Alkane Questions in English

(A) Petroleum is a complex mixture of a very large number of different hydrocarbons.
The most commonly found molecules in petroleum are $alkanes$ (paraffins),$cycloalkanes$ (naphthenes),and $aromatic$ hydrocarbons.
Among these,$aliphatic$ hydrocarbons (specifically alkanes and cycloalkanes) constitute the major portion of crude oil.
Therefore,petroleum consists mainly of aliphatic hydrocarbons.

(A) . Petroleum ether is a mixture of volatile aliphatic hydrocarbons,primarily pentane and hexane. It is widely used as a non-polar solvent for fats,oils,waxes,varnishes,and rubber in laboratory and industrial processes.

(A) . In alkanes,the carbon atoms are $sp^3$ hybridized.
Due to $sp^3$ hybridization,the geometry is tetrahedral,and the bond angle is $109.5^o$.

(D) The process described is the $Kolbe's$ electrolysis method.
In this method,a concentrated aqueous solution of sodium or potassium salts of saturated carboxylic acids is subjected to electrolysis.
The reaction is as follows:
$2CH_3COONa + 2H_2O \xrightarrow{\text{Electrolysis}} CH_3-CH_3 + 2CO_2 + 2NaOH + H_2$

(C) The halogenation of alkanes is a chain reaction that proceeds via a free-radical mechanism in the presence of sunlight $(h\nu)$ or heat.
The reaction involves three steps:
$1$. Initiation: The halogen molecule $(X_2)$ undergoes homolytic fission to form halogen radicals $(X\cdot)$.
$2$. Propagation: The halogen radical abstracts a hydrogen atom from the alkane $(R-H)$ to form an alkyl radical $(R\cdot)$ and $HX$. The alkyl radical then reacts with another halogen molecule to form the alkyl halide $(R-X)$ and regenerates a halogen radical.
$3$. Termination: The reaction stops when two radicals combine to form a stable molecule.
Since the reaction proceeds through the formation of free radicals,it is classified as a free-radical substitution reaction.
The reaction sequence is as follows:
$CH_4 + Cl_2 \rightarrow CH_3Cl + HCl$ (chloromethane)
$CH_3Cl + Cl_2 \rightarrow CH_2Cl_2 + HCl$ (dichloromethane)
$CH_2Cl_2 + Cl_2 \rightarrow CHCl_3 + HCl$ (trichloromethane)
$CHCl_3 + Cl_2 \rightarrow CCl_4 + HCl$ (tetrachloromethane)

(D) The Wurtz reaction involving a mixture of two different alkyl halides ($R-I$ and $R'-I$) leads to the formation of three alkanes ($R-R$,$R'-R'$,and $R-R'$).
$C_2H_5I + 2Na + C_2H_5I \xrightarrow{\text{Dry Ether}} C_2H_5-C_2H_5 \text{ (Butane)} + 2NaI$
$C_3H_7I + 2Na + C_3H_7I \xrightarrow{\text{Dry Ether}} C_3H_7-C_3H_7 \text{ (Hexane)} + 2NaI$
$C_2H_5I + 2Na + C_3H_7I \xrightarrow{\text{Dry Ether}} C_2H_5-C_3H_7 \text{ (Pentane)} + 2NaI$
Thus,a mixture of three alkanes is formed.

(D) The number of isomeric monobromo derivatives depends on the number of non-equivalent hydrogen atoms in the alkane.
$1$. $CH_4$ (Methane) has only one type of $H$ atom,yielding $1$ derivative.
$2$. $CH_3-CH_3$ (Ethane) has only one type of $H$ atom,yielding $1$ derivative.
$3$. $C(CH_3)_4$ (Neopentane) has only one type of $H$ atom,yielding $1$ derivative.
$4$. $CH_3-CH_2-CH_3$ (Propane) has two types of non-equivalent $H$ atoms (terminal $CH_3$ and central $CH_2$),yielding two isomeric monobromo derivatives: $1$-bromopropane $(CH_3-CH_2-CH_2Br)$ and $2$-bromopropane $(CH_3-CH(Br)-CH_3)$.

(A) Kerosene is a thin,clear liquid formed from hydrocarbons obtained from the fractional distillation of petroleum between $150^{\circ} C$ and $275^{\circ} C$.
It is composed of carbon chains that typically contain between $6$ and $16$ carbon atoms per molecule.
Kerosene's major components are branched and straight-chain $Alkanes$ and naphthenes (cycloalkanes),which normally account for at least $70 \, \%$ by volume.

(D) Iso-octane is a branched alkane ($2,2,4$-trimethylpentane) used as a reference standard to measure the antiknock quality of gasoline.
It is assigned an octane rating of $100$,while $n$-heptane is assigned $0$.
Mixing iso-octane with petrol helps in reducing the knocking tendency of the fuel,thereby acting as an antiknock agent.

(A) . Methane $(CH_4)$ is the primary component of natural gas,typically accounting for $70-90\%$ of its composition.

(D) Propane is obtained from propene by catalytic hydrogenation.
In this reaction,propene reacts with hydrogen in the presence of a metal catalyst like $Ni$,$Pt$,or $Pd$ to form propane:
$CH_3-CH=CH_2 + H_2 \xrightarrow{Ni/Pt/Pd} CH_3-CH_2-CH_3$.

(D) . $CH_3CH_2Cl + KOH (\text{alc.}) \to CH_2=CH_2 + KCl + H_2O$.
In the presence of alcoholic $KOH$,dehydrohalogenation occurs,resulting in the formation of an alkene (ethene) rather than an alkane.

(B) . Liquefied petroleum gas $(LPG)$ is primarily a mixture of propane $(C_3H_8)$ and butane $(C_4H_{10})$,along with smaller amounts of ethane $(C_2H_6)$.
Among the given options,$C_4H_{10} + C_3H_8$ represents the primary components of $LPG$.

(D) $CH_4 + O_2 \xrightarrow{\Delta, \text{Limited supply of air}} C + 2H_2O$
Carbon black contains $98-99 \%$ carbon. It is obtained by the incomplete combustion or thermal decomposition of methane $(CH_4)$ in a limited supply of air. It is widely used in making black ink,paints,and shoe polishes.

(B) Tetraethyl lead $(Pb(C_2H_5)_4)$ is used as an anti-knocking agent in petrol.
It increases the octane number of the fuel,thereby improving its quality and preventing engine knocking.

(A) The boiling point of alkanes depends on the surface area of the molecule.
As the branching increases,the surface area decreases,which leads to weaker van der Waals forces of attraction.
$n-$hexane $(C_6H_{14})$ is a straight-chain alkane with the largest surface area among the given options,resulting in the strongest van der Waals forces.
Therefore,$n-$hexane has the highest boiling point.

(B) Knocking is the sudden and irregular burning of the fuel mixture,which causes jerks against the piston and gives rise to a violent sound. This phenomenon is known as knocking.

(A) Petroleum is a complex mixture of a very large number of different hydrocarbons.
The most abundant components found in petroleum are $Alkanes$ (paraffins),followed by cycloalkanes (naphthenes) and aromatic hydrocarbons.
Therefore,it is primarily considered a mixture of alkanes.

(B) The correct answer is $(B)$ $n-$octane.
$1.$ Boiling point depends on molecular mass. $A$ higher molecular mass leads to a higher boiling point.
$2.$ For isomers with the same molecular mass,the boiling point depends on the surface area. Straight-chain (linear) alkanes have a larger surface area compared to branched alkanes,leading to stronger van der Waals forces and thus a higher boiling point.
$3.$ Among the given options,$n-$octane $(C_8H_{18})$ has the highest molecular mass compared to $n-$butane $(C_4H_{10})$ and is a straight-chain isomer compared to iso-octane and $2, 2, 3, 3-$tetramethylbutane,which are branched.

(C) The bromine water test is used to detect the presence of unsaturation (double or triple bonds) in a compound.
Unsaturated hydrocarbons like $Acetylene$ $(C_2H_2)$,$Propene$ $(C_3H_6)$,and $Propyne$ $(C_3H_4)$ react with bromine to form addition products,thereby decolourising the reddish-brown bromine solution.
$Ethane$ $(C_2H_6)$ is a saturated hydrocarbon (alkane) and does not undergo addition reactions with bromine under normal conditions.
Therefore,it does not decolourise the bromine solution.

(B) The reaction of anhydrous sodium acetate with sodalime $(NaOH + CaO)$ is a decarboxylation reaction.
The chemical equation is: $CH_3COONa + NaOH \xrightarrow{CaO} CH_4 + Na_2CO_3$.
Thus,the product formed is methane $(CH_4)$.

(B) The octane number of a fuel is defined as the percentage of iso-octane ($2,2,4$-trimethylpentane) in a mixture of iso-octane and $n$-heptane that has the same knocking characteristics as the fuel being tested.
Since the sample contains $81\%$ iso-octane,its octane number is $81$.

(C) The reaction is known as $Kolbe's \text{ electrolysis}$.
The chemical equation is:
$2CH_3COOK(aq) + 2H_2O(l) \xrightarrow{\text{Electrolysis}} CH_3-CH_3(g) + 2CO_2(g) + 2KOH(aq) + H_2(g)$.
At the anode,ethane and carbon dioxide are produced,while at the cathode,hydrogen gas and potassium hydroxide are formed.

(A) The boiling point of alkanes increases with an increase in the number of carbon atoms.
This is because the surface area of the molecule increases as the chain length increases,which leads to stronger $Van \ der \ Waals$ forces of attraction between the molecules.

(C) Cracking is the process in which high molecular weight hydrocarbons (alkanes) are heated at high temperatures in the presence or absence of a catalyst to break down into smaller,lower molecular weight hydrocarbons (alkanes,alkenes,etc.). This is a key process in petroleum refining.

(C) $n-$heptane is defined as the zero point of the octane rating scale.
$2,2,4-$Trimethylpentane has an octane rating of $100$,whereas $n-$heptane has an octane rating of $0$.
The octane number is based on an arbitrary scale which is a measure of its ability to resist knocking as it burns on combustion.

(C) The reaction of sodium propanoate with sodalime $(NaOH + CaO)$ is a decarboxylation reaction.
The chemical equation is: $CH_3CH_2COONa + NaOH \xrightarrow{CaO} C_2H_6 + Na_2CO_3$.
Thus,the product formed is ethane $(C_2H_6)$.

(A) Bromination of alkanes is a free radical substitution reaction.
Bromine is highly selective and prefers to substitute the hydrogen on the more stable radical intermediate.
In $n$-butane $(CH_3-CH_2-CH_2-CH_3)$,the secondary $(2^{\circ})$ carbon radical is more stable than the primary $(1^{\circ})$ carbon radical.
Therefore,$2$-bromobutane $(CH_3-CH_2-CH(Br)-CH_3)$ is the chief product.

(C) The cracking of $n-$butane $(CH_3-CH_2-CH_2-CH_3)$ involves the breaking of $C-C$ bonds at high temperatures.
Specifically,the cracking of $n-$butane can yield methane $(CH_4)$ and propene $(CH_3-CH=CH_2)$ as follows:
$CH_3-CH_2-CH_2-CH_3 \xrightarrow{\text{Cracking}} CH_4 + CH_3-CH=CH_2$.

(A) The members of the homologous series of alkanes are not necessarily straight-chain compounds; they can also be branched-chain alkanes (e.g.,isobutane).
All members of a homologous series differ from each other by a $-CH_2-$ unit and share similar chemical properties,but they do not have to be exclusively straight-chained.

(A) Cracking is the process of breaking down large hydrocarbon molecules into smaller,more volatile ones by heating them at high temperatures in the absence of air.
For example,liquid hexane $(C_6H_{14})$ can be cracked into smaller gaseous hydrocarbons like butane $(C_4H_{10})$ and ethene $(C_2H_4)$:
$C_6H_{14} \xrightarrow{\Delta} C_4H_{10} + C_2H_4$.

(A) The reaction of alkyl halides with $Zn$ metal to form alkanes is known as the Frankland reaction.
$(I)$ Wurtz reaction: $2 RX + 2 Na \xrightarrow{\text{ether}} R-R + 2 NaX$
$(II)$ Kolbe's reaction: $2 RCOONa + 2 H_2O \xrightarrow{\text{electrolysis}} R-R + 2 CO_2 + H_2 + 2 NaOH$
$(III)$ Frankland reaction: $2 RX + Zn \rightarrow R-R + ZnX_2$
$(IV)$ Cannizzaro reaction: This is a disproportionation reaction of aldehydes lacking $\alpha$-hydrogen atoms.

(D) Hydrocarbons with $1$ to $4$ carbon atoms are gases at $STP$.
Hydrocarbons with $5$ to $17$ carbon atoms are liquids at $STP$.
Therefore,$n-$pentane $(C_5H_{12})$ is a liquid at $STP$.

(A) Alkanes are non-polar hydrocarbons and are insoluble in water,which is a polar solvent.
Water molecules are held together by hydrogen bonding,whereas alkanes only have weak van der Waals forces.
Therefore,the statement that large number alkanes are soluble in water is false.

(A) The Fischer-Tropsch process is used in the manufacturing of synthetic petrol.
This process involves a collection of chemical reactions that convert a mixture of carbon monoxide $(CO)$ and hydrogen $(H_2)$ into liquid hydrocarbons.
In the standard industrial implementation,the feedstocks ($CO$ and $H_2$) are produced from coal,natural gas,or biomass through a process known as gasification.

(A) The Wurtz reaction involves the coupling of two alkyl halide molecules in the presence of sodium metal to form a symmetrical alkane with an even number of carbon atoms.
$2R-X + 2Na \rightarrow R-R + 2NaX$.
Since the reaction requires at least two alkyl groups to join,the smallest alkane that can be formed is ethane $(C_2H_6)$.
Therefore,methane $(CH_4)$ cannot be prepared by the Wurtz reaction because it contains only one carbon atom.

(B) The octane number is defined as the percentage by volume of iso-octane in a mixture of iso-octane and $n-$heptane that exhibits the same antiknocking properties as the fuel being tested.
Given that the fuel contains $25 \%$ $n-$heptane and $75 \%$ iso-octane,the octane number is equal to the percentage of iso-octane.
Therefore,the octane number is $75$.

(A) Step $1$: Dehalogenation of $2,3-$dibromo$-3-$methylpentane with zinc dust $(Zn)$ leads to the formation of an alkene via elimination of $ZnBr_2$.
$CH_3-CH(Br)-C(Br)(CH_3)-CH_2-CH_3 + Zn \to CH_3-CH=C(CH_3)-CH_2-CH_3 + ZnBr_2$.
Step $2$: The resulting alkene ($3-$methylpent$-2-$ene) is heated with $HI$ in the presence of red phosphorus $(Red \ P)$. Red phosphorus and $HI$ act as a strong reducing agent,which reduces the alkene to the corresponding alkane.
$CH_3-CH=C(CH_3)-CH_2-CH_3 + 2HI \xrightarrow{Red \ P} CH_3-CH_2-CH(CH_3)-CH_2-CH_3$ ($3-$methylpentane).
Thus,the final product is $3-$methylpentane.

(A) The reaction of sodium propionate with soda lime $(NaOH + CaO)$ is a decarboxylation reaction.
The reaction is: $CH_3CH_2COONa + NaOH \xrightarrow[\Delta]{CaO} CH_3CH_3 + Na_2CO_3$.
Thus,the main product formed is ethane $(CH_3CH_3)$.

(D) Gasoline,also known as petrol,is a complex mixture of hydrocarbons.
It primarily consists of alkanes,cycloalkanes,and aromatic hydrocarbons.
The alkane component in gasoline typically contains carbon atoms in the range of $C_7 - C_{12}$.

(A) The rotation about a $C-C$ bond is determined by the steric hindrance of the substituents attached to the carbon atoms.
$A$) $CH_3-CH_3$ (Ethane) has only small hydrogen atoms,allowing free rotation.
$B$) $CH_2=CH_2$ (Ethylene) has a double bond,which restricts rotation.
$C$) $CH \equiv CH$ (Acetylene) has a triple bond,which restricts rotation.
$D$) $C_2Cl_6$ (Hexachloroethane) has bulky chlorine atoms that cause significant steric hindrance,making rotation difficult.
Therefore,ethane has the least hindered rotation.

(B) The given reaction is an example of $Cracking$ (or $Pyrolysis$).
$C_{10}H_{22} \xrightarrow{900 \ K} C_4H_8 + C_6H_{14}$
In this process,higher alkanes are broken down into smaller alkanes and alkenes by heating at high temperatures in the absence of air. Both $Cracking$ and $Pyrolysis$ are correct terms for this process,but $Cracking$ is the standard term used for alkanes in this context. Since both are provided as options,$Cracking$ is the most specific industrial term.

(D) The structure of $2,2,3-$trimethylpentane is $CH_3-C(CH_3)_2-CH(CH_3)-CH_2-CH_3$.
We identify the types of carbon atoms based on their environment:
$1$. Primary $(1^{\circ})$ carbons: The terminal $CH_3$ groups at positions $1, 2, 2, 3, 5$ are not all equivalent.
$2$. The $CH_3$ attached to $C-2$ are equivalent.
$3$. The $CH_3$ attached to $C-3$ is different.
$4$. The $CH_3$ at the end of the chain $(C-5)$ is different.
$5$. The $CH_2$ at $C-4$ is unique.
$6$. The $CH$ at $C-3$ is unique.
$7$. The quaternary carbon at $C-2$ is unique.
Counting the distinct chemical environments for carbon atoms,we find $4$ types:
Type $1$: $CH_3$ groups attached to $C-2$.
Type $2$: $CH_3$ group attached to $C-3$.
Type $3$: $CH_3$ group at $C-5$.
Type $4$: The $CH_2$ group at $C-4$.
Type $5$: The $CH$ group at $C-3$.
Type $6$: The quaternary carbon at $C-2$.
Wait,let us re-evaluate:
$C-1$ ($CH_3$ on $C-2$): $1^{\circ}$
$C-2$ (Quaternary): $4^{\circ}$
$C-3$ (Tertiary): $3^{\circ}$
$C-4$ (Secondary): $2^{\circ}$
$C-5$ (Primary): $1^{\circ}$
$CH_3$ on $C-2$: $1^{\circ}$
$CH_3$ on $C-3$: $1^{\circ}$
There are $4$ types of carbon atoms based on degree of substitution $(1^{\circ}, 2^{\circ}, 3^{\circ}, 4^{\circ})$.

(B) The correct answer is $(b)$.
In neopentane $(CH_3-C(CH_3)_2-CH_3)$,all $12$ hydrogen atoms are chemically equivalent because the molecule is highly symmetrical.
Therefore,the replacement of any of these hydrogen atoms by a chlorine atom results in the formation of only one unique monosubstituted product,which is $1-$chloro$-2,2-$dimethylpropane.

(D) $1$. Decarboxylation of sodium acetate: $CH_3COONa + NaOH \xrightarrow{CaO} CH_4 + Na_2CO_3$
$2$. Reduction of iodomethane: $CH_3I + 2[H] \xrightarrow{Zn/HCl} CH_4 + HI$
$3$. Hydrolysis of aluminium carbide: $Al_4C_3 + 12H_2O \to 3CH_4 + 4Al(OH)_3$
Since all three reactions produce $CH_4$,the correct option is $D$.

(D) Paraffin waxes are mixtures of solid alkanes.
Alkanes with carbon numbers ranging from $C_{20}$ to $C_{40}$ are typically solids at room temperature.
Among the given options,$C_{20}H_{42}$ (icosane) is the only compound that exists as a solid at room temperature,while $C_3H_8$,$C_8H_{18}$,and $C_4H_{10}$ are gases or liquids.

(B) $Alkanes$ do not give addition reactions because a multiple bond is absent.

(D) Natural gas is primarily composed of methane $(CH_4)$,along with smaller amounts of other alkanes such as ethane $(C_2H_6)$,propane $(C_3H_8)$,and butane $(C_4H_{10})$.
Therefore,the correct option is $D$.

(D) In the $Wurtz$ reaction,alkyl halides react with $Na$ in the presence of dry ether to form alkanes.
When a mixture of $CH_3I$ and $C_2H_5I$ is used,three different coupling reactions occur:
$1. CH_3I + 2Na + ICH_3 \rightarrow CH_3-CH_3 (Ethane) + 2NaI$
$2. C_2H_5I + 2Na + IC_2H_5 \rightarrow C_2H_5-C_2H_5 (Butane) + 2NaI$
$3. CH_3I + 2Na + IC_2H_5 \rightarrow CH_3-C_2H_5 (Propane) + 2NaI$
Therefore,the product is a mixture of $Ethane$,$Propane$,and $Butane$.

(D) Vapor phase nitration of alkanes involves heating the alkane with nitric acid at high temperatures ($400$ to $500^{\circ}C$).
Due to the free radical mechanism,the $C$-$C$ bonds are cleaved,resulting in a mixture of nitroalkanes.
For propane $(CH_3CH_2CH_3)$,the products include $1$-nitropropane,$2$-nitropropane,nitroethane,and nitromethane.
Therefore,all of these are obtained.