Class 12 Chemistry Chapter 6 Haloalkanes And Haloarenes

Key Concept: Reaction Mechanism Analysis, Stereochemical Considerations

c) Achiral product formation
[Solution Description]
In the addition of HBr to vinyl chloride ($CH_2=CHCl$), Markovnikov’s rule applies, where the hydrogen atom from HBr adds to the carbon with more hydrogens, and bromine adds to the carbon with fewer hydrogens. This results in 1-bromo-2-chloroethane. Since there are no chiral centers formed in this reaction, the product does not exhibit stereoisomerism. Thus, the result is achiral, meaning there is no stereochemical outcome.

Key Concept: Ambident Nucleophiles

c) Nitroethane
[Solution Description] Sodium nitrite $(NaNO_2)$ acts as an ambident nucleophile, meaning it can attack through either the nitrogen or oxygen atom. In this case, when reacting with ethyl chloride, the nitrogen atom tends to be the site of attack, resulting in the formation of nitroethane $(C_2H_5Cl)$
.

Key Concept: Reaction Conditions

c) Concentrated $HNO_3$ and concentrated $H_2SO_4$, heat
[Solution Description]
To nitrate chlorobenzene, a typical electrophilic substitution reaction, drastic conditions are required compared to benzene due to the deactivating nature of chlorine through its -I effect. This requires concentrated $HNO_3$ and concentrated $H_2SO_4$ with heating.

Key Concept: Simple Definition

c) To donate fluoride ions for replacing other halogens
[Solution Description] In the Swarts Reaction, metallic fluorides such as AgF serve as fluoride donors. They help replace halogen atoms (such as Cl or Br) with fluorine in alkyl halides to form alkyl fluorides.

Key Concept: Carbocation Stability

d) Tertiary
[Solution Description] Tertiary carbocations are the most stable in SN1 reactions due to hyperconjugation and inductive effects. The stability increases with the number of alkyl groups attached to the positively charged carbon atom, which helps disperse the positive charge.

Key Concept: Nature of C-X Bond

a) Haloalkanes are more reactive due to longer bond length.

[Solution Description]

In haloalkanes, carbon is $sp^3$ hybridised while in haloarenes it is $sp^2$ hybridised. The $sp^2$ hybridized carbon has more s-character than $sp^3$, leading to a shorter C-Cl bond length in haloarenes (169 pm) compared to haloalkanes (177 pm), making the C-Cl bond in haloarenes less reactive and harder to break.

Key Concept: Isomer Identification

a) 1-Bromobutane

[Solution Description] A structural isomer has the same molecular formula but a different connectivity between atoms. The molecular formula of 1-Bromo-2-methylpropane is $$C_4H_9Br$$. An isomer could rearrange the bromine or the methyl group. Given options are checked which have the same formula:

– 1-Bromobutane ($C_4H_9Br$)

– Isopropyl bromide ($C_3H_7Br$)

– Neopentyl bromide ($C_5H_{11}Br$)

– Cyclobutyl bromide ($C_4H_7Br$)

Thus, the compound with the same formula as 1-Bromo-2-methylpropane is 1-Bromobutane.

Key Concept: Prefix Usage

c) 1,3-
[Solution Description] In the naming of dihalogen derivatives like dichlorobenzenes, the prefix ‘m-‘ (meta) indicates that the two chlorine atoms are on carbons separated by one carbon atom, specifically at positions 1 and 3 on a benzene ring.

Key Concept: Hybridisation Difference

b) $sp^2$

[Solution Description] In haloarenes, the carbon atom bonded directly to the halogen is part of the aromatic ring structure. The carbons in the aromatic rings are $sp^2$ hybridised because they form three sigma bonds, one with each adjacent carbon and one with the hydrogen or substitute group. Therefore, the carbon attached to the halogen in haloarenes is $sp^2$ hybridized.

Key Concept: Reactivity Order

a) $$(CH_3)_3CCl > C_6H_5CH_2Cl > C_3H_7Br > CH_3Cl$$
[Solution Description]
In SN1 reactions, the stability of the carbocation formed is crucial. The tert-butyl chloride ($(CH_3)_3CCl$) forms a tertiary carbocation which is highly stable. Benzyl chloride ($C_6H_5CH_2Cl$) forms a resonance-stabilized benzyl cation. Propyl bromide ($C_3H_7Br$) forms a secondary carbocation, while methyl chloride ($CH_3Cl$) forms a primary carbocation, which is least stable.
Therefore, the order of reactivity is:

$$(CH_3)_3CCl > C_6H_5CH_2Cl > C_3H_7Br > CH_3Cl$$.

Key Concept: Phosphorus Halides

c) Assertion is true, but Reason is false.

[Solution Description] Phosphorus tribromide ($PBr_3$) is used in the preparation of alkyl bromides from alcohols. It is typically generated in situ by the reaction between red phosphorus and bromine, which does not require any additional catalyst. The assertion correctly states that $PBr_3$ is generated in situ. However, the reason given is incorrect because no catalyst is involved in this process. Thus, the Assertion is true, but the Reason is false.

Key Concept: Reaction Identification

d) Fittig reaction

[Solution Description] The reaction described involves the formation of a compound by joining two aryl groups through the use of sodium in dry ether. This is characteristic of the Fittig Reaction, where aryl halides react with sodium to produce biaryl compounds.

Key Concept: Basic Concept Understanding

c) Haloalkanes have high solubility in organic solvents due to similar intermolecular forces.
[Solution Description] The solubility of haloalkanes in water is low, but they dissolve well in organic solvents due to similar intermolecular forces.

Key Concept: Molecular Mass Impact

c) 1-Iodopropane
[Solution Description] Boiling points increase with molecular mass. Among the options, 1-Chlorobutane $(C_4H_9Cl)$, 1-Bromopropane $(C_3H_7Br)$, 1-Iodopropane $(C_3H_7I)$, and 1-Fluorobutane $C_4H_9F$, the compound with the highest molecular mass is 1-Iodopropane due to iodine being heavier than other halogens present in the options.

Key Concept: Reaction Identification

b) Reaction with potassium iodide
[Solution Description] The replacement of the diazonium group by iodine without using cuprous halide is achieved through the use of potassium iodide. This specific Sandmeyer reaction does not require any cuprous halide for the iodine substitution.

Key Concept: Reactivity Order

b) $CH_3–F$ < $CH_3–Cl$ < $CH_3–Br$ < $CH_3–I$ [Solution Description] Reactivity of haloalkanes increases as the bond becomes weaker. The bond strength decreases from $C–F$ to $C–I$ due to decreasing bond enthalpy. Thus, the order of increasing reactivity is $CH_3–F$ < $CH_3–Cl$ < $CH_3–Br$ < $CH_3–I$.

Key Concept: Reaction Speed

b) Slower than benzene
[Solution Description] Electrophilic substitution reactions in haloarenes are slower compared to benzene. This is because the halogen’s inductive effect withdraws electrons, making the aromatic ring less reactive.

Key Concept: Health Impact Analysis

a) Chloroform
[Solution Description] Chloroform is known to be metabolized into phosgene in the human body. Phosgene is a toxic compound that can cause severe liver damage upon chronic exposure.

Key Concept: Simple Application

d) Refrigerants
[Solution Description] Iodoform is utilized in the production of certain chemicals, including refrigerants and propellants for aerosol cans due to its chemical properties. It is involved in synthesizing chlorofluorocarbons and similar compounds.

Key Concept: Catalyst Requirement

d) $ZnCl_2$
[Solution Description] Primary and secondary alcohols require a catalyst to react with HCl. The catalyst used is $ZnCl_2$, which helps in the formation of alkyl chlorides from these alcohols.

Key Concept: Reaction Condition

c) Dry conditions
[Solution Description] The Swarts Reaction is usually conducted under dry conditions to minimize side reactions that may occur due to moisture.

Key Concept: Prefix Identification

b) vic-
[Solution Description] When both halogen atoms are on adjacent carbons, the term ‘vicinal’ is used. Therefore, the prefix “vic-” is used.

Key Concept: Complex Reaction Pathways, Stereochemistry

d) Assertion is false, but Reason is true.

[Solution Description] Free radical chlorination of propane involves substituting hydrogen atoms with chlorine. Propane has primary hydrogen at C-1 and secondary hydrogen at C-2. The energy required to break a primary C-H bond is more than breaking a secondary C-H bond because secondary radicals are more stable due to hyperconjugation and inductive effects. Thus, secondary hydrogens are preferentially replaced, leading to more 2-chloropropane than 1-chloropropane. The reason correctly explains why different products form in varying proportions, but not in equal amounts as asserted.

Therefore, while the assertion is false, the reason provided for product formation is true.

Key Concept: Electrophilic Substitution Conditions

b) Drastic conditions such as high temperature or pressure.
[Solution Description]
Electrophilic substitution reactions in haloarenes require drastic conditions due to deactivation by the halogen atoms. Halogens are both ortho and para directing but deactivate the benzene ring overall, necessitating harsher conditions for successful substitution.

Key Concept: Real-world Application, Reaction Mechanism, Environmental Impact

c) Controlled stability permits selective transformations.
[Solution Description] Aryl chlorides are often utilized in pharmaceutical syntheses due to their ability to engage in specific reactions while offering sufficient stability to withstand various synthetic conditions. Despite their environmental impact stemming from long-term stability, their characteristic low reactivity toward nucleophilic substitution (due to resonance stabilization of the C—Cl bond) provides controlled reaction pathways necessary for complex drug construction. This stability allows precise manipulations essential in generating active pharmaceutical ingredients like chloramphenicol.

Key Concept: Reaction Mechanism

c) To prevent moisture interference
[Solution Description] The Fittig reaction involves the coupling of aryl halides using sodium metal. Sodium is highly reactive with water, so it is essential to use anhydrous conditions to prevent any side reactions between sodium and moisture. Dry ether acts as a solvent that keeps the reaction medium free from water.

Key Concept: Optical Activity Basics

b) Dextrorotatory
[Solution Description] A compound that rotates plane-polarized light to the right (clockwise) is termed as dextrorotatory, denoted by the symbol ($+$).

Key Concept: Reaction Outcome

a) Ethane
[Solution Description]
In the Wurtz reaction, two molecules of methyl iodide will react with sodium in dry ether. The sodium provides electrons, facilitating the coupling of two $\text{CH}_3$ groups to form ethane $(\text{C}_2\text{H}_6).$

Reaction: $$2 \text{CH}_3\text{I} + 2 \text{Na} \xrightarrow{\text{dry ether}} \text{C}_2\text{H}_6 + 2 \text{NaI}$$.
Thus, the product formed is ethane.

Key Concept: Environmental Impact, Real-world Application

c) 1,2-Dichloro-4-(trichloromethyl)benzene
[Solution Description] Polychlorinated biphenyls (PCBs) are known for their persistence in the environment and capacity to bioaccumulate in the food chain, leading to severe ecological and health impacts. Compound 1,2-dichloro-4-(trichloromethyl)benzene has multiple chlorine atoms, similar to PCBs, making it environmentally concerning.

Key Concept: Density Comparison

d) $CCl_4$

[Solution Description] Bromo, iodo, and polychloro derivatives tend to be denser than water due to their higher atomic masses. Among the options given, carbon tetrachloride $(CCl_4)$ is known to be denser than water.

Key Concept: Basic Identification

b) $CH_2=CHCH_2$Br
[Solution Description] To identify an allylic halide, we need to look for a compound where a halogen atom is attached to the carbon adjacent to a double bond. The general structure for an allylic halide is $RCH=CHCH_2X$, where $X$ is a halogen.

Key Concept: Complex Structure Analysis, Environmental Chemistry

d) Perhaloalkane
[Solution Description] The given compound is a polyhalogenated compound due to multiple halogen atoms. Each carbon atom in the chain is fully substituted with chlorine or fluorine making it a perfluorocarbon (PFC). Such compounds are often stable and can persist in the environment contributing to ozone layer depletion and acting as greenhouse gases. In terms of haloalkane classification, because every carbon is bonded to multiple halogens and there’s no hydrogen atom directly attached, it’s neither primary, secondary nor tertiary but rather a perhaloalkane.

Key Concept: Markovnikov’s Rule

b) 2-bromobutane
[Solution Description] Markovnikov’s rule states that in the addition of HX to an alkene, the hydrogen atom attaches to the carbon with the greatest number of hydrogen atoms already present, while the halogen attaches to the other carbon. For but-1-ene, the hydrogen from HBr will attach to the terminal $CH_2$ group, and bromine will attach to the adjacent carbon, forming 2-bromobutane.

Key Concept:  Reaction Pathway Analysis, Solvent Effects

d) The reaction rate increases due to less nucleophile-solvent interaction.
[Solution Description]
SN2 reactions are favored by polar aprotic solvents because such solvents do not stabilize anions as effectively as polar protic solvents, allowing the nucleophile to remain more reactive. DMSO is a common polar aprotic solvent, which enhances the reactivity of nucleophiles like $CN^-$. This increases the efficiency of the backside attack on the electrophilic carbon in $CH_3I$, leading to a faster reaction rate.

Therefore, using DMSO will accelerate the SN2 reaction significantly compared to a polar protic solvent.

Key Concept: Stereochemical Implications, Mechanism Comparison

c) Single enantiomer of 3-methyl-2-butyl cyanide (inverted)
[Solution Description] 2-Bromo-3-methylbutane can undergo both $SN1$ and $SN2$ reactions depending on the reaction conditions. In a protic solvent, the $SN1$ mechanism predominates, forming a carbocation intermediate that results in racemization at the chiral center. This would produce a mix of enantiomers in equal proportions: 3-methyl-2-butyl cyanide. Conversely, in a polar aprotic solvent, the $SN2$ pathway is favored, leading to inversion of configuration at the chiral center, producing only one enantiomer of 3-methyl-2-butyl cyanide.

Key Concept: Basic Properties

c) As a cleaning fluid
[Solution Description] Historically, Carbon Tetrachloride has been used as a cleaning fluid due to its ability to dissolve greases and oils effectively.

Key Concept: Resonance and Reactivity

c) It increases reactivity by stabilizing intermediates.
[Solution Description] The presence of a nitro group at the ortho or para position significantly increases the reactivity of haloarenes in nucleophilic substitution reactions. This is because the nitro group stabilizes the carbanion intermediate through resonance, lowering the energy barrier for the reaction and facilitating the process.

Key Concept: Natural Occurrence

c) Chloramphenicol

[Solution Description] Chloramphenicol is a naturally occurring antibiotic that contains chlorine. It is known for its broad-spectrum activity and is used to treat various bacterial infections. Other options listed are either synthetic compounds or do not fit the description of a natural antibiotic containing chlorine.

Key Concept: Real-World Application, Reaction Mechanism Analysis

d) Freon
[Solution Description] Freons, also known as chlorofluorocarbons (CFCs), are known to cause significant damage to the ozone layer through chlorine radicals formed during decomposition in the upper atmosphere. Other options like dichloromethane and chloroform are less likely linked to such extensive ozone depletion effects.

Key Concept: Reaction Conditions

c) Oxidizing agent
[Solution Description] In the iodination of benzene, iodine reacts with benzene only if an oxidizing agent is present. This is because hydrogen iodide (HI), which is formed during the reaction, needs to be removed. An oxidizing agent converts HI into $I_2$ and $H_2O$, ensuring the forward direction of the reaction.

Key Concept: Reaction Pathway Analysis, Advanced Mechanism

b) Use of potassium tert-butoxide in dimethyl sulfoxide at elevated temperature
[Solution Description]
To determine the favored reaction pathway, we need to consider factors such as the type of halide and base used. In this case, 2-bromo-3-methylbutane is a secondary alkyl halide, which can undergo both elimination (E2) and substitution (SN2) reactions. However, bulky bases typically hinder substitution pathways in favor of elimination. Additionally, high temperatures also promote elimination over substitution.

Given these considerations, using a strong, bulky base like tert-butoxide in conjunction with elevated temperature will favor the E2 mechanism due to steric hindrance and increased energy promoting elimination.

The correct answer is option b) because it uses a bulky base and heat.

Key Concept: Mechanism Analysis, Electron Withdrawing Group Effect

a) It decreases the electron density on the aromatic ring.
[Solution Description] The $-NO_2$ group is an electron-withdrawing group, which increases the positive charge character on the aromatic ring, thereby making haloarenes more susceptible to attack by nucleophiles. This occurs because the withdrawal of electrons from the ring stabilizes the transition state in nucleophilic aromatic substitution, making the reaction more favorable.

Key Concept: Chemical Composition

c) $CCl_2F_2$
[Solution Description] The chemical formula for Freon 12 can be determined by identifying the molecular structure as a chlorofluorocarbon compound typically represented as $CCl_2F_2$. This means it contains two chlorine atoms and two fluorine atoms attached to a central carbon atom.

Key Concept: Grignard and Halides

b) Hydrocarbons
[Solution Description]
Grignard reagents can convert halides into hydrocarbons through a substitution reaction where the halogen atom is replaced by the alkyl or aryl group present in the Grignard reagent. For example, reacting ethyl magnesium bromide with methyl chloride will replace the chloride with the ethyl group, producing ethane and releasing magnesium chloride as byproducts.

Key Concept: Competing Reactions

c) Strong, bulky base and high temperature
[Solution Description]
For an E2 elimination reaction to be favored over an SN2 substitution reaction in a primary alkyl halide, a strong and bulky base is needed. A bulky base hinders backside attack necessary for SN2 while facilitating deprotonation required for E2. The presence of heat also favors elimination over substitution.

Key Concept: Radical Reactions, Long-term Impact

a) By initiating radical chain reactions that deplete the ozone, leading to increased UV radiation reaching Earth
[Solution Description]
Freons, which contain chlorine atoms, are released into the atmosphere where they eventually reach the stratosphere. Here, ultraviolet (UV) radiation causes them to release chlorine radicals through photodissociation. The reaction can be expressed as follows:

$$CF_2Cl_2 \rightarrow CF_2Cl + Cl$$

These chlorine radicals react with ozone $(O_3)$, causing its depletion by converting $O_3$ molecules into oxygen $(O_2)$:

$$Cl + O_3 \rightarrow ClO + O_2$$

$$ClO + O \rightarrow Cl + O_2$$

This cycle continues, leading to significant depletion of the ozone layer. The ozone layer is vital for blocking harmful UV radiation from reaching Earth’s surface. A long-term consequence of this depletion is an increase in skin cancers and cataracts among humans, as well as environmental impacts such as reduced crop yields due to increased UV exposure.

Key Concept: Density and Molecular Structure

b) Presence of heavy halogen atoms
[Solution Description] The density of haloalkanes is influenced by the presence of heavy halogen atoms such as bromine (Br), iodine (I), or multiple chlorine (Cl) atoms. These elements have higher atomic masses compared to hydrogen, which they replace in hydrocarbons, thus increasing the overall molecular mass and consequently the density.

Key Concept: Electrophilic vs. Nucleophilic Reactions, Mechanism Analysis

c) Assertion is true, but Reason is false.

[Solution Description] Although both haloarenes and haloalkanes can participate in substitution reactions, haloarenes exhibit greater reactivity towards electrophilic substitutions due to their ability to stabilize positive charge through resonance in an aromatic system. In electrophilic substitution reactions, the aromatic ring acts as a site for electrophiles to attack due to electron-rich $\pi$-clouds resulting from resonance stabilization. However, the reason given is incorrect because it incorrectly attributes this increased reactivity to enhanced susceptibility to nucleophilic attacks, which are instead hindered by resonance stabilization that prevents the formation of phenyl cations—a requirement for nucleophilic substitution processes like $S_N1$. Therefore, the assertion is true but the reason does not correctly explain the assertion.

Key Concept: Solvent Role

c) Acetone
[Solution Description]
The Finkelstein reaction requires a polar aprotic solvent to enhance the nucleophilicity of iodide ions by not solvating them well, unlike polar protic solvents that would hydrogen bond with them. Acetone is often used because it is a polar aprotic solvent and increases the reactivity of the iodide ion in the $SN2$ mechanism.

Key Concept: Chemical Stability

d) It is fat-soluble
[Solution Description] DDT is chemically stable and fat-soluble, which prevents it from being metabolized rapidly in animal tissues.

Key Concept: Reaction Products

a) Ortho-bromobenzenesulfonic acid and para-bromobenzenesulfonic acid
[Solution Description] Bromobenzene is a type of haloarene where the bromine atom is ortho and para-directing due to resonance, despite its deactivating effect. When bromobenzene undergoes sulphonation with fuming sulfuric acid, the electrophile (sulfonyl group) preferentially attacks the ortho and para positions relative to the bromine atom. Hence, the major products formed will be ortho-bromobenzenesulfonic acid and para-bromobenzenesulfonic acid.

Key Concept: Mechanism Understanding

b) Dry ether as solvent
[Solution Description] In the Wurtz-Fittig reaction, sodium acts as a reducing agent and dry ether is used as a solvent to facilitate the reaction between alkyl halides and aryl halides to form an alkylarene.

Key Concept: Boiling Point Trends, Transition State Characteristics

b) 2-Chlorobutane
[Solution Description] Boiling points are influenced by both molecular weight and the degree of branching within a molecule. A higher molecular mass typically increases boiling points due to increased Van der Waals forces, while more branching tends to decrease the boiling point as it reduces surface area available for intermolecular interactions. Comparing these haloalkanes, 2-Chlorobutane has the highest molecular weight with relatively low branching among these options, leading to stronger intermolecular attractions and thus the highest boiling point.

Key Concept: Halogen Exchange

c) Assertion is true, but Reason is false.

[Solution Description]

The assertion is correct as alkyl iodides are prepared using the Finkelstein reaction. However, the reason provided is incorrect because the Finkelstein reaction involves the use of sodium iodide (NaI) in dry acetone, not concentrated hydrochloric acid. Concentrated hydrochloric acid is unrelated to this reaction.

Key Concept: Reaction Pathway Analysis, Reaction Condition Impact

d) Octane
[Solution Description]
The Wurtz reaction involves the coupling of two alkyl halides using sodium in dry ether to form an alkane with double the carbon atoms. For 1-bromobutane $(C_4H_9Br)$, the process can be outlined as follows:

1. Two molecules of $C_4H_9Br$ will react with Na to eliminate 2NaBr.
2. The remaining butyl groups couple together forming octane $(C_8H_{18}).$

Thus, the chemical equation is:
$$2\, \text{C}_4\text{H}_9\text{Br} + 2\, \text{Na} \rightarrow \text{C}_8\text{H}_{18} + 2\, \text{NaBr}$$
Therefore, the product formed is octane.

Key Concept: Resonance in Haloarenes, Environmental Impact

a) Chlorobenzene; Higher environmental impact
[Solution Description]
Chlorobenzene and benzyl chloride differ significantly due to resonance effects. Chlorobenzene exhibits extensive resonance stabilization because the chlorine atom can participate in delocalizing its electron pair with the aromatic ring. This resonance makes chlorobenzene less reactive in nucleophilic substitution reactions.

On the other hand, benzyl chloride does not benefit from such resonance stabilization and is thus more reactive. Regarding reactivity towards ozone, polyhalogenated benzenes, like chlorobenzene, have been known to contribute to ozone depletion when they break down into respective radical species and release halogens that catalyze ozone destruction. Benzyl chloride typically does not exhibit this behavior.

Therefore, although chlorobenzene is stable owing to resonance, it has a higher potential to affect the ozone layer negatively than benzyl chloride.

Key Concept: Environmental Impact, Real-World Application

d) Freon
[Solution Description] Among the given options, Freon is known as a chlorofluorocarbon (CFC), which contributes significantly to the depletion of the ozone layer. CFCs are stable enough to reach the stratosphere, where they release chlorine atoms upon exposure to ultraviolet radiation. These chlorine atoms catalyze the breakdown of ozone into oxygen molecules, thus depleting the ozone layer. Other compounds listed have lesser or different environmental impacts and do not contribute significantly to ozone depletion.

Key Concept: Structural Isomer Identification

b) 2-Bromobutane

[Solution Description] 1-bromo-2-methylpropane has the chemical formula $C_4H_9Br$. Structural isomers have the same molecular formula but different connectivity between atoms. Drawing out possible structures, we find:

– 1-Bromo-2-methylpropane: $CH_3CH(Br)CH(CH_3)CH_3$

– 2-Bromobutane: $CH_3CH(Br)CH_2CH_3$

– 1-Bromobutane: $BrCH_2CH_2CH_2CH_3$

– 1-Bromo-2,2-dimethylpropane: $C(CH_3)_2(Br)CH_3$

The correct structural isomer from the options given must have a different structure but the same molecular formula as 1-bromo-2-methylpropane. Here, 2-bromobutane has this property.

Key Concept: Classification of Polyhalogen Compounds

c) Chloroform ($CHCl_3$)

[Solution Description] A polyhalogen compound contains more than one halogen atom in its molecule. Among the given options, chloroform ($CHCl_3$) contains three chlorine atoms and therefore qualifies as a polyhalogen compound.

Key Concept: Safety Measures

b) To avoid oxidation to phosgene
[Solution Description] Chloroform is stored in closed dark-colored bottles to prevent its oxidation into phosgene, a highly toxic compound. Exposure to light can accelerate this oxidation process, hence the need for such storage measures.

Key Concept: Health Effects

c) Liver
[Solution Description] Chronic exposure to chloroform vapours may lead to damage to the liver and kidneys due to its toxic nature when inhaled over prolonged periods.

Key Concept: Reaction Mechanism

d) Condensation
[Solution Description] The steps involved in free radical halogenation are initiation, propagation, and termination. In the initiation step, radicals are created usually by splitting halogen molecules using UV light or heat. During the propagation step, these radicals react with alkanes to create alkyl radicals and hydrogen halides. In the termination step, remaining radicals combine to form stable products. Condensation is not part of this process.

Key Concept: Resonance Stabilization

c) It stabilizes the carbanion
[Solution Description] Resonance plays a crucial role in stabilizing the carbanion formed during nucleophilic attack on haloarenes when $-NO_2$ groups are present at ortho and para positions. This stabilization makes the reaction more favorable.

Key Concept: Stability and Reactivity

d) $(CH_3)_3CBr$
[Solution Description] In E1 reactions, the rate-determining step involves the formation of a carbocation. Therefore, more stable carbocations lead to faster reactions. Among the options, $(CH_3)_3CBr$ forms a tertiary carbocation, which is the most stable and thus most reactive in an E1 mechanism.

Key Concept: Energy Requirement

b) Energy to break existing hydrogen bonds
[Solution Description]
Dissolving haloalkanes in water requires energy to break existing hydrogen bonds between water molecules. This is necessary for accommodating haloalkane molecules among the water molecules. Unlike ionic or covalent interactions, the new interactions formed do not compensate for this energy input.

Key Concept: Reaction Mechanism

c) It forms an alkyl chloride
[Solution Description] In this reaction, known as the Lucas test, primary alcohols react with HCl in the presence of anhydrous zinc chloride to form alkyl chlorides. The mechanism involves the formation of a carbocation intermediate, which undergoes nucleophilic substitution by chloride ion resulting in the formation of an alkyl chloride through an SN1 pathway. However, in primary alcohols, the reaction can proceed via an SN2 mechanism due to the instability of the primary carbocation.

Key Concept: Environmental Impact

c) Bioaccumulates in aquatic food chains, affecting top predators

[Solution Description] DDT is highly toxic to fish and other aquatic organisms due to its persistence and bioaccumulation in water bodies. It affects their nervous system, leading to mortality or sub-lethal effects such as reproductive and developmental issues.

Key Concept: Basic Naming

a) 1-Bromo-2-chloropropane
[Solution Description] The structure $CH_3CHBrCH_2Cl$ has a bromine and a chlorine atom attached to a propane chain. According to IUPAC naming conventions, we must number the carbon atoms in such a way that the substituents receive the lowest possible numbers. Starting from the end nearest to the substituent gives us 1-Bromo-2-chloropropane.

Key Concept: Exposure Symptoms

d) Severe skin burns
[Solution Description] Low levels of dichloromethane exposure tend to cause mild symptoms like impaired hearing, dizziness, and nausea. Skin burns are more severe and typically associated with higher concentrations or direct contact, rather than low-level exposure.