Alkane Questions in English

(B) $n$-Butane is $CH_3-CH_2-CH_2-CH_3$.
Monochlorination can occur at the $1^{st}$ or $2^{nd}$ carbon.
$1.$ $CH_3-CH_2-CH_2-CH_2-Cl$ ($1$-chlorobutane) is achiral,resulting in $1$ isomer.
$2.$ $CH_3-CH(Cl)-CH_2-CH_3$ ($2$-chlorobutane) is chiral,resulting in $2$ stereoisomers ($R$ and $S$ configurations).
Total number of monochloro products = $1 + 2 = 3$.

(A) The correct answer is $A$. To maximize the yield of the monohalogenated product $(CH_3Cl)$,the reaction should be carried out with an excess of the alkane $(CH_4)$.
By keeping the concentration of $CH_4$ high relative to $Cl_2$,the probability of the chlorine radical $(Cl^{\bullet})$ colliding with a $CH_4$ molecule is significantly higher than its collision with a $CH_3Cl$ molecule.
If $Cl_2$ were in excess,the chlorine radicals would react with the already formed $CH_3Cl$ to produce polyhalogenated products like $CH_2Cl_2$ via the following steps:
$Cl^{\bullet} + CH_3Cl \to {^{\bullet}}CH_2Cl + HCl$
${^{\bullet}}CH_2Cl + Cl_2 \to CH_2Cl_2 + Cl^{\bullet}$
Therefore,a high ratio of $CH_4$ to $Cl_2$ is required.

(C) The reactant is $2$-methylbutane (isopentane). Monochlorination yields four structural isomers:
$1.$ $Cl-CH_2-CH(CH_3)-CH_2-CH_3$ ($1$-chloro-$2$-methylbutane): The $C2$ carbon is chiral,so it exists as a pair of enantiomers ($2$ stereoisomers).
$2.$ $CH_3-CCl(CH_3)-CH_2-CH_3$ ($2$-chloro-$2$-methylbutane): This molecule is achiral,so it exists as $1$ stereoisomer.
$3.$ $CH_3-CH(CH_3)-CHCl-CH_3$ ($2$-chloro-$3$-methylbutane): The $C3$ carbon is chiral,so it exists as a pair of enantiomers ($2$ stereoisomers).
$4.$ $CH_3-CH(CH_3)-CH_2-CH_2-Cl$ ($1$-chloro-$3$-methylbutane): This molecule is achiral,so it exists as $1$ stereoisomer.
Total number of monochloro products including stereoisomers = $2 + 1 + 2 + 1 = 6$.

(C) For isomeric alkanes,the heat of combustion is inversely proportional to their stability.
As branching increases,the stability of the alkane increases.
Stability order: Neopentane > Isopentane > $n$-pentane.
Therefore,the order of heat of combustion is: $n$-pentane > Isopentane > Neopentane,which is $(iii) > (ii) > (i)$.
Heat of combustion $\propto \frac{1}{\text{Stability}}$

(D) The reaction is a free radical bromination of $2$-bromopentane. The rate-determining step involves the abstraction of a hydrogen atom to form a free radical. The most stable free radical is formed at the $C-3$ position. Since the chiral center at $C-2$ is not involved in the reaction,its configuration remains unchanged. The $C-3$ carbon becomes a new chiral center upon bromination,leading to two diastereomers. The configuration at $C-2$ is fixed as $S$ in the starting material. Therefore,the products must retain the $S$ configuration at $C-2$. The option showing a change in configuration at $C-2$ (i.e.,$2R, 3S$) is not possible.

(B) The combustion reaction of ethane is: $2C_2H_6(g) + 7O_2(g) \rightarrow 4CO_2(g) + 6H_2O(l)$.
In ethane $(C_2H_6)$,the carbon atoms are $sp^3$ hybridized.
In the product carbon dioxide $(CO_2)$,the carbon atom is $sp$ hybridized.
Therefore,the hybridization changes from $sp^3$ to $sp$.

(A) The conversion of $n$-alkanes to branched-chain alkanes in the presence of anhydrous $AlCl_3$ and $HCl$ gas is known as isomerization.
In this reaction,$n$-butane is converted into its isomer,isobutane ($2$-methylpropane).

(C) Soda-lime de-carboxylation involves heating sodium acetate with soda-lime $(NaOH + CaO)$.
The reaction is: $CH_3COONa + NaOH \xrightarrow[\Delta]{CaO} CH_4 + Na_2CO_3$.
This method is suitable for the laboratory preparation of pure methane.

(C) The molecular weight of the alkane is $86$.
General formula for alkane: $C_nH_{2n+2} = 86$
$12n + 2n + 2 = 86$
$14n = 84$
$n = 6$
The alkane is $C_6H_{14}$.
$2$,$3$-dimethylbutane $(CH_3-CH(CH_3)-CH(CH_3)-CH_3)$ has two types of equivalent hydrogen atoms:
$1.$ Twelve $CH_3$ hydrogens are equivalent (on the four methyl groups).
$2.$ Two $CH$ hydrogens are equivalent (on the two tertiary carbons).
Thus,it gives only two monobromo derivatives.

(A) The reactivity of cycloalkanes towards addition reactions is governed by ring strain.
According to Baeyer's strain theory,the ring strain decreases as the ring size increases from $3$ to $5$ members.
$(I)$ Cyclopropane has the highest ring strain ($60^{\circ}$ bond angle),making it the most unstable and most reactive towards addition reactions.
$(II)$ Cyclobutane has intermediate ring strain ($90^{\circ}$ bond angle).
$(III)$ Cyclopentane has the lowest ring strain ($108^{\circ}$ bond angle),making it the most stable and least reactive towards addition reactions.
Therefore,the order of reactivity is $I > II > III$.

(B) The reaction is an example of an intramolecular Wurtz reaction.
Sodium $(Na)$ acts as a reducing agent,providing electrons $(e^{\ominus})$ to the substrate.
The mechanism involves the following steps:
$1$. An electron $(e^{\ominus})$ attacks the carbon attached to the bromine atom,leading to the departure of the bromide ion $(Br^{\ominus})$ and the formation of a radical.
$2$. Another electron $(e^{\ominus})$ attacks the carbon attached to the chlorine atom,leading to the departure of the chloride ion $(Cl^{\ominus})$ and the formation of a second radical.
$3$. The two radicals then combine to form a new $C-C$ bond,resulting in the formation of bicyclobutane.
Therefore,the correct product $(A)$ is bicyclobutane.

(C) Methyl bromide $(CH_3-Br)$ can be reduced to methane $(CH_4)$ using a reducing agent like $Zn/HCl$ or $LiAlH_4$ in one step.
Methyl bromide $(CH_3-Br)$ can be converted to ethane $(CH_3-CH_3)$ using the $Wurtz$ reaction $(2CH_3-Br + 2Na \xrightarrow{dry \ ether} CH_3-CH_3 + 2NaBr)$ in one step.
Therefore,$CH_3-Br$ is the suitable reactant.

(D) For an alkane to exist in an enantiomeric form,it must possess at least one chiral carbon atom (a carbon atom bonded to four different groups).
Let the chiral carbon be $C^*$. The four groups attached to $C^*$ must be different.
In an alkane,the smallest groups are hydrogen $(-H)$,methyl $(-CH_3)$,ethyl $(-CH_2CH_3)$,and propyl $(-CH_2CH_2CH_3)$.
Summing the carbon atoms in these groups: $1$ (for the chiral carbon) $+ 1$ (methyl) $+ 2$ (ethyl) $+ 3$ (propyl) $= 7$ carbon atoms.
Therefore,the smallest alkane capable of exhibiting enantiomerism is $3$-methylhexane,which has $7$ carbon atoms.

(B) Bromination is highly selective for the most stable radical intermediate.
$(P)$ Cyclohexane gives bromocyclohexane ($2^o$ halide).
$(Q)$ Methylcyclobutane gives $1-$bromo$-1-$methylcyclobutane ($3^o$ halide).
$(R)$ Ethylbenzene gives $1-$bromo$-1-$phenylethane (benzylic halide,which is more stable than $2^o$ alkyl halide,but often classified as $2^o$ in terms of carbon substitution).
$(S)$ $3-$Methylhexane gives $3-$bromo$-3-$methylhexane ($3^o$ halide).
$(T)$ Decalin gives a $3^o$ halide at the bridgehead position.
$(U)$ Butane gives $2-$bromobutane ($2^o$ halide).
Therefore,$P, R, U$ are the reactions where $2^o$ (or equivalent) halides are the major products.

(B) The free radical chlorination of an alkane $(A)$ is a chain reaction that can lead to polychlorination if $Cl_2$ is present in excess.
To maximize the yield of the monochloro product,the concentration of the alkane $(A)$ must be kept high relative to $Cl_2$.
By adding reactant $(A)$ in excess,the probability of a $Cl^{\bullet}$ radical colliding with an unreacted molecule of $(A)$ is significantly higher than colliding with a monochloro product molecule,thereby minimizing further chlorination.

(A) The reaction of $n$-butane with $Br_2$ in the presence of $hv$ (light) proceeds via a free radical mechanism.
The abstraction of a secondary hydrogen atom is more favorable than a primary hydrogen atom due to the greater stability of the $2^o$ alkyl radical.
The intermediate $CH_3-CH^{\bullet}-CH_2-CH_3$ is $sp^2$-hybridized and planar.
Attack of $Br_2$ on either face of the planar radical leads to the formation of both enantiomers of $2$-bromobutane in equal amounts.
Therefore,the major product is a racemic mixture.

(C) The free-radical chlorination of methane proceeds through three stages: initiation,propagation,and termination.
Step $(1)$ is the initiation step where $Cl_2$ undergoes homolytic cleavage to form chlorine radicals.
Propagation steps are those where a radical reacts with a molecule to form a new molecule and a new radical.
Step $(3)$ is a propagation step: $Cl^{\bullet} + CH_4 \to CH_3^{\bullet} + HCl$.
Step $(5)$ is a propagation step: $CH_3^{\bullet} + Cl_2 \to CH_3Cl + Cl^{\bullet}$.
Therefore,the propagation steps are $(3)$ and $(5)$.

(B) The reaction of methylcyclohexane with $Br_2/hv$ involves free radical substitution.
Methylcyclohexane has different types of hydrogen atoms available for substitution:
$1$. The hydrogen on the methyl group (primary carbon).
$2$. The hydrogen on the tertiary carbon of the cyclohexane ring (at the position where the methyl group is attached).
$3$. The hydrogens on the secondary carbons of the cyclohexane ring at the ortho position.
$4$. The hydrogens on the secondary carbons of the cyclohexane ring at the meta position.
$5$. The hydrogen on the secondary carbon of the cyclohexane ring at the para position.
Thus,there are $5$ distinct positions where bromine can substitute,leading to $5$ structural isomers. Excluding stereoisomers,the total number of monobromo products is $5$.

(A) The bromine radical $(Br^{\bullet})$ is highly selective in hydrogen abstraction reactions.
It preferentially abstracts the hydrogen atom that leads to the formation of the most stable free radical.
Comparing the positions:
- Hydrogen '$a$' is on a tertiary carbon atom. Removal of this hydrogen results in a $3^{\circ}$ free radical.
- Hydrogen '$b$' is on a secondary carbon atom.
- Hydrogen '$c$' is on a primary carbon atom.
- Hydrogen '$d$' is on a secondary carbon atom.
Since the $3^{\circ}$ free radical is the most stable among the options,the bromine radical will most readily abstract the hydrogen at position '$a$'.

(C) The reactivity of hydrogens towards free radical chlorination depends on the stability of the intermediate radical formed after hydrogen abstraction.
Stability of radicals follows the order: $3^o > 2^o > 1^o$.
Additionally,the $-I$ effect of the fluorine atom destabilizes the radical formed on the adjacent carbons,with the effect decreasing as the distance from the $F$ atom increases.
For $CH_3(a)-CH_2(b)-CH_2(c)-CH_2(d)-F$:
- Hydrogen $d$ is on the carbon attached to $F$ (most destabilized by $-I$ effect).
- Hydrogen $c$ is on the $\beta$-carbon.
- Hydrogen $b$ is on the $\gamma$-carbon.
- Hydrogen $a$ is on the $\delta$-carbon (least affected by $-I$ effect).
However,considering the radical stability,$b$ and $c$ are $2^o$ radicals,while $a$ and $d$ are $1^o$ radicals.
Comparing the $-I$ effect,the reactivity order is $b > c > a > d$.

(C) The reaction of cyclopentane with excess $Cl_2$ under $h\nu$ leads to the formation of various dichloro derivatives.
$1$. $1,1$-dichlorocyclopentane: $1$ isomer.
$2$. $1,2$-dichlorocyclopentane: This has $2$ chiral centers,so it exists as $3$ stereoisomers ($cis$ (meso) and $trans$ (enantiomeric pair)).
$3$. $1,3$-dichlorocyclopentane: This also has $2$ chiral centers,so it exists as $3$ stereoisomers ($cis$ (meso) and $trans$ (enantiomeric pair)).
Total number of dichloro products = $1 + 3 + 3 = 7$.

(A) The reaction is a free radical bromination.
The most stable radical is formed at the benzylic position $(Ph-\dot{C}H-CH(D)-CH_3)$ due to resonance stabilization with the phenyl ring.
Bromination at this position creates a new chiral center.
Since the starting molecule already contains a chiral center at the carbon atom bonded to deuterium $(D)$,the formation of a second chiral center results in a pair of diastereomers.

(C) The reaction of isobutane with ethene in the presence of $HF$ at low temperature is an alkylation reaction.
$1$. The mechanism involves the formation of a tert-butyl carbocation $(CH_3)_3C^+$ by the protonation of isobutane.
$2$. This carbocation adds to the ethene molecule to form a new carbocation $(CH_3)_3C-CH_2-CH_2^+$.
$3$. Through a series of rearrangements ($1$,$2$-hydride shift followed by $1,2-$methyl shift),a more stable carbocation is formed.
$4$. Finally,hydride abstraction from another isobutane molecule yields the major product,$2$,$2$-dimethylbutane $(CH_3-C(CH_3)_2-CH_2-CH_3)$.

(B) $(I)$ Reaction with $HBr$ in the presence of peroxide follows anti-Markovnikov addition to form $1-$bromopropane.
$(II)$ Reaction with $HBr$ in $CCl_4$ follows Markovnikov's addition to form $2-$bromopropane.
$(III)$ Free radical bromination of propane $(CH_3-CH_2-CH_3)$ with $Br_2$ under $hv$ (light) preferentially forms $2-$bromopropane due to the higher stability of the secondary free radical intermediate.
$(IV)$ Reaction with $Br_2$ in $CCl_4$ results in the formation of $1,2-$dibromopropane.
Therefore,reactions $(II)$ and $(III)$ result in the formation of $2-$bromopropane.

(D) The reaction of an alkane with diazomethane $(CH_2N_2)$ in the presence of heat or light involves the generation of a carbene intermediate $(:CH_2)$.
The carbene undergoes an insertion reaction into the $C-H$ bonds of the alkane.
For $n-$pentane $(CH_3-CH_2-CH_2-CH_2-CH_3)$,there are three types of $C-H$ bonds:
$1.$ Insertion into the primary $C-H$ bond ($C_1$ or $C_5$) gives $n-$hexane $(CH_3-CH_2-CH_2-CH_2-CH_2-CH_3)$.
$2.$ Insertion into the secondary $C-H$ bond at $C_2$ (or $C_4$) gives $2-$methylpentane $(CH_3-CH(CH_3)-CH_2-CH_2-CH_3)$.
$3.$ Insertion into the secondary $C-H$ bond at $C_3$ gives $3-$methylpentane $(CH_3-CH_2-CH(CH_3)-CH_2-CH_3)$.
Therefore,all the listed products are obtained.

(D) The given reaction is the bromination of toluene in the presence of light $(h\nu)$.
This reaction proceeds via a free radical mechanism where a hydrogen atom on the methyl group of toluene is replaced by a bromine atom.
Therefore,it is a $Free \ radical \ substitution$ reaction.

(A) $CH_3MgBr$ is a Grignard reagent,which acts as a strong base.
Methanol $(CH_3OH)$ contains an acidic hydrogen atom attached to the oxygen atom.
When $CH_3MgBr$ reacts with $CH_3OH$,the methyl group $(CH_3^-)$ abstracts the acidic proton from the methanol to form methane $(CH_4)$ gas.
The reaction is: $CH_3MgBr + CH_3OH \rightarrow CH_4 \uparrow + Mg(OCH_3)Br$.

(D) The general combustion reaction for an alkane $C_nH_{2n+2}$ is:
$C_nH_{2n+2} + (\frac{3n+1}{2})O_2 \to nCO_2 + (n+1)H_2O$
According to Avogadro's law,at constant temperature and pressure,the volume of a gas is directly proportional to the number of moles.
Therefore,the ratio of volumes is equal to the ratio of stoichiometric coefficients:
$\frac{V_{alkane}}{V_{O_2}} = \frac{1}{\frac{3n+1}{2}} = \frac{2}{3n+1}$
Given $V_{alkane} = 5 \, L$ and $V_{O_2} = 25 \, L$:
$\frac{5}{25} = \frac{2}{3n+1}$
$\frac{1}{5} = \frac{2}{3n+1}$
$3n+1 = 10$
$3n = 9$
$n = 3$
Since $n=3$,the alkane is propane $(C_3H_8)$.

(D) The bromination of alkanes follows a free radical mechanism.
Bromine is highly selective and preferentially replaces the hydrogen atom on the carbon that forms the most stable radical.
In $2-$methylbutane $(CH_3-CH(CH_3)-CH_2-CH_3)$,the $C2$ position is a tertiary $(3^\circ)$ carbon.
The $3^\circ$ radical is significantly more stable than $2^\circ$ or $1^\circ$ radicals.
Thus,the major product is $2-$bromo$-2-$methylbutane.

(A) The photobromination of $2-$methylbutane involves the formation of a free radical intermediate.
$CH_3-CH(CH_3)-CH_2-CH_3 \xrightarrow[Br_2]{hv} CH_3-C(Br)(CH_3)-CH_2-CH_3$ ($2-$bromo$-2-$methylbutane).
The reactivity of hydrogen atoms towards free radical substitution follows the order: $3^{\circ} > 2^{\circ} > 1^{\circ}$.
Since $2-$methylbutane has one tertiary $(3^{\circ})$ hydrogen atom at the $C-2$ position,the abstraction of this hydrogen leads to the most stable tertiary free radical.
Therefore,$2-$bromo$-2-$methylbutane is the major product.

(C) Photochlorination proceeds via a free radical mechanism. The ease of replacement of a hydrogen atom depends on the stability of the resulting free radical intermediate.
$1$. $H_b$ is a benzylic hydrogen. The resulting radical is resonance-stabilized by the benzene ring,making it the most stable radical.
$2$. $H_d$ is a secondary $(2^{\circ})$ alkyl hydrogen. The resulting radical is a secondary alkyl radical,which is more stable than primary radicals.
$3$. $H_c$ is a primary $(1^{\circ})$ alkyl hydrogen. The resulting radical is a primary alkyl radical.
$4$. $H_a$ is an aromatic hydrogen attached directly to the benzene ring. Replacing this requires breaking a strong $sp^2$ $C-H$ bond,which is very difficult under standard photochlorination conditions.
Therefore,the stability order of the radicals is: Benzylic > $2^{\circ}$ alkyl > $1^{\circ}$ alkyl > Aromatic.
The order of ease of replacement is $H_b > H_d > H_c > H_a$.

(D) The Wurtz reaction involves the coupling of two alkyl halide molecules in the presence of sodium metal to form a symmetric alkane.
When a single alkyl halide is used,the resulting alkane must have an even number of carbon atoms and a symmetric structure formed by joining two identical alkyl groups.
For example,$2,3-$Dimethylbutane can be synthesized from $2-$chloropropane:
$2CH_3-CH(Cl)-CH_3 + 2Na \xrightarrow{\text{dry ether}} CH_3-CH(CH_3)-CH(CH_3)-CH_3 + 2NaCl$
$n-$Butane,$n-$Pentane,and $2-$Methylbutane are either asymmetric or have an odd number of carbon atoms,making them unsuitable for synthesis from a single alkyl halide in high yield.

(D) compound gives only one monochloro product if all its hydrogen atoms are equivalent.
$(A)$ $2,2-$dimethylpropane $(C(CH_3)_4)$ has all $12$ hydrogen atoms equivalent,so it gives only one monochloro product.
$(B)$ $2,2,3,3-$tetramethylbutane $((CH_3)_3C-C(CH_3)_3)$ has all $18$ hydrogen atoms equivalent,so it gives only one monochloro product.
$(C)$ Cyclopentane $(C_5H_{10})$ has all $10$ hydrogen atoms equivalent,so it gives only one monochloro product.
$(D)$ $2-$methylpropane $(CH_3-CH(CH_3)-CH_3)$ has two types of hydrogen atoms (primary and tertiary),so it gives two different monochloro products: $1-$chloro-$2-$methylpropane and $2-$chloro-$2-$methylpropane.

(A) The molecular formula $C_5H_{10}$ corresponds to a degree of unsaturation of $1$ $(5 - 10/2 + 1 = 1)$.
Cycloalkanes are saturated cyclic hydrocarbons with the general formula $C_nH_{2n}$.
For $C_5H_{10}$,the possible cycloalkanes are:
$1$. Cyclopentane
$2$. Methylcyclobutane
$3$. $1,1$-Dimethylcyclopropane
$4$. $1,2$-Dimethylcyclopropane (exists as cis and trans isomers)
$5$. Ethylcyclopropane
Counting the stereoisomers ($cis$ and $trans$ for $1,2$-dimethylcyclopropane),the total number of distinct cycloalkanes is $5$.

(B) The reaction of a Grignard reagent $(PhMgCl)$ with water (hydrolysis) involves the protonation of the carbanion part of the reagent.
The phenyl group $(Ph^-)$ acts as a strong base and abstracts a proton $(H^+)$ from the water molecule $(H_2O)$.
The reaction is as follows:
$PhMgCl + H_2O \rightarrow C_6H_6 + Mg(OH)Cl$
Here,$PhMgCl$ reacts with $H_2O$ to form benzene $(C_6H_6)$ and magnesium hydroxychloride $(Mg(OH)Cl)$.
Therefore,the product is benzene.

(C) The correct option is $(C)$.
In Kolbe's electrolysis of sodium acetate $(CH_3COONa)$,the acetate ions $(CH_3COO^{-})$ migrate to the anode.
At the anode,they undergo oxidation to form ethane $(CH_3-CH_3)$ and carbon dioxide $(CO_2)$.
The reaction at the anode is: $2CH_3COO^{-} \rightarrow CH_3-CH_3 + 2CO_2 + 2e^{-}$.

(C) The structure of $n-$butane is $CH_3-CH_2-CH_2-CH_3$.
Chlorination of $n-$butane can occur at the $C-1$ position or the $C-2$ position.
Substitution at $C-1$ gives $1-$chlorobutane $(CH_3-CH_2-CH_2-CH_2Cl)$.
Substitution at $C-2$ gives $2-$chlorobutane $(CH_3-CH_2-CHCl-CH_3)$.
Since $2-$chlorobutane has a chiral center,it exists as a pair of enantiomers ($d$ and $l$ forms).
Therefore,the total number of distinct monochloro-products (including stereoisomers) is $3$ ($1-$chlorobutane,$d-2-$chlorobutane,and $l-2-$chlorobutane).

(B) The reaction of a Grignard reagent $(R-MgBr)$ with $D_2O$ is a standard method for the deuteration of alkanes.
In this reaction,the nucleophilic carbon atom of the Grignard reagent attacks the deuterium atom of $D_2O$,replacing the $MgBr$ group with a deuterium atom $(D)$.
The starting material is $3$-methylcyclohexylmagnesium bromide.
Upon treatment with $D_2O$,the $MgBr$ group at the $1$-position is replaced by $D$.
Thus,the product $(A)$ is $1$-deuterio$-3-$methylcyclohexane.

(A) The reaction of a carboxylic acid with sodalime $(NaOH + CaO)$ is known as decarboxylation.
In this process,the $-COOH$ group is removed and replaced by a hydrogen atom.
For cyclohexanecarboxylic acid,the reaction is:
$\text{Cyclohexanecarboxylic acid} + NaOH/CaO \rightarrow \text{Cyclohexane} + Na_2CO_3$.
Thus,the product is cyclohexane.

(D) Homolytic bond dissociation energy $(BDE)$ is inversely proportional to the stability of the free radical formed after the bond breaks.
Stability of free radicals follows the order: $3^\circ > 2^\circ > 1^\circ$.
In the given molecule (isopentane):
$1.$ $C_2-H$ is a $3^\circ$ bond (forms a $3^\circ$ radical,which is the most stable).
$2.$ $C_3-H$ is a $2^\circ$ bond (forms a $2^\circ$ radical).
$3.$ $C_1-H$ and $C_4-H$ are $1^\circ$ bonds (form $1^\circ$ radicals,which are the least stable).
Among the $1^\circ$ bonds,the $C_4$ radical is slightly more stable than the $C_1$ radical because $C_4$ has two $\beta$-hydrogens available for hyperconjugation,while $C_1$ has only one.
Therefore,the $BDE$ order is: $C_1-H > C_4-H > C_3-H > C_2-H$.

(B) The reaction of ethane $(CH_3-CH_3)$ with bromine $(Br_2)$ leads to the substitution of hydrogen atoms by bromine atoms. The possible substitution products are:
$1$. $CH_3-CH_2Br$ (Bromoethane)
$2$. $CH_3-CHBr_2$ ($1$,$1$-Dibromoethane)
$3$. $CH_2Br-CH_2Br$ ($1$,$2$-Dibromoethane)
$4$. $CH_3-CBr_3$ ($1$,$1$,$1$-Tribromoethane)
$5$. $CH_2Br-CHBr_2$ ($1$,$1$,$2$-Tribromoethane)
$6$. $CHBr_2-CHBr_2$ ($1$,$1$,$2$,$2$-Tetrabromoethane)
$7$. $CHBr_2-CBr_3$ (Pentabromoethane)
$8$. $CBr_3-CBr_3$ (Hexabromoethane)
There are a total of $8$ possible substitution products.

(D) Grignard reagents $(R-MgX)$ react with compounds containing active hydrogen (like alcohols,$R'-OH$) to form alkanes $(R-H)$ and magnesium alkoxides $(R'-OMgX)$.
The reaction of isobutyl magnesium bromide with absolute alcohol (ethanol) is as follows:
$CH_3-CH(CH_3)-CH_2-MgBr + CH_3-CH_2-OH \rightarrow CH_3-CH(CH_3)-CH_3 + CH_3-CH_2-OMgBr$.
The products are isobutane and magnesium ethoxide bromide.

(B) The reaction involves the abstraction of a hydrogen atom by a bromine radical $(\dot{Br})$.
Stability of free radicals follows the order: $3^{\circ} > 2^{\circ} > 1^{\circ}$.
In the given molecule $CH_3-CH(D)-CH(CH_3)-CH_3$,there are different types of hydrogen atoms available for abstraction.
The abstraction of a hydrogen atom from the $C-3$ position leads to the formation of a $3^{\circ}$ free radical,$CH_3-CH(D)-\dot{C}(CH_3)-CH_3$,which is the most stable intermediate.
Since the $C-H$ bond energy is lower than the $C-D$ bond energy,the hydrogen atom is abstracted preferentially over the deuterium atom.
Thus,the major product $X$ is $CH_3-CH(D)-\dot{C}(CH_3)-CH_3$.

(B) The starting material is $2$-methylbutane,$CH_3-CH(CH_3)-CH_2-CH_3$.
Chlorination with $Cl_2/hv$ can occur at four different types of hydrogen atoms:
$1$. $CH_3-CH(CH_3)-CH_2-CH_2-Cl$ ($1$-chloro$-2-$methylbutane,achiral)
$2$. $CH_3-CH(CH_3)-CHCl-CH_3$ ($2$-chloro$-2-$methylbutane,chiral,gives $d$ and $l$ enantiomers)
$3$. $CH_3-C(Cl)(CH_3)-CH_2-CH_3$ ($2$-chloro$-2-$methylbutane,achiral)
$4$. $ClCH_2-CH(CH_3)-CH_2-CH_3$ ($1$-chloro$-3-$methylbutane,chiral,gives $d$ and $l$ enantiomers)
Total number of isomers $N = 1 + 2 + 1 + 2 = 6$.
Fractional distillation separates compounds based on boiling points. Enantiomers ($d$ and $l$ pairs) have identical boiling points and cannot be separated by fractional distillation.
Thus,the number of fractions $M = 4$ (corresponding to the four constitutional isomers).

(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 $(R-H + I_2 \rightleftharpoons R-I + H-I)$. The $H-I$ formed is a strong reducing agent and reduces $R-I$ back to $R-H$. To shift the equilibrium forward,an oxidizing agent like $HIO_3$ or $HNO_3$ is used to consume $H-I$ $(5H-I + HIO_3 \rightarrow 3I_2 + 3H_2O)$. Thus,option $C$ is correct.

(A) The reaction of isobutane with $Cl_2$ in the presence of sunlight (hv) proceeds via a free radical mechanism.
Isobutane has $9$ primary $(1^o)$ hydrogen atoms and $1$ tertiary $(3^o)$ hydrogen atom.
The relative reactivity of $1^o$,$2^o$,and $3^o$ hydrogen atoms towards chlorination is approximately $1 : 3.8 : 5$.
For $1^o$ hydrogen atoms: $\text{Number of } H \times \text{relative rate} = 9 \times 1 = 9$.
For $3^o$ hydrogen atom: $\text{Number of } H \times \text{relative rate} = 1 \times 5 = 5$.
Since the product of the number of hydrogen atoms and their relative reactivity is higher for the $1^o$ position $(9 > 5)$,the $1^o$ alkyl chloride (isobutyl chloride) is the major product.

(A) The free radical chlorination of alkanes is a chain reaction that often leads to a mixture of products due to polyhalogenation.
To obtain the mono-substituted product $C_2H_5Cl$ in the best yield,it is necessary to minimize further chlorination.
By taking an excess of ethane $(C_2H_6)$,the probability of a chlorine radical colliding with an ethane molecule is significantly higher than its collision with a $C_2H_5Cl$ molecule.
Therefore,$C_2H_6 \text{ (excess)} + Cl_2 \xrightarrow{UV}$ is the condition that favors the formation of $C_2H_5Cl$ as the major product.

(C) The reaction is a Corey-House synthesis,which is used to prepare alkanes by coupling an alkyl halide with a lithium dialkylcuprate (Gilman reagent).
Step $1$: Formation of Gilman reagent $(A)$.
$2(CH_3)_2CHBr + 4Li \rightarrow 2(CH_3)_2CHLi + 2LiBr$
$2(CH_3)_2CHLi + CuI \rightarrow [(CH_3)_2CH]_2CuLi + LiI$
So,$(A)$ is the Gilman reagent $[(CH_3)_2CH]_2CuLi$.
Step $2$: Coupling reaction.
$[(CH_3)_2CH]_2CuLi + (CH_3)_2CHCH_2Br \rightarrow (CH_3)_2CH-CH_2CH(CH_3)_2 + (CH_3)_2CHCu + LiBr$
The product $B$ is $2,4$-dimethylpentane,which is $(CH_3)_2CHCH_2CH(CH_3)_2$.

(A) The product of the Wurtz reaction is $2,3$-dimethylbutane $(CH_3-CH(CH_3)-CH(CH_3)-CH_3)$. This indicates that the starting alkyl bromide $R'-Br$ is isopropyl bromide $(CH_3-CH(Br)-CH_3)$.
Step $1$: $2 CH_3-CH(Br)-CH_3 + 2 Na \xrightarrow{ether} CH_3-CH(CH_3)-CH(CH_3)-CH_3 + 2 NaBr$.
Step $2$: $CH_3-CH(Br)-CH_3 + Mg \xrightarrow{ether} CH_3-CH(MgBr)-CH_3$ $(D)$.
Step $3$: $CH_3-CH(MgBr)-CH_3 + H_2O \to CH_3-CH_2-CH_3$ $(E)$ $+ Mg(OH)Br$.
Therefore,$D$ is isopropyl magnesium bromide and $E$ is propane.

(C) In the chlorination of isobutane $(CH_3-CH(CH_3)-CH_3)$,there are $9$ primary $(1^\circ)$ hydrogens and $1$ tertiary $(3^\circ)$ hydrogen.
The relative reactivity of $Cl$ radicals with $1^\circ, 2^\circ, 3^\circ$ hydrogens is approximately $1 : 3.8 : 5$.
Relative yield of $1^\circ$ product = $9 \times 1 = 9$.
Relative yield of $3^\circ$ product = $1 \times 5 = 5$.
Since $9 > 5$,the $1^\circ$ alkyl halide ($1$-chloro-$2$-methylpropane) is the major product.