon chain, while chain isomerism arises from variations in the carbon skeleton.

Key Concept: Aromaticity Criteria

c) Alternating single and double bonds within a plane

[Solution Description] Benzene’s aromatic stability arises from planarity, complete delocalization of its π electrons, and adherence to Hückel’s rule of having (4n+2) π electrons. This allows it to maintain a conjugated system that increases its energetic stability. A non-planar structure would disrupt this conjugation.

Key Concept: Chemical Properties, Environmental Considerations

c), Non-polarity

[Solution Description] Hydrocarbons are non-polar compounds, which makes them excellent solvents for dissolving other non-polar substances. Their relatively low reactivity ensures they do not participate in unwanted side reactions during usage. Furthermore, many hydrocarbons have moderate boiling points, facilitating easy removal or recovery post-application. However, aromatic hydrocarbons might pose higher health risks, necessitating careful selection based on environmental guidelines.

Key Concept: Electrophilic Addition, Reaction Mechanisms

b) 2-Chloro-2-methylpropane

[Solution Description]

In the electrophilic addition reaction of HCl to 2-methylpropene, the first step involves the protonation of the double bond to form a more stable carbocation. The carbocation formed from 2-methylpropene is a tertiary carbocation due to the positive charge being on the carbon atom bonded to three other carbon atoms. This stability leads to the major formation of 2-chloro-2-methylpropane:

$$\text{C}_3\text{H}_6 + \text{HCl} \rightarrow \text{C}_4\text{H}_9\text{Cl}$$

where C$_4$H$_9$Cl corresponds to 2-chloro-2-methylpropane.

Thus, the major product is 2-chloro-2-methylpropane.

Key Concept: Double Bond Characteristics

c) Alkenes’ pi bonds have mobile electrons making them reactive.

[Solution Description] Alkenes contain a carbon-carbon double bond consisting of both a sigma ($\sigma$) and pi ($\pi$) bond. The pi bond, formed by lateral overlapping of 2p orbitals, contains mobile electrons that make alkenes susceptible to electrophilic attacks. This increased electron density in the pi bond is less stable than the sigma bond in alkanes, making alkenes more reactive.

Key Concept: Reaction Pathway, Catalyst Role

b) AlC$l_3$ facilitates carbocation formation leading to rearrangement; main product is 2-methylbutane

[Solution Description] In isomerisation, AlC$l_3$ acts as a Lewis acid facilitating carbocation formation from alkanes, allowing rearrangement to form branched-chain alkanes. For n-pentane, the reaction yields 2-methylbutane as a major product following the rearrangement facilitated by the carbocation intermediate.

Key Concept: Energy Consideration, Mechanism Comparison

b) Due to stronger H-Cl bond energy

[Solution Description]

The peroxide effect is specific to HBr because it involves breaking the H-Br bond, which has a lower bond energy (363.7 kJ mol${^{-1}}$) compared to the H-Cl bond (430.5 kJ mol${^{-1}}$). The higher bond energy of H-Cl makes it energetically unfavorable to produce chlorine radicals under similar conditions. Hence, no free radical mechanism occurs with HCl, and thus, no peroxide effect.

Steps:

– Bond energy consideration: $\text{H-Br bond energy} = 363.7 \text{kJ mol}^{-1}$

– Compare with: $\text{H-Cl bond energy} = 430.5 \text{kJ mol}^{-1}$

Conclusion: Higher bond energy for H-Cl prevents the peroxide effect.

Key Concept: Real-world Application, Reaction Pathway

c) Use excess benzene

[Solution Description]

To optimize the formation of ethylbenzene, one effective strategy is to use excess benzene. This minimizes the formation of polyalkylated products because the high concentration of benzene dilutes the chance of further alkylation once ethylbenzene is formed, thus favoring mono-alkylation over multiple substitutions.

Key Concept: Reaction Outcome

b) Aldehydes or ketones

[Solution Description]

Ozonolysis of alkenes typically results in the formation of aldehydes or ketones. The alkene’s double bond is cleaved to produce these carbonyl compounds, depending on the substituents attached to the original double bond.

Key Concept: Hybridization

c) From sp to sp${^2}$

[Solution Description] In ethyne, each carbon atom is sp hybridized, allowing formation of a linear structure with a triple bond. Upon adding bromine, it breaks into a single bond changing the hybridization of the carbon from sp to sp${^2}$.

Key Concept: Industrial Application

b) Preparation of oil gas from kerosene

[Solution Description] Pyrolysis is widely used in the industry for transforming kerosene oil or petrol into gaseous fuels like oil gas or petrol gas. This process is an essential aspect of converting liquid hydrocarbons into valuable gaseous products.

Key Concept: Reaction Pathway Analysis, Reaction Condition Optimization

c) $C_2H_6 \xrightarrow{(CH_3COO)Mn/\Delta} CH_3COOH$

[Solution Description]

Under controlled oxidation, ethane can be oxidized to ethanoic acid using a specific catalyst and temperature condition. The process is carried out as follows:

1. Use the catalyst $(CH_3COO)Mn$.

2. The reaction requires heating ($\Delta$).

This setup ensures that the ethane ($C_2H_6$) is converted to ethanoic acid ($CH_3COOH$) efficiently without over-oxidation to carbon dioxide or other by-products.

Key Concept: Reaction Pathway, Intermediate Stability

d) 1,2-Dibromoethane

[Solution Description]

Acetylene ($C_2H_2$) undergoes radical addition of HBr in the presence of peroxide. The antimarkovnikov rule applies, leading to initial formation of vinyl bromide ($CH_2=CHBr$). Further addition of HBr forms 1,2-dibromoethane ($CH_2Br-CH_2Br$).

Key Concept: Radical Chain Length, Product Distribution

a) Concentration of chlorine

[Solution Description]

The chain length in a radical halogenation reaction refers to the number of times a propagation cycle occurs before termination. In the chlorination of propane, factors affecting chain length include reactant concentration, temperature, and inhibitors. Among these, the concentration of reactants, particularly chlorine, directly influences how many times the propagation cycle can occur before termination happens. Higher concentrations result in longer chain lengths, facilitating more complete conversion to various chlorinated products.

For chloropropanes, their distribution depends on which hydrogen atom is substituted, influenced by chain length and reactivity preference (tertiary < secondary < primary for chlorination).

Key Concept: Reaction Temperature

c) 333 K

[Solution Description] The specific condition required for the addition of water to alkynes involves warming them at 333 K along with mercuric sulphate and dilute sulphuric acid. This temperature facilitates the conversion of alkynes to their corresponding carbonyl compounds.

Key Concept: Reaction Conditions, Real-world Application

a) High temperature and pressure with a nickel catalyst

[Solution Description] Hydrogenation of benzene to form cyclohexane requires breaking the stable aromatic structure. This process typically demands high temperatures and pressures along with a suitable catalyst such as nickel. Under these harsh conditions, the double bonds within the benzene ring can be saturated with hydrogen atoms, converting it to cyclohexane.

Key Concept: Preparation Methods

b) Hydrogenation

[Solution Description] The conversion of ethene (an alkene) to ethane (an alkane) involves adding hydrogen across the double bond, a process known as hydrogenation. This requires a catalyst such as platinum or palladium.

Key Concept: Hybridization and Bonding

b) sp$^2$

[Solution Description] A carbon-carbon double bond involves one sigma bond formed by sp$^2$ hybridized orbitals and one pi bond formed by unhybridized p orbitals overlapping sideways. Therefore, sp$^2$ hybridization is involved.

Key Concept: Real-world Application, Synthesis Strategy

d) Use of Br$_2$ in water

[Solution Description] Phenol contains an -OH group that directs incoming groups to the ortho and para positions. To synthesize p-bromophenol, bromination should be performed under conditions that favor para substitution. The -OH group can increase electron density through resonance, making the para position more reactive compared to the ortho position when using dilute bromine water (Br$_2$). Therefore, the correct reagent sequence starts with Br$_2$.

Key Concept: Real-World Solubility, Complex Boiling Point Analysis

c) Sodium chloride will dissolve entirely in ethanol, not mixing with octane.

[Solution Description] Sodium chloride is ionic and will not dissolve in a non-polar solvent like octane but can dissolve in a polar solvent like ethanol. However, given the mixed nature of the solvent, it will primarily dissolve in ethanol due to its ionic nature.

Key Concept: Mechanistic Comparison, Reaction Pathway

b) Arenium ion

[Solution Description]

The mechanism of nitration involves three main steps:

1. Generation of the electrophile ($NO_2^+$).

2. Formation of the arenium ion after the electrophile attacks the benzene ring.

3. Removal of a proton from the arenium ion to restore the aromaticity of the benzene ring.

In step 2, the formation of an arenium ion is crucial as it temporarily disrupts the aromaticity of benzene while allowing for further transformations. Ultimately, the arenium ion plays a pivotal role in step 3 by donating a proton, thus restoring the aromatic character of the molecule.

Thus, the arenium ion is the key intermediate that facilitates proton removal and helps in regaining aromatic stability.

Key Concept: Basic Reaction

b) Toluene

[Solution Description]

Benzene undergoes Friedel-Crafts alkylation with methyl chloride in the presence of anhydrous AlCl$_3$. An electrophilic substitution occurs where a methyl group replaces a hydrogen atom on the benzene ring, forming toluene (methylbenzene).

Key Concept: Reaction Mechanism, Reaction Efficiency

b) Decrease temperature to favor selective radical formation

[Solution Description] In the case of chlorination of propane, the formation of 2-chloropropane involves the abstraction of a secondary hydrogen atom because secondary radicals are more stable than primary ones due to hyperconjugation and inductive effects. Chlorine prefers reacting with more substituted carbon atoms. To maximize the yield of 2-chloropropane, it is crucial to control the reaction such that propagation steps favor secondary radical formation. This can be achieved by controlling the concentration of chlorine radicals and the temperature to minimize excessive reactivity which might lead to competing reactions.

Key Concept: Comparative Analysis

b) Ozonolysis cleaves double bonds completely, KMnO$_4$ adds across the double bond

[Solution Description] Ozonolysis involves breaking the C=C double bond entirely to give smaller molecules like aldehydes or ketones after reductive workup, whereas oxidation with KMnO$_4$ adds oxygen across the double bond without breaking the carbon chain to form glycols (diols).

Key Concept: Resonance Structures

c) The two Kekulé structures with alternating single and double bonds.

[Solution Description] Benzene is a resonance hybrid primarily described by the two Kekulé structures, which have alternating single and double bonds. These structures allow for delocalization of π electrons across all carbon atoms in the ring.

Key Concept: Reaction Mechanism Details, Reaction Pathway

c) By generating the electrophile needed for the reaction

[Solution Description]

In the Friedel-Crafts alkylation of benzene, anhydrous AlC$l_3$ acts as a Lewis acid. It reacts with the alkyl halide reagent to generate a complex that forms a more potent electrophile (an alkyl cation, R⊕) suitable for attacking the aromatic ring. This interaction between AlC$l_3$ and the alkyl halide results in the polarization and weakening of the carbon-halogen bond, promoting the generation of the carbocation. Once the carbocation is formed, it can attack the benzene ring to form a sigma-complex (arenium ion), which eventually loses a proton to restore the aromaticity of the benzene ring. AlC$l_3$ thereby plays a crucial role in generating the active electrophile and completing the reaction mechanism.

Key Concept: Reaction Kinetics, Catalyst Comparison

b) Combine palladium and platinum under moderate conditions

[Solution Description]

When designing a catalytic system for optimizing the hydrogenation rate of ethyne to ethane, it is important to choose a catalyst effective in both adsorbing hydrogen and facilitating its addition to ethyne. Platinum and palladium are highly selective towards hydrogenation reactions and are efficient for such processes.

Using a combination of these metals can enhance the hydrogenation rate due to their complementary properties: platinum has excellent hydrogen absorption capabilities, while palladium efficiently facilitates the transfer of hydrogen atoms to the substrate.

Nickel is another viable alternative; however, it generally requires more extreme conditions than platinum or palladium for similar efficiency. Therefore, blending platinum with palladium under moderate pressure and temperature optimizes the reaction kinetics effectively.

Key Concept: Real-world Application, Non-Straightforward Path

b) Open-air burning of household waste in rural areas.

[Solution Description] Incomplete combustion of coal, often seen in traditional power plants or domestic heating systems, releases large quantities of soot containing polynuclear hydrocarbons. These particles can become airborne, increasing atmospheric concentrations and posing health risks through inhalation. Other options involve less release or more controlled environments.

Key Concept: Position Isomerism

a) Pent-2-yne

[Solution Description] Position isomers have the same molecular formula but differ in the position of the functional group.

– Pent-1-yne has the triple bond starting at the first carbon.

– The only option with the triple bond at another position while maintaining five carbons is pent-2-yne.

Key Concept: Reaction Mechanism Comparison, Advanced Concept Integration

b) Determines whether the alkene is cis or trans

[Solution Description] During dehalogenation, the stereochemistry of the starting material influences the configuration of the alkene formed. Vicinal dibromides with anti-stereochemistry lead to trans-alkenes while syn-stereochemistry leads to cis-alkenes after dehalogenation. This is because the removal of bromine atoms occurs in a concerted manner.

For example, if we have $Br–CH_2–CH_2–Br$ in an anti conformation, the bromine atoms are removed across the plane, forming a trans-alkene. Conversely, for syn conformation, the same process results in a cis-alkene.

Key Concept: Reaction Conditions, Mechanistic Details

b) Basic medium with zinc

[Solution Description]

Vicinal dihalides require basic conditions to effectively remove the two halogen atoms from adjacent carbon atoms. This process typically involves the use of zinc in a reaction that favors elimination to form an alkene.

Key Concept: Industrial Process Analysis, Reaction Optimization

c) Improving catalyst performance

[Solution Description] To analyze and optimize the reaction, we consider Le Chatelier’s principle, where increasing pressure or changing temperature can affect the equilibrium position. Here, the reaction involves one mole of methane ($CH_4$) reacting with steam ($H_2O$), resulting in more moles of products than reactants. According to Le Chatelier’s principle, reducing pressure would favor the side with more moles.

For this endothermic reaction:

$$CH_4(g) + H_2O(g) \rightarrow CO(g) + 3H_2(g)$$

Lowering the pressure would move the equilibrium towards the right, enhancing $H_2$ production. However, given industrial constraints on cost and equipment, the other significant option is improving catalytic activity by using an effective catalyst. Thus, increasing catalyst efficiency would be the best practical approach for maximizing hydrogen yield.

Key Concept: Advanced Projection Analysis, Isomerism Complexity

c) 2-Pentene

[Solution Description] To exhibit both chain and geometrical isomerism, the compound must have different connectivity for chain isomers and double bonds for geometrical isomers. Among the given options, only a compound like 2-pentene can show both isomerisms: it can form pent-2-ene or 2-methylbut-2-ene as chain isomers, and exist in cis-trans forms due to the C=C bond.

Key Concept: Advanced Hybridization, Reaction Mechanisms

a) Increased s character in sp hybridized carbon enhances electronegativity, increasing acidity.

[Solution Description]

The sp hybridization in ethyne results in carbon having a 50% s character, giving it a higher electronegativity. This increases the ability of the carbon to stabilize negative charge when the hydrogen is removed as a proton, thereby making the hydrogen atoms more acidic. Compared to ethane, where carbon is sp$^3$ hybridized with only 25% s character, this makes hydrogen atoms in ethyne much more acidic, facilitating reactions like forming monosodium and disodium ethynide.