Alkane Questions in English

(C) Saturated hydrocarbons (alkanes) exhibit chain isomerism due to the difference in the arrangement of carbon atoms in the chain.
This occurs because of the $(C)$ formation of branches in the chain of $C$ atoms.
For example,$n$-butane $(\text{CH}_3-\text{CH}_2-\text{CH}_2-\text{CH}_3)$ and isobutane $(\text{CH}_3-\text{CH}(\text{CH}_3)-\text{CH}_3)$ are isomers.

(C) The photochemical chlorination of an alkane proceeds via a free radical mechanism.
In the initiation step,the $Cl-Cl$ bond undergoes homolytic fission (homolysis) in the presence of light $(h
u)$ to generate chlorine free radicals $(Cl^{\bullet})$.

(C) . Marsh gas,natural gas,and coal gas contain $CH_4$,but producer gas is a mixture of $CO$ and $N_2$.
$\underset{\text{Coke, red hot}}{2C} + \underbrace{O_2 + 4N_2}_{\text{Air}}$ $\longrightarrow \underbrace{2CO + 4N_2}_{\text{Producer gas}}$

(C) The octane number is a measure of the anti-knock properties of a fuel. $n$-heptane is assigned an octane number of $0$,while $2,2,4$-trimethylpentane (Iso-octane) is assigned an octane number of $100$. Aromatic hydrocarbons and branched alkanes generally have higher octane numbers than straight-chain alkanes. Among the given options,Iso-octane has the highest octane number $(100)$.

(B) The Wurtz reaction involves the coupling of alkyl halides in the presence of sodium metal to form higher alkanes.
When a mixture of ethyl iodide $(C_2H_5I)$ and $n-$propyl iodide $(C_3H_7I)$ is used,the following products are formed:
$1$. Self-coupling of ethyl iodide: $C_2H_5-C_2H_5$ ($n-$butane).
$2$. Self-coupling of $n-$propyl iodide: $C_3H_7-C_3H_7$ ($n-$hexane).
$3$. Cross-coupling of ethyl iodide and $n-$propyl iodide: $C_2H_5-C_3H_7$ ($n-$pentane).
Therefore,$n-$propane is not formed in this reaction.

(C) The dichlorination of propane $(CH_3-CH_2-CH_3)$ leads to the formation of various isomers based on the position of chlorine atoms.
The possible isomers are:
$1,1$-dichloropropane
$1,2$-dichloropropane
$1,3$-dichloropropane
$2,2$-dichloropropane
Thus,the mixture contains $4$ isomers.

(A) Cyclopropane reacts with $Br_2$ to form an open-chain compound,$1,3-dibromopropane$.
This occurs because cyclopropane experiences significant ring strain due to its small bond angles of $60^{\circ}$,which deviate greatly from the ideal tetrahedral angle of $109.5^{\circ}$.
Consequently,the ring undergoes a ring-opening reaction with bromine,whereas larger cycloalkanes like cyclopentane and cyclohexane typically undergo substitution reactions rather than ring-opening.

(C) Heating of alkanes with fuming sulphuric acid or oleum at high temperature,which forms sulphonic acid,is called Sulphonation.
$H_2S_2O_7$ (Oleum) $\rightarrow H_2SO_4 + SO_3$
$R-H + SO_3 \rightarrow R-SO_2OH$

(A) . As the number of branches increases,the surface area of the molecule decreases.
This reduction in surface area leads to weaker $Vander \ Waals$ forces of attraction between the molecules.
Consequently,the boiling point $(B.P.)$ of branched-chain alkanes is lower than that of their corresponding straight-chain isomers.

(A) Candles consist of paraffin wax,which is a mixture of solid alkanes.
These are generally saturated hydrocarbons with a carbon chain length ranging from $C_{20}$ to $C_{40}$ (i.e.,higher saturated hydrocarbons).

(B) The reaction $CH_4 + Cl_2 \xrightarrow{uv \ light} CH_3Cl + HCl$ involves the replacement of a hydrogen atom in methane $(CH_4)$ by a chlorine atom $(Cl)$.
This type of reaction,where one atom or group of atoms in a molecule is replaced by another atom or group,is known as a substitution reaction.
Specifically,this is a free-radical substitution reaction initiated by $UV$ light.

(B) The conversion of $n$-butane to isobutane is an isomerization reaction.
This reaction is catalyzed by anhydrous $AlCl_3$ in the presence of $HCl$ gas.
$CH_3CH_2CH_2CH_3 \xrightarrow[\Delta]{Anhyd. AlCl_3 / HCl} CH_3-CH(CH_3)-CH_3$
Thus,the correct reagent is $AlCl_3$.

(A) The reaction of aluminium carbide $(Al_4C_3)$ with water $(H_2O)$ is a hydrolysis reaction that produces methane $(CH_4)$ gas and aluminium 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 $A$.

(C) The carbon-carbon bond distance depends on the hybridization of the carbon atoms.
In $Ethyne$ $(C_2H_2)$,the $C-C$ bond is a triple bond with $sp$ hybridization,having a bond length of $1.20 \ \mathring{A}$.
In $Ethene$ $(C_2H_4)$,the $C-C$ bond is a double bond with $sp^2$ hybridization,having a bond length of $1.34 \ \mathring{A}$.
In $Benzene$ $(C_6H_6)$,the $C-C$ bond length is $1.39 \ \mathring{A}$ due to resonance.
In $Ethane$ $(C_2H_6)$,the $C-C$ bond is a single bond with $sp^3$ hybridization,having a bond length of $1.54 \ \mathring{A}$.
Therefore,the maximum carbon-carbon bond distance is found in $Ethane$.

(A) The reaction $2RCOOK + 2H_2O \xrightarrow{\text{Electrolysis}} R-R + 2CO_2 + 2KOH + H_2$ is known as the Kolbe's electrolysis method.
This method is used for the preparation of alkanes (hydrocarbons) with an even number of carbon atoms.
It proceeds via a free radical mechanism and provides good yields of the symmetric alkane product $(R-R)$.

(C) Petroleum is a mixture of various hydrocarbons that can be separated by fractional distillation based on their boiling points.
According to the fractional distillation process:
$1$. Fractions with lower molecular mass and lower boiling points vaporize and are collected at the top of the column at lower temperatures.
$2$. Fractions with higher molecular mass and higher boiling points are collected at the bottom at higher temperatures.
Comparing the given options:
- Gasoline (Petrol) has a boiling point range of approximately $40-175 \ ^\circ C$.
- Kerosene has a boiling point range of approximately $175-250 \ ^\circ C$.
- Diesel oil has a boiling point range of approximately $250-350 \ ^\circ C$.
- Heavy oil (Fuel oil) has a boiling point above $350 \ ^\circ C$.
Therefore,among the given options,Gasoline has the lowest boiling point.

(B) The correct answer is $(B)$.
Decarboxylation of sodium acetate with soda lime $(NaOH + CaO)$ yields methane.
The reaction is: $CH_3COONa + NaOH \xrightarrow{CaO} CH_4 + Na_2CO_3$.

(A) According to Baeyer's strain theory,the angle strain in cycloalkanes is related to the deviation of the bond angle from the ideal tetrahedral angle of $109.5^o$.
$Cyclopropane$ has a bond angle of $60^o$,which is the largest deviation from the ideal tetrahedral angle,making it the most strained and most reactive cycloalkane.
Therefore,the correct option is $(A)$.

(C) Alkenes $(C_2H_4)$ and alkynes $(C_2H_2)$ undergo electrophilic addition reactions with $Cl_2$ even in the dark.
$CH_3CHO$ reacts with $Cl_2$ via substitution or addition depending on conditions.
$CH_4$ (methane) is an alkane and requires ultraviolet light or high temperature to initiate free radical substitution with $Cl_2$.
In the dark,there is no energy source to generate chlorine radicals,so $CH_4$ does not react with $Cl_2$.

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

(C) The reduction of alkyl halides (specifically methyl chloride) with zinc and hydrochloric acid produces methane.
$CH_3-Cl + 2H \xrightarrow{Zn/HCl} CH_4 + HCl$
$Wurtz$ reaction and $Kolbe's$ reaction are generally used to prepare alkanes with an even number of carbon atoms or higher alkanes,while hydrogenation of alkenes requires at least two carbon atoms to form ethane.

(D) Stability in alkanes is generally governed by the heat of combustion per $CH_2$ group and the degree of branching.
For straight-chain alkanes,as the chain length increases,the stability per carbon atom increases slightly due to the decrease in the relative contribution of end-group effects.
Among the given options,$n$-butane $(C_4H_{10})$ has the longest chain and the highest molecular weight,making it the most stable compared to methane $(CH_4)$,ethane $(C_2H_6)$,and propane $(C_3H_8)$.

(D) Ozonolysis is a reaction characteristic of unsaturated compounds containing multiple bonds (alkenes and alkynes).
$A$) Ethene $(CH_2=CH_2)$ is an alkene and forms an ozonide.
$B$) Propyne $(CH_3-C\equiv CH)$ is an alkyne and forms an ozonide.
$C$) Propene $(CH_3-CH=CH_2)$ is an alkene and forms an ozonide.
$D$) Propane $(CH_3-CH_2-CH_3)$ is a saturated alkane and does not contain any multiple bonds,therefore it does not undergo ozonolysis to form an ozonide.

(C) The reaction between an alkyl halide and sodium metal in the presence of dry ether is known as the $Wurtz$ reaction.
$2C_2H_5I + 2Na \xrightarrow{\text{Dry Ether}} C_4H_{10} + 2NaI$
Here,ethyl iodide $(C_2H_5I)$ reacts with sodium to form $n$-butane $(C_4H_{10})$.

(B) The process of breaking down higher hydrocarbons into lower hydrocarbons by the application of heat is known as $Cracking$ or $Pyrolysis$.
This process involves the cleavage of $C-C$ and $C-H$ bonds in alkanes to produce smaller,more volatile hydrocarbons.

(A) Successive members of the alkane homologous series differ in their molecular formula by an extra methylene group ($-CH_2-$ unit) inserted in the carbon chain.

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

(C) The correct answer is $CH_3Br$.
$1$. Methane $(CH_4)$ can be prepared by the reduction of methyl bromide $(CH_3Br)$ using $LiAlH_4$: $CH_3Br + H_2 \xrightarrow{LiAlH_4} CH_4 + HBr$.
$2$. Ethane $(CH_3-CH_3)$ can be prepared by the Wurtz reaction of methyl bromide $(CH_3Br)$ with sodium metal in dry ether: $2CH_3Br + 2Na \xrightarrow{\text{dry ether}} CH_3-CH_3 + 2NaBr$.

(D) . Photochemical chlorination of an alkane takes place via a free radical mechanism,which is initiated by the homolysis of the $Cl-Cl$ bond.
$Cl_2 \xrightarrow{hv} Cl^{\bullet} + Cl^{\bullet}$
$CH_3-CH_3 + Cl^{\bullet} \to CH_3CH_2^{\bullet} + HCl$

(C) Among alkanes,the boiling point increases with an increase in molecular weight.
For isomeric alkanes,straight-chain alkanes have a higher boiling point than branched alkanes because they have a larger surface area,leading to stronger van der Waals forces.
Comparing the given options:
$(A)$ $n-$butane $(C_4H_{10})$ has the lowest molecular weight.
$(B)$,$(C)$,and $(D)$ are isomers with the molecular formula $C_8H_{18}$.
Among these,$n-$octane is a straight-chain alkane,while iso-octane and $2, 2, 3, 3-$tetramethyl butane are branched.
Therefore,$n-$octane has the highest boiling point.

(C) The complete combustion of alkanes like $CH_4$ in the presence of excess oxygen results in the formation of carbon dioxide and water vapor.
The balanced chemical equation is:
$CH_4(g) + 2O_2(g) \to CO_2(g) + 2H_2O(l)$

(C) The knocking tendency of fuels is inversely proportional to their octane number. Straight chain alkanes have the lowest octane number and thus the highest knocking tendency. Among the given options,straight chain alkanes exhibit the highest knocking.

(A) The reaction of aluminum carbide $(Al_{4}C_{3})$ with water produces methane $(CH_{4})$ gas.
The chemical equation is: $Al_{4}C_{3} + 12H_{2}O \to 3CH_{4} + 4Al(OH)_{3}$.
In contrast,calcium carbide $(CaC_{2})$ reacts with water to produce ethyne $(C_{2}H_{2})$,while $SiC$ and $B_{4}C$ are inert to water under normal conditions.

(D) $2,2-$dimethyl propane,Pentane,and $2-$methyl butane are isomers with the molecular formula $C_5H_{12}$.
$2,2-$dimethyl butane has the molecular formula $C_6H_{14}$.
Therefore,$2,2-$dimethyl butane is the odd one out as it belongs to the $C_6$ series while others belong to the $C_5$ series.

(B) The reaction of alkanes with bromine in the presence of sunlight (or $UV$ light) proceeds via a free radical mechanism.
Bromination is highly selective compared to chlorination.
The reactivity order of hydrogen atoms towards free radical substitution is $3^\circ > 2^\circ > 1^\circ$.
In $2-$methylbutane $(CH_3-CH(CH_3)-CH_2-CH_3)$,the hydrogen atom on the $C-2$ carbon is a tertiary $(3^\circ)$ hydrogen.
Therefore,the abstraction of the $3^\circ$ hydrogen is the fastest,leading to the formation of the most stable tertiary free radical.
Consequently,the major product is $2-$bromo$-2-$methylbutane: $CH_3-C(Br)(CH_3)-CH_2-CH_3$.

(B) $2,3$-dimethylbutane has the structure $CH_3-CH(CH_3)-CH(CH_3)-CH_3$.
It contains $2$ types of equivalent hydrogen atoms: $12$ primary hydrogens (on the four $-CH_3$ groups) and $2$ tertiary hydrogens (on the two $-CH$ groups).
Therefore,it can form only $2$ monochlorinated isomers.

(A) Sabatier and Senderen's reaction involves the catalytic hydrogenation of unsaturated hydrocarbons (alkenes or alkynes) to form alkanes.
The general reaction is: $C_nH_{2n} + H_2 \xrightarrow{Ni} C_nH_{2n+2}$ or $C_nH_{2n-2} + 2H_2 \xrightarrow{Ni} C_nH_{2n+2}$.
Since the simplest alkene is ethene $(C_2H_4)$ and the simplest alkyne is ethyne $(C_2H_2)$,the minimum number of carbon atoms in the resulting alkane is $2$.
Therefore,methane $(CH_4)$ cannot be prepared by this method because it contains only one carbon atom.

(D) The octane number is a measure of the anti-knocking property of a fuel. The order of octane number is: $\text{Straight chain alkanes} < \text{Branched chain alkanes} < \text{Olefins} < \text{Cycloalkanes} < \text{Aromatic compounds}$.
Since straight chain paraffins (alkanes) have the lowest octane number,they have the highest tendency to knock in an internal combustion engine.
Therefore,the correct option is $D$.

(A) saturated cyclic compound (cycloalkane) follows the general formula $C_nH_{2n}$.
For $n = 3$,the formula is $C_3H_{2(3)} = C_3H_6$,which represents cyclopropane.
Cyclopropane is a saturated cyclic hydrocarbon because all carbon-carbon bonds are single bonds.
Therefore,the correct option is $A$.

(C) Paraffins are non-polar compounds. The intermolecular forces are weak Van der Waals forces. As the molecular mass increases,the Van der Waals forces increase. Hence,the boiling point increases.

(A) The reactivity order of halogens in free radical halogenation is $F_2 > Cl_2 > Br_2 > I_2$.
Fluorine is the most reactive halogen due to its small size and high electronegativity,making the reaction highly exothermic and often explosive.

(B) Paraffins (alkanes) are non-polar hydrocarbons.
According to the principle of "like dissolves like",non-polar substances are soluble in non-polar solvents.
Benzene $(C_6H_6)$ is a non-polar organic solvent,whereas water and methanol are polar solvents.
Therefore,paraffins are soluble in benzene.

(C) The reaction $CH_2 = CH_2 + H_2 \xrightarrow[250 - 300^{\circ}C]{Ni} CH_3 - CH_3$ is known as the Sabatier-Senderens reaction.
This process involves the catalytic hydrogenation of unsaturated hydrocarbons (alkenes or alkynes) in the presence of a metal catalyst like $Ni$,$Pt$,or $Pd$ at elevated temperatures to form alkanes.

(A) The correct answer is $A$.
$1$. Kolbe's electrolysis of concentrated aqueous solutions of sodium or potassium salts of saturated monocarboxylic acids yields higher alkanes at the anode.
$2$. In the case of salts of dicarboxylic acids (e.g.,succinic acid salts),the electrolysis yields alkenes at the anode.
$3$. Thus,depending on the starting material (monocarboxylic vs. dicarboxylic acid salts),Kolbe's reaction can yield both alkanes and alkenes.
$4$. For monocarboxylic acid: $2CH_3COONa + 2H_2O \xrightarrow{\text{electrolysis}} CH_3-CH_3 + 2CO_2 + H_2 + 2NaOH$.
$5$. For dicarboxylic acid (e.g.,sodium succinate): $NaOOC-CH_2-CH_2-COONa + 2H_2O \xrightarrow{\text{electrolysis}} CH_2=CH_2 + 2CO_2 + H_2 + 2NaOH$.

(A) The correct answer is $A$.
Members of a homologous series are not isomers of each other because they have different molecular formulas and different chain lengths.
Members of the same homologous series share the same general formula ($C_nH_{2n+2}$ for alkanes) and exhibit similar chemical properties.
The difference between any two successive members is a $-CH_2-$ group,which corresponds to a mass difference of $14 \text{ u}$ $(12 + 2 = 14)$.

(A) To determine the type of carbon atoms,we identify how many other carbon atoms are attached to each carbon:
$1^o$ (Primary): Carbon attached to $1$ other carbon.
$2^o$ (Secondary): Carbon attached to $2$ other carbons.
$3^o$ (Tertiary): Carbon attached to $3$ other carbons.
$4^o$ (Quaternary): Carbon attached to $4$ other carbons.
In the structure $CH_3-CH(CH_3)-C(CH_3)_2-CH_2-CH(CH_3)-CH_2-CH_3$:
- Primary $(1^o)$ carbons: The $6$ terminal $CH_3$ groups.
- Secondary $(2^o)$ carbons: The $2$ $CH_2$ groups.
- Tertiary $(3^o)$ carbons: The $2$ $CH$ groups.
- Quaternary $(4^o)$ carbon: The central $C$ atom attached to $4$ carbons.
Thus,the count is $6$ primary,$2$ secondary,$2$ tertiary,and $1$ quaternary carbon.

(A) The free radical chlorination of alkanes is a chain reaction that continues until all hydrogen atoms are replaced by chlorine atoms if $Cl_2$ is in excess.
To obtain the mono-substituted product $(C_2H_5Cl)$ as the major product,the alkane $(C_2H_6)$ must be taken in excess.
Therefore,the reaction $C_2H_6$ (excess) $+ Cl_2 \xrightarrow{UV \text{ Light}} C_2H_5Cl + HCl$ provides the maximum yield of the mono-chloro product.

(A) The correct option is $A$.
In the Fischer-Tropsch process,a mixture of carbon monoxide $(CO)$ and hydrogen $(H_2)$,known as water gas,is used as the raw material.
The reaction is: $nCO + (2n+1)H_2 \xrightarrow{\text{Co or Ni, heat}} C_nH_{2n+2} + nH_2O$.
This process produces a mixture of hydrocarbons,including petrol.

(A) . Hydroboration-oxidation of propene yields propan$-1-$ol,not propane.
$CH_3-CH=CH_2 \xrightarrow{B_2H_6, H_2O_2, OH^-} CH_3-CH_2-CH_2OH$.
$B$. Reduction of $1$-iodopropane with $HI/P$ gives propane: $CH_3CH_2CH_2I + HI \xrightarrow{P} CH_3CH_2CH_3 + I_2$.
$C$. Wurtz reaction of $1$-chloropropane with $Na$ gives $n$-hexane: $2CH_3CH_2CH_2Cl + 2Na \rightarrow CH_3CH_2CH_2CH_2CH_2CH_3 + 2NaCl$.
Since both $B$ and $C$ also do not produce propane as the primary product (Wurtz gives hexane,$HI/P$ gives propane),the question implies which reaction fails to yield propane. However,$B$ yields propane. Thus,$A$ and $C$ do not yield propane. Given the options,$A$ is the most direct answer for a reaction that yields an alcohol instead of an alkane.