Alkane Questions in English

(D) The structure of isopentane is $CH_3-CH(CH_3)-CH_2-CH_3$.
In this molecule,there is only one secondary $(2^{\circ})$ carbon atom (the $-CH_2-$ group).
Therefore,$1 \ mol$ of isopentane contains $1 \ mol$ of secondary carbon atoms.
Thus,$n \ mol$ of isopentane contains $n \ mol$ of secondary carbon atoms.

(D) The general equation for the combustion of alkanes is $C_n H_{2n+2} + (\frac{3n+1}{2}) O_2 \rightarrow n CO_2 + (n+1) H_2O + \text{Heat}$.
Greater the value of $n$ (number of carbon atoms),greater will be the amount of heat produced upon combustion.
Comparing the given alkanes: Methane $(n=1)$,Ethane $(n=2)$,Propane $(n=3)$,and Butane $(n=4)$.
Since Butane has the highest number of carbon atoms $(n=4)$,it releases the maximum energy upon combustion.

(D) The reaction given is the Wurtz reaction,where an alkyl halide reacts with sodium in the presence of dry ether to form a symmetric alkane.
$2R^{\prime}-X + 2Na \xrightarrow{\text{ether}} R^{\prime}-R^{\prime} + 2NaX$
Given the product is $2,3-\text{dimethylbutane}$,which has the structure $(CH_3)_2CH-CH(CH_3)_2$.
This indicates that the alkyl group $R^{\prime}$ must be an isopropyl group,$(CH_3)_2CH-$.
Therefore,$R^{\prime}-X$ is isopropyl halide,$(CH_3)_2CH-X$.

(D) The reaction of sodium acetate $(CH_3COONa)$ with sodalime $(NaOH + CaO)$ is known as decarboxylation.
When sodium acetate is heated with sodalime,it loses a molecule of $CO_2$ to form methane $(CH_4)$ and sodium carbonate $(Na_2CO_3)$.
The chemical equation is: $CH_3COONa + NaOH \xrightarrow{CaO, \Delta} CH_4 + Na_2CO_3$.
Thus,the product obtained is methane.

(C) The given reaction is the free radical chlorination of $n$-butane.
This reaction proceeds via a free radical mechanism initiated by ultraviolet $(UV)$ light.
$CH_3-CH_2-CH_2-CH_3 + Cl_2 \xrightarrow{UV \ light} CH_3-CH_2-CH_2-CH_2-Cl + CH_3-CH_2-CHCl-CH_3 + HCl$.
Therefore,the correct reagent is $Cl_2 / UV$ light.

(D) The given reaction is a Wurtz reaction,where two molecules of an alkyl halide $(R'-Cl)$ react with sodium in the presence of dry ether to form a symmetric alkane.
The reaction is: $2R'-Cl + 2Na \xrightarrow{\text{Dry ether}} R'-R' + 2NaCl$.
The product formed is $2,3-\text{dimethylbutane}$,which has the structure $CH_3-CH(CH_3)-CH(CH_3)-CH_3$.
By splitting this symmetric alkane at the center,we get two identical fragments of $CH_3-CH(CH_3)-$. Thus,the alkyl group $R'$ is the isopropyl group,$CH_3-CH(CH_3)-$.
Therefore,$R'-Cl$ is isopropyl chloride.

(C) Mono-chlorination involves the replacement of a hydrogen atom with a chlorine atom in the presence of light $(h\nu)$.
For a compound to yield only one mono-chlorination product,all hydrogen atoms in the molecule must be equivalent.
$A$. $n-$Butane $(CH_3-CH_2-CH_2-CH_3)$ has two types of hydrogen atoms,yielding two products.
$B$. iso-pentane $((CH_3)_2CH-CH_2-CH_3)$ has four types of hydrogen atoms,yielding four products.
$C$. neo-pentane $((CH_3)_4C)$ has all $12$ hydrogen atoms equivalent (attached to the same type of carbon),yielding only one product: $1-$chloro-$2,2-$dimethylpropane.
$D$. $n-$pentane $(CH_3-CH_2-CH_2-CH_2-CH_3)$ has three types of hydrogen atoms,yielding three products.
Therefore,the correct answer is $C$.

(B) The structure of isopentane is $CH_3-CH(CH_3)-CH_2-CH_3$.
There are $4$ distinct types of hydrogen atoms in isopentane:
$1$. Primary hydrogens on the terminal $CH_3$ group attached to the $CH$ group.
$2$. Tertiary hydrogen on the $CH$ group.
$3$. Secondary hydrogens on the $CH_2$ group.
$4$. Primary hydrogens on the terminal $CH_3$ group attached to the $CH_2$ group.
Since there are $4$ different types of hydrogen atoms,replacing any one of them with a chlorine atom will result in $4$ different monochloro structural isomers.
Therefore,the correct answer is $4$.

(A) The complete reaction sequence involved is as follows:
$CH_3COONa + NaOH \xrightarrow[\Delta]{CaO} CH_4 + Na_2CO_3$
(Sodium ethanoate) (Methane $X$)
$2CH_3COONa + 2H_2O \xrightarrow{\text{Electrolysis}} CH_3-CH_3 + 2CO_2 + H_2 + 2NaOH$
(Ethane $Y$)
Thus,the compounds $X$ and $Y$ are methane and ethane respectively.

(D) This is $Kolbe's$ electrolytic method which produces alkanes.
$2 RCOONa + 2 H_2O \xrightarrow{\text{Electrolysis}} R-R + 2 NaOH + H_2 + 2 CO_2$
Steps of reaction:
$2 CH_3COONa \rightleftharpoons 2 CH_3COO^{-} + 2 Na^{+}$
$2 H_2O \rightleftharpoons 2 OH^{-} + 2 H^{+}$
At anode:
$2 CH_3COO^{-} - 2 e^{-}$ $\rightarrow [2 CH_3COO]$ $\rightarrow CH_3-CH_3 + 2 CO_2$
At cathode:
$2 H^{+} + 2 e^{-}$ $\rightarrow [2 H]$ $\rightarrow H_2$

(D) The neutralization reaction is: $CH_3COOH + NaOH \rightarrow CH_3COONa + H_2O$.
Moles of $CH_3COOH = 0.1 \ M \times 0.1 \ L = 0.01 \ mol$.
Thus,$0.01 \ mol$ of $CH_3COONa$ is formed.
Kolbe's electrolysis reaction is: $2CH_3COONa + 2H_2O \rightarrow C_2H_6 + 2CO_2 + H_2 + 2NaOH$.
According to the stoichiometry,$2 \ mol$ of $CH_3COONa$ produces $1 \ mol$ of $C_2H_6$.
Therefore,$0.01 \ mol$ of $CH_3COONa$ produces $0.005 \ mol$ of $C_2H_6$.
Volume of $C_2H_6$ at $STP = 0.005 \ mol \times 22400 \ mL/mol = 112 \ mL$.

(D) In an ideal $sp^3$ hybridized carbon atom,the bond angle is $109^{\circ} 28^{\prime}$.
In cyclobutane,the carbon atoms form a square,so the internal bond angle is $90^{\circ}$.
The total deviation from the tetrahedral angle is $109^{\circ} 28^{\prime} - 90^{\circ} = 19^{\circ} 28^{\prime}$.
According to Baeyer's strain theory,the angle strain is half of this deviation.
$\text{Angle strain} = \frac{1}{2} \times (109^{\circ} 28^{\prime} - 90^{\circ}) = \frac{19^{\circ} 28^{\prime}}{2} = 9^{\circ} 44^{\prime}$.

(A) The bond angle in cyclopropane is calculated as $\theta = \frac{180(n-2)}{n} = \frac{180(3-2)}{3} = 60^{\circ}$.
Angle strain is defined as $\alpha = \frac{1}{2} (109^{\circ} 28^{\prime} - \theta)$.
Substituting the value of $\theta = 60^{\circ}$:
$\alpha = \frac{1}{2} (109^{\circ} 28^{\prime} - 60^{\circ}) = \frac{1}{2} (49^{\circ} 28^{\prime}) = 24^{\circ} 44^{\prime}$.

(A) When $1,4$-dibromopentane is treated with sodium metal,an intramolecular Wurtz reaction occurs.
This leads to the formation of a four-membered ring with a methyl substituent.
The reaction is represented as:
$CH_3-CH(Br)-CH_2-CH_2-CH_2-Br + 2Na \rightarrow \text{Methylcyclobutane} + 2NaBr$.
Thus,the product formed is methyl cyclobutane.

(D) When $\omega$-dihalides (terminal dihalides) react with sodium metal in the presence of dry ether,an intramolecular Wurtz reaction occurs,leading to the formation of a cyclic hydrocarbon.
For example,$1, 4-$dibromobutane reacts with $2Na$ to form cyclobutane and $2NaBr$ as a byproduct.
The reaction is: $Br-CH_2-CH_2-CH_2-CH_2-Br + 2Na \rightarrow \text{Cyclobutane} + 2NaBr$.

(A) This reaction is an example of the $Wurtz$ reaction,where alkyl halides react with sodium metal in the presence of dry ether to form higher alkanes.
$2 CH_{3}CH_{2}CH_{2}Cl + 2 Na \xrightarrow{\text{dry ether}} CH_{3}CH_{2}CH_{2}CH_{2}CH_{2}CH_{3} + 2 NaCl$
As $n$-propyl chloride $(CH_{3}CH_{2}CH_{2}Cl)$ is used,the product formed is $n$-hexane $(CH_{3}CH_{2}CH_{2}CH_{2}CH_{2}CH_{3})$.

(C) The chlorination of $methane$ is carried out by treating it with chlorine in the presence of ultraviolet light or by heating the reaction mixture to $520-670 \ K$.
$CH_{4} + Cl_{2} \xrightarrow{hv \ or \ \Delta} CH_{3}Cl + HCl$
This $methyl \ chloride$ reacts with metallic $Na$ in the presence of dry ether to form a symmetrical alkane with a higher number of carbon atoms $(ethane)$.
$2CH_{3}Cl + 2Na \xrightarrow{\text{dry ether}} CH_{3}-CH_{3} + 2NaCl$
This reaction is called the $Wurtz$ reaction.

(A) The reaction proceeds as follows:
$1$. Oxidation of isopentane ($2$-methylbutane) with $KMnO_4$ occurs at the tertiary carbon atom to form $2-$methylbutan$-2-$ol $(X)$.
$2$. Dehydration of $2-$methylbutan$-2-$ol using an acid catalyst follows Saytzeff's rule to form the more substituted and stable alkene,$2$-methylbut$-2-$ene,as the major product $(Y)$.

(B) In Wurtz reaction, two molecules of bromoethane react with $2$ molecules of sodium to form $n$-butane:
$2CH_3CH_2Br + 2Na \rightarrow CH_3CH_2CH_2CH_3 + 2NaBr$
Decarboxylation of the sodium salt of a carboxylic acid with sodalime $(NaOH + CaO)$ removes one carbon atom as $Na_2CO_3$ to produce an alkane with one less carbon atom than the original acid.
To obtain $n$-butane ($4$ carbons), the sodium salt must be derived from a pentanoic acid ($5$ carbons).
The reaction is:
$CH_3CH_2CH_2CH_2COONa + NaOH \xrightarrow{CaO, \Delta} CH_3CH_2CH_2CH_3 + Na_2CO_3$
Thus, $X$ is pentanoic acid, $CH_3(CH_2)_3COOH$.

(A) The electrolysis of sodium salts of carboxylic acids is known as Kolbe's electrolysis.
For sodium butanoate $(CH_3CH_2CH_2COONa)$,the reaction is:
$2CH_3CH_2CH_2COONa + 2H_2O \rightarrow CH_3CH_2CH_2-CH_2CH_2CH_3 + 2CO_2 + H_2 + 2NaOH$.
The hydrocarbon formed is hexane $(C_6H_{14})$,which contains $6$ carbon atoms.

(A) The molecular formula $C_5H_{12}$ corresponds to a molecular mass of $72 \ g/mol$.
Among the isomers of pentane,$2, 2-$dimethylpropane $(neopentane)$ has all $12$ hydrogen atoms equivalent.
Therefore,it yields only one monochloro derivative upon photochlorination.
Further substitution leads to two distinct dichloro derivatives.
The structure of $A$ is $CH_3-C(CH_3)_2-CH_3$.

(C) The reaction of a mixture of two different alkyl halides with $Na$ metal in dry ether is known as the Wurtz reaction.
When a mixture of ethyl bromide $(C_2H_5Br)$ and $n$-propyl bromide $(C_3H_7Br)$ reacts with $Na$ in dry ether,three different alkanes are formed due to the coupling of alkyl radicals:
$1$. Coupling of two ethyl radicals: $C_2H_5-C_2H_5$ ($n$-butane)
$2$. Coupling of two $n$-propyl radicals: $C_3H_7-C_3H_7$ ($n$-hexane)
$3$. Cross-coupling of one ethyl and one $n$-propyl radical: $C_2H_5-C_3H_7$ ($n$-pentane)
Thus,a total of $3$ different alkanes are formed.

(A) The given reaction is: $2 CH_3 CH_2 Br + 2 Na \xrightarrow{\text{Dry ether}} CH_3 CH_2 CH_2 CH_3 + 2 NaBr$.
This reaction involves the coupling of two alkyl halide molecules in the presence of sodium metal and dry ether to form a higher alkane.
This specific reaction is known as the Wurtz reaction.
The product formed is $n$-butane $(CH_3 CH_2 CH_2 CH_3)$.

(A) In reaction $I$,$n$-pentyl bromide $(CH_3CH_2CH_2CH_2CH_2Br)$ reacts with $Zn/H^+$ (a reducing agent),which replaces the $Br$ atom with an $H$ atom to form pentane $(CH_3CH_2CH_2CH_2CH_3)$ as product $P$.
In reaction $II$,$n$-pentyl bromide reacts with $Na$ in the presence of dry ether,which is the Wurtz reaction. This reaction couples two alkyl groups to form a symmetric alkane,resulting in decane $(CH_3(CH_2)_8CH_3)$ as product $Q$.

(D) In the free radical chlorination of methane $(CH_4)$:
$1.$ **Initiation step $(X)$:** $A$ neutral molecule breaks into radicals,usually in the presence of $UV$ light or heat. Here,$d) Cl_2 \rightarrow 2\dot{Cl}$ is the initiation step.
$2.$ **Propagation steps:** $A$ radical reacts with a neutral molecule to produce another radical and a neutral molecule. Steps $(c)$ and $(e)$ are propagation steps.
$3.$ **Termination steps $(Y)$:** Two radicals combine to form a neutral molecule,ending the chain reaction. Steps $(a), (b),$ and $(f)$ are termination steps.
Thus,$X = \{d\}$ and $Y = \{a, b, f\}$.

(D) In Kolbe's electrolysis,the reaction is $2CH_3CH_2COONa + 2H_2O \xrightarrow{\text{electrolysis}} CH_3CH_2CH_2CH_2CH_3 + 2CO_2 + H_2 + 2NaOH$.
Statement-$I$: Sodium propionate $(CH_3CH_2COONa)$ undergoes electrolysis to form $n$-butane $(CH_3CH_2CH_2CH_3)$,not $n$-hexane. Thus,statement-$I$ is incorrect.
Statement-$II$: During the electrolysis of sodium salts of carboxylic acids,$CO_2$ is evolved at the anode and $H_2$ is evolved at the cathode. Thus,statement-$II$ is correct.

(B) The Wurtz reaction involves the coupling of alkyl halides in the presence of sodium and dry ether.
When a mixture of ethyl iodide $(C_2H_5I)$ and $n$-propyl iodide $(C_3H_7I)$ is used,the following coupling reactions occur:
$1$. $C_2H_5I + C_2H_5I \xrightarrow{Na/dry ether} C_4H_{10}$ ($n$-butane)
$2$. $C_3H_7I + C_3H_7I \xrightarrow{Na/dry ether} C_6H_{14}$ ($n$-hexane)
$3$. $C_2H_5I + C_3H_7I \xrightarrow{Na/dry ether} C_5H_{12}$ ($n$-pentane)
Thus,$n$-butane,$n$-hexane,and $n$-pentane are formed,while propane is not formed.

(C) In Kolbe's electrolysis,the reaction is as follows:
$2CH_3CH_2COONa + 2H_2O \rightarrow CH_3CH_2CH_2CH_3 + 2CO_2 + H_2 + 2NaOH$
At the anode,the propanoate ion $(CH_3CH_2COO^-)$ undergoes oxidation to form an alkyl radical $(CH_3CH_2^\bullet)$,which then dimerizes to form $n$-butane $(C_4H_{10})$ along with the release of $CO_2$.
At the cathode,water is reduced to form hydrogen gas $(H_2)$ and hydroxide ions $(OH^-)$.
Therefore,the products at the anode and cathode are $C_4H_{10}$ and $H_2$ respectively.

(A) The boiling point of isomeric alkanes decreases with increasing branching due to a decrease in the exposed surface area,which leads to weaker van der Waals forces.
$n$-Hexane $(i)$ is a straight-chain alkane with the largest surface area,hence it has the highest boiling point.
$2$-methylpentane $(ii)$ has one branch,reducing its surface area compared to $n$-hexane.
$2,3$-dimethylbutane $(iii)$ has two branches,resulting in the most compact structure and the smallest surface area,hence it has the lowest boiling point.
Therefore,the correct order is $i > ii > iii$.

(C) In Kolbe's electrolysis,sodium acetate $(CH_3COONa)$ undergoes electrolysis to produce ethane $(C_2H_6)$.
The dissociation is: $CH_3COONa \rightleftharpoons CH_3COO^{-} + Na^{+}$.
At the anode,the acetate ion loses an electron to form an acetoxy radical,which then decarboxylates to form a methyl free radical $(CH_3^{\bullet})$. Two methyl radicals combine to form ethane.
At the cathode,water is reduced to produce hydrogen gas $(H_2)$ and hydroxide ions $(OH^{-})$.
Therefore,the methyl free radical is formed at the anode,not the cathode. Thus,Assertion $(A)$ is true,but Reason $(R)$ is false.

(C) cycloalkane is a saturated hydrocarbon that contains one ring structure.
For an open-chain alkane,the general formula is $C_n H_{2n+2}$.
When a ring is formed,two hydrogen atoms are removed from the chain to connect the ends,resulting in the loss of $2$ hydrogen atoms.
Therefore,the general formula for a cycloalkane becomes $C_n H_{2n+2-2} = C_n H_{2n}$.

(C) Kolbe electrolysis of $CH_3COONa$ produces $C_2H_6$ and $CO_2$ at the anode.
$2CH_3COONa + 2H_2O \xrightarrow{\text{electrolysis}} CH_3-CH_3 (A) + 2CO_2 (B) + H_2 + 2NaOH$
When $B$ $(CO_2)$ is heated with $O_2$ in the presence of $(CH_3COO)_2Mn$,it is not the standard reaction; however,the question implies the oxidation of $CH_3-CH_3$ $(A)$. The combustion of ethane is:
$2C_2H_6 + 7O_2 \xrightarrow{(CH_3COO)_2Mn} 4CO_2 (C) + 6H_2O (D)$
$C$ $(CO_2)$ reacts with $NaOH$ to form sodium carbonate $(Na_2CO_3)$ or sodium bicarbonate $(NaHCO_3)$,which are salts.
Thus,$A$ is $C_2H_6$ and $D$ is $H_2O$.

(C) $(i)$ Petrol and $CNG$ are cleaner fuels compared to diesel or coal,thus they cause less pollution. This statement is correct.
$(ii)$ Alkanes containing a tertiary hydrogen atom can be oxidized to tertiary alcohols using $KMnO_4$. This statement is correct.
$(iii)$ Kolbe's electrolytic method involves the electrolysis of sodium or potassium salts of carboxylic acids to produce alkanes. Methane $(CH_4)$ cannot be prepared by this method because the reaction requires at least two carbon atoms to form an alkane. This statement is incorrect.
$(iv)$ Alkyl chlorides undergo reduction with $Zn$ and dilute $HCl$ to form alkanes $(R-Cl + 2[H] \xrightarrow{Zn/HCl} R-H + HCl)$. This statement is correct.
Therefore,the correct statements are $(i)$,$(ii)$,and $(iv)$.

(D) The reaction $CH_3-CH_2-COONa \xrightarrow{NaOH + CaO / \Delta} CH_3-CH_3 + Na_2CO_3$ is a decarboxylation reaction,where $X = NaOH + CaO / \Delta$ and $Y = CH_3-CH_3$.
The reaction $2CH_3COONa + 2H_2O \xrightarrow{\text{Kolbe's electrolysis}} CH_3-CH_3 + 2CO_2 + H_2 + 2NaOH$ involves $Z = CH_3COONa$ as the reactant for Kolbe's electrolysis to produce ethane.

(A) The isomerisation of $n$-alkanes to branched chain alkanes is a process used to improve the octane rating of fuel.
This reaction is typically carried out in the presence of a Lewis acid catalyst,such as anhydrous $AlCl_3$ combined with $HCl$ gas.
Therefore,the correct catalyst is anhydrous $AlCl_3 / HCl$.

(D) The controlled catalytic oxidation of ethane with air at high pressure in the presence of manganese acetate $(CH_3 COO)_2 Mn$ yields acetic acid as the final product.
The chemical equation is:
$2 CH_3-CH_3 + 3 O_2 \xrightarrow[\Delta]{(CH_3 COO)_2 Mn} 2 CH_3 COOH + 2 H_2 O$
Therefore,'$Q$' is $CH_3 COOH$.

(D) The reaction of methane with iodine is a reversible process: $CH_4 + I_2 \rightleftharpoons CH_3I + HI$.
Since $HI$ is a strong reducing agent,it reduces $CH_3I$ back to $CH_4$.
To drive the reaction in the forward direction,an oxidizing agent like $HNO_3$ is added to consume $HI$ and convert it back into $I_2$: $2HNO_3 + 2HI \longrightarrow 2NO_2 + 2H_2O + I_2$.
Therefore,$I_2$ and $HNO_3$ are the suitable reagents.

(D) The partial oxidation of methane $(CH_4)$ under different conditions yields different products.
$1$. When $CH_4$ is oxidized with $O_2$ in the presence of $Mo_2O_3$ at high temperature $(\Delta)$,it forms formaldehyde ($HCHO$ or $CH_2O$):
$CH_4 + O_2 \xrightarrow{Mo_2O_3 / \Delta} HCHO + H_2O$
Thus,$X = CH_2O$.
$2$. When $CH_4$ is oxidized with $O_2$ in the presence of $Cu$ at $523 \ K$ and $100 \ atm$,it forms methanol $(CH_3OH)$:
$2CH_4 + O_2 \xrightarrow{Cu / 523 \ K / 100 \ atm} 2CH_3OH$
Thus,$Y = CH_3OH$.
Therefore,$X = CH_2O$ and $Y = CH_3OH$.

(D) The reaction $CH_3-CH_2-CH_3 \xrightarrow{Br_2, h\nu} CH_3-CH(Br)-CH_3$ is a free radical substitution reaction.
In the propagation step,a bromine radical abstracts a hydrogen atom from the secondary carbon of propane to form a secondary propyl radical $(CH_3-\dot{C}H-CH_3)$.
This secondary free radical is more stable than a primary free radical,making this the preferred pathway.

(C) The reaction sequence is as follows:
$1$. $2-$methyl$-2-$bromopropane reacts with $Mg$ in the presence of dry ether to form a Grignard reagent,$X$,which is $2-$methyl$-2-$propylmagnesium bromide: $(CH_3)_3CBr + Mg \xrightarrow{\text{Dry ether}} (CH_3)_3CMgBr (X)$.
$2$. The Grignard reagent $X$ then reacts with water $(H_2O)$ to undergo hydrolysis,yielding $2-$methyl propane $(Z)$ and magnesium hydroxybromide: $(CH_3)_3CMgBr + H_2O \rightarrow (CH_3)_3CH (Z) + Mg(OH)Br$.
$3$. Therefore,$Z$ is $2-$methyl propane.

(D) The reaction sequence is as follows:
$1$. $C_2H_4 + HBr \rightarrow CH_3CH_2Br$ $(W)$
$2$. $W + Mg / \text{dry ether} \rightarrow CH_3CH_2MgBr$
$3$. $CH_3CH_2MgBr + CO_2 \xrightarrow{H_3O^+} CH_3CH_2COOH$ $(X)$
$4$. $CH_3CH_2COOH$ undergoes Kolbe's electrolysis $(Y)$ to form $CH_3CH_2CH_2CH_3$ $(Z)$.
$5$. $W + Na / \text{dry ether}$ (Wurtz reaction) also forms $CH_3CH_2CH_2CH_3$ $(Z)$.
Thus,$Y$ is electrolysis and $Z$ is $CH_3CH_2CH_2CH_3$.

(D) The chemical reaction between aluminum carbide and heavy water is given by:
$Al_4 C_3 + 12 D_2 O \rightarrow 4 Al(OD)_3 + 3 CD_4$
Aluminum carbide $(Al_4 C_3)$ reacts with $D_2 O$ to produce aluminum deuteroxide $(Al(OD)_3)$ and deutero-methane $(CD_4)$.
Therefore,'$X$' is $CD_4$.

(B) The reaction of iso-pentane ($2$-methylbutane) with $KMnO_4$ is an oxidation reaction that occurs at the tertiary carbon atom,which is the most reactive site for oxidation in alkanes. This leads to the formation of $2$-methylbutan-$2$-ol as the major product $X$.
$CH_3-CH(CH_3)-CH_2-CH_3 \xrightarrow{KMnO_4} CH_3-C(OH)(CH_3)-CH_2-CH_3$ $(X)$
Next,the dehydration of $2$-methylbutan-$2$-ol $(X)$ in the presence of $20 \% H_3PO_4$ at $358 \ K$ follows Saytzeff's rule,which states that the more substituted alkene is the major product.
$CH_3-C(OH)(CH_3)-CH_2-CH_3 \xrightarrow[358 \ K]{20 \% H_3PO_4} CH_3-C(CH_3)=CH-CH_3$ $(Y)$ + $H_2O$
Thus,$X$ is $2$-methylbutan-$2$-ol and $Y$ is $2$-methylbut-$2$-ene.

(B) The controlled oxidation of methane under different conditions yields different products.
$1$. When methane is heated with $Cu$ at $523 \ K$ and $100 \ atm$ pressure, it gives methanol: $2CH_4 + O_2 \xrightarrow{Cu / 523 \ K / 100 \ atm} 2CH_3OH$.
$2$. When methane is heated with $Mo_2O_3$, it gives methanal: $CH_4 + O_2 \xrightarrow{Mo_2O_3 / \Delta} HCHO + H_2O$.
Therefore, $X = Cu$ and $Y = Mo_2O_3$.

(A) The reaction $C_2H_6 + \frac{3}{2}O_2 \xrightarrow[(CH_3COO)_2Mn]{\Delta} CH_3COOH$ produces $X = CH_3COOH$ (acetic acid).
The reaction $CH_3COOH + NaOH \rightarrow CH_3COONa + H_2O$ produces $Y = CH_3COONa$ (sodium acetate).
Kolbe's electrolysis of aqueous sodium acetate $(CH_3COONa)$ is represented as:
$2CH_3COONa + 2H_2O \xrightarrow{\text{electrolysis}} CH_3-CH_3 + 2CO_2 + H_2 + 2NaOH$.
At the anode,the products formed are ethane $(C_2H_6)$ and carbon dioxide $(CO_2)$.
Thus,$P$ and $Q$ are $C_2H_6$ and $CO_2$ respectively.

(C) The reaction given is a decarboxylation reaction,also known as the soda-lime decarboxylation.
When a sodium salt of a carboxylic acid is heated with soda lime $(NaOH + CaO)$,it loses a molecule of $CO_2$ to form an alkane with one carbon atom less than the parent carboxylic acid.
$CH_3-CH_2-CO_2^{\ominus} Na^{\oplus} + NaOH \xrightarrow{CaO, \Delta} CH_3-CH_3 + Na_2CO_3$
Here,sodium propanoate ($3$ carbons) reacts to form ethane ($2$ carbons).
Therefore,$Z$ is ethane.

(B) Step $1$: Reaction of acetic acid with metallic sodium:
$2CH_3COOH + 2Na \rightarrow 2CH_3COONa + H_2 \uparrow$
Here,$X$ is sodium acetate $(CH_3COONa)$.
Step $2$: Decarboxylation of sodium acetate with sodalime $(NaOH + CaO)$:
$CH_3COONa + NaOH \xrightarrow{CaO, \Delta} CH_4 + Na_2CO_3$
Here,$Y$ is methane $(CH_4)$.