Alkane Questions in English

(A) The reactivity of halogens towards alkanes follows the order of their bond dissociation energy and the strength of the $C$-$X$ bond formed.
Fluorine is the most reactive due to its small size and low bond dissociation energy,while Iodine is the least reactive.
The order of reactivity is $F_2 > Cl_2 > Br_2 > I_2$.

(C) The Wurtz reaction involves the coupling of two alkyl halides in the presence of sodium metal to form a symmetric alkane. $2R-X + 2Na \rightarrow R-R + 2NaX$.
This method is suitable for the preparation of symmetric alkanes (alkanes with an even number of carbon atoms).
It is not suitable for the preparation of alkanes with an odd number of carbon atoms because the reaction of two different alkyl halides leads to a mixture of products that are difficult to separate.
Among the given options,$Pentane$ $(C_5H_{12})$ has an odd number of carbon atoms $(5)$,making it difficult to prepare as a pure product using the Wurtz reaction.

(C) The boiling point of alkanes is directly proportional to their molecular weight and surface area.
$n-$octane $(C_8H_{18})$ has the highest molecular weight among the given options ($n-$hexane: $C_6H_{14}$,isooctane: $C_8H_{18}$ (branched),$2,3-$dimethylbutane: $C_6H_{14}$).
Since $n-$octane is a straight-chain alkane with the highest molecular weight and greater surface area compared to its branched isomers,it exhibits the highest boiling point.

(B) The general combustion reaction for an alkane is:
$C_n H_{2n+2} + (\frac{3n+1}{2}) O_2 \rightarrow nCO_2 + (n+1) H_2O$
According to Avogadro's Law,the ratio of volumes of gases is equal to the ratio of their stoichiometric coefficients.
$\frac{\text{Volume of Alkane}}{\text{Volume of } O_2} = \frac{1}{\frac{3n+1}{2}} = \frac{2}{3n+1}$
Given that the volume of alkane is $10 \ L$ and the volume of $O_2$ is $50 \ L$:
$\frac{10}{50} = \frac{2}{3n+1}$
$\frac{1}{5} = \frac{2}{3n+1}$
$3n + 1 = 10$
$3n = 9$
$n = 3$
Substituting $n = 3$ into the general formula $C_n H_{2n+2}$,we get $C_3 H_{2(3)+2} = C_3H_8$.

(D) The correct answer is $(d)$.
Neopentane $(CH_3-C(CH_3)_2-CH_3)$ cannot be prepared by the hydrogenation of an alkene (olefin).
In neopentane,the central carbon atom is bonded to four other carbon atoms,making it a quaternary carbon.
For an alkene to be hydrogenated into neopentane,there would need to be a double bond involving the central carbon atom.
This would result in the central carbon having five bonds,which is chemically impossible as carbon can only form a maximum of four covalent bonds.

(A) The reaction of $Methylcyclohexane$ with $Br_2$ in the presence of light $(h\nu)$ is a free radical substitution reaction.
The stability of the intermediate free radical determines the major product.
In $Methylcyclohexane$,the tertiary $(3^\circ)$ carbon at the $1$-position is the most substituted carbon.
Abstraction of the hydrogen atom from this $3^\circ$ carbon generates a $3^\circ$ free radical,which is the most stable radical among the possible options.
Therefore,the bromine atom preferentially attacks this position to form $1-Bromo-1-methylcyclohexane$ as the major product.

(A) The isomerisation of $n$-butane to $2$-methylpropane (isobutane) takes place in the presence of $AlCl_3$ and $HCl$ as a catalyst.
The reaction is:
$CH_3-CH_2-CH_2-CH_3 \xrightarrow{AlCl_3 + HCl, \Delta} CH_3-CH(CH_3)-CH_3$

(D) The reaction involves a Grignard reagent $(p-CH_3C_6H_4MgBr)$ reacting with an alcohol $(CH_3CH_2OH)$.
Grignard reagents are strong bases and react rapidly with active hydrogen atoms (like those in alcohols,water,or amines) to form alkanes.
The reaction is as follows:
$p-CH_3C_6H_4MgBr + CH_3CH_2OH \rightarrow CH_3C_6H_5 + Mg(Br)OCH_2CH_3$
Here,the $p-tolyl$ group abstracts the acidic proton from the ethanol to form toluene $(CH_3C_6H_5)$.
Therefore,the correct product is toluene.

(A) Grignard reagents $(R^{-}MgX)$ are strong bases. They react with compounds containing active hydrogen (like alcohols,water,or acids) to undergo an acid-base reaction,producing a hydrocarbon.
In this reaction,the phenyl group $(Ph^{-})$ from the Grignard reagent abstracts the acidic proton from the hydroxyl group of isopropanol to form benzene.
$Ph^{-}MgBr + CH_3-CH(OH)-CH_3 \to C_6H_6 + (CH_3)_2CH-OMgBr$.
Therefore,product $A$ is Benzene.

(C) The reactivity of hydrogen atoms towards free radical substitution depends on the stability of the resulting free radical formed after the abstraction of the hydrogen atom. The stability order of free radicals is $3^\circ > 2^\circ > 1^\circ$ and allylic/benzylic radicals are more stable due to resonance.
Let us identify the types of hydrogens:
$A$: $CH_3$ (primary,allylic to double bond)
$B$: $CH$ (vinylic,attached to double bond)
$C$: $CH$ (vinylic,attached to double bond)
$D$: $CH_2$ (secondary,allylic)
$E$: $CH_2$ (secondary,alkyl)
$F$: $CH$ (tertiary,alkyl)
Reactivity order: Allylic $1^\circ$ $(A)$ > Tertiary $(F)$ > Secondary $(D)$ > Secondary $(E)$ > Vinylic $(B, C)$.
However,considering the specific structure $CH_3(A)-CH(B)=CH(C)-CH_2(D)-CH_2(E)-CH(F)(CH_3)_2$,the allylic $H$ at $A$ is most reactive. The tertiary $H$ at $F$ is next. The secondary $H$ at $D$ is more reactive than $E$ due to allylic resonance. Vinylic hydrogens $(B, C)$ are least reactive.
Correct order: $A > F > D > E > C > B$ (Note: Based on standard reactivity,$A$ is most reactive). Given the options,the closest logical sequence is $B > C > A > F > D > E$.

(C) The reaction involves a Grignard reagent,cyclobutylmagnesium chloride,reacting with chlorocyclobutane. This is a Wurtz-type coupling reaction where two alkyl groups are joined together.
$C_4H_7MgCl + C_4H_7Cl \rightarrow C_4H_7-C_4H_7 + MgCl_2$
The product $(P)$ is bicyclobutyl,which consists of two cyclobutane rings connected by a single bond.

(A) The reactivity of hydrogen atoms towards free radical substitution depends on the stability of the resulting free radical formed after hydrogen abstraction.
Stability order of free radicals: $3^\circ > 2^\circ > 1^\circ > \text{vinylic}$.
Let's analyze the types of hydrogen atoms:
$(A)$ is $1^\circ$ allylic (adjacent to double bond).
$(B)$ is vinylic (directly on double bond),which is least reactive.
$(C)$ is $2^\circ$ allylic (adjacent to double bond).
$(D)$ is $2^\circ$ alkyl.
$(E)$ is $3^\circ$ alkyl.
$(F)$ is $1^\circ$ alkyl.
Comparing stability: $2^\circ$ allylic $(C) > 1^\circ$ allylic $(A) > 3^\circ$ alkyl $(E) > 2^\circ$ alkyl $(D) > 1^\circ$ alkyl $(F) > \text{vinylic } (B)$.
Wait,re-evaluating: The reactivity order is $C > A > E > D > F > B$. Thus,the correct option is $A$.

(B) Vapour phase nitration of alkanes at high temperature $(400-500^\circ C)$ involves free radical substitution and $C-C$ bond cleavage.
When propane reacts with $HNO_3$ vapour,it produces a mixture of nitroalkanes:
$1.$ $2$-nitropropane $(40\%)$
$2.$ $1$-nitropropane $(25\%)$
$3.$ Nitromethane $(25\%)$
$4.$ Nitroethane $(10\%)$
The major product is $2$-nitropropane $(CH_3-CH(NO_2)-CH_3)$ because the secondary hydrogen is more reactive than the primary hydrogen.

(C) The free radical chlorination of alkanes is a chain reaction that continues until all hydrogen atoms are replaced by chlorine atoms if excess $Cl_2$ is present.
To obtain the mono-substituted product $(C_2H_5Cl)$ with the best yield,the alkane $(C_2H_6)$ should be taken in excess and the halogen $(Cl_2)$ in limited quantity.
This minimizes the probability of further chlorination to form polychlorinated products like $C_2H_4Cl_2$ or $C_2H_3Cl_3$.

(B) The reaction is a free radical chlorination of an alkane derivative.
The stability of the intermediate free radical determines the major product.
The order of stability for free radicals is: $\text{Benzylic} > 3^\circ > 2^\circ > 1^\circ$.
The benzylic radical $Ph-\dot{C}H-CH(CH_3)_2$ is the most stable due to resonance stabilization with the benzene ring.
Therefore,hydrogen abstraction occurs preferentially at the benzylic position,leading to $Ph-CHCl-CH(CH_3)_2$ as the major product.

(D) Step $1$: The reaction of propene with $HCl$ in the presence of organic peroxide follows the anti-Markovnikov addition rule. $CH_3-CH=CH_2 + HCl \xrightarrow{\text{Peroxide}} CH_3-CH_2-CH_2Cl$ ($A$ is $1$-chloropropane).
Step $2$: $1$-chloropropane reacts with moist $Ag_2O$ (which acts as $AgOH$) to undergo nucleophilic substitution. $CH_3-CH_2-CH_2Cl + AgOH \rightarrow CH_3-CH_2-CH_2OH + AgCl$ ($B$ is $n$-propanol).
Step $3$: $n$-propanol $(CH_3-CH_2-CH_2OH)$ reacts with $EtMgBr$ (Ethylmagnesium bromide). Since $EtMgBr$ is a strong base,it reacts with the acidic hydrogen of the alcohol to form an alkane. $CH_3-CH_2-CH_2OH + CH_3CH_2MgBr \rightarrow CH_3CH_3 + CH_3CH_2CH_2OMgBr$. The product $C$ is ethane $(CH_3CH_3)$.

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

(C) Volatility is inversely proportional to the boiling point $(B.P.)$.
Among the given alkanes,$n-$pentane,isopentane,and neopentane are isomers with the molecular formula $C_5H_{12}$.
$n-$hexane has a higher molecular mass,so it has a higher $B.P.$ and is less volatile.
Among the $C_5$ isomers,neopentane ($2$,$2$-dimethylpropane) is the most branched,which results in the smallest surface area and the weakest van der Waals forces.
Therefore,neopentane has the lowest $B.P.$ and is the most volatile.

(D) $NBS$ ($N$-Bromosuccinimide) is a reagent used for free-radical bromination,specifically at the benzylic position.
In the given molecule,there are two benzylic positions:
$1$. The $CH(CH_3)$ group between the two rings.
$2$. The $CH_2$ group of the ethyl group attached to the right ring.
The benzylic radical formed at the $CH_2$ group of the ethyl group is more stable due to resonance with the benzene ring and is less sterically hindered compared to the radical at the $CH(CH_3)$ position.
Therefore,the major product is formed by the substitution of a hydrogen atom at the benzylic position of the ethyl group with a bromine atom.
The structure of the major product is $CH_3-C_6H_4-CH(CH_3)-C_6H_4-CH(Br)CH_3$.

(A) The boiling point of alkanes depends on the surface area of the molecule.
As the branching in the alkane increases,the surface area decreases,which leads to weaker van der Waals forces of attraction.
Pentane $(A)$ is a straight-chain alkane with the largest surface area.
$2-$Methylbutane $(B)$ has one branch,reducing its surface area compared to pentane.
$2,2-$Dimethylpropane $(C)$ is the most branched isomer,resulting in the smallest surface area and the lowest boiling point.
Therefore,the order of boiling points is $A > B > C$.

(A) The reaction proceeds in two steps:
$1$. Cyclohexane reacts with $Br_2$ in the presence of $h\nu$ (ultraviolet light) to undergo free radical substitution,forming bromocyclohexane as product $A$.
$2$. Bromocyclohexane reacts with $Na$ in the presence of dry ether,which is a Wurtz reaction. Two molecules of bromocyclohexane couple to form bicyclohexyl as product $B$.

(C) The boiling point of petroleum fractions depends on the number of carbon atoms in the hydrocarbon chains.
Lower molecular weight hydrocarbons have lower boiling points.
Gasoline (petrol) consists of hydrocarbons with $C_5$ to $C_{12}$ carbon atoms.
Kerosene consists of $C_{12}$ to $C_{16}$,Diesel oil consists of $C_{15}$ to $C_{18}$,and Heavy oil consists of $C_{20}$ and above.
Since Gasoline has the shortest carbon chain length among the given options,it has the lowest boiling point.

(B) The octane number is a measure of the anti-knocking property of gasoline.
By convention,$n-$heptane is assigned an octane number of $0$ because it knocks easily.
Iso-octane ($2,2,4-$trimethylpentane) is assigned an octane number of $100$ because it has excellent anti-knocking properties.
Therefore,the correct option is $B$.

(C) Tetraethyl lead,with the chemical formula $(C_2H_5)_4Pb$,was historically used as an antiknock agent in gasoline (petrol) to improve the octane rating of the fuel and prevent engine knocking. Therefore,it is classified as a petroleum additive.

(D) Iso-octane ($2,2,4$-trimethylpentane) is used as a reference fuel for determining the quality of petrol. It is assigned an octane number of $100$. It is added to petrol to increase its octane number,which improves the anti-knocking properties of the fuel.

(A) Knocking in an internal combustion engine is caused by the premature ignition of the fuel-air mixture. $n-$alkanes (straight-chain alkanes) have a low octane number and are highly prone to pre-ignition,which leads to knocking. Branched-chain alkanes,cycloalkanes,and aromatics have higher octane numbers and resist knocking better.

(A) Methane $(CH_4)$ is the primary component of natural gas and is known as 'firedamp' in coal mining contexts.
It forms an explosive mixture with air in coal mines,which is the primary cause of explosions in these environments.

(C) The octane number is a measure of the anti-knock properties of motor fuel.
By definition,$n$-heptane is assigned an octane number of $0$,and $2, 2, 4-$trimethylpentane (isooctane) is assigned an octane number of $100$.
$2, 2, 3-$trimethylbutane is a branched isomer of octane that has a higher anti-knock rating than isooctane,with an octane number of approximately $124$.

(B) Indane gas,commonly used for domestic cooking,is a commercial mixture of $isobutane$ and $propane$.

(B) Tetraethyl lead $(TEL)$ was historically added to petrol as an antiknock agent.
However,its combustion leads to the deposition of lead oxide in the engine.
To prevent this,ethylene dibromide $(C_2H_4Br_2)$ is added to the petrol.
It reacts with lead oxide to form volatile lead bromide $(PbBr_2)$,which is expelled from the engine through the exhaust.

(B) The octane number of a fuel is defined as the percentage of isooctane in a mixture of isooctane and $n$-heptane that has the same knocking characteristics as the fuel.
Since the fuel has the same knocking property as a mixture of $70\%$ isooctane and $30\%$ $n$-heptane,its octane number is $70$.

(A) The octane number is a measure of the anti-knocking property of a fuel.
By definition,$n-$heptane is assigned an octane number of $0$ because it has very poor anti-knocking properties.
Iso-octane ($2,2,4-$trimethylpentane) is assigned an octane number of $100$ because it has excellent anti-knocking properties.
Therefore,the correct answer is $n-$heptane.

(B) The octane number is a measure of the anti-knocking property of a fuel.
By definition,$n$-heptane is assigned an octane number of $0$ because it has very poor anti-knocking properties and causes significant knocking in engines.
Iso-octane ($2,2,4$-trimethylpentane) is assigned an octane number of $100$.
Since $n$-heptane has the lowest defined octane number among standard reference fuels,it is the correct answer.

(B) Marsh gas is formed by the anaerobic decomposition of organic matter in waterlogged areas like marshes and swamps.
It primarily consists of methane $(CH_4)$,which is the simplest alkane.
Therefore,the correct option is $B$.

(C) The photochemical chlorination of alkanes is a free radical substitution reaction.
It is initiated by the homolytic cleavage (homolysis) of the $Cl-Cl$ bond in the presence of $UV$ light or heat,which generates chlorine free radicals $(Cl^{\bullet})$.

(C) The electrolysis of an aqueous solution of potassium acetate is known as the Kolbe's electrolysis reaction.
The reaction is: $2CH_3COOK + 2H_2O \rightarrow CH_3-CH_3 + 2CO_2 + H_2 + 2KOH$.
During this process,ethane $(CH_3-CH_3)$ is produced at the anode.

(D) The reaction between methane $(CH_4)$ and chlorine $(Cl_2)$ in the presence of sunlight is a free radical substitution reaction.
This reaction proceeds through the following steps:
$CH_4 + Cl_2 \xrightarrow{h\nu} CH_3Cl + HCl$
$CH_3Cl + Cl_2 \xrightarrow{h\nu} CH_2Cl_2 + HCl$
$CH_2Cl_2 + Cl_2 \xrightarrow{h\nu} CHCl_3 + HCl$
$CHCl_3 + Cl_2 \xrightarrow{h\nu} CCl_4 + HCl$
Additionally,ethane $(CH_3CH_3)$ can be formed as a minor byproduct due to the coupling of two methyl radicals $(CH_3^{\bullet} + CH_3^{\bullet} \rightarrow CH_3CH_3)$.
Propane $(CH_3CH_2CH_3)$ contains three carbon atoms,whereas methane only contains one carbon atom. Therefore,propane cannot be formed in this reaction.

(B) The reagent $Red\, P + HI$ is a powerful reducing agent that can reduce various functional groups like halides,alcohols,aldehydes,ketones,and carboxylic acids directly into their corresponding alkanes.
$Zn - Hg / HCl$ (Clemmensen reduction) is specific to carbonyl compounds.
$LiAlH_4$ is a reducing agent but typically reduces carbonyls to alcohols,not alkanes.

(D) The electrolysis of a mixture of sodium salts of carboxylic acids is known as the Kolbe electrolysis reaction.
For a mixture of sodium acetate $(CH_3COONa)$ and sodium propionate $(CH_3CH_2COONa)$,the following reactions occur at the anode:
$1$. $CH_3COO^-$ $\rightarrow CH_3COO^{\bullet} + e^-$ $\rightarrow CH_3^{\bullet} + CO_2$
$2$. $CH_3CH_2COO^-$ $\rightarrow CH_3CH_2COO^{\bullet} + e^-$ $\rightarrow CH_3CH_2^{\bullet} + CO_2$
The free radicals formed ($CH_3^{\bullet}$ and $CH_3CH_2^{\bullet}$) combine in three possible ways:
- $CH_3^{\bullet} + CH_3^{\bullet} \rightarrow CH_3-CH_3$ (Ethane)
- $CH_3CH_2^{\bullet} + CH_3CH_2^{\bullet} \rightarrow CH_3CH_2-CH_2CH_3$ (Butane)
- $CH_3^{\bullet} + CH_3CH_2^{\bullet} \rightarrow CH_3-CH_2CH_3$ (Propane)
Thus,the mixture of products obtained is Ethane,Propane,and Butane.

(C) Grignard reagents $(RMgX)$ act as strong bases when they react with compounds containing active hydrogen atoms (like alcohols,water,or amines).
In this reaction,$CH_3MgBr$ reacts with $C_2H_5OH$ to form methane $(CH_4)$ and magnesium ethoxy bromide $(Mg(OC_2H_5)Br)$.
The reaction is: $CH_3MgBr + C_2H_5OH \rightarrow CH_4 + Mg(OC_2H_5)Br$.
Therefore,the product is methane $(CH_4)$.

(D) The Wurtz reaction involves the coupling of two alkyl halide molecules to form a symmetric alkane: $2R-X + 2Na \rightarrow R-R + 2NaX$.
Given the product is $4, 5-$diethyl octane,which has the structure $CH_3-CH_2-CH_2-CH(C_2H_5)-CH(C_2H_5)-CH_2-CH_2-CH_3$.
Splitting this molecule at the center (between $C_4$ and $C_5$),we get two identical fragments: $CH_3-CH_2-CH_2-CH(C_2H_5)-$.
Replacing the radical with a bromine atom,we get the alkyl bromide $(X)$ as $CH_3-CH_2-CH_2-CH(Br)-CH_2-CH_3$.
This corresponds to $3-$bromohexane,which is $CH_3-CH_2-CH_2-CH(Br)CH_2CH_3$.

(B) The $photochemical \ chlorination$ of methane is a free radical chain reaction.
Oxygen $(O_2)$ acts as an inhibitor in this process because it reacts with the free radical intermediates (like $CH_3^{\bullet}$) to form stable peroxy radicals.
This effectively terminates the chain reaction,thereby inhibiting the reaction for some time until the oxygen is consumed.

(A) Alkanes with an even number of carbon atoms pack better in the crystal lattice due to their symmetrical structure,leading to stronger intermolecular forces of attraction.
Consequently,alkanes with an even number of carbon atoms have higher melting points compared to their immediate neighbors with an odd number of carbon atoms.
Therefore,alkanes with an odd number of carbon atoms have lower melting points.

(B) Paraffin refers to alkanes,which are non-polar hydrocarbons. According to the principle of 'like dissolves like',non-polar substances dissolve in non-polar solvents. Among the given options,$Benzene$ $(C_6H_6)$ is a non-polar organic solvent,while water,methanol,and saline water are polar solvents. Therefore,paraffin is soluble in $Benzene$.

(A) The reaction of $C_2H_5I$ with $Na$ in the presence of dry ether is the Wurtz reaction:
$2C_2H_5I + 2Na \xrightarrow{\text{dry ether}} C_4H_{10} + 2NaI$.
In this reaction,side products like ethene $(C_2H_4)$ and ethane $(C_2H_6)$ can be formed due to disproportionation or radical abstraction.
Among the given options,$C_2H_4$ is a possible impurity formed via the elimination reaction of the ethyl radical or ethyl iodide.

(D) The reactivity of alkanes towards free radical substitution reactions depends on the stability of the intermediate free radical formed during the reaction.
In the given alkanes,the stability of the free radical formed by the abstraction of a hydrogen atom follows the order: $3^\circ > 2^\circ > 1^\circ$.
$CH_4$ is a primary alkane,$CH_3CH_2CH_3$ has $2^\circ$ hydrogens,$CH_3(CH_2)_2CH_3$ has $2^\circ$ hydrogens,and $(CH_3)_3CH$ (isobutane) has a $3^\circ$ hydrogen.
The abstraction of the $3^\circ$ hydrogen from isobutane leads to the formation of a tertiary butyl radical,which is the most stable among the options provided.
Therefore,$(CH_3)_3CH$ is the most reactive alkane.

(B) The iodination of alkanes is a reversible reaction: $CH_4 + I_2 \rightleftharpoons CH_3I + HI$.
To drive the reaction in the forward direction,the byproduct $HI$ must be removed.
This is achieved by using an oxidizing agent like $HIO_3$ (iodic acid) or $HNO_3$,which oxidizes $HI$ to $I_2$ $(5HI + HIO_3 \rightarrow 3I_2 + 3H_2O)$.
Therefore,an oxidizing agent is necessary.

(B) In the $Wurtz$ reaction,alkyl halides react with sodium metal in the presence of dry ether to form alkanes.
However,the reaction proceeds via a free radical mechanism.
Sometimes,these free radicals undergo disproportionation,where one radical abstracts a hydrogen atom from another,resulting in the formation of an alkane and an alkene.
Therefore,the formation of an alkene is a common side reaction due to the disproportionation of free radicals.

(A) The halogenation of alkanes follows the reactivity order $F_2 > Cl_2 > Br_2 > I_2$.
Chlorination is highly exothermic and less selective,whereas bromination is endothermic and highly selective.
Due to the higher activation energy required for the abstraction of hydrogen by a bromine radical compared to a chlorine radical,the rate of bromination is significantly slower than that of chlorination.