Alkane Questions in English

(C) For saturated hydrocarbons (alkanes),the boiling point increases with an increase in the molecular mass or the number of carbon atoms due to an increase in the magnitude of van der Waals forces.
Given boiling points are:
$A = -102\,^{\circ}C$
$B = -43.44\,^{\circ}C$
$C = -0.6\,^{\circ}C$
Since the boiling point of $C$ is the highest,it must contain the highest number of carbon atoms.

(A) The boiling point of alkanes depends on the surface area and branching.
$n$-Butane is a straight-chain alkane with a higher boiling point than its branched isomer,isobutane.
Isobutane ($2$-methylpropane) has more branching,which reduces the surface area and thus the van der Waals forces,resulting in a lower boiling point compared to $n$-butane.
Among the given options,isobutane has the lowest boiling point due to its compact branched structure.

(A) Alkanes are non-polar hydrocarbons. According to the principle of 'like dissolves like',non-polar substances are insoluble in polar solvents like water. Therefore,long-chain alkanes are insoluble in water. Statement $A$ is incorrect.

(A) The reactivity of alkanes towards free radical substitution depends on the stability of the intermediate free radical formed during the reaction.
The order of stability of free radicals is: $3^\circ > 2^\circ > 1^\circ > \text{methyl radical}$.
$CH_4$ forms a methyl radical,which is the least stable.
$(CH_3)_3CH$ forms a $3^\circ$ radical,which is the most stable.
$CH_3CH_2CH_3$ forms a $2^\circ$ radical.
$CH_3CH_3$ forms a $1^\circ$ radical.
Since the reactivity is directly proportional to the stability of the radical formed,$CH_4$ is the least reactive.

(C) The bond length of a $C-H$ bond depends on the hybridization of the carbon atom.
As the $s$-character in the hybrid orbital increases,the electronegativity of the carbon atom increases,pulling the electrons closer to the nucleus and shortening the bond length.
$C_2H_2$ has $sp$ hybridization ($50\% \ s$-character),$C_2H_4$ has $sp^2$ hybridization ($33.3\% \ s$-character),and $C_2H_6$ has $sp^3$ hybridization ($25\% \ s$-character).
Since $sp^3$ hybridized carbon has the least $s$-character,the $C-H$ bond in $C_2H_6$ (ethane) is the longest.

(C) The substrate is isobutane,which has $9$ equivalent $1^o$ hydrogens and $1$ $3^o$ hydrogen.
Relative rate for $3^o$ product $(a)$: $1 \times 1600 = 1600$.
Relative rate for $1^o$ product $(b)$: $9 \times 1 = 9$.
Total relative rate = $1600 + 9 = 1609$.
Percentage of product $(a) = (1600 / 1609) \times 100 \approx 99.4\%$.
Percentage of product $(b) = (9 / 1609) \times 100 \approx 0.6\%$.
Thus,the yields are $99.4\%$ and $0.6\%$ respectively.

(A) The reaction of sodium acetate $(CH_3COONa)$ with sodalime $(NaOH + CaO)$ is known as decarboxylation.
The chemical equation is: $CH_3COONa + NaOH \xrightarrow{CaO, \Delta} CH_4 + Na_2CO_3$.
Thus,the product obtained is methane $(CH_4)$.

(C) The octane number of a fuel is defined as the percentage by volume of isooctane ($2,2,4$-trimethylpentane) in a mixture of isooctane and $n$-heptane that has the same knocking characteristics as the fuel being tested.
Since the octane number is $40$,it means the mixture contains $40\%$ isooctane and $(100 - 40) = 60\% \ n$-heptane by volume.
Therefore,the correct composition is $60\% \ n$-heptane and $40\%$ isooctane.

(B) The halogenation of alkanes (e.g.,chlorination) proceeds via a free radical chain mechanism.
This process involves three steps: initiation,propagation,and termination.
In the presence of light $(h\nu)$ or heat,the halogen molecule undergoes homolytic cleavage to form free radicals,which then react with the alkane.

(A) The physical state of alkanes depends on the number of carbon atoms in the chain.
Alkanes with $C_1$ to $C_4$ (methane to butane) are gases at room temperature.
Alkanes with $C_5$ to $C_{17}$ (pentane to heptadecane) are liquids at room temperature.
Since pentane has $5$ carbon atoms $(C_5H_{12})$,it is a liquid at room temperature.

(C) mono-chlorinated product is formed when a hydrogen atom is replaced by a chlorine atom. If all hydrogen atoms in an alkane are equivalent,only one mono-chlorinated product is possible.
In $(CH_3)_4C$ (neopentane),all $12$ hydrogen atoms are equivalent because the molecule is highly symmetrical.
Therefore,replacing any of these $12$ hydrogen atoms with a chlorine atom results in the same product,$1$-chloro-$2,2$-dimethylpropane.
Other options like propane,butane,and pentane have non-equivalent hydrogen atoms,leading to multiple isomeric mono-chlorinated products.

(A) The boiling point of alkanes depends on the surface area of the molecule.
Straight-chain alkanes have a larger surface area compared to branched-chain alkanes,leading to stronger van der Waals forces of attraction.
Among the given options,$n$-octane $(C_8H_{18})$ is a straight-chain alkane with the highest molecular weight and the largest surface area.
$n$-butane has a lower molecular weight,while isooctane and neopentane are branched,which reduces their surface area and boiling points.
Therefore,$n$-octane has the highest boiling point.

(D) The reaction between an alkyl halide $(R-X)$ and a dialkyl copper lithium reagent $(R'_2CuLi)$,known as the Corey-House synthesis,results in the formation of an alkane $(R-R')$.
This reaction is used to prepare alkanes with a specific carbon skeleton.

(D) Isomerism in alkanes requires the presence of at least $4$ carbon atoms to form branched structures.
For $n=1, 2, 3$,only straight-chain alkanes exist (methane,ethane,propane).
For $n=4$,we have $n$-butane $(CH_3-CH_2-CH_2-CH_3)$ and isobutane $(CH_3-CH(CH_3)-CH_3)$,which are structural isomers.
Therefore,the smallest alkane exhibiting isomerism is butane,which contains $4$ carbon atoms.

(D) The density of cyclohexane is approximately $0.779 \ g/cm^3$ at $20 \ ^\circ C$,whereas the density of water is $1.00 \ g/cm^3$.
Since the density of cyclohexane is lower than that of water,it floats on the surface of water when added.

(B) The reaction of $n$-butane with bromine at high temperatures (like $130^\circ C$) or under $UV$ light proceeds via a free-radical substitution mechanism.
In $n$-butane $(CH_3-CH_2-CH_2-CH_3)$,there are two types of hydrogen atoms: primary $(1^\circ)$ and secondary $(2^\circ)$.
The secondary hydrogen atoms are more reactive towards free-radical substitution because the resulting secondary free radical is more stable than the primary free radical.
Therefore,the major product is $2$-bromobutane,which is $CH_3CH_2CH(Br)CH_3$.

(A) The reaction of an alkane with chlorine in the presence of ultraviolet light is a free radical substitution reaction.
To obtain only one monochloroalkane,all the hydrogen atoms in the alkane must be equivalent.
In $Neopentane$ $(2,2-dimethylpropane)$,all $12$ hydrogen atoms are equivalent because the molecule is highly symmetrical.
Therefore,substitution of any hydrogen atom by a chlorine atom results in the same product,$1-chloro-2,2-dimethylpropane$.

(B) The given reaction is $CH_4 + Cl_2 \xrightarrow{UV} CH_3Cl + HCl$.
In this reaction,one hydrogen atom of methane $(CH_4)$ is replaced by a chlorine atom $(Cl)$ to form chloromethane $(CH_3Cl)$.
Such a reaction where an atom or group of atoms is replaced by another atom or group is known as a substitution reaction.
Therefore,this is a free radical substitution reaction.

(D) $LPG$ (Liquefied Petroleum Gas) is primarily a mixture of hydrocarbons,consisting mainly of $n$-butane and isobutane,along with smaller amounts of propane. Among the given options,isobutane is a major component of $LPG$.

(A) Petroleum is a complex mixture of various hydrocarbons,primarily consisting of aliphatic hydrocarbons (alkanes,cycloalkanes) and some aromatic hydrocarbons. Aliphatic alcohols are not major components of crude petroleum. Therefore,the most accurate description among the choices is that it contains aliphatic and aromatic hydrocarbons. However,since the question asks for the main constituents,aliphatic hydrocarbons are the most abundant. Given the options,$(A)$ is the most appropriate answer.

(A) Petroleum is a complex mixture of hydrocarbons,primarily consisting of alkanes (paraffins),cycloalkanes,and aromatic hydrocarbons. Among these,alkanes are the most abundant components in crude oil.

(A) The reaction of alkanes with bromine in the presence of sunlight (or $UV$ light) is a free radical substitution reaction.
Bromination is highly selective compared to chlorination.
The reactivity order of $C-H$ bonds towards free radical substitution is $3^\circ > 2^\circ > 1^\circ$.
In $2$-methylbutane $(CH_3-CH(CH_3)-CH_2-CH_3)$,the $3^\circ$ hydrogen atom is present at the $C-2$ position.
Therefore,the bromine radical will abstract the $3^\circ$ hydrogen to form the most stable tertiary free radical,which then reacts with $Br_2$ to form $2$-bromo-$2$-methylbutane as the major product.

(A) The hydrolysis of aluminum carbide $(Al_4C_3)$ yields methane gas $(CH_4)$ and aluminum hydroxide $(Al(OH)_3)$.
The chemical reaction is: $Al_4C_3 + 12H_2O \rightarrow 3CH_4 + 4Al(OH)_3$.

(C) At $STP$,$1 \ mole$ of any gas occupies $22.4 \ L$ volume.
Given: Weight = $11 \ g$,Volume = $5.6 \ L$.
Number of moles = $\frac{\text{Given Volume}}{\text{Molar Volume at STP}} = \frac{5.6 \ L}{22.4 \ L/mol} = 0.25 \ mol$.
Molar mass $(M)$ = $\frac{\text{Weight}}{\text{Moles}} = \frac{11 \ g}{0.25 \ mol} = 44 \ g/mol$.
For an alkane with formula $C_nH_{2n+2}$,the molar mass is $12n + 1(2n+2) = 14n + 2$.
Setting $14n + 2 = 44$,we get $14n = 42$,so $n = 3$.
Therefore,the molecular formula is $C_3H_8$.

(C) The reaction of $CH_4$ with $Cl_2$ in the presence of sunlight follows a free radical substitution mechanism.
The stepwise chlorination produces: $CH_4$ $\rightarrow CH_3Cl$ $\rightarrow CH_2Cl_2$ $\rightarrow CHCl_3$ $\rightarrow CCl_4$.
Additionally,the coupling of methyl free radicals $(CH_3^{\bullet} + CH_3^{\bullet} \rightarrow CH_3-CH_3)$ leads to the formation of ethane $(CH_3CH_3)$.
Propane $(CH_3CH_2CH_3)$ cannot be formed in this reaction because it requires a higher number of carbon atoms than are present in the reactant $CH_4$.

(D) Stability of alkanes increases with an increase in the number of carbon atoms due to increased van der Waals forces and molecular complexity. Among the given options,$n$-butane $(C_4H_{10})$ has the highest number of carbon atoms,making it the most stable.

(A) To obtain $C_2H_5Cl$ as the major product,the alkane $(C_2H_6)$ must be taken in excess to prevent further chlorination of the product.
$C_2H_6 \text{ (excess)} + Cl_2 \xrightarrow{UV \text{ Light}} C_2H_5Cl \text{ (Major product)} + HCl$

(D) The general formula for alkanes is $C_nH_{2n+2}$.
For option $A$,$C_5H_8$ follows $C_nH_{2n-2}$ (alkyne).
For option $B$,$C_8H_6$ does not follow the general alkane formula.
For option $C$,$C_9H_{10}$ does not follow the general alkane formula.
For option $D$,$C_7H_{16}$ follows the formula $C_nH_{2n+2}$ where $n=7$,so $2(7)+2 = 16$. Thus,$C_7H_{16}$ is an alkane.

(A) The structure of $2$-methylbutane is $CH_3-CH(CH_3)-CH_2-CH_3$.
Monochlorination can occur at four different positions,resulting in four constitutional isomers:
$1$. $1$-chloro-$2$-methylbutane (chiral center at $C2$),
$2$. $2$-chloro-$2$-methylbutane (achiral),
$3$. $2$-chloro-$3$-methylbutane (chiral center at $C2$ and $C3$),
$4$. $1$-chloro-$3$-methylbutane (achiral).
Among these,$1$-chloro-$2$-methylbutane and $2$-chloro-$3$-methylbutane possess chiral centers and exist as pairs of enantiomers.
Thus,there are $2$ pairs of enantiomers.

(B) The reaction involves the abstraction of a hydrogen atom by a bromine radical $(Br^\bullet)$.
The stability of the resulting alkyl radical determines the major product.
The starting material is $2-deuterio-3-methylbutane$.
Abstraction of a hydrogen atom from the $C-3$ position leads to a tertiary $(3^\circ)$ radical,which is more stable than a secondary $(2^\circ)$ or primary $(1^\circ)$ radical.
Structure $B$ represents the radical formed at the $C-3$ position: $H_3C-CH(D)-C^\bullet(CH_3)-CH_3$.

(C) The molecular mass of the hydrocarbon is $72 \ u$. The general formula for an alkane is $C_nH_{2n+2}$.
$12n + 1(2n+2) = 72 \implies 14n = 70 \implies n = 5$.
Thus,the hydrocarbon is pentane $(C_5H_{12})$.
The isomers of pentane are $n$-pentane,isopentane,and neopentane.
Neopentane ($2,2$-dimethylpropane) has all $12$ hydrogen atoms equivalent.
Therefore,upon reaction with a halogen,it produces only one mono-substituted alkyl halide.

(A) quaternary carbon atom is a carbon atom bonded to four other carbon atoms.
To form a quaternary carbon,we need a central carbon atom bonded to four methyl groups $(-CH_3)$.
The structure is $CH_3-C(CH_3)_2-CH_3$,which is $2,2$-dimethylpropane (neopentane).
Counting the total number of carbon atoms: $1$ (central) $+ 4$ (methyl groups) $= 5$ carbon atoms.
Thus,the minimum number of carbon atoms required is $5$.

(A) The starting material is $2$-methylbutane $(CH_3-CH(CH_3)-CH_2-CH_3)$.
When $2$-methylbutane reacts with $Cl_2/h\nu$,it undergoes free radical chlorination.
The possible monochlorinated products are:
$1$. $1$-chloro-$2$-methylbutane
$2$. $2$-chloro-$2$-methylbutane
$3$. $2$-chloro-$3$-methylbutane
$4$. $1$-chloro-$3$-methylbutane
Considering stereoisomers:
- $1$-chloro-$2$-methylbutane has $1$ chiral center ($2$ enantiomers).
- $2$-chloro-$2$-methylbutane is achiral ($1$ isomer).
- $2$-chloro-$3$-methylbutane has $2$ chiral centers ($2$ pairs of enantiomers = $4$ stereoisomers).
- $1$-chloro-$3$-methylbutane has $1$ chiral center ($2$ enantiomers).
Total number of isomers $(N)$ = $2 + 1 + 4 + 2 = 9$ (Wait,let's re-evaluate: $1$-chloro-$2$-methylbutane $(2)$,$2$-chloro-$2$-methylbutane $(1)$,$2$-chloro-$3$-methylbutane ($2$ diastereomers,each with $2$ enantiomers = $4$),$1$-chloro-$3$-methylbutane $(2)$. Total = $9$).
However,based on standard textbook problems for $2$-methylbutane,the constitutional isomers are $4$. The total number of stereoisomers is $6$. Fractional distillation separates compounds based on boiling points. Stereoisomers (enantiomers) have the same boiling point and cannot be separated by fractional distillation. Thus,the number of fractions $(F)$ is $4$ (the constitutional isomers).
Therefore,$N = 6$ (if only considering chiral centers) or $9$. Given the options,$N=6$ and $F=4$ is the standard answer.

(D) Chirality in an alkane requires the presence of a chiral carbon atom,which is a carbon atom bonded to four different groups.
For an alkane $(C_nH_{2n+2})$,the simplest chiral molecule is $3$-methylhexane,which has the formula $C_7H_{16}$.
Therefore,the minimum number of carbon atoms required to exhibit chirality in an alkane is $7$.

(C) The structure of $2,3-$dimethylbutane is $(CH_3)_2CH-CH(CH_3)_2$.
It has two types of equivalent hydrogen atoms: the $12$ primary hydrogens on the four methyl groups and the $2$ tertiary hydrogens on the $C_2$ and $C_3$ carbons.
Therefore,substitution of these hydrogens leads to only two distinct monochlorinated products.

(C) The structure of $2$-methylbutane is $CH_3-CH(CH_3)-CH_2-CH_3$.
Monochlorination can occur at four different types of hydrogen atoms:
$1$. At $C_1$ (terminal methyl): $Cl-CH_2-CH(CH_3)-CH_2-CH_3$ (Achiral).
$2$. At $C_2$ (tertiary carbon): $CH_3-CCl(CH_3)-CH_2-CH_3$ (Chiral,as $C_2$ is bonded to four different groups: $-CH_3, -Cl, -CH_2CH_3, -H$).
$3$. At $C_3$ (secondary carbon): $CH_3-CH(CH_3)-CHCl-CH_3$ (Chiral,as $C_3$ is bonded to four different groups: $-CH_3, -Cl, -H, -CH_2CH_3$).
$4$. At $C_4$ (terminal methyl): $CH_3-CH(CH_3)-CH_2-CH_2Cl$ (Achiral).
Thus,there are $2$ chiral products obtained.

(A) The given reaction is $2CH_3Br + 2Na \to CH_3-CH_3 + 2NaBr$.
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 is a characteristic reaction known as the Wurtz reaction.

(C) The electrolysis of an aqueous solution of potassium acetate $(CH_3COOK)$ is known as the Kolbe's electrolysis reaction.
The reaction proceeds as follows:
$2CH_3COOK + 2H_2O \rightarrow CH_3-CH_3 + 2CO_2 + 2KOH + H_2$
At the anode,ethane $(C_2H_6)$ and carbon dioxide $(CO_2)$ are produced,while at the cathode,hydrogen gas $(H_2)$ and potassium hydroxide $(KOH)$ are formed.
Thus,the main organic product obtained is ethane $(C_2H_6)$.

(A) The electrolysis of sodium or potassium salts of carboxylic acids is known as the Kolbe electrolysis reaction,which is an effective method for the preparation of alkanes (hydrocarbons).
The reaction is: $2RCOOK + 2H_2O \xrightarrow{\text{electrolysis}} R-R + 2CO_2 + H_2 + 2KOH$.
Therefore,option $A$ is the correct reaction for obtaining a good yield of hydrocarbons.

(C) Cracking is the process of breaking down long-chain hydrocarbons into smaller,more useful hydrocarbons.
Petrol (gasoline) consists of higher alkanes,and upon cracking,it yields a mixture of smaller alkanes and alkenes,such as $CH_4$ and $C_3H_6$.

(A) Cetane is chemically hexadecane,which is an alkane with $16$ carbon atoms.
Its chemical formula is $CH_3(CH_2)_{14}CH_3$.

(B) The reaction $CH_4 + Cl_2 \xrightarrow{h\nu} CH_3Cl + HCl$ is a free-radical substitution reaction.
In this reaction,the $Cl-Cl$ bond undergoes homolytic fission in the presence of light $(h\nu)$ to generate chlorine free radicals $(Cl^{\bullet})$.
These radicals then attack the methane molecule to substitute a hydrogen atom,which is a characteristic step of free-radical substitution.

(D) The structure of $2, 2-$dimethylbutane is $CH_3-C(CH_3)_2-CH_2-CH_3$.
In this molecule:
$1.$ Primary $(1^\circ)$ carbons are the terminal carbons and the methyl branches. There are $4$ such carbons,having $12$ primary hydrogens.
$2.$ Secondary $(2^\circ)$ carbon is the $-CH_2-$ group at position $3$. It has $2$ secondary hydrogens.
$3.$ Quaternary $(4^\circ)$ carbon is at position $2$,which has no hydrogens.
Therefore,the number of secondary hydrogens is $2$.

(A) The general formula for an alkane is $C_nH_{2n+2}$.
Given molecular weight is $72$.
$12n + 1(2n+2) = 72 \implies 14n + 2 = 72 \implies 14n = 70 \implies n = 5$.
So,the molecular formula is $C_5H_{12}$.
An alkane that forms only one monochlorinated product must have all its hydrogen atoms equivalent.
In $2,2-$dimethylpropane (neopentane),$(CH_3)_4C$,all $12$ hydrogen atoms are equivalent.
Therefore,it forms only one monochlorinated product upon reaction with $Cl_2$ in the presence of sunlight.

(A) The Wurtz reaction involves the coupling of alkyl halides with sodium metal in the presence of dry ether to form higher alkanes.
Alkyl fluorides $(R-F)$ are generally unreactive in the Wurtz reaction due to the high bond dissociation energy of the $C-F$ bond.
Therefore,$C_2H_5F$ does not undergo the Wurtz reaction under standard conditions.
The correct option is $A$.

(B) Lower hydrocarbons exist in the gaseous state,while higher ones exist in the liquid or solid state.
On cracking or pyrolysis,a hydrocarbon with a higher molecular mass breaks down into a mixture of hydrocarbons having lower molecular masses.
Therefore,by the process of cracking,a liquid hydrocarbon can be converted into a mixture of gaseous hydrocarbons.

(B) The molecular mass of the hydrocarbon is $72 \ u$. The general formula for alkanes is $C_nH_{2n+2}$.
$12n + 1(2n+2) = 72 \implies 14n = 70 \implies n = 5$.
Thus,the hydrocarbon is pentane $(C_5H_{12})$.
Among the isomers of pentane (n-pentane,isopentane,and neopentane),neopentane $(2,2-\text{dimethylpropane})$ has all $12$ hydrogen atoms equivalent.
Therefore,upon mono-halogenation,it yields only one mono-substituted alkyl halide.
$(CH_3)_4C + Cl_2 \xrightarrow{h\nu} (CH_3)_3C-CH_2Cl + HCl$.

(C) $2$-methylbutane is $CH_3-CH(CH_3)-CH_2-CH_3$.
Monochlorination yields the following products:
$1.$ $Cl-CH_2-CH(CH_3)-CH_2-CH_3$ ($1$-chloro-$2$-methylbutane): This molecule has a chiral center at $C_2$,so it exists as $2$ enantiomers.
$2.$ $CH_3-CCl(CH_3)-CH_2-CH_3$ ($2$-chloro-$2$-methylbutane): This molecule is achiral.
$3.$ $CH_3-CH(CH_3)-CHCl-CH_3$ ($2$-chloro-$3$-methylbutane): This molecule has a chiral center at $C_3$,so it exists as $2$ enantiomers.
$4.$ $CH_3-CH(CH_3)-CH_2-CH_2Cl$ ($1$-chloro-$3$-methylbutane): This molecule is achiral.
Total chiral compounds = $2 + 2 = 4$.

(D) The hydrocarbon $C_5H_{12}$ is an alkane. Alkanes do not react with chlorine in the dark but undergo free-radical substitution in the presence of sunlight.
$n$-pentane $(CH_3-CH_2-CH_2-CH_2-CH_3)$ has three types of equivalent hydrogens,yielding three monochloro isomers.
Isopentane $(CH_3-CH(CH_3)-CH_2-CH_3)$ has four types of equivalent hydrogens,yielding four monochloro isomers.
Neopentane $(C(CH_3)_4)$ has only one type of equivalent hydrogen,yielding only one monochloro isomer.
Since the question asks for a hydrocarbon that gives six monochloro compounds,and none of the standard isomers of $C_5H_{12}$ produce six,there might be a misunderstanding in the question's premise or it refers to a different hydrocarbon. However,based on the options provided,none are correct.

(A) The free radical chlorination of ethane follows a chain mechanism consisting of initiation,propagation,and termination steps.
The propagation steps are:
Step $(i)$: The chlorine radical abstracts a hydrogen atom from ethane to form an ethyl radical and hydrogen chloride: $\dot{Cl} + CH_3-CH_3 \rightarrow CH_3-C\dot{H}_2 + HCl$.
Step $(ii)$: The ethyl radical reacts with a chlorine molecule to form ethyl chloride and regenerate the chlorine radical: $CH_3-C\dot{H}_2 + Cl_2 \rightarrow CH_3-CH_2Cl + \dot{Cl}$.
Thus,option $A$ represents the correct propagation steps.