Alkane Questions in English

(D) The chemical formula for propane is $C_3H_8$.
The structural formula is $CH_3-CH_2-CH_3$.
In this structure,there are $2$ $C-C$ single bonds and $8$ $C-H$ single bonds.
Each single bond represents one shared pair of electrons.
Total shared pairs of electrons = $2 + 8 = 10$.

(A) The complete combustion of methane $(CH_4)$ in the presence of excess oxygen $(O_2)$ produces carbon dioxide $(CO_2)$ and water $(H_2O)$.
The balanced chemical equation is:
$CH_4(g) + 2O_2(g) \to CO_2(g) + 2H_2O(l)$

(C) Soda lime is a mixture of sodium hydroxide $(NaOH)$ and calcium oxide $(CaO)$.
It is typically prepared in a ratio of $3:1$ $(NaOH:CaO)$.

(A) The hydrolysis of aluminum carbide $(Al_4C_3)$ produces methane gas $(CH_4)$ and aluminum hydroxide $(Al(OH)_3)$.
The balanced chemical equation is:
$Al_4C_3 + 12H_2O \to 3CH_4 + 4Al(OH)_3$
Therefore,the correct option is $A$.

(A) When $Cl_2$ is exposed to $UV$ light,it undergoes homolytic fission.
$Cl-Cl \xrightarrow{UV} Cl^{\bullet} + Cl^{\bullet}$
This process results in the formation of chlorine free radicals $(Cl^{\bullet})$.

(C) The correct answer is $(C)$.
The structure of $(CH_3)_3CCH_2CH(CH_3)_2$ is $CH_3-C(CH_3)_2-CH_2-CH(CH_3)-CH_3$.
$1$. $A$ quaternary $(4^o)$ carbon is present at the central position bonded to four other carbons.
$2$. $A$ tertiary $(3^o)$ carbon is present in the $CH$ group.
$3$. $A$ secondary $(2^o)$ carbon is present in the $CH_2$ group.
$4$. Primary $(1^o)$ carbons are present at the terminal methyl groups.

(C) The structure of the compound $(CH_3)_2CHCH_2C(CH_3)_3$ can be expanded as $CH_3-CH(CH_3)-CH_2-C(CH_3)_3$.
$A$ tertiary carbon atom is a carbon atom bonded to three other carbon atoms.
In the given structure:
$1$. The carbon atom at the second position from the left is bonded to two $CH_3$ groups and one $CH_2$ group,making it a tertiary carbon ($3^{\circ}$ carbon).
$2$. The carbon atom at the fourth position is bonded to three $CH_3$ groups and one $CH_2$ group,making it a quaternary carbon ($4^{\circ}$ carbon).
$3$. All other carbon atoms are either primary $(1^{\circ})$ or secondary $(2^{\circ})$.
Therefore,there is only $1$ tertiary carbon atom in the molecule.

(B) The general formula for alkanes is $C_nH_{2n+2}$.
For $n = 9$,the formula becomes $C_9H_{2(9)+2} = C_9H_{20}$.
According to $IUPAC$ nomenclature,an alkane with $9$ carbon atoms is named as Nonane.

(B) Alkanes are saturated hydrocarbons containing only single bonds between carbon atoms. The general formula for acyclic alkanes is $C_nH_{2n+2}$,where $n$ represents the number of carbon atoms.

(B) The general formula for an alkene (containing one double bond) is $C_nH_{2n}$.
The general formula for an alkane (saturated hydrocarbon) is $C_nH_{2n+2}$.
Therefore,the correct pair is $C_nH_{2n}$ and $C_nH_{2n+2}$.

(C) Paraffins are another name for alkanes,which are saturated hydrocarbons. The general formula for alkanes is $C_nH_{2n + 2}$,where $n$ is the number of carbon atoms.

(C) cycloalkane is a saturated hydrocarbon that contains one ring structure.
For an open-chain alkane,the general formula is $C_nH_{2n+2}$.
When a ring is formed,two hydrogen atoms are removed to create the bond between the carbon atoms,resulting in the general formula $C_nH_{2n}$.

(D) primary hydrogen atom is one that is attached to a primary carbon atom (a carbon atom bonded to only one other carbon atom).
$A$. Isobutene $(CH_2=C(CH_3)_2)$ contains both primary and secondary hydrogens.
$B$. $2,3-$Dimethylbutane contains primary and tertiary hydrogens.
$C$. Cyclohexane contains only secondary hydrogens.
$D$. Neopentane $(C(CH_3)_4)$ has a central quaternary carbon atom bonded to four methyl groups. All $12$ hydrogen atoms are attached to primary carbon atoms,making them all primary hydrogen atoms.

(D) The structure of isopentane is $CH_3-CH(CH_3)-CH_2-CH_3$.
It contains three $CH_3$ groups,one $CH_2$ group,and one $CH$ group.
All carbon atoms in isopentane are bonded to at least one hydrogen atom.
Therefore,the statement that it has a carbon which is not bonded to hydrogen is false.

(C) The bond dissociation energy of a $C-H$ bond depends on the stability of the resulting alkyl radical formed after homolytic cleavage.
Stability of alkyl radicals follows the order: $3^o > 2^o > 1^o > \text{methyl}$.
Since the $3^o$ radical is the most stable,the energy required to break the $3^o$ $C-H$ bond is the lowest.
Therefore,the tertiary $(3^o)$ $C-H$ bond has the lowest bond dissociation energy.

(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 bond length decreases.
In $C_2H_6$ (ethane),the carbon is $sp^3$ hybridized ($25\% \ s$-character).
In $C_2H_4$ (ethene),the carbon is $sp^2$ hybridized ($33.3\% \ s$-character).
In $C_2H_2$ (ethyne),the carbon is $sp$ hybridized ($50\% \ s$-character).
Since $sp^3$ hybridization has the lowest $s$-character,the $C-H$ bond in $C_2H_6$ is the longest.

(D) In a free radical chain reaction,the chain termination step involves the combination of two radicals to form a stable molecule,thereby ending the chain propagation.
Step $5$ $(Br^{\bullet} + Br^{\bullet} \to Br_2)$ is the chain termination step because two free radicals combine to form a stable $Br_2$ molecule,and no new radicals are generated.

(D) The reaction of $CH_4$ with $Cl_2$ in the presence of sunlight $(h\nu)$ is a halogenation reaction.
This process proceeds via a free radical chain mechanism.
In this reaction,a hydrogen atom of methane is replaced by a chlorine atom to form methyl chloride $(CH_3Cl)$.
Therefore,it is classified as a free radical substitution reaction.

(D) The halogenation of alkanes,including bromination,proceeds via a free radical mechanism. This reaction typically occurs in the presence of light $(h\nu)$ or at elevated temperatures,which facilitates the homolytic cleavage of the halogen molecule to generate free radicals.

(A) To determine the number of monochloro isomers,we count the number of non-equivalent hydrogen atoms in the molecule.
$1$. For $n-$butane $(CH_3-CH_2-CH_2-CH_3)$,there are two sets of equivalent hydrogens: the terminal $CH_3$ groups and the internal $CH_2$ groups. Thus,it gives two monochloro products: $1-$chlorobutane and $2-$chlorobutane.
$2$. For $2,4-$dimethyl pentane,there are three sets of equivalent hydrogens.
$3$. For Benzene,all six hydrogen atoms are equivalent,giving only one monochloro product (chlorobenzene).
$4$. For $1-$methyl propane (isobutane),there are two sets of equivalent hydrogens,but the terminal $CH_3$ groups are equivalent to each other,and the central $CH$ is unique. However,the question asks for the specific case of $n-$butane as the standard answer for two isomers.

(D) Structural isomerism in alkanes requires a minimum of $4$ carbon atoms to form branched structures.
For $n=1$ (methane),$n=2$ (ethane),and $n=3$ (propane),only one structural arrangement is possible.
For $n=4$ (butane),two isomers exist: $n$-butane $(CH_3-CH_2-CH_2-CH_3)$ and isobutane $(CH_3-CH(CH_3)-CH_3)$.
Therefore,the correct answer is $4$.

(D) The resistance to bond rotation depends on the bond order.
$Ethane$ $(CH_3-CH_3)$ contains only $C-C$ single $(\sigma)$ bonds,which allow free rotation.
$Ethylene$ $(CH_2=CH_2)$ has a double bond,$Acetylene$ $(CH \equiv CH)$ has a triple bond,and $Hexachloroethane$ $(CCl_3-CCl_3)$ has significant steric hindrance due to bulky chlorine atoms.
Therefore,$Ethane$ exhibits the minimum resistance to bond rotation.

(A) The rotation about a $C-C$ bond is most free in $Ethane$ $(CH_3-CH_3)$ because it contains a single sigma bond.
$Ethylene$ $(CH_2=CH_2)$ has a double bond,and $Ethyne$ $(CH \equiv CH)$ has a triple bond,both of which restrict rotation due to the presence of pi bonds.
$Hexachloroethane$ $(CCl_3-CCl_3)$ has bulky chlorine atoms that cause significant steric hindrance,making rotation difficult.
Therefore,the least hindered rotation is observed in $Ethane$.

(A) For a compound to yield only two isomeric monochloro derivatives,it must have only two types of non-equivalent hydrogen atoms.
$(A)$ $2-$methyl propane $(CH_3-CH(CH_3)-CH_3)$: It has two types of hydrogen atoms (primary and tertiary),leading to two monochloro derivatives: $1-$chloro-$2-$methylpropane and $2-$chloro-$2-$methylpropane.
$(B)$ $n-$pentane: It has three types of hydrogen atoms,yielding three isomers.
$(C)$ Benzene: It has only one type of hydrogen atom,yielding only one monochloro derivative (chlorobenzene).
$(D)$ $2, 4-$dimethyl pentane: It has four types of hydrogen atoms,yielding four isomers.
Therefore,the correct option is $A$.

(B) The molecular formula $C_7H_{16}$ corresponds to an alkane $(C_nH_{2n+2})$.
To find the number of structural isomers,we draw the carbon skeletons:
$1$. $n$-heptane: $CH_3-CH_2-CH_2-CH_2-CH_2-CH_2-CH_3$
$2$. $2$-methylhexane: $CH_3-CH(CH_3)-CH_2-CH_2-CH_2-CH_3$
$3$. $3$-methylhexane: $CH_3-CH_2-CH(CH_3)-CH_2-CH_2-CH_3$
$4$. $3$-ethylpentane: $CH_3-CH_2-CH(CH_2CH_3)-CH_2-CH_3$
$5$. $2,2$-dimethylpentane: $CH_3-C(CH_3)_2-CH_2-CH_2-CH_3$
$6$. $2,3$-dimethylpentane: $CH_3-CH(CH_3)-CH(CH_3)-CH_2-CH_3$
$7$. $2,4$-dimethylpentane: $CH_3-CH(CH_3)-CH_2-CH(CH_3)-CH_3$
$8$. $3,3$-dimethylpentane: $CH_3-CH_2-C(CH_3)_2-CH_2-CH_3$
$9$. $2,2,3$-trimethylbutane: $CH_3-C(CH_3)_2-CH(CH_3)-CH_3$
There are a total of $9$ structural isomers for $C_7H_{16}$.

(D) The general formula for an alkane is $C_nH_{2n+2}$.
For option $D$,$C_7H_{16}$,we have $n = 7$.
Substituting $n = 7$ into the formula gives $C_7H_{2(7)+2} = C_7H_{14+2} = C_7H_{16}$.
Therefore,$C_7H_{16}$ represents an alkane.

(A) The boiling point of an alkane depends on the surface area of the molecule. $A$ larger surface area leads to stronger van der Waals forces,resulting in a higher boiling point.
Straight-chain isomers have a larger surface area compared to branched-chain isomers. As the number of branches increases,the molecule becomes more spherical,reducing the surface area and consequently lowering the boiling point.
Comparing the pentane isomers:
$n-$pentane (straight chain) has the largest surface area.
iso-pentane (one branch) has a smaller surface area than $n-$pentane.
neo-pentane (two branches) has the smallest surface area.
Therefore,the decreasing order of boiling points is $n-$pentane > iso-pentane > neo-pentane.

(A) The Wurtz reaction involves the coupling of two alkyl halides in the presence of sodium metal to form a symmetrical alkane.
To prepare $n-$hexane $(C_6H_{14})$,we require two $n-$propyl groups.
Therefore,the reaction is: $2CH_3CH_2CH_2Br + 2Na \xrightarrow{\text{dry ether}} CH_3CH_2CH_2CH_2CH_2CH_3 + 2NaBr$.
Thus,the reactant is $n-$propyl bromide.

(B) The conversion of sodium acetate to ethane is achieved by the Kolbe's electrolytic method.
In this process,the electrolysis of an aqueous solution of sodium acetate yields ethane at the anode.
The reaction is as follows:
$2CH_3COONa + 2H_2O \xrightarrow{\text{Electrolysis}} CH_3-CH_3 + 2CO_2 + 2NaOH + H_2$

(C) $1,2$-dibromoethane is used along with tetraethyllead $(TEL)$ in antiknock compositions.
It reacts with the lead oxides formed during combustion to produce volatile lead bromide $(PbBr_2)$,which is then removed through the exhaust,preventing lead oxide deposition on engine parts.

(B) Kerosene is a mixture of hydrocarbons obtained from the fractional distillation of petroleum.
It typically consists of hydrocarbons containing $11$ to $16$ carbon atoms per molecule $(C_{11} - C_{16})$.
The main constituents are saturated straight-chain and branched-chain hydrocarbons (paraffins) and cyclic hydrocarbons (naphthenes).
Therefore,the correct range is $C_{11} - C_{16}$.

(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 to form a higher alkane.
This specific type of reaction is known as the $Wurtz$ reaction.

(C) Alkanes with a tertiary hydrogen atom can be oxidised by strong oxidising agents like $KMnO_4$ to form tertiary alcohols.
Isobutane $(CH_3)_3CH$ contains a tertiary hydrogen atom.
$(CH_3)_3CH \xrightarrow{KMnO_4} (CH_3)_3COH$ (tert-butyl alcohol).
Therefore,the correct option is $(C)$.

(A) The boiling point is inversely proportional to the branching in the compound.
More branching results in a lower boiling point and higher volatility.
Among the given isomers of $C_5H_{12}$,$2,2-$dimethylpropane is the most branched,hence it has the lowest boiling point and is the most volatile.

(C) The Wurtz reaction involves the coupling of two alkyl halides in the presence of sodium metal and dry ether to form a higher alkane.
The general reaction is: $2RX + 2Na \xrightarrow[\text{Dry ether}]{} R-R + 2NaX$.
Therefore,the correct reagent is $Na$ / Dry ether.

(D) The octane number is a measure of the anti-knocking property of a fuel.
$2,2,4-$trimethylpentane,commonly known as iso-octane,is assigned an octane number of $100$ and is used as a reference standard.
Straight-chain alkanes like $n-$hexane,$n-$heptane,and $n-$pentane have much lower octane numbers compared to branched alkanes.
Therefore,$2,2,4-$trimethylpentane has the highest octane number among the given options.

(B) Reaction $(a)$ involves Grignard reagent formation followed by hydrolysis: $CH_3-CH_2-CH_2-Cl$ $\xrightarrow{Mg} CH_3-CH_2-CH_2-MgCl$ $\xrightarrow{H_2O} CH_3-CH_2-CH_3$ (Propane).
Reaction $(b)$ involves $CH_3-COCl$ reacting with $CH_3MgX$ to form acetone $(CH_3-CO-CH_3)$ or $2$-methylpropan-$2$-ol $((CH_3)_3COH)$,not propane.
Reaction $(c)$ is hydroboration followed by protonolysis: $CH_3-CH=CH_2$ $\xrightarrow{B_2H_6} (CH_3-CH_2-CH_2)_3B$ $\xrightarrow{CH_3COOH} CH_3-CH_2-CH_3$ (Propane).
Reaction $(d)$ is the reduction of an alcohol using $P/HI$: $CH_3-CH(OH)-CH_3 \xrightarrow{P/HI} CH_3-CH_2-CH_3$ (Propane).

(B) The reaction of alkanes with $Br_2$ in the presence of light (free radical substitution) depends on the number of non-equivalent hydrogen atoms.
$A$. But-$2$-ene: Contains different types of hydrogens (allylic and vinylic),leading to multiple products.
$B$. $2,2$-dimethylpropane (neopentane): All $12$ hydrogen atoms are equivalent. Replacing any one of them with $Br$ results in the same product,$1$-bromo-$2,2$-dimethylpropane. Thus,it gives only one monobrominated product.
$C$. But-$1$-yne: Contains different types of hydrogens,leading to multiple products.
$D$. Butan-$2$-ol: Contains different types of hydrogens,leading to multiple products.
Therefore,the correct answer is $B$.

(A) Kerosene is a mixture of hydrocarbons with higher boiling points compared to petrol.
Due to its higher boiling point,it is less volatile,which makes it safer to store and handle as a fuel compared to more volatile substances like petrol.

(B) When $n$-alkanes are heated with anhydrous $AlCl_3$ and $HCl$ (or $HBr$) gas,they undergo isomerization to form branched-chain alkanes.
$n$-butane isomerizes to isobutane ($2$-methylpropane).
$CH_3-CH_2-CH_2-CH_3 \xrightarrow[HBr]{AlCl_3} CH_3-CH(CH_3)-CH_3$

(B) Ethane $(C_2H_6)$ is a saturated hydrocarbon (alkane).
$(A)$ It undergoes free radical substitution with chlorine to form chloroethane.
$(B)$ Catalytic hydrogenation is a process used for unsaturated hydrocarbons (alkenes or alkynes) to convert them into alkanes. Since ethane is already saturated,it cannot be catalytically hydrogenated.
$(C)$ Complete combustion of alkanes produces $CO_2$ and $H_2O$.
$(D)$ Ethane $(C_2H_6)$ and iso-butane $(C_4H_{10})$ belong to the same homologous series (alkanes),differing by two $CH_2$ units,so they are homologues.
Therefore,the statement that is not true is $(B)$.

(A) The correct answer is $A$. Branched chain hydrocarbons are more desirable in commercial gasoline because they have higher octane ratings compared to their straight-chain isomers,which helps in reducing engine knocking.

(D) The reaction of $CH_4$ with $Cl_2$ in the presence of sunlight follows a free radical substitution mechanism.
The successive chlorination products are $CH_3Cl$, $CH_2Cl_2$, $CHCl_3$, and $CCl_4$.
Additionally, the coupling of methyl free radicals $(\cdot CH_3 + \cdot CH_3 \to CH_3-CH_3)$ leads to the formation of ethane $(CH_3CH_3)$.
Propane $(CH_3CH_2CH_3)$ is not formed in this reaction because it requires a higher carbon chain length than what is available from the reactants.

(C) The boiling point of an alkane depends on the surface area and the strength of the van der Waals forces.
As the number of carbon atoms increases,the molecular weight and surface area increase,leading to stronger van der Waals forces and a higher boiling point.
$n$-heptane $(C_7H_{16})$ has $7$ carbon atoms,$n$-butane $(C_4H_{10})$ has $4$ carbon atoms,neopentane $(C_5H_{12})$ has $5$ carbon atoms,and isobutane $(C_4H_{10})$ has $4$ carbon atoms.
Since $n$-heptane has the longest carbon chain and the highest molecular weight,it exhibits the strongest intermolecular forces.
The boiling points are: $n$-heptane $(98.4 \ ^\circ C)$,neopentane $(9.5 \ ^\circ C)$,$n$-butane $(-1 \ ^\circ C)$,and isobutane $(-11.7 \ ^\circ C)$.
Therefore,$n$-heptane has the highest boiling point.

(C) Aluminium carbide $(Al_4C_3)$ reacts with water to produce methane $(CH_4)$ gas.
The chemical reaction is:
$Al_4C_3 + 12H_2O \longrightarrow 3CH_4 + 4Al(OH)_3$

(D) $PCl_5$ reacts with compounds containing $-OH$ groups or carbonyl oxygen atoms to form chloro-derivatives.
$1. CH_3OH + PCl_5 \rightarrow CH_3Cl + POCl_3 + HCl$
$2. CH_3COOH + PCl_5 \rightarrow CH_3COCl + POCl_3 + HCl$
$3. CH_3CHO + PCl_5 \rightarrow CH_3CHCl_2 + POCl_3$
$C_2H_6$ (ethane) is an alkane,which is chemically inert towards $PCl_5$ under normal conditions because it lacks functional groups like $-OH$ or $C=O$ that can be replaced by chlorine atoms.

(C) $Hexane$ is an alkane,which is also known as paraffin due to its low reactivity.
Alkanes are non-polar and do not react with or dissolve in concentrated $H_2SO_4$.
In contrast,$Ethylene$ (alkene) reacts via addition,$Benzene$ undergoes sulfonation,and $Aniline$ (base) reacts to form a salt,making them soluble in concentrated $H_2SO_4$.

(A) . $CH_3I + 2H \xrightarrow{Zn/HCl} CH_4 + HI$ (Reduction of methyl iodide gives methane).
$2CH_3I + 2Na \xrightarrow{Dry \ Ether} CH_3-CH_3 + 2NaI$ (Wurtz reaction of methyl iodide gives ethane).

(D) Paraffin wax is a mixture of solid hydrocarbons of the general formula $C_nH_{2n+2}$.
These are long-chain alkanes,which are saturated hydrocarbons.
Therefore,the correct answer is $D$.

(D) 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 group): $Cl-CH_2-CH(CH_3)-CH_2-CH_3$. This molecule has a chiral center at $C-2$,so it exists as a pair of enantiomers.
$2$. At $C-2$: $CH_3-CCl(CH_3)-CH_2-CH_3$. This molecule is achiral.
$3$. At $C-3$: $CH_3-CH(CH_3)-CHCl-CH_3$. This molecule has a chiral center at $C-3$,so it exists as a pair of enantiomers.
$4$. At $C-4$: $CH_3-CH(CH_3)-CH_2-CH_2Cl$. This molecule is achiral.
Thus,there are $2$ chiral products,each forming an enantiomeric pair. Therefore,the total number of enantiomeric pairs is $2$.