Alkane Questions in English

(N/A) An aqueous solution of sodium or potassium salt of a carboxylic acid on electrolysis gives an alkane containing an even number of carbon atoms.
Example: Electrolysis of sodium acetate yields ethane.
$2 CH_{3}COO^{-}Na^{+} + 2 H_{2}O \xrightarrow{\text{Electrolysis}} CH_{3}-CH_{3} + 2 CO_{2} + H_{2} + 2 NaOH$
Mechanism:
$a$. In aqueous solution,the salt dissociates:
$2 CH_{3}COO^{-}Na^{+} \rightleftharpoons 2 CH_{3}COO^{-}_{(aq)} + 2 Na^{+}_{(aq)}$
$b$. At the Anode (Oxidation):
$(i)$ $2 CH_{3}COO^{-} \rightarrow 2 CH_{3}COO^{\bullet} + 2 e^{-}$
$(ii)$ $2 CH_{3}COO^{\bullet} \rightarrow 2 \dot{C}H_{3} + 2 CO_{2} \uparrow$
$(iii)$ $\dot{C}H_{3} + \dot{C}H_{3} \rightarrow CH_{3}-CH_{3} \uparrow$ (Ethane)
$c$. At the Cathode (Reduction):
Water is reduced instead of $Na^{+}$ ions:
$2 H_{2}O + 2 e^{-} \rightarrow 2 OH^{-} + H_{2} \uparrow$
Overall reaction: $2 CH_{3}COO^{-}Na^{+} + 2 H_{2}O \rightarrow CH_{3}-CH_{3} + 2 CO_{2} + H_{2} + 2 NaOH$

(N/A) As the molecular weight of an alkane increases,both its boiling point and melting point generally increase. This is due to the following reasons:
$1$. Alkanes contain $C-C$ and $C-H$ covalent bonds. Due to the small difference in electronegativity between carbon $(2.5)$ and hydrogen $(2.1)$,alkanes are essentially non-polar molecules.
$2$. Consequently,the intermolecular forces present are weak van der Waals forces.
$3$. As the molecular weight increases,the molecular size and surface area of the alkane increase. $A$ larger surface area leads to stronger van der Waals forces of attraction between the molecules.
$4$. Stronger intermolecular forces require more energy to overcome,resulting in higher boiling and melting points.

Molecular formula	Name	Molecular weight	$B.P. (K)$	$M.P. (K)$
$CH_4$	Methane	$16$	$111.0$	$90.5$
$C_2H_6$	Ethane	$30$	$184.4$	$101.0$
$C_3H_8$	Propane	$44$	$230.9$	$85.3$
$C_4H_{10}$	$n$-Butane	$58$	$272.4$	$134.6$
$C_5H_{12}$	$n$-Pentane	$72$	$309.1$	$143.3$
$C_6H_{14}$	Hexane	$86$	$341.9$	$178.5$
$C_7H_{16}$	Heptane	$100$	$371.4$	$182.4$
$C_8H_{18}$	Octane	$114$	$398.7$	$216.2$
$C_9H_{20}$	Nonane	$128$	$423.8$	$222.0$
$C_{10}H_{22}$	Decane	$142$	$447.1$	$243.3$
$C_{20}H_{42}$	Icosane	$282$	$615.0$	$236.2$

(N/A) Alkanes are non-polar molecules held together by weak van der Waals forces.
The strength of these van der Waals forces increases with the increase in molecular size and surface area as the number of carbon atoms increases.
For lower alkanes ($C_{1}$ to $C_{4}$),the molecular mass is small,and the van der Waals forces are too weak to hold the molecules together in a liquid or solid state at room temperature $(298 \ K)$,hence they exist as gases.
As the number of carbon atoms increases ($C_{5}$ to $C_{17}$),the molecular mass and surface area increase,leading to stronger van der Waals forces,which are sufficient to hold the molecules together in the liquid state at room temperature.

(N/A) $(i)$ In alkanes,weak van der Waals forces exist; therefore,at normal temperature,alkanes up to $C_{4}$ are gases,and $C_{5}$ to $C_{13}$ are liquids.
$(ii)$ As the molecular weight of alkanes increases,the van der Waals forces increase,which leads to an increase in both boiling and melting points.
$(iii)$ In branched isomers of alkanes,the molecular shape becomes more spherical,which reduces the surface area available for van der Waals interactions,resulting in a lower boiling point.
$(iv)$ In isomers of alkanes,branching leads to a more spherical shape,which allows for better packing in the crystal lattice,thereby increasing the melting point.

(N/A) $(i)$ Alkanes are colourless and odourless.
$(ii)$ Alkanes are insoluble in water but soluble in non-polar organic solvents.
$(iii)$ The boiling points of alkanes increase with an increase in molecular weight due to the increase in van der Waals forces.
$(iv)$ For isomeric alkanes,the boiling point decreases with branching because the surface area decreases,leading to weaker intermolecular forces.
$(v)$ Melting points do not show a regular trend; however,alkanes with an even number of carbon atoms generally have higher melting points than those with an odd number of carbon atoms due to better packing in the crystal lattice.

(N/A) For isomers of pentane $(C_5H_{12})$,as the number of branches increases,the boiling point decreases.
Explanation:
$1$. As the number of branched chains increases,the molecule tends to attain a more spherical shape.
$2$. $A$ spherical shape results in a smaller surface area of contact between the molecules.
$3$. Due to the smaller surface area,the magnitude of intermolecular van der Waals forces of attraction decreases.
$4$. Consequently,less energy is required to overcome these forces,leading to a decrease in the boiling point.

The chlorination of methane is a free radical substitution reaction that occurs in the presence of ultraviolet light $(hv)$ or at high temperatures $(573-773 \ K)$. The hydrogen atoms of methane are successively replaced by chlorine atoms.
$(i) \ CH_4 + Cl_2 \xrightarrow{hv \text{ or } 573-773 \ K} CH_3Cl + HCl$
$(ii) \ CH_3Cl + Cl_2 \xrightarrow{hv} CH_2Cl_2 + HCl$
$(iii) \ CH_2Cl_2 + Cl_2 \xrightarrow{hv} CHCl_3 + HCl$
$(iv) \ CHCl_3 + Cl_2 \xrightarrow{hv} CCl_4 + HCl$
The overall reaction is: $CH_4$ $\xrightarrow{Cl_2, hv, -HCl} CH_3Cl$ $\xrightarrow{Cl_2, hv, -HCl} CH_2Cl_2$ $\xrightarrow{Cl_2, hv, -HCl} CHCl_3$ $\xrightarrow{Cl_2, hv, -HCl} CCl_4$.

(N/A) In alkanes,one or more hydrogen atoms can be substituted with a halogen,nitro group,or sulphonic acid group. These reactions are known as substitution reactions.
$(a)$ Nitration: The hydrogen atom of an alkane is substituted with a $-NO_{2}$ group.
$(b)$ Sulphonation: The hydrogen atom of an alkane is substituted with a $-SO_{3}H$ group.
$(c)$ Halogenation: The hydrogen atom of an alkane is replaced with a halogen ($F$,$Cl$,$Br$,$I$). The reactivity order of halogens is: $F_{2} > Cl_{2} > Br_{2} > I_{2}$.
$(d)$ Fluorination: This reaction is highly exothermic and violent,making it difficult to control. Example: $CH_{4} + F_{2} \xrightarrow{N_{2} \text{ or } Ar} CH_{3}F + HF$.
$(e)$ Iodination: This reaction is very slow and reversible. It can be driven to completion in the presence of oxidizing agents like $HIO_{3}$ or $HNO_{3}$.
$CH_{4} + I_{2} \rightleftharpoons CH_{3}I + HI$
$HIO_{3} + 5HI \rightarrow 3I_{2} + 3H_{2}O$
$(f)$ Chlorination and Bromination: These are controlled reactions. If chlorine is present in excess,all hydrogen atoms are substituted with $-Cl$ atoms.
$CH_{4}$ $\xrightarrow{h\nu, Cl_{2}, -HCl} CH_{3}Cl$ $\xrightarrow{h\nu, Cl_{2}, -HCl} CH_{2}Cl_{2}$ $\xrightarrow{h\nu, Cl_{2}, -HCl} CHCl_{3}$ $\xrightarrow{h\nu, Cl_{2}, -HCl} CCl_{4}$
Example: $CH_{3}-CH_{3} + Cl_{2} \xrightarrow{h\nu} CH_{3}-CH_{2}Cl + HCl$ (Ethane $\rightarrow$ Chloroethane).
These reactions occur at temperatures between $573 \ K$ and $773 \ K$ or in the presence of diffused sunlight or $UV$ light.

(N/A) The chlorination of methane proceeds via a free radical chain mechanism,which involves three main steps: initiation,propagation,and termination.
$1$. Initiation: The reaction begins with the homolytic cleavage of the $Cl-Cl$ bond in the presence of light $(hv)$ or heat to produce chlorine free radicals:
$Cl-Cl \xrightarrow{hv} 2\dot{C}l$
$2$. Propagation: The chlorine free radical abstracts a hydrogen atom from methane to form a methyl free radical $(dot{C}H_3)$:
$CH_4 + \dot{C}l \rightarrow \dot{C}H_3 + HCl$
The methyl free radical then reacts with another $Cl_2$ molecule to form chloromethane $(CH_3Cl)$ and regenerate a chlorine free radical:
$\dot{C}H_3 + Cl-Cl \rightarrow CH_3Cl + \dot{C}l$
$3$. Termination: The reaction chain is terminated when two free radicals combine. The formation of ethane $(CH_3-CH_3)$ occurs when two methyl free radicals collide and combine:
$\dot{C}H_3 + \dot{C}H_3 \rightarrow CH_3-CH_3$

(N/A) The mechanism involves a chain reaction known as free radical substitution.
$1)$. Chain initiation: The reaction is initiated by $UV$ light,which breaks a chlorine molecule into two chlorine free radicals: $Cl_2 \xrightarrow{h\nu} 2Cl \cdot$
$2)$. Chain propagation: These steps keep the reaction going.
Step $A$: $CH_4 + Cl \cdot \rightarrow \cdot CH_3 + HCl$
Step $B$: $\cdot CH_3 + Cl_2 \rightarrow CH_3Cl + Cl \cdot$
$3)$. Chain termination: These reactions remove free radicals from the system.
$2Cl \cdot \rightarrow Cl_2$
$\cdot CH_3 + Cl \cdot \rightarrow CH_3Cl$
$\cdot CH_3 + \cdot CH_3 \rightarrow CH_3-CH_3$

(N/A) $1$. Substitution reactions: Alkanes undergo free radical substitution,such as halogenation (Fluorination,Chlorination,Bromination,and Iodination).
$2$. Combustion: Alkanes react with oxygen to produce $CO_2$,$H_2O$,and heat.
$3$. Controlled oxidation: Alkanes can be oxidized to alcohols,aldehydes,or carboxylic acids using specific catalysts.
$4$. Isomerization: $n$-alkanes are converted into branched-chain isomers in the presence of $AlCl_3$ and $HCl$.
$5$. Aromatization: Alkanes with $6$ or more carbon atoms are converted into benzene and its homologs at high temperature and pressure in the presence of catalysts like $V_2O_5$ or $Cr_2O_3$.
$6$. Reaction with steam: Alkanes react with steam in the presence of a nickel catalyst to form carbon monoxide and hydrogen (steam reforming).
$7$. Pyrolysis (Cracking): Decomposition of alkanes into smaller hydrocarbons by heating in the absence of air.

(N/A) Complete combustion of alkane: Alkanes on heating in the presence of air or dioxygen are completely oxidized to carbon dioxide and water with the evolution of a large amount of heat.
The general reaction for the combustion of an alkane is:
$C_{n}H_{2n+2} + (\frac{3n+1}{2}) O_{2} \rightarrow n CO_{2} + (n+1) H_{2}O + \text{Heat}$
Examples:
$(i)$ Methane: $CH_{4(g)} + 2 O_{2(g)} \rightarrow CO_{2(g)} + 2 H_{2}O_{(l)}$ ; $\Delta_{c} H^{\ominus} = -890 \ kJ \ mol^{-1}$
$(ii)$ Butane: $C_{4}H_{10(g)} + \frac{13}{2} O_{2(g)} \rightarrow 4 CO_{2(g)} + 5 H_{2}O_{(l)}$ ; $\Delta_{c} H^{\ominus} = -2875.84 \ kJ \ mol^{-1}$
Due to the evolution of a large amount of heat,alkanes are used as fuels in homes,vehicles,and industries.
$(b)$ Incomplete combustion of alkane: During incomplete combustion of alkanes with an insufficient amount of air or dioxygen,'carbon black' is formed.

Complete combustion occurs in the presence of excess oxygen,producing carbon dioxide and water.
For Heptane $(C_{7}H_{16})$:
$C_{7}H_{16(l)} + 11 O_{2(g)} \rightarrow 7 CO_{2(g)} + 8 H_{2}O_{(l)}$
For Nonane $(C_{9}H_{20})$:
$C_{9}H_{20(l)} + 14 O_{2(g)} \rightarrow 9 CO_{2(g)} + 10 H_{2}O_{(l)}$
Incomplete combustion occurs in the presence of limited oxygen,producing carbon black $(C)$ and water.
For Heptane $(C_{7}H_{16})$:
$C_{7}H_{16(l)} + 4 O_{2(g)} \rightarrow 7 C_{(s)} + 8 H_{2}O_{(l)}$
For Nonane $(C_{9}H_{20})$:
$C_{9}H_{20(l)} + 5 O_{2(g)} \rightarrow 9 C_{(s)} + 10 H_{2}O_{(l)}$

(N/A) $n$-Alkanes,upon heating in the presence of anhydrous aluminium chloride $(AlCl_3)$ and hydrogen chloride $(HCl)$ gas,undergo isomerization to form branched-chain alkanes.
For example,$n$-hexane isomerizes to a mixture of $2$-methylpentane and $3$-methylpentane as shown below:
$CH_3(CH_2)_4CH_3 \xrightarrow[HCl_{(g)}]{Anhy. AlCl_3} CH_3CH_2CH(CH_3)CH_2CH_3 + CH_3CH(CH_3)CH_2CH_2CH_3$
($n$-Hexane) $\rightarrow$ ($3$-methylpentane) + ($2$-methylpentane)

(N/A) $n$-Alkanes having six or more carbon atoms,on heating to $773 \ K$ at $10-20 \ \text{atm}$ pressure in the presence of oxides of vanadium $(V_2O_5)$,molybdenum $(Mo_2O_3)$,or chromium $(Cr_2O_3)$ supported over alumina $(Al_2O_3)$,undergo dehydrogenation and cyclization to form benzene and its homologues. This reaction is known as aromatization or reforming.
Example-$1$: $n$-Hexane $\xrightarrow[10-20 \ \text{atm}]{773 \ K, \ Cr_2O_3/V_2O_5/Mo_2O_3}$ Benzene
Example-$2$: $n$-Heptane $\xrightarrow[10-20 \ \text{atm}]{773 \ K, \ Cr_2O_3/V_2O_5/Mo_2O_3}$ Toluene

(A) The controlled oxidation of methane in the presence of a copper catalyst at $523 \ K$ and $100 \ atm$ pressure yields methanol $(CH_3OH)$.
The chemical reaction is:
$2 CH_4 + O_2 \xrightarrow{Cu, 523 \ K, 100 \ atm} 2 CH_3OH$

Definition: Higher alkanes,on heating to a high temperature,decompose into lower alkanes,alkenes,etc. Such a decomposition reaction into smaller fragments by the application of heat is called pyrolysis or cracking.
General reaction:
$Higher \ alkane \xrightarrow[Pyrolysis]{Temperature} Lower \ alkane + Alkene \ mixture$
Example: Pyrolysis of hexane $(C_6H_{14})$ at $773 \ K$:
$C_6H_{14} \xrightarrow{773 \ K} C_6H_{12} + H_2$ $(i)$
$C_6H_{14} \xrightarrow{773 \ K} C_4H_8 + C_2H_6$ $(ii)$
$C_6H_{14} \xrightarrow{773 \ K} C_3H_6 + C_2H_4 + CH_4$ $(iii)$
Mechanism: Pyrolysis of alkanes is believed to be a free radical reaction. The preparation of oil gas or petrol gas from kerosene oil or petrol involves the principle of pyrolysis. For example,dodecane $(C_{12}H_{26})$,a constituent of kerosene oil,on heating to $973 \ K$ in the presence of platinum,palladium,or nickel,gives a mixture of heptane and pentene.
$C_{12}H_{26} \xrightarrow[973 \ K]{Pd, Pt, Ni} C_7H_{16} + C_5H_{10} + \text{Other products}$

(N/A) The reaction of methane with chlorine in the presence of ultraviolet light $(hv)$ or at high temperatures $(573-773 \ K)$ proceeds via a free radical mechanism,leading to the successive replacement of hydrogen atoms by chlorine atoms:
$CH_4$ $\xrightarrow{hv, Cl_2, -HCl} CH_3Cl$ $\xrightarrow{hv, Cl_2, -HCl} CH_2Cl_2$ $\xrightarrow{hv, Cl_2, -HCl} CHCl_3$ $\xrightarrow{hv, Cl_2, -HCl} CCl_4$

(N/A) The reaction of methane with iodine is reversible:
$CH_4 + I_2 \rightleftharpoons CH_3I + HI$
To drive this reaction in the forward direction,an oxidizing agent like $HIO_3$ is used to remove $HI$:
$5HI + HIO_3 \rightarrow 3I_2 + 3H_2O$
The controlled chlorination of ethane is a free radical substitution reaction:
$CH_3-CH_3 + Cl_2 \xrightarrow{h\nu} CH_3-CH_2Cl + HCl$

(N/A) The complete combustion (oxidation) of alkanes produces carbon dioxide and water.
$1$. For methane $(CH_4)$:
$CH_{4(g)} + 2O_{2(g)} \rightarrow CO_{2(g)} + 2H_2O_{(l)}$
$\Delta_{c}H^{\ominus} = -890 \ kJ \ mol^{-1}$
$2$. For butane $(C_4H_{10})$:
$C_4H_{10(g)} + \frac{13}{2}O_{2(g)} \rightarrow 4CO_{2(g)} + 5H_2O_{(l)}$
$\Delta_{c}H^{\ominus} = -2875.84 \ kJ \ mol^{-1}$

(N/A) $1$. Complete oxidation of alkane:
$C_{n}H_{2n+2} + (\frac{3n+1}{2})O_{2} \rightarrow nCO_{2} + (n+1)H_{2}O$
$2$. Incomplete combustion of methane:
$CH_{4(g)} + O_{2(g)} \xrightarrow{\text{Incomplete combustion}} C_{(s)} + 2H_{2}O_{(l)}$
In the incomplete combustion of methane,carbon black (soot) is produced.

$1$. Isomerization: $n-\text{hexane}$ on heating with anhydrous $AlCl_3$ and $HCl_{(g)}$ at $573 \ K$ undergoes isomerization to form a mixture of $2-\text{methylpentane}$ and $3-\text{methylpentane}$.
$CH_3(CH_2)_4CH_3$ $\xrightarrow[anhydrous \ AlCl_3, HCl_{(g)}]{573 \ K} CH_3-CH(CH_3)-CH_2-CH_2-CH_3 + CH_3-CH_2-CH(CH_3)-CH_2-CH_3$
$2$. Aromatization: $n-\text{hexane}$ when heated to $773 \ K$ under $10-20 \ atm$ pressure in the presence of $V_2O_5$,$Cr_2O_3$ or $Mo_2O_3$ supported over alumina,gets dehydrogenated and cyclized to benzene.
$CH_3(CH_2)_4CH_3 \xrightarrow[773 \ K, 10-20 \ atm]{Cr_2O_3/V_2O_5/Mo_2O_3} C_6H_6 (\text{Benzene}) + 4H_2$

(N/A) $(i)$ $n$-heptane undergoes aromatization in the presence of $Cr_2O_3$,$V_2O_5$,or $Mo_2O_3$ at $773 \ K$ and $10-20 \ atm$ to form toluene $(C_6H_5CH_3)$:
$C_7H_{16} \xrightarrow[773 \ K, 10-20 \ atm]{Cr_2O_3/V_2O_5/Mo_2O_3} C_6H_5CH_3 + 4H_2$
(ii) Methane undergoes controlled oxidation in the presence of copper catalyst at $523 \ K$ and $100 \ atm$ to form methanol:
$2CH_4 + O_2 \xrightarrow[523 \ K, 100 \ atm]{Cu} 2CH_3OH$

(A) The controlled oxidation of alkanes in the presence of specific catalysts leads to different products.
$1. CH_4 + O_2 \xrightarrow{Mn_2O_3, \Delta} HCHO + H_2O$ (Formaldehyde is formed).
$2. 2CH_3CH_3 + 3O_2 \xrightarrow{(CH_3COO)_2Mn, \Delta} 2CH_3COOH + 2H_2O$ (Acetic acid is formed).

(N/A) Pyrolysis is the thermal decomposition of higher alkanes into smaller alkanes,alkenes,and hydrogen at high temperatures.
For $n$-hexane $(C_6H_{14})$ at $773 \ K$:
$(i)$ $C_6H_{14} \xrightarrow{773 \ K} C_6H_{12} + H_2$
(ii) $C_6H_{14} \xrightarrow{773 \ K} C_4H_8 + C_2H_6$
(iii) $C_6H_{14} \xrightarrow{773 \ K} C_3H_6 + C_2H_4 + CH_4$
For kerosene (represented by dodecane,$C_{12}H_{26}$) at $973 \ K$ in the presence of $Pd, Pt, Ni$:
$C_{12}H_{26} \xrightarrow{973 \ K, Pd/Pt/Ni} C_7H_{16} + C_5H_{10} + \text{Other products}$

(N/A) $(i)$ The reaction of methane with steam is used for the industrial production of dihydrogen:
$CH_{4(g)} + H_{2}O_{(g)} \xrightarrow[Ni]{1273 \ K} CO_{(g)} + 3H_{2(g)}$
(ii) Oxidation of $2$-methylpropane $(isobutane)$ with potassium permanganate $(KMnO_4)$ yields $2$-methylpropan-$2$-ol ($tert$-butyl alcohol):
$(CH_{3})_{3}CH \xrightarrow{[O]} (CH_{3})_{3}COH$

(N/A) Hexane undergoes different chemical reactions depending on the conditions provided:
$(a)$ Aromatization: $n$-Hexane,when heated to $773 \ K$ under $10-20 \ atm$ pressure in the presence of $Cr_2O_3/V_2O_5/Mo_2O_3$ catalysts,undergoes cyclization and dehydrogenation to form benzene.
$(b)$ Pyrolysis (Thermal Cracking): When $C_6H_{14}$ is heated to $773 \ K$,it undergoes thermal decomposition to produce a mixture of smaller hydrocarbons and hydrogen gas,such as:
$(i)$ $C_6H_{14} \rightarrow C_6H_{12} + H_2$
(ii) $C_6H_{14} \rightarrow C_4H_8 + C_2H_6$
(iii) $C_6H_{14} \rightarrow C_3H_6 + C_2H_4 + CH_4$ (Note: The original image showed $CH_3$ radical,which is corrected to $CH_4$ for balanced stoichiometry).
$(c)$ Isomerization: $n$-Hexane,when heated with anhydrous $AlCl_3$ and $HCl$,undergoes isomerization to form branched-chain alkanes like $2$-methylpentane and $3$-methylpentane.

(N/A) The chemical reactions for the preparation of products from methane $(CH_4)$ are as follows:
$(i)$ $CH_4 + Cl_2 \xrightarrow{h\nu} CH_3Cl + HCl$ (Methyl chloride)
$(ii)$ $CH_4 + O_2(g) \xrightarrow{\text{Incomplete combustion}} C(s) + 2H_2O$ (Carbon black)
$(iii)$ $CH_4 + 2O_2(g) \xrightarrow{\text{Complete combustion}} CO_2(g) + 2H_2O(l)$ (Carbon dioxide)
$(iv)$ $2CH_4 + O_2 \xrightarrow{Cu, 523 K, 100 \ atm} 2CH_3OH$ (Methanol)
$(v)$ $CH_4 + O_2 \xrightarrow{Mo_2O_3, \Delta} HCHO + H_2O$ (Methanal)
$(vi)$ & $(vii)$ $CH_4 + H_2O(g) \xrightarrow{Ni, \Delta} CO + 3H_2$ (Carbon monoxide and Dihydrogen)
$(viii)$ $CH_4 + I_2 \xrightarrow{HIO_3} CH_3I + HI$ (Methyl iodide)
$(ix)$ $CH_4 + Br_2 \xrightarrow{h\nu} CH_3Br + HBr$ (Methyl bromide)

(N/A) The chemical reactions of methane $(CH_4)$ are as follows:
$1$. Halogenation: $CH_4 + Cl_2 \xrightarrow{hv} CH_3Cl + HCl$ (Methyl chloride)
$2$. Incomplete combustion: $2CH_4 + 3O_2 \xrightarrow{} 2C_{(s)} + 4H_2O$ (Carbon black)
$3$. Complete combustion: $CH_4 + 2O_2 \xrightarrow{} CO_{2(g)} + 2H_2O_{(l)}$ (Carbon dioxide)
$4$. Controlled oxidation: $2CH_4 + O_2 \xrightarrow{Cu, 523 K, 100 atm} 2CH_3OH$ (Methanol)
$5$. Controlled oxidation: $CH_4 + O_2 \xrightarrow{Mo_2O_3, \Delta} HCHO + H_2O$ (Methanal)
$6$. Reaction with steam: $CH_4 + H_2O_{(g)} \xrightarrow{Ni, \Delta} CO + 3H_2$ (Carbon monoxide and Dihydrogen)
$7$. Iodination: $CH_4 + I_2 \xrightarrow{HIO_3 + 5HI_3 + 3I_2 + 3H_2O} CH_3I + HI$ (Methyl iodide)
$8$. Bromination: $CH_4 + Br_2 \xrightarrow{hv} CH_3Br + HBr$ (Methyl bromide)

(N/A) $(i)$ Reduction of methyl chloride: $CH_{3}Cl + H_{2} \xrightarrow{Zn, H^{+}} CH_{4} + HCl$
$(ii)$ Decarboxylation of acetic acid:
$CH_{3}COOH + NaOH \rightarrow CH_{3}COONa + H_{2}O$
$CH_{3}COONa + NaOH \xrightarrow{CaO, \Delta} CH_{4} + Na_{2}CO_{3}$

(N/A) $(i)$ Hydrogenation of ethene: $CH_{2}=CH_{2} + H_{2} \xrightarrow{Pt/Pd/Ni} CH_{3}-CH_{3}$
$(ii)$ Reduction of ethyl chloride: $CH_{3}CH_{2}Cl + H_{2} \xrightarrow{Zn, H^{+}} CH_{3}-CH_{3} + HCl$
$(iii)$ Wurtz reaction: $2CH_{3}Br + 2Na \xrightarrow{\text{dry ether}} CH_{3}-CH_{3} + 2NaBr$
$(iv)$ Decarboxylation of sodium propanoate: $CH_{3}CH_{2}COONa + NaOH \xrightarrow{CaO, \Delta} CH_{3}-CH_{3} + Na_{2}CO_{3}$
$(v)$ Kolbe's electrolysis: $2CH_{3}COONa + 2H_{2}O \xrightarrow{\text{electrolysis}} CH_{3}-CH_{3} + 2CO_{2} + H_{2} + 2NaOH$

(N/A) The preparation of ethane $(CH_3-CH_3)$ from the given compounds is as follows:
$(i)$ From Ethene:
Ethene undergoes catalytic hydrogenation in the presence of $Pt/Pd/Ni$ to form ethane.
$CH_2=CH_2 + H_2 \xrightarrow{Pt/Pd/Ni} CH_3-CH_3$
$(ii)$ From Ethyl chloride:
Ethyl chloride undergoes reduction with $Zn$ and $H^+$ to form ethane.
$CH_3CH_2Cl + H_2 \xrightarrow{Zn, H^+} CH_3-CH_3 + HCl$

(N/A) $(i)$ Preparation of ethane from Bromo-methane (Wurtz reaction):
$2CH_3Br + 2Na \xrightarrow{\text{dry ether}} CH_3-CH_3 + 2NaBr$
$(ii)$ Preparation of ethane from Propanoic acid (Decarboxylation):
$CH_3CH_2COOH + NaOH \rightarrow CH_3CH_2COONa + H_2O$
$CH_3CH_2COONa + NaOH \xrightarrow{CaO, \Delta} CH_3-CH_3 + Na_2CO_3$

(N/A) $(i)$ From Acetic acid $(CH_3COOH)$:
$2CH_3COOH + 2NaOH \rightarrow 2CH_3COONa + 2H_2O$
$2CH_3COONa + 2H_2O \xrightarrow{\text{Electrolysis}} CH_3-CH_3 + 2CO_2 + H_2 + 2NaOH$ (Kolbe's Electrolysis)
$(ii)$ From Methane $(CH_4)$:
$2CH_4 + 2Cl_2 \xrightarrow{h\nu} 2CH_3Cl + 2HCl$
$2CH_3Cl + 2Na \xrightarrow{\text{dry ether}} CH_3-CH_3 + 2NaCl$ (Wurtz Reaction)
$(iii)$ From Acetate ion $(CH_3COO^{-})$:
$2CH_3COO^{-}$ $\xrightarrow{-2e^{-}} 2CH_3COO^{\bullet}$ $\rightarrow 2CH_3^{\bullet} + 2CO_2$
$2CH_3^{\bullet} \rightarrow CH_3-CH_3$ (Kolbe's Electrolysis mechanism)

(N/A) The given reaction sequence is for the preparation of alkanes via decarboxylation.
Step $1$: Propanoic acid $(CH_3CH_2COOH)$ reacts with $NaOH$ to form sodium propanoate $(CH_3CH_2COONa)$ and water.
Thus,$(A) = CH_3CH_2COOH$ (Propanoic acid) and $(B) = CH_3CH_2COONa$ (Sodium propanoate).
Step $2$: Sodium propanoate undergoes decarboxylation with soda-lime $(NaOH + CaO)$ upon heating to yield ethane $(CH_3CH_3)$.

(A) The reaction sequence is as follows:
$CH_3CH_3$ $\xrightarrow{Cl_2, h\nu, -HCl} CH_3CH_2Cl$ $\xrightarrow{Zn, H^+, +H_2, -HCl} CH_3CH_3$
Here,$(A)$ is ethane $(CH_3CH_3)$ and $(B)$ is chloroethane $(CH_3CH_2Cl)$.

(N/A) Alkanes can have an infinite number of conformations due to rotation around $C-C$ single bonds.
Reason: $A$ sigma bond is present between two carbon atoms in an alkane.
The electron distribution of the sigma molecular orbital is symmetrical around the internuclear axis of the $C-C$ bond,which is not disturbed by rotation about its axis.
This rotation results in different spatial arrangements of atoms,which can interconvert into one another.

(N/A) Alkanes can have an infinite number of conformations by rotation around $C-C$ single bonds. However,it should be noted that rotation around a $C-C$ single bond is not completely free. It is hindered by a small energy barrier of $1-20 \ kJ \ mol^{-1}$ due to weak repulsive interactions between the adjacent bonds. Such a type of repulsive interaction is called torsional strain.

(N/A) $(1)$ The reaction of $1$-chlorobutane with $H_2$ in the presence of $Zn, H^{+}$ is a reduction reaction that yields $n$-butane:
$CH_3CH_2CH_2CH_2Cl + H_2 \xrightarrow{Zn, H^{+}} CH_3CH_2CH_2CH_3 + HCl$
$(2)$ The reaction of $n$-butane with anhydrous $AlCl_3$ and $HCl$ at high temperature is an isomerisation reaction that yields $2$-methylpropane (isobutane):
$CH_3CH_2CH_2CH_3 \xrightarrow[HCl, \Delta]{Anhy. AlCl_3} CH_3-CH(CH_3)-CH_3$

(N/A) The Wurtz reaction is a method for preparing hydrocarbons from alkyl halides.
Sodium metal is highly reactive and reacts with alkyl halides to form hydrocarbons containing an even number of carbon atoms.
$2RX + 2Na \xrightarrow{\text{dry ether}} R-R + 2NaX$
The reaction in which alkyl halide compounds are treated with sodium metal in the presence of dry ether to form higher alkanes is known as the Wurtz reaction.
Example:
$2CH_3Br + 2Na \xrightarrow{\text{dry ether}} CH_3-CH_3 + 2NaBr$

(N/A) The molecular formula of the hydrocarbon with molecular mass $72 \ g \ mol^{-1}$ is $C_5H_{12}$ $(5 \times 12 + 12 \times 1 = 72)$.
$C_5H_{12}$ has three isomers: $n$-pentane,isopentane,and neopentane.
Neopentane $(2,2-\text{dimethylpropane})$ has all $12$ hydrogen atoms equivalent,so it gives only one monochloro derivative upon photochlorination.
Further chlorination of this monochloro derivative yields two dichloro derivatives: $1,3-\text{dichloro}-2,2-\text{dimethylpropane}$ and $1,1-\text{dichloro}-2,2-\text{dimethylpropane}$.
Thus,the hydrocarbon is neopentane,with the structure: $CH_3-C(CH_3)_2-CH_3$.

(N/A) Decarboxylation by sodalime: Carboxylic acids lose carbon dioxide to form hydrocarbons when their sodium salts are heated with sodalime ($NaOH$ and $CaO$ in the ratio of $3:1$). The reaction is known as decarboxylation.
$\underset{\text{Sodium carboxylate}}{RCOONa}$ $\xrightarrow[\text{Sodalime, Heat}]{NaOH \text{ and } CaO} \underset{\text{Hydrocarbon}}{RH} + Na_{2}CO_{3}$
In this decarboxylation reaction,due to the removal of $CO_{2}$,the number of carbon atoms decreases by one compared to the original acid salt.
$(b)$ Decarboxylation by Kolbe electrolysis: Alkali metal salts of carboxylic acids also undergo decarboxylation on electrolysis of their aqueous solutions and form hydrocarbons having twice the number of carbon atoms present in the alkyl group of the acid. The reaction is known as Kolbe electrolysis.
$2RCOO^{-}Na^{+} + 2H_{2}O \xrightarrow{\text{Electrolysis}} R-R + 2CO_{2} + 2NaOH + H_{2} \uparrow$
Examples of sodalime decarboxylation:
$(i)$ $CH_{3}COONa + NaOH \xrightarrow{\text{Sodalime}, \Delta} CH_{4} + Na_{2}CO_{3}$
$(ii)$ $\text{Sodium benzoate} \xrightarrow{\text{Sodalime}, \Delta, -Na_{2}CO_{3}} \text{Benzene}$
$(iii)$ $\text{Sodium phthalate} \xrightarrow{\text{Sodalime}, \Delta, -Na_{2}CO_{3}} \text{Benzene}$
$(iv)$ $\text{Sodium salicylate} \xrightarrow{\text{Sodalime}, \Delta, -Na_{2}CO_{3}} \text{Phenol}$

(N/A) The conversion of methane $(CH_4)$ to ethene $(CH_2=CH_2)$ proceeds through the following steps:
$Step \ 1: CH_4 + Cl_2 \xrightarrow{h\nu} CH_3Cl + HCl$ (Chlorination)
$Step \ 2: 2CH_3Cl + 2Na \xrightarrow{dry \ ether} CH_3-CH_3 + 2NaCl$ (Wurtz reaction)
$Step \ 3: CH_3-CH_3 + Cl_2 \xrightarrow{h\nu} CH_3-CH_2Cl + HCl$ (Chlorination)
$Step \ 4: CH_3-CH_2Cl + alc.KOH \rightarrow CH_2=CH_2 + KCl + H_2O$ (Dehydrohalogenation)
Thus,the unknown compound $X$ is methane $(CH_4)$.

(N/A) Alkanes experience intermolecular Van der Waals forces. The stronger the force,the higher the boiling point of the alkane.
As branching increases,the surface area of the molecule decreases,which results in a smaller area of contact between molecules.
Consequently,the Van der Waals forces decrease,which can be overcome at a relatively lower temperature. Hence,the boiling point of an alkane chain decreases with an increase in branching.
For example,consider the isomers of pentane $(C_5H_{12})$:
$1$. $n$-pentane (straight chain): Highest surface area,strongest Van der Waals forces,highest boiling point $(309.1 \ K)$.
$2$. iso-pentane ($2$-methylbutane): One branch,lower surface area,lower boiling point $(301.0 \ K)$.
$3$. neo-pentane ($2$,$2$-dimethylpropane): Two branches,spherical shape,smallest surface area,lowest boiling point $(282.5 \ K)$.
Therefore,the order of boiling point is: $n$-pentane $>$ iso-pentane $>$ neo-pentane.

$2CH_3-CH_3 + 2Br_2 \xrightarrow{hv} 2CH_3-CH_2Br + 2HBr$
$2CH_3-CH_2Br + 2Na \xrightarrow{\text{dry ether}} CH_3-CH_2-CH_2-CH_3 + 2NaBr$
Ethane $\rightarrow$ Bromination $\rightarrow$ Bromoethane $\rightarrow$ Wurtz reaction $\rightarrow$ $n$-Butane

(N/A) The conversion from $methane$ to $ethane$ can be performed in two steps:
$1$. Bromination of $methane$ to form $methyl$ $bromide$:
$CH_4 + Br_2 \xrightarrow{h\nu} CH_3Br + HBr$
$2$. $Wurtz$ reaction:
$2CH_3Br + 2Na \xrightarrow{dry \ ether} CH_3-CH_3 + 2NaBr$

(N/A) The boiling point of alkanes increases with an increase in molecular mass due to an increase in van der Waals forces. For isomers,the boiling point decreases with increased branching because the surface area decreases,leading to weaker intermolecular forces.
$(i)$ Methane $(CH_4)$ < Ethane $(C_2H_6)$ < Propane $(C_3H_8)$ < Butane $(C_4H_{10})$ < Pentane $(C_5H_{12})$.
$(ii)$ Neo-pentane (most branched) < Iso-pentane < $n$-pentane (straight chain).
$(iii)$ Ethane $(C_2H_6)$ < Propane $(C_3H_8)$ < $2$-methyl propane $(C_4H_{10})$ < $n$-butane $(C_4H_{10})$. Note: $n$-butane has a higher boiling point than its branched isomer $2$-methyl propane.
$(iv)$ $2,2$-dimethyl propane $(C_5H_{12})$ < $2,2$-dimethyl butane $(C_6H_{14})$ < $2$-methyl pentane $(C_6H_{14})$.

(B) The molecule $(CH_3)_3CH$ is isobutane.
In this molecule,there are three terminal $-CH_3$ groups. The carbon atoms in these groups are primary $(1^{\circ})$ carbons,so the hydrogen atoms attached to them are $1^{\circ}$ hydrogen atoms.
The central carbon atom is bonded to three other carbon atoms,making it a tertiary $(3^{\circ})$ carbon. The hydrogen atom attached to this central carbon is therefore a $3^{\circ}$ hydrogen atom.
Thus,the molecule contains both $1^{\circ}$ and $3^{\circ}$ hydrogen atoms.

(N/A) The saturated hydrocarbon with $2$ carbon atoms is ethane,which has the molecular formula $C_{2}H_{6}$.
The structural formula is $CH_{3}-CH_{3}$.
In ethane,all $6$ hydrogen atoms are attached to primary $(1^{\circ})$ carbon atoms,making them all equivalent primary $(1^{\circ})$ hydrogen atoms.

(A) In $(i)$ Hydrogenation,an alkene or alkyne is converted to an alkane without changing the number of carbon atoms.
In $(ii)$ Alkyl halide reduction $(R-X + 2[H] \rightarrow R-H + HX)$,the number of carbon atoms remains the same.
In $(iii)$ Wurtz reaction,the number of carbon atoms doubles.
In $(iv)$ Decarboxylation $(R-COONa + NaOH \rightarrow R-H + Na_2CO_3)$,the number of carbon atoms decreases by $1$.
In $(v)$ Kolbe reaction,the number of carbon atoms in the product is $2n-2$ (where $n$ is the number of carbons in the acid salt).
Therefore,the carbon number does not change in $(i)$ and $(ii)$.