Alkane Questions in English

(C) The combustion reaction of an alkane is: $C_nH_{2n+2} + \frac{3n+1}{2} O_2 \to nCO_2 + (n+1) H_2O$
The reaction of $CO_2$ with $NaOH$ to form $Na_2CO_3$ is: $CO_2 + 2NaOH \to Na_2CO_3 + H_2O$
Given $300 \ mL$ of $2 \ N \ NaOH$ solution. Since the $n$-factor for $NaOH$ is $1$,$Normality = Molarity$. Thus,moles of $NaOH = 0.3 \ L \times 2 \ M = 0.6 \ mol$.
From the stoichiometry of the second reaction,$2 \ mol$ of $NaOH$ react with $1 \ mol$ of $CO_2$. Therefore,moles of $CO_2 = \frac{0.6}{2} = 0.3 \ mol$.
From the combustion reaction,$1 \ mol$ of $C_nH_{2n+2}$ produces $n \ mol$ of $CO_2$. Thus,moles of alkane = $\frac{0.3}{n}$.
Given the mass of the alkane is $4.3 \ g$ and its molar mass is $(14n + 2) \ g/mol$,we have: $\frac{4.3}{14n + 2} = \frac{0.3}{n}$.
Solving for $n$: $4.3n = 0.3(14n + 2) \implies 4.3n = 4.2n + 0.6 \implies 0.1n = 0.6 \implies n = 6$.
Thus,the alkane is $C_6H_{14}$.

(C) In Kolbe's electrolysis of sodium pivalate $((CH_3)_3CCOONa)$,the tert-butyl radical $((CH_3)_3C \cdot)$ is formed.
This radical undergoes disproportionation to form isobutylene $(CH_2=C(CH_3)-CH_3)$ and isobutane $(CH_3-CH(CH_3)-CH_3)$,or dimerization to form $2,2,3,3-tetramethylbutane$.
Methane $(CH_4)$ is not a product of this reaction.
Therefore,option $C$ is incorrect.

(D) $1$. Cyclohexane reacts with $Br_2$ in the presence of $h\nu$ (free radical substitution) to form bromocyclohexane $(A)$.
$2$. Bromocyclohexane $(A)$ reacts with $Na$ in dry ether (Wurtz reaction) to form bicyclohexyl $(B)$.
$3$. Bicyclohexyl $(B)$ reacts with $Br_2$ in the presence of $h\nu$ to form $1-$bromobicyclohexyl $(C)$.
$4$. $1-$bromobicyclohexyl $(C)$ undergoes hydrolysis with $H_2O$ to form $1-$hydroxybicyclohexyl (or $1-$cyclohexylcyclohexanol) $(D)$.
$5$. $1-$cyclohexylcyclohexanol $(D)$ undergoes acid-catalyzed dehydration with $conc. H_2SO_4$ to form the major product $(E)$,which is bicyclohexylidene (or dicyclohexylidene).

(A) The product ratio is calculated as: $(\text{Number of } H \text{ atoms}) \times (\text{Selectivity ratio})$.
For Propane $(CH_3-CH_2-CH_3)$:
$1^o H = 6$,$2^o H = 2$.
Ratio = $(6 \times 1) : (2 \times 3.8) = 6 : 7.6$. Thus,option $A$ is correct.
For $2,3-$Dimethylbutane:
$1^o H = 12$,$3^o H = 2$.
Ratio = $(12 \times 1) : (2 \times 5) = 12 : 10 = 6 : 5$. Option $C$ is incorrect.
For $2,4-$Dimethylpentane:
$1^o H = 12$,$2^o H = 2$,$3^o H = 2$.
Ratio = $(12 \times 1) : (2 \times 3.8) : (2 \times 5) = 12 : 7.6 : 10$. Option $D$ is incorrect.

(C) To obtain only two monochlorinated products,the isomer must have only two types of equivalent hydrogen atoms.
In $2,3-$dimethylbutane,the structure is $(CH_3)_2CH-CH(CH_3)_2$.
The two tertiary $H$ atoms are equivalent,and the twelve primary $H$ atoms (in four $CH_3$ groups) are equivalent.
Thus,substitution at these two distinct positions yields only two monochlorinated isomers.

(D) For an alkane to form only one type of mono-chlorinated product,all hydrogen atoms in the molecule must be equivalent.
In $2,2,3,3$-tetramethylbutane,all $18$ hydrogen atoms are equivalent because the molecule is highly symmetrical.
Therefore,mono-chlorination of $2,2,3,3$-tetramethylbutane yields only one isomer of $C_8H_{17}Cl$.

(A) In Kolbe's electrolysis,the electrolysis of an aqueous solution of the potassium salt of a carboxylic acid produces a hydrocarbon (alkane) at the anode.
The reaction is: $2RCOOK + 2H_2O \xrightarrow{\text{Electrolysis}} R-R + 2CO_2 + H_2 + 2KOH$.
Thus,option $A$ is the correct reaction.

(B) In the presence of sunlight,the reaction proceeds via a free radical mechanism.
Bromination is highly selective,and the stability of the intermediate free radical determines the major product.
The $3^{\circ}$ free radical is more stable than $1^{\circ}$ or $2^{\circ}$ free radicals.
Therefore,the hydrogen atom attached to the $C-2$ carbon is replaced by bromine to form $2-$bromo$-2-$methylbutane as the major product.

(A) Photochlorination involves the substitution of a hydrogen atom with a chlorine atom. Diastereomers are stereoisomers that are not mirror images of each other. For a compound to produce diastereomers upon chlorination,it must already contain at least one chiral center,and the chlorination must occur at a position that creates a second chiral center.
$1$. Consider $3-$methylhexane: The structure is $CH_3-CH_2-CH(CH_3)-CH_2-CH_2-CH_3$. It has a chiral center at $C3$. Upon photochlorination,substitution at different positions (like $C2$ or $C4$) creates a new chiral center,resulting in diastereomeric pairs.
$2$. $2-$chlorobutane already has a chiral center,but chlorination would create a dichlorobutane derivative which is not the smallest hydrocarbon precursor.
$3$. $1-$bromopropane and $1-$bromo$-3-$methylbutane do not satisfy the requirement of being the smallest hydrocarbon capable of generating diastereomers through this specific mechanism.
Thus,$3-$methylhexane is the smallest alkane that produces diastereomers upon photochlorination.

(A) $2,3-$dimethylbutane $(CH_3-CH(CH_3)-CH(CH_3)-CH_3)$ contains two types of equivalent hydrogen atoms: primary $(CH_3)$ and tertiary $(CH)$.
Substitution of a hydrogen atom from the primary carbon leads to $1-$chloro-$2,3-$dimethylbutane.
Substitution of a hydrogen atom from the tertiary carbon leads to $2-$chloro-$2,3-$dimethylbutane.
Since there are only two distinct types of hydrogen atoms,it yields exactly two isomeric monochloro derivatives.

(A) The boiling point of alkanes depends on the surface area of the molecule.
As the branching in an alkane increases,its surface area decreases,which leads to weaker van der Waals forces of attraction.
Therefore,the boiling point decreases as branching increases.
Comparing the isomers of pentane $(C_5H_{12})$:
$1$. $n$-pentane (straight chain) has the maximum surface area and thus the highest boiling point.
$2$. Isopentane ($2$-methylbutane) has one branch.
$3$. Neopentane ($2$,$2$-dimethylpropane) has two branches (most branched).
Among the given options,$n$-pentane is a straight-chain isomer with no branching,giving it the largest surface area and the highest boiling point.

(B) The simplest hydrocarbon with two $3^o$ and one $2^o$ carbon atom is $2,3-$dimethylbutane.
Structure: $(CH_3)_2CH-CH(CH_3)_2$.
In this structure:
- The two $CH$ groups are $3^o$ carbon atoms (attached to three other carbons).
- The $CH_2$ group is not present,but wait,the question asks for one $2^o$ carbon atom.
- Let us re-evaluate: $2,4-$dimethylpentane has the structure $CH_3-CH(CH_3)-CH_2-CH(CH_3)-CH_3$.
- Here,there are two $3^o$ carbon atoms (at positions $2$ and $4$) and one $2^o$ carbon atom (at position $3$).
- Counting the $1^o$ carbon atoms (terminal $CH_3$ groups): there are four $1^o$ carbon atoms.
Therefore,the number of $1^o$ carbon atoms is $4$.

(D) racemic mixture is formed when a chiral center is generated during a reaction.
$n-$pentane $(CH_3CH_2CH_2CH_2CH_3)$ upon chlorination gives $2-$chloropentane,which has a chiral carbon at the $C-2$ position.
Isopentane $(CH_3CH(CH_3)CH_2CH_3)$ upon chlorination at the $C-2$ position gives $2-$chloro-$2-$methylbutane (achiral) but chlorination at the $C-3$ position gives $2-$methyl-$3-$chlorobutane,which has a chiral carbon at the $C-3$ position.
Since both $n-$pentane and isopentane can form products with chiral centers via free radical halogenation,they can lead to the formation of racemic mixtures.

(A) The reaction shown is a free radical halogenation of an alkane (cyclohexane).
The reactivity order of halogens towards free radical substitution is $F_2 > Cl_2 > Br_2 > I_2$.
Fluorine $(F_2)$ is the most reactive halogen due to its high electronegativity and the low bond dissociation energy of the $F-F$ bond,making the reaction with $F_2$ extremely fast and often explosive.
Therefore,the fastest rate of reaction is observed with $F_2$.

(A) The reagent $Br_2 + H_2O$ (bromine water) is used to test for unsaturation (alkenes and alkynes). Compounds containing double or triple bonds will decolorize the reddish-brown color of bromine water.
$1$. $Al_4C_3 + 12H_2O \rightarrow 3CH_4 + 4Al(OH)_3$. Methane $(CH_4)$ is a saturated alkane and does not react with bromine water.
$2$. $Mg_2C_3 + 4H_2O \rightarrow CH_3-C \equiv CH + 2Mg(OH)_2$. Propyne is an alkyne and will decolorize bromine water.
$3$. $CaC_2 + 2H_2O \rightarrow HC \equiv CH + Ca(OH)_2$. Ethyne is an alkyne and will decolorize bromine water.
$4$. $Ag_2C_2$ is a metal acetylide that is generally insoluble and does not undergo simple hydrolysis to yield an unsaturated hydrocarbon product in this context.
Comparing the options,$Al_4C_3$ produces methane,which is saturated and will not decolorize bromine water.

(A) The structure of $2$-methylpentane is $CH_3-CH(CH_3)-CH_2-CH_2-CH_3$.
There are $5$ distinct types of hydrogen atoms in the molecule:
$1$. The $CH_3$ group attached to $C_2$ (position $1$).
$2$. The $CH$ group at $C_2$ (position $2$).
$3$. The $CH_2$ group at $C_3$ (position $3$).
$4$. The $CH_2$ group at $C_4$ (position $4$).
$5$. The $CH_3$ group at $C_5$ (position $5$).
Since there are $5$ non-equivalent hydrogen atoms,replacing any one of them with a chlorine atom will result in a unique monochlorinated product.
Note that the two $CH_3$ groups at the $C_2$ position are equivalent.
Therefore,the total number of monochlorinated products is $5$.

(C) The reactant is $3$-methylpentane. Let us identify the number of hydrogen atoms of each type:
$1^o$ hydrogens: There are $9$ such hydrogens ($3$ at each terminal methyl group).
$2^o$ hydrogens: There are $4$ such hydrogens ($2$ at $C_2$ and $2$ at $C_4$).
$3^o$ hydrogens: There is $1$ such hydrogen (at $C_3$).
The relative rates of formation are calculated as (Number of $H$ atoms) $\times$ (Relative reactivity):
For $1^o$ products: $9 \times 1 = 9$
For $2^o$ products: $4 \times 4 = 16$
For $3^o$ products: $1 \times 6 = 6$
Total relative rate = $9 + 16 + 6 = 31$.
The product shown is a $2^o$ chloro derivative formed by replacing a hydrogen at $C_2$ or $C_4$. The relative amount of this specific product is $16/31$.

(B) The Wurtz reaction involves the coupling of two alkyl halides in the presence of sodium metal to form a symmetrical alkane.
It is most effective for the synthesis of symmetrical alkanes with an even number of carbon atoms.
Among the given options,$n-butane$ $(CH_3-CH_2-CH_2-CH_3)$ is a symmetrical alkane that can be efficiently prepared by the Wurtz reaction using ethyl chloride $(CH_3-CH_2-Cl)$ and sodium metal.

(A) The reaction between a Gilman reagent (lithium dialkylcuprate) and an alkyl halide is known as the Corey-House synthesis.
The general reaction is: $R_2CuLi + R'-X \rightarrow R-R' + R-Cu + LiX$.
In this specific case,the Gilman reagent is $(CH_3-CH_2)_2CuLi$ (lithium diethylcuprate) and the alkyl halide is $CH_3-Cl$ (methyl chloride).
The ethyl group from the Gilman reagent replaces the chlorine atom on the methyl chloride to form propane:
$(CH_3-CH_2)_2CuLi + CH_3-Cl \rightarrow CH_3-CH_2-CH_3 + CH_3-CH_2-Cu + LiCl$.
The major product is propane $(CH_3-CH_2-CH_3)$.

(B) The reaction of a carboxylic acid with soda lime $(NaOH + CaO)$ is known as decarboxylation.
In this reaction,the $-COOH$ group is removed as $Na_2CO_3$ and replaced by a hydrogen atom.
The reactant is $3$-methylbutanoic acid,$CH_3-CH(CH_3)-CH_2-COOH$.
Upon heating with soda lime,the $-COOH$ group is eliminated,resulting in the formation of isobutane $(CH_3-CH(CH_3)-CH_3)$.

(C) To determine the number of monochlorinated products,we need to count the number of non-equivalent hydrogen atoms in the molecule.
$3-$Methylhexane has the structure $CH_3-CH_2-CH(CH_3)-CH_2-CH_2-CH_3$.
Let's identify the distinct types of hydrogen atoms:
$1$. The $CH_3$ group at the end of the chain (position $1$ in the image).
$2$. The $CH_2$ group adjacent to the $CH_3$ (position $2$).
$3$. The $CH$ group with the methyl branch (position $3$).
$4$. The $CH_3$ group attached to the $CH$ (position $7$).
$5$. The $CH_2$ group next to the $CH$ (position $4$).
$6$. The $CH_2$ group next to that (position $5$).
$7$. The terminal $CH_3$ group at the other end (position $6$).
Since there are $7$ non-equivalent hydrogen atoms,$3-$Methylhexane yields $7$ different monochlorinated products.

(B) The oxidation of alkanes under controlled conditions yields different products:
$A$. $(CH_3)_3CH + O_2 \xrightarrow{KMnO_4} (CH_3)_3COH$ (tert-butyl alcohol) matches with $ii$.
$B$. $2CH_4 + O_2 \xrightarrow[100 \ atm]{Cu/523 \ K} 2CH_3OH$ (methanol) matches with $iii$.
$C$. $CH_4 + O_2 \xrightarrow[\Delta]{Mo_2O_3} HCHO + H_2O$ (methanal) matches with $iv$.
$D$. $CH_3-CH_3 + \frac{3}{2}O_2 \xrightarrow{(CH_3COO)_2Mn} CH_3COOH + H_2O$ (ethanoic acid) matches with $i$.
Thus,the correct matching is $A-ii, B-iii, C-iv, D-i$.

(D) Rotation is restricted around double and triple bonds due to the presence of $\pi$ bonds.
In $C_2H_4$ $(CH_2=CH_2)$,there is a $C=C$ double bond.
In $C_2H_2$ $(CH \equiv CH)$,there is a $C \equiv C$ triple bond.
In $H_2O_2$ $(H-O-O-H)$,although there is a single bond,the lone pairs on oxygen atoms cause repulsion,leading to a restricted rotation in its non-planar 'open book' structure.
In $C_2H_6$ $(CH_3-CH_3)$,there is a $C-C$ single bond,which allows free rotation around the bond axis.

(A) $i$. The conversion of $n$-butane to $2$-methylpropane is an example of isomerization,which is categorized under $C$. Isomerism.
$ii$. The reaction $CH_4 + Cl_2 \xrightarrow{hv} CH_3Cl + HCl$ is a classic example of free radical substitution,which is $A$.
$iii$. The reaction $R-COONa \xrightarrow{\text{soda-lime}} R-H$ is the decarboxylation of sodium salts of carboxylic acids,which is $D$.
$iv$. The reaction $R-X + Na \xrightarrow{\text{Ether}} R-R$ is the Wurtz reaction,which is $B$.
Therefore,the correct matching is $i-C, ii-A, iii-D, iv-B$.

(A) When methane reacts with concentrated nitric acid at high temperatures (vapor phase nitration),it undergoes nitration to form nitromethane.
$CH_4 + HNO_3 \xrightarrow[400^{\circ}C]{\Delta} CH_3NO_2 + H_2O$

(C) The peroxide effect (Kharasch effect) is only observed with $HBr$. $HCl$ and $HI$ do not show the peroxide effect because the $H-Cl$ bond is too strong for the radical chain propagation step,and the addition of $I\cdot$ to the alkene is endothermic. Therefore,$HCl$ adds to propene according to Markovnikov's rule to give $2$-chloropropane $(CH_3-CHCl-CH_3)$,making the product in option $(c)$ incorrect.

(A) The reactant is isopropylcyclopentane. To find the number of structural mono-chloro products,we identify all non-equivalent hydrogen atoms in the molecule.
$1$. The isopropyl group has a tertiary hydrogen at the junction $(CH)$ and six equivalent primary hydrogens on the two methyl groups $(CH_3)$. This gives $2$ products.
$2$. The cyclopentane ring has five carbons. The carbon attached to the isopropyl group is a tertiary carbon $(CH)$. The two adjacent carbons $(CH_2)$ are equivalent,and the two carbons further away $(CH_2)$ are also equivalent.
$3$. Thus,the distinct positions for substitution are:
- Tertiary $H$ on the isopropyl group ($1$ product).
- Primary $H$ on the methyl groups ($1$ product).
- Tertiary $H$ on the cyclopentane ring ($1$ product).
- Secondary $H$ on the carbons adjacent to the isopropyl group ($1$ product).
- Secondary $H$ on the carbons further from the isopropyl group ($1$ product).
Total structural isomers = $1 + 1 + 1 + 1 + 1 = 5$.

(A) $1$. The first step is an intramolecular Wurtz reaction: $Br-(CH_2)_5-Br \xrightarrow{Na, \text{D.E.}} \text{cyclopentane} (A)$.
$2$. The second step is free radical bromination: $\text{cyclopentane} \xrightarrow{Br_2, \Delta} \text{bromocyclopentane} (B)$.
$3$. The third step is dehydrohalogenation using alcoholic $KOH$: $\text{bromocyclopentane} \xrightarrow{\text{alc. KOH}, \Delta} \text{cyclopentene} (C)$.

(D) In the Hunsdiecker reaction,silver salts of carboxylic acids react with $Br_2$ in $CCl_4$ to form alkyl bromides.
$CH_3-COOAg + Br_2 \xrightarrow{CCl_4, \Delta} CH_3-Br + CO_2 + AgBr$
Methyl bromide $(CH_3-Br)$ is an alkyl halide,not an alkane.
In other options:
$(a)$ Wurtz reaction produces butane ($C_4H_{10}$,an alkane).
$(b)$ Kolbe's electrolysis produces ethane ($C_2H_6$,an alkane).
$(c)$ Decarboxylation produces methane ($CH_4$,an alkane).

(A) Isobutane $CH_3-CH(CH_3)-CH_3$ contains $9$ primary $(1^o)$ hydrogens and $1$ tertiary $(3^o)$ hydrogen.
Relative reactivity: $1^o H : 3^o H = 1 : 1600$.
Relative amount of product $(A)$ (from $3^o H$) $= 1 \times 1600 = 1600$.
Relative amount of product $(B)$ (from $1^o H$) $= 9 \times 1 = 9$.
Total relative amount $= 1600 + 9 = 1609$.
$\% \text{ yield of } (A) = \frac{1600}{1609} \times 100 \approx 99.4\%$.
$\% \text{ yield of } (B) = \frac{9}{1609} \times 100 \approx 0.6\%$.

(C) The given reaction is a free radical halogenation of alkanes.
$1$. The reactivity order of halogens is $F_2 > Cl_2 > Br_2 > I_2$. Thus,option $A$ is incorrect.
$2$. The reaction proceeds via a free radical substitution mechanism,not electrophilic substitution. Thus,option $B$ is incorrect.
$3$. Iodination is a reversible reaction. To drive the reaction forward and increase the yield of alkyl iodide,an oxidizing agent like $HIO_3$ or $HNO_3$ is used to consume the $HI$ produced. Thus,option $C$ is correct.

(B) The given reaction is a Wurtz reaction,where an alkyl halide reacts with sodium metal in the presence of dry ether to form a symmetrical alkane.
The reaction is:
$2 C_6H_{11}CH_2Br + 2 Na \xrightarrow{\text{Dry ether}} C_6H_{11}CH_2-CH_2C_6H_{11} + 2 NaBr$
Here,two molecules of (cyclohexylmethyl) bromide couple together to form $1,2-$dicyclohexylethane.

(B) The hydrolysis reactions of the given carbides are as follows:
$Al_4C_3 + 12H_2O \to 4Al(OH)_3 + 3CH_4$ (Methane is produced)
$Be_2C + 4H_2O \to 2Be(OH)_2 + CH_4$ (Methane is produced)
$Mg_2C_3 + 4H_2O \to 2Mg(OH)_2 + C_3H_4$ (Propyne is produced)
$CaC_2 + 2H_2O \to Ca(OH)_2 + C_2H_2$ (Acetylene is produced)
Therefore,$CH_4$ gas is obtained from the hydrolysis of $Al_4C_3$ and $Be_2C$ only.

(A) $1$. The reaction of cyclohexane with $Br_2$ in the presence of $h\nu$ ($UV$ light) is a free-radical substitution reaction,which produces bromocyclohexane as product $(A)$.
$2$. The reaction of bromocyclohexane with $Na$ in the presence of dry ether is a Wurtz reaction,which involves the coupling of two alkyl groups to form a higher alkane.
$3$. Two molecules of bromocyclohexane react with $2Na$ to form bicyclohexyl as product $(B)$.

(A) The bond length between carbon atoms depends on the hybridization of the carbon atoms involved.
$1$. In $CH_3-CH_3$ (ethane), both carbons are $sp^3$ hybridized. The $C-C$ bond is a single bond with a bond length of approximately $154 \ pm$.
$2$. In $CH_2=CH_2$ (ethene), carbons are $sp^2$ hybridized. The $C=C$ bond length is $134 \ pm$.
$3$. In $CH\equiv CH$ (ethyne), carbons are $sp$ hybridized. The $C\equiv C$ bond length is $120 \ pm$.
$4$. In $CH_2=C=CH_2$ (allene), the central carbon is $sp$ hybridized and terminal carbons are $sp^2$ hybridized. The $C-C$ bond length is approximately $131 \ pm$.
Comparing these, the $C-C$ single bond in ethane $(sp^3-sp^3)$ has the maximum bond length.

(B) The boiling point of a compound depends on its molecular weight and intermolecular forces.
In option $B$,$CH_3-CH_2-CH_2-CH_3$ (Butane) has a higher molecular weight than $CH_3-CH_2-CH_3$ (Propane),leading to stronger London dispersion forces and a higher boiling point.
In option $A$,the second compound $(CH_3-CH(OH)-CH_3)$ can form hydrogen bonds,whereas the first $(CH_3-CH_2-O-CH_3)$ cannot,so the second has a higher boiling point.
In option $C$,the second compound is an alcohol,which has higher boiling points due to hydrogen bonding compared to the alkane.
In option $D$,the second compound $(CH_3-CH_2-CH_2-Cl)$ has a higher molecular weight and dipole-dipole interactions compared to the alkane,resulting in a higher boiling point.

(A) The heat of hydrogenation per $CH_2$ unit is directly proportional to the ring strain present in the cycloalkane.
Cyclopropane has the highest ring strain due to significant angle strain ($60^{\circ}$ bond angles compared to the ideal $109.5^{\circ}$),making it highly unstable.
Upon hydrogenation,the ring opens to form a stable alkane,releasing a large amount of energy.
The order of heat of combustion ($H$.$O$.$C$.) per $CH_2$ unit is: cyclopropane $>$ cyclobutane $>$ cyclopentane $>$ cyclohexane.
Therefore,the reaction involving cyclopropane releases the maximum amount of heat per $CH_2$ unit.

(C) The enthalpy change of a free radical halogenation reaction depends on the bond dissociation energies of the reactants and products.
$(i)$ Chlorination is generally more exothermic than bromination because the $C-Cl$ bond formed is stronger than the $C-Br$ bond,and the $H-Cl$ bond formed is stronger than the $H-Br$ bond.
$(ii)$ Among the given options,the chlorination of alkanes is more exothermic than bromination.
$(iii)$ Comparing the chlorination of propane and isobutane,the reaction with isobutane is more exothermic because it leads to the formation of a more stable $3^{\circ}$ alkyl chloride compared to the $2^{\circ}$ alkyl chloride formed from propane,due to the higher stability of the $3^{\circ}$ free radical intermediate.
Therefore,the reaction of isobutane with $Cl_2$ is the most exothermic.

(C) The rotational barrier is determined by steric hindrance and the electronic nature of the bond.
$A$: Cyclopentane has a cyclic structure,and rotation about a $C-C$ bond is restricted by the ring strain.
$B$: Buta$-1,3-$diene has partial double-bond character between $C_2$ and $C_3$ due to conjugation,leading to a high rotational barrier.
$C$: $n$-Butane $(CH_3-CH_2-CH_2-CH_3)$ has a simple $sp^3-sp^3$ single bond with minimal steric hindrance compared to the other options,resulting in the lowest rotational energy barrier.
$D$: $2,3-$Dimethylbutane has significant steric repulsion between the bulky methyl groups during rotation,leading to a higher barrier than $n$-butane.

(A) is correct. In $cis-1,2-dimethylcyclopropane$,the two methyl groups are on the same side of the ring,leading to significant van der Waals' repulsion (steric strain) between them,which makes the $cis$ isomer less stable than the $trans$ isomer.
$B$ is incorrect because cyclohexane has the lowest heat of combustion per $CH_2$ group due to its lack of angle strain.
$C$ is incorrect because the principal source of strain in the boat conformation of cyclohexane is torsional strain (eclipsing interactions) and flagpole interactions,not angle strain.
$D$ is incorrect because the principal source of strain in the gauche conformation of butane is van der Waals' repulsion (steric strain) between the methyl groups,not torsional strain.

(B) The given compound $Ph-CH=NO_2H$ is an aci-nitro compound.
Nitro-aci-nitro tautomerism is a type of tautomerism where a nitro compound $(R-CH_2-NO_2)$ exists in equilibrium with its aci-form $(R-CH=NO_2H)$.
In this case,the aci-nitro form $Ph-CH=NO_2H$ tautomerizes to the stable nitro form $Ph-CH_2-NO_2$.
Therefore,the isomer $(x)$ is $Ph-CH_2-NO_2$.

(B) The structure provided is $n$-pentane $(CH_3-CH_2-CH_2-CH_2-CH_3)$.
Mono-chlorination of $n$-pentane can occur at three non-equivalent carbon positions:
$1$. At the $C_1$ position (terminal carbon): $CH_3-CH_2-CH_2-CH_2-CH_2Cl$ ($1$-chloropentane).
$2$. At the $C_2$ position: $CH_3-CH_2-CH_2-CHCl-CH_3$ ($2$-chloropentane).
$3$. At the $C_3$ position: $CH_3-CH_2-CHCl-CH_2-CH_3$ ($3$-chloropentane).
Note that $2$-chloropentane and $3$-chloropentane are chiral molecules,meaning each exists as a pair of enantiomers ($R$ and $S$ forms).
However,fractional distillation separates compounds based on their boiling points. Enantiomers have identical boiling points and cannot be separated by fractional distillation.
Therefore,the distinct products with different boiling points are:
- $1$-chloropentane
- $2$-chloropentane (racemic mixture)
- $3$-chloropentane (racemic mixture)
This results in $3$ distinct fractions.

(C) Grignard reagents $(RMgX)$ react with compounds containing active hydrogen atoms (hydrogen atoms attached to electronegative atoms like $O, S, N$) to produce alkanes.
The reaction is: $R-MgX + R'-H \rightarrow R-H + Mg(R')X$.
In the given compound $HO-CH_2-CH(OH)-CH_2-SH$,there are three such active hydrogen atoms (two in $-OH$ groups and one in the $-SH$ group).
Thus,$3$ moles of $CH_3MgBr$ will react with $1$ mole of the compound to release $3$ moles of methane $(CH_4)$.
Therefore,$x = 3$.

(C) When $1,3-$dibromopropane $(Br-CH_2-CH_2-CH_2-Br)$ is heated with zinc dust in an ether solvent,an intramolecular dehalogenation reaction occurs.
Zinc acts as a reducing agent and removes the two bromine atoms from the terminal carbons.
The two terminal carbon atoms then bond together to form a three-membered ring.
The reaction is: $Br-CH_2-CH_2-CH_2-Br + Zn \xrightarrow{\text{ether}} \text{cyclopropane} + ZnBr_2$.
Thus,the product formed is cyclopropane.

(D) For an alkane to give only one monohalogenated product,all hydrogen atoms in the molecule must be equivalent.
$1$. $2-$methylbutane has different types of hydrogen atoms,leading to multiple monohalogenated products.
$2$. $2,2-$dimethylpropane (neopentane) has $12$ equivalent hydrogen atoms attached to the four equivalent methyl groups. Thus,it gives only one monohalogenated product.
$3$. Cyclopentane has $10$ equivalent hydrogen atoms in a cyclic structure. Thus,it also gives only one monohalogenated product.
Therefore,both $(b)$ and $(c)$ are correct.

(B) $2CH_3-CH_2-Cl \xrightarrow[Dry \ ether]{Na} CH_3-CH_2-CH_2-CH_3 + 2NaCl$
The Wurtz reaction is most effective for the preparation of symmetrical alkanes with an even number of carbon atoms.
In the case of $n$-butane,it is formed by the coupling of two ethyl radicals $(CH_3-CH_2 \cdot)$,making it the ideal product for this reaction.

(B) The molecular weight of hydrocarbon $A = 2 \times V.D. = 2 \times 36 = 72$.
For an alkane with formula $C_nH_{2n+2}$,$12n + 2n + 2 = 72$ $\Rightarrow 14n = 70$ $\Rightarrow n = 5$.
The molecular formula is $C_5H_{12}$.
Among the isomers of $C_5H_{12}$,neo-pentane $(CH_3-C(CH_3)_2-CH_3)$ has all $12$ hydrogen atoms equivalent.
Therefore,it forms only one monochloro substitution product.

(B) The Wurtz reaction involves the coupling of alkyl halides in the presence of sodium and dry ether.
When a mixture of two different alkyl halides ($R-X$ and $R'-X$) is used,a mixture of three possible alkanes is formed: $R-R$,$R'-R'$,and $R-R'$.
Here,the reactants are ethyl iodide $(C_2H_5I)$ and $n$-propyl iodide $(C_3H_7I)$.
The possible products are:
$1$. $C_2H_5-C_2H_5$ ($n$-butane)
$2$. $C_3H_7-C_3H_7$ ($n$-hexane)
$3$. $C_2H_5-C_3H_7$ ($n$-pentane)
Propane $(C_3H_8)$ is not formed in this reaction.

(B) The reactant is $2$-methylbutane $(CH_3-CH(CH_3)-CH_2-CH_3)$. Monochlorination can occur at four different positions:
$1.$ At $C1$: $ClCH_2-CH(CH_3)-CH_2-CH_3$ ($1$-chloro-$2$-methylbutane). Here,$C2$ becomes a chiral center.
$2.$ At $C2$: $CH_3-C(Cl)(CH_3)-CH_2-CH_3$ ($2$-chloro-$2$-methylbutane). No chiral center is formed.
$3.$ At $C3$: $CH_3-CH(CH_3)-CHCl-CH_3$ ($2$-chloro-$3$-methylbutane). Here,$C3$ becomes a chiral center.
$4.$ At $C4$: $CH_3-CH(CH_3)-CH_2-CH_2Cl$ ($1$-chloro-$3$-methylbutane). No chiral center is formed.
Therefore,the total number of unique chiral centers generated is $2$.

(D) $1$. $CH_3Cl \xrightarrow{Zn/H^{+}} CH_4$ (Reduction by metal-acid).
$2$. $CH_3Cl \xrightarrow{LiAlH_4} CH_4$ (Reduction by strong reducing agent).
$3$. $CH_3Cl$ $\xrightarrow{Mg, \text{ dry ether}} CH_3MgCl$ $\xrightarrow{H_2O} CH_4$ (Grignard reagent formation followed by hydrolysis).
Since all three methods successfully convert chloromethane to methane,the correct option is $D$.