Alkane Questions in English

(B) The reaction of $n$-butane with bromine in the presence of light is a free radical substitution reaction.
At high temperatures like $725\,^oC$,the reaction is highly selective.
Bromination is more selective than chlorination,and the reactivity order of hydrogen atoms is $3^\circ > 2^\circ > 1^\circ$.
In $n$-butane $(CH_3-CH_2-CH_2-CH_3)$,the $2^\circ$ hydrogen atoms are more reactive than the $1^\circ$ hydrogen atoms.
Therefore,the major product formed is $2$-bromobutane,which is $CH_3CH_2CH(Br)CH_3$.

(C) The reaction of alkyl halides with sodium in dry ether is known as the Wurtz reaction. When a mixture of two different alkyl halides (e.g.,$R-X$ and $R'-X$) is used,the reaction produces a mixture of three alkanes.
In this case,the reactants are ethyl bromide $(C_2H_5Br)$ and $n$-propyl bromide $(C_3H_7Br)$.
The possible products are:
$1$. Coupling of two ethyl groups: $C_2H_5-C_2H_5$ ($n$-butane)
$2$. Coupling of two $n$-propyl groups: $C_3H_7-C_3H_7$ ($n$-hexane)
$3$. Cross-coupling of ethyl and $n$-propyl groups: $C_2H_5-C_3H_7$ ($n$-pentane)
Thus,a mixture of three alkanes is obtained.

(C) The reaction is a Corey-House synthesis (alkylation of organocopper reagents).
Lithium di-$3-$pentylcuprate is $(CH_3CH_2CH(CH_2CH_3))_2CuLi$.
Ethyl bromide is $CH_3CH_2Br$.
The reaction involves the coupling of the alkyl group from the cuprate with the alkyl group from the halide:
$(CH_3CH_2CH(CH_2CH_3))_2CuLi + CH_3CH_2Br \rightarrow CH_3CH_2CH(CH_2CH_3)CH_2CH_3 + CH_3CH_2Cu + LiBr$.
The product is $CH_3CH_2CH(CH_2CH_3)CH_2CH_3$.
Naming the structure: The longest chain is $6$ carbons long (hexane),with an ethyl group at position $3$.
Thus,the structure is $3-$ethylhexane.
Wait,checking the carbon count: $3-$ethylhexane has $8$ carbons $(C_8H_{18})$.
Let's re-evaluate: The reactant is lithium di-$3-$pentylcuprate ($C_5H_{11}$ group) and ethyl bromide ($C_2H_5$ group).
$C_5H_{11} + C_2H_5 = C_7H_{16}$.
The structure formed is $CH_3CH_2CH(CH_2CH_3)CH_2CH_3$ which is $3-$ethylpentane.
$3-$ethylpentane has $7$ carbons $(C_7H_{16})$.

(D) The reduction of methyl halides (like $CH_3Cl$ or $CH_3I$) with zinc and hydrochloric acid $(Zn/HCl)$ or lithium aluminium hydride $(LiAlH_4)$ is a standard laboratory method for preparing methane. Among the given options,none represent the standard laboratory preparation method (like the reduction of methyl halides or the decarboxylation of sodium acetate). Therefore,the correct choice is $D$.

(C) Pure methane $(CH_4)$ is produced by the decarboxylation of sodium acetate with sodalime $(NaOH + CaO)$.
The reaction is: $CH_3COONa + NaOH \xrightarrow{CaO, \Delta} CH_4 + Na_2CO_3$.
Other methods like Wurtz reaction are used for higher alkanes (even number of carbon atoms),and Kolbe's electrolysis of sodium acetate produces ethane.

(B) To determine the number of monochloro derivatives,we identify the number of non-equivalent hydrogen atoms in the molecule.
$1$. $2-$methylpropane $(CH_3-CH(CH_3)-CH_3)$: It has two types of hydrogen atoms (primary and tertiary),so it gives two monochloro derivatives.
$2$. $n-$butane $(CH_3-CH_2-CH_2-CH_3)$: It has two types of hydrogen atoms (terminal and internal),so it gives two monochloro derivatives.
$3$. Benzene $(C_6H_6)$: All hydrogen atoms are equivalent,so it gives only one monochloro derivative.
$4$. $2,4-$dimethylpentane: It has three types of hydrogen atoms,so it gives three monochloro derivatives.
Since both $2-$methylpropane and $n-$butane yield two isomers,the question implies a standard selection. However,$n-$butane is the most common textbook example for this specific property.

(B) The general formula for an alkane is $C_nH_{2n+2}$.
Given molecular mass = $86 \ g/mol$.
$12n + 1(2n+2) = 86$
$14n + 2 = 86$
$14n = 84$
$n = 6$.
The hydrocarbon is hexane $(C_6H_{14})$.
The structural isomers of hexane are:
$1$. $n$-hexane $(CH_3-CH_2-CH_2-CH_2-CH_2-CH_3)$
$2$. $2$-methylpentane $(CH_3-CH(CH_3)-CH_2-CH_2-CH_3)$
$3$. $3$-methylpentane $(CH_3-CH_2-CH(CH_3)-CH_2-CH_3)$
$4$. $2,2$-dimethylbutane $(CH_3-C(CH_3)_2-CH_2-CH_3)$
$5$. $2,3$-dimethylbutane $(CH_3-CH(CH_3)-CH(CH_3)-CH_3)$
Total number of structural isomers = $5$.

(C) The given compound is $3-$ethyl$-2-$methylhexane.
First,determine the molecular formula by counting the total number of carbon atoms.
The parent chain is hexane ($6$ carbons),plus an ethyl group ($2$ carbons) and a methyl group ($1$ carbon).
Total carbons = $6 + 2 + 1 = 9$.
The general formula for an alkane is $C_nH_{2n+2}$. For $n=9$,the formula is $C_9H_{20}$.
An isomer must have the same molecular formula.
Among the options,$n-$nonane $(C_9H_{20})$ is the straight-chain isomer of $3-$ethyl$-2-$methylhexane.

(A) Neopentane is $2,2$-dimethylpropane,which has the structure $C(CH_3)_4$.
All $12$ hydrogen atoms in neopentane are equivalent because the molecule has a highly symmetrical tetrahedral structure.
Replacing any one of these $12$ hydrogen atoms with a bromine atom results in the same product,$1$-bromo-$2,2$-dimethylpropane.
Therefore,there is only $1$ possible monobromo isomer for neopentane.

(A) The restriction of rotation around a $C-C$ bond depends on the bond order and the steric hindrance caused by substituents.
$1$. $Ethane$ $(CH_3-CH_3)$ has a $C-C$ single bond,allowing free rotation.
$2$. $Ethene$ $(CH_2=CH_2)$ has a $C=C$ double bond,which restricts rotation.
$3$. $Acetylene$ $(CH \equiv CH)$ has a $C \equiv C$ triple bond,which further restricts rotation.
$4$. $Hexachloroethane$ $(CCl_3-CCl_3)$ has a $C-C$ single bond,but the large $Cl$ atoms cause significant steric hindrance,restricting rotation compared to $Ethane$.
Therefore,$Ethane$ has the least restricted rotation.

(C) An alkane is optically active if it contains at least one chiral carbon atom (a carbon atom bonded to four different groups).
$2-$ Methyl pentane: $CH_3-CH(CH_3)-CH_2-CH_2-CH_3$ (No chiral carbon).
$3-$ Methyl pentane: $CH_3-CH_2-CH(CH_3)-CH_2-CH_3$ (No chiral carbon).
$3-$ Methyl hexane: $CH_3-CH_2-CH(CH_3)-CH_2-CH_2-CH_3$. The $C_3$ atom is bonded to $-H$,$-CH_3$,$-CH_2CH_3$,and $-CH_2CH_2CH_3$. Since all four groups are different,it is chiral.
$3-$ Methyl hexane has $7$ carbon atoms,which is the lowest number among the options provided that exhibits optical activity.

(B) The boat conformation of cyclohexane is less stable than the chair conformation due to two main factors:
$1$. Torsional strain caused by eclipsing interactions of bonds on adjacent carbons.
$2$. Steric strain (van der Waals repulsion) between the two hydrogen atoms at the $C-1$ and $C-4$ positions,which are known as flagpole hydrogens.
The flagpole-flagpole interaction is the most significant source of instability in the boat conformation.

(A) According to $Baeyer's$ strain theory,the angular strain is given by the formula: $\text{Angle strain} = \frac{1}{2} (109^\circ 28' - \text{bond angle})$.
For $Cyclopentane$,the bond angle is $108^\circ$,which is very close to the tetrahedral angle $(109^\circ 28')$,resulting in low strain.
However,among the given options,$Cyclopentane$ has the smallest ring size compared to $Cyclohexane$,$Cyclooctane$,and $Cyclodecane$.
According to $Baeyer$,the deviation from the tetrahedral angle increases as the ring size deviates from the ideal $5$-membered ring (which he incorrectly assumed to be the most stable).
In the context of standard chemistry problems regarding $Baeyer$ strain,$Cyclopentane$ is often cited for its specific strain characteristics compared to larger rings.

(D) The reaction of an alkane with $Br_2$ in the presence of $hv$ (ultraviolet light) proceeds via a free radical substitution mechanism.
In the first step,the bromine molecule undergoes homolytic cleavage to form bromine radicals $(Br^{\bullet})$.
These bromine radicals then abstract a hydrogen atom from the alkane to form an alkyl radical intermediate.
For $2-\text{methylbutane}$ $(CH_3-CH(CH_3)-CH_2-CH_3)$,the abstraction of a hydrogen atom from the $C-2$ position leads to a tertiary free radical,while abstraction from the $C-3$ position leads to a secondary free radical.
The formation of $2-\text{bromo}-3-\text{methylbutane}$ involves the secondary free radical intermediate formed at the $C-3$ position.

(B) The reaction of alkyl halides with sodium in the presence of dry ether is known as the Wurtz reaction.
When a mixture of two different alkyl halides ($R-X$ and $R'-X$) is used,it results in a mixture of three alkanes:
$1$. $CH_3I + CH_3I + 2Na \xrightarrow{\text{dry ether}} CH_3-CH_3 + 2NaI$ (Ethane)
$2$. $C_2H_5I + C_2H_5I + 2Na \xrightarrow{\text{dry ether}} C_2H_5-C_2H_5 + 2NaI$ (n-Butane)
$3$. $CH_3I + C_2H_5I + 2Na \xrightarrow{\text{dry ether}} CH_3-C_2H_5 + 2NaI$ (Propane)
Thus,a total of $3$ different alkanes are formed as products.

(C) When $1, 3-$dibromopropane $(Br-CH_2-CH_2-CH_2-Br)$ is treated with metallic zinc $(Zn)$,an intramolecular dehalogenation reaction occurs.
This reaction results in the formation of a three-membered ring,which is $Cyclopropane$ $(C_3H_6)$,along with the byproduct $ZnBr_2$.

(B) Ethylene dichloride is $Cl-CH_2-CH_2-Cl$.
When it reacts with aqueous $KOH$,it undergoes nucleophilic substitution where the chlorine atoms are replaced by hydroxyl groups.
The reaction is: $Cl-CH_2-CH_2-Cl + 2KOH(aq) \rightarrow HO-CH_2-CH_2-OH + 2KCl$.
The product formed is ethylene glycol $(CH_2(OH)-CH_2OH)$.

(A) $1$. The first step is the Wurtz reaction: $2CH_3-CH(CH_3)-CH_2-Cl + 2Na \xrightarrow{\text{dry ether}} CH_3-CH(CH_3)-CH_2-CH_2-CH(CH_3)-CH_3 + 2NaCl$. The product $X$ is $2,5-\text{dimethylhexane}$.
$2$. The second step is the free radical monobromination of $2,5-\text{dimethylhexane}$ with $Br_2/h\nu$.
$3$. The structure of $2,5-\text{dimethylhexane}$ is $CH_3-CH(CH_3)-CH_2-CH_2-CH(CH_3)-CH_3$. It has primary,secondary,and tertiary hydrogens. The reactivity order for free radical substitution is $3^\circ > 2^\circ > 1^\circ$.
$4$. The tertiary hydrogens are at the $C-2$ and $C-5$ positions. Replacing a tertiary hydrogen leads to the most stable free radical intermediate,resulting in the major product: $2-\text{bromo}-2,5-\text{dimethylhexane}$.
$5$. Reviewing the options,the provided structures in the images correspond to different isomers. Based on the standard reaction path for $2,5-\text{dimethylhexane}$,the major product is $2-\text{bromo}-2,5-\text{dimethylhexane}$. Given the options provided,option $A$ represents $2-\text{bromo}-2,3-\text{dimethylbutane}$,which is not the product of this specific reaction. However,assuming the question intended to show $2,5-\text{dimethylhexane}$ as $X$,the major product is $2-\text{bromo}-2,5-\text{dimethylhexane}$.

(A) When glycerol $(CH_2OH-CHOH-CH_2OH)$ is heated with excess hydriodic acid $(HI)$,it undergoes reduction.
First,the hydroxyl groups are replaced by iodine atoms to form $1,2,3-triiodopropane$.
This compound is unstable and loses iodine to form allyl iodide $(CH_2=CH-CH_2I)$.
Further reaction with $HI$ leads to the formation of $2-iodopropane$ $(CH_3-CH(I)-CH_3)$.
Finally,the reduction of $2-iodopropane$ with $HI$ yields $2-iodopropane$ which further reduces to propane $(CH_3-CH_2-CH_3)$.

(A) Pyrolysis (or cracking) is the process of breaking down higher alkanes into lower alkanes and alkenes by heating them in the absence of air.
For example: $C_6H_{14} \xrightarrow[\Delta]{\text{Pyrolysis}} C_2H_4 + C_4H_{10}$
This is the most effective industrial method for obtaining hydrocarbons with a lower carbon number from higher ones.

(A) The Wurtz reaction involves the coupling of two alkyl halide molecules to form a symmetric alkane with an even number of carbon atoms.
$CH_4$ (methane) contains only one carbon atom.
Since the Wurtz reaction requires at least two alkyl groups to combine,it is impossible to form a single-carbon alkane like methane.
Therefore,$CH_4$ cannot be prepared by this method.

(A) The chlorination of $CH_4$ is a free radical substitution reaction.
In the dark,there is no energy source (like $h\nu$ or high temperature) to initiate the homolytic cleavage of the $Cl-Cl$ bond to produce $Cl$ free radicals.
Therefore,the reaction does not occur in the dark.
Sunlight provides the necessary energy to initiate the formation of $Cl$ free radicals,which then propagate the reaction.
Thus,both the Assertion and Reason are correct,and the Reason is the correct explanation of the Assertion.

(B) For an alkane to produce only one mono-chloro product,all hydrogen atoms in the molecule must be equivalent.
Neopentane ($2,2-$dimethylpropane) has the structure $C(CH_3)_4$.
In neopentane,all $12$ hydrogen atoms are attached to equivalent primary carbon atoms.
Therefore,substitution of any of these $12$ hydrogen atoms by a chlorine atom results in the same product,$1-$chloro$-2,2-$dimethylpropane.
Other options like $n-$pentane,isopentane,and $2,2-$dimethylbutane contain non-equivalent hydrogen atoms,leading to a mixture of isomeric mono-chloro products.

(D) The hydrocarbon $(A)$ is methane $(CH_4)$.
$CH_4$ reacts with $Br_2$ in the presence of light $(hv)$ via free radical substitution to form methyl bromide $(CH_3Br)$.
$CH_4 + Br_2 \xrightarrow{hv} CH_3Br + HBr$
Methyl bromide $(CH_3Br)$ undergoes the Wurtz reaction in the presence of sodium $(Na)$ and dry ether to form ethane $(CH_3-CH_3)$.
$2CH_3Br + 2Na \xrightarrow{\text{dry ether}} CH_3-CH_3 + 2NaBr$
Ethane $(CH_3-CH_3)$ is a gaseous hydrocarbon containing two carbon atoms,which is less than four carbon atoms.

(A) The reaction sequence is as follows:
$1$. $A$ reacts with $Br_2, h\nu$ to form a brominated product.
$2$. Treatment with $KOH$ (alc.) causes dehydrohalogenation to form an alkene.
$3$. Ozonolysis $(O_3, (CH_3)_2S)$ of the alkene cleaves the ring to form a dialdehyde.
$4$. Intramolecular aldol condensation followed by dehydration with $NaOH (aq) + \Delta$ yields the final product,which is cyclopent$-1-$ene$-1-$carbaldehyde.
$5$. Working backwards from the final product,the precursor dialdehyde is hexane$-1,6-$dial. The alkene formed after step (ii) is cyclohexene. The brominated product is bromocyclohexane. Therefore,$A$ must be cyclohexane.

(N/A) $(i) \ CH_{3}-CH_{2}-CH_{2}-CH_{2}-CH_{2}-CH_{3}$ : $n$-Hexane
$(ii) \ CH_{3}-CH(CH_{3})-CH_{2}-CH_{2}-CH_{3}$ : $2$-Methylpentane
$(iii) \ CH_{3}-CH_{2}-CH(CH_{3})-CH_{2}-CH_{3}$ : $3$-Methylpentane
$(iv) \ CH_{3}-CH(CH_{3})-CH(CH_{3})-CH_{3}$ : $2,3$-Dimethylbutane
$(v) \ CH_{3}-C(CH_{3})_{2}-CH_{2}-CH_{3}$ : $2,2$-Dimethylbutane

(N/A) $(i)$ The parent chain is heptane ($7$ carbons). Substituents are four methyl groups at positions $3, 4, 4,$ and $5$. The structure is: $CH_3-CH_2-CH(CH_3)-C(CH_3)_2-CH(CH_3)-CH_2-CH_3$
$(ii)$ The parent chain is hexane ($6$ carbons). Substituents are two methyl groups at positions $2$ and $5$. The structure is: $CH_3-CH(CH_3)-CH_2-CH_2-CH(CH_3)-CH_3$

(N/A) To prepare propane $(CH_3CH_2CH_3)$ by decarboxylation,the sodium salt of butanoic acid $(CH_3CH_2CH_2COONa)$ is required.
The chemical equation for the reaction is:
$CH_3CH_2CH_2COONa + NaOH \xrightarrow{CaO, \Delta} CH_3CH_2CH_3 + Na_2CO_3$

(N/A) The chlorination of methane proceeds via a free radical chain mechanism,which involves three steps:
Step $1$: Initiation: The reaction begins with the homolytic cleavage of the $Cl-Cl$ bond: $Cl-Cl \xrightarrow{hv} 2\overset{\centerdot }{Cl}$.
Step $2$: Propagation: Chlorine free radicals attack methane molecules to generate methyl radicals: $CH_{4} + \overset{\centerdot }{Cl} \to \overset{\centerdot }{CH_{3}} + HCl$. These methyl radicals then react with $Cl_{2}$ to form methyl chloride: $\overset{\centerdot }{CH_{3}} + Cl-Cl \to CH_{3}Cl + \overset{\centerdot }{Cl}$.
Step $3$: Termination: The chain reaction terminates when free radicals combine. Ethane is formed by the combination of two methyl radicals: $\overset{\centerdot }{CH_{3}} + \overset{\centerdot }{CH_{3}} \to CH_{3}-CH_{3}$ (Ethane). Thus,ethane is obtained as a minor by-product during the termination step.

(N/A) $1^{\circ}$ carbon atoms are bonded to only one other carbon atom. In the given structure,there are five $1^{\circ}$ carbon atoms (the terminal methyl groups),which are bonded to a total of $15 \ H$ atoms.
$2^{\circ}$ carbon atoms are bonded to two other carbon atoms. In the given structure,there are two $2^{\circ}$ carbon atoms (the methylene groups $-CH_2-$),which are bonded to a total of $4 \ H$ atoms.
$3^{\circ}$ carbon atoms are bonded to three other carbon atoms. In the given structure,there is one $3^{\circ}$ carbon atom (the methine group $-CH-$),which is bonded to $1 \ H$ atom.

(N/A) Alkanes are saturated open-chain hydrocarbons containing carbon-carbon single bonds.
$Methane$ $(CH_{4})$ is the first member of this family. If you replace one hydrogen atom of $Methane$ with a $-CH_{3}$ group,the resulting hydrocarbon is $Ethane$ $(C_{2}H_{6})$.
By continuing this process of replacing a hydrogen atom with a $-CH_{3}$ group,we can derive higher alkanes. For example,replacing a hydrogen in $Ethane$ leads to $Propane$ $(C_{3}H_{8})$,and further substitution leads to $Butane$ $(C_{4}H_{10})$ and its isomers like $Isobutane$.

(N/A) The general formula for alkanes is $C_{n}H_{2n+2}$, where $n$ is the number of carbon atoms and $(2n+2)$ is the number of hydrogen atoms in the molecule. For example, butane has $4$ carbon atoms and $10$ hydrogen atoms, so its formula is $C_{4}H_{10}$.
Bonding and structure: Methane has a tetrahedral structure. The carbon atom lies at the center, and the four hydrogen atoms lie at the four corners of a regular tetrahedron.
$(i)$ In methane, the $H-C-H$ bond angle is $109.5^{\circ}$.
(ii) In methane, the $C-H$ bond length is $112 \text{ pm}$.
(iii) The $C-H$ bonds are formed by the head-on overlapping of $sp^{3}$ hybrid orbitals of carbon and $1s$ orbitals of hydrogen atoms.
In alkanes, tetrahedra are joined together where $C-C$ and $C-H$ bond lengths are $154 \text{ pm}$ and $112 \text{ pm}$ respectively. All these are sigma $(\sigma)$ bonds. The $C-C$ sigma bond is formed by $sp^{3}-sp^{3}$ overlap, and the $C-H$ sigma bond is formed by $sp^{3}-1s$ overlap.

(N/A) The simplest cycloalkane is cyclopropane $(C_3H_6)$.
It consists of a three-membered carbon ring where each carbon is bonded to two hydrogen atoms.
Since all positions in the ring are equivalent,it has only one possible structure.
Name: Cyclopropane
Formula: $C_3H_6$
Structure: $A$ triangle with a $CH_2$ group at each vertex.

(C) The given molecule is $2$-methylbutane,which has the structure $(CH_3)_2CHCH_2CH_3$.
There are four distinct types of hydrogen atoms in this molecule:
$1$. Primary hydrogens on the terminal methyl groups attached to the $CH$ group.
$2$. The tertiary hydrogen on the $CH$ group.
$3$. The secondary hydrogens on the $CH_2$ group.
$4$. The primary hydrogens on the terminal methyl group attached to the $CH_2$ group.
Replacing each type of hydrogen with a chlorine atom gives the following four structural isomers:
$(i) (CH_3)_2CHCH_2CH_2Cl$
$(ii) (CH_3)_2CHCH(Cl)CH_3$
$(iii) (CH_3)_2C(Cl)CH_2CH_3$
$(iv) CH_3CH(CH_2Cl)CH_2CH_3$
Thus,there are $4$ possible monochloro structural isomers.

(N/A) $(i)$ To have a single monochloride,there should be only one type of $H$ atom in the isomer of the alkane of the molecular formula $C_{5}H_{12}$. This is because replacement of any $H$ atom leads to the formation of the same product. The isomer is neopentane ($2,2-$dimethylpropane).
$CH_{3}-C(CH_{3})_{2}-CH_{3}$ (neopentane)
$(ii)$ To have three isomeric monochlorides,the isomer of the alkane of the molecular formula $C_{5}H_{12}$ should contain three different types of $H$ atoms. Therefore,the isomer is $n-$pentane. It can be observed that there are three types of $H$ atoms labelled as $a, b,$ and $c$ in $n-$pentane.
$CH_{3}^{c}-CH_{2}^{b}-CH_{2}^{a}-CH_{2}^{b}-CH_{3}^{c}$ ($n-$pentane)
$(iii)$ To have four isomeric monochlorides,the isomer of the alkane of the molecular formula $C_{5}H_{12}$ should contain four different types of $H$ atoms. Therefore,the isomer is $2-$methylbutane. It can be observed that there are four types of $H$ atoms labelled as $a, b, c,$ and $d$ in $2-$methylbutane.
$CH_{3}^{a}-CH^{b}(CH_{3}^{a})-CH_{2}^{c}-CH_{3}^{d}$ ($2-$methylbutane)

(N/A) hydrocarbon with the molecular formula $C_{5}H_{10}$ belongs to the group with a general molecular formula $C_{n}H_{2n}$.
Therefore,it may either be an alkene or a cycloalkane.
Since the hydrocarbon does not react with chlorine in the dark,it cannot be an alkene (alkenes undergo addition reactions with chlorine even in the dark).
Thus,it must be a cycloalkane.
Further,the hydrocarbon gives a single monochloro compound,$C_{5}H_{9}Cl$,by reacting with chlorine in bright sunlight.
Since a single monochloro compound is formed,all hydrogen atoms in the hydrocarbon must be equivalent.
In cyclopentane,all $10$ hydrogen atoms are equivalent due to the symmetry of the ring.
Hence,the hydrocarbon is cyclopentane.

(N/A) In branched chain hydrocarbons,the alkyl group is present as a branch.
$\rightarrow$ Alkyl groups $(-R)$: The removal of one hydrogen atom from the molecule of an alkane gives an alkyl group.
$R-H \xrightarrow{-H} R-$
$\Rightarrow$ Nomenclature: $(\text{Alkane} - \text{ane} + \text{yl} = \text{Alkyl})$. These small alkyl groups are found in branched chain hydrocarbons. In a branched chain compound,small chains of carbon atoms are attached at one or more carbon atoms of the parent chain. The small carbon chains (branches) are called alkyl groups.
For example:
In a four-carbon containing parent chain,the $-CH_3$ (Methyl) alkyl group is in the branch.
In a five-carbon containing parent chain,the $-CH_3$ and $-CH_2CH_3$ groups are in the branch.
$(B)$ Nomenclature of alkyl group: In order to name such compounds,the name of the alkyl groups is prefixed to the name of the parent alkane.
$\rightarrow$ General formula of alkyl group: $C_nH_{2n+1}$
- Name of alkyl group: $(\text{Alkane} - \text{ane} + \text{yl}) = \text{Alkyl}$. One $H$ atom is less in an alkyl group than in the corresponding alkane.

Alkane (Formula,Name)	Alkyl (Formula,Name,Short Name)
$CH_4$,Methane	$CH_3-$,Methyl,$Me$
$C_2H_6$,Ethane	$CH_3CH_2-$,Ethyl,$Et$
$C_3H_8$,Propane	$C_3H_7-$,Propyl,$Pr$
$C_4H_{10}$,Butane	$C_4H_9-$,Butyl,$Bu$
$C_5H_{12}$,Pentane	$C_5H_{11}-$,Pentyl,$Pent$
$C_{10}H_{22}$,Decane	$C_{10}H_{21}-$,Decyl,$Dec$

(N/A) The general formula for the alkane series is $C_nH_{2n+2}$,where $n$ represents the number of carbon atoms. Based on this,the series from $n=1$ to $n=10$ is as follows:

$C$ atoms	Name	Formula
$1$	Methane	$CH_4$
$2$	Ethane	$C_2H_6$
$3$	Propane	$C_3H_8$
$4$	Butane	$C_4H_{10}$
$5$	Pentane	$C_5H_{12}$
$6$	Hexane	$C_6H_{14}$
$7$	Heptane	$C_7H_{16}$
$8$	Octane	$C_8H_{18}$
$9$	Nonane	$C_9H_{20}$
$10$	Decane	$C_{10}H_{22}$

(N/A) An $ \text{alkyl group} $ is a univalent radical derived from an alkane by the removal of one hydrogen atom.
The general formula for an alkane is $ C_nH_{2n+2} $.
By removing one hydrogen atom,the general formula for an alkyl group becomes $ C_nH_{2n+1} $.
These groups are often represented by the symbol $ R $.
For example,removing one hydrogen atom from methane $( CH_4 )$ results in a methyl group $( -CH_3 )$,and removing one hydrogen atom from ethane $( C_2H_6 )$ results in an ethyl group $( -C_2H_5 )$.

(N/A) In organic chemistry,the prefix $n-$ (short for normal) is used to denote a straight-chain alkane where all carbon atoms are linked in a continuous,unbranched chain.
This means that no carbon atom is attached to more than two other carbon atoms.
For example,$CH_3-CH_2-CH_2-CH_3$ is called $n$-butane.

(N/A) $1$. $Alkane$: An alkane is a saturated hydrocarbon with the general formula $C_nH_{2n+2}$. It is a stable,neutral molecule. For example,$CH_4$ is methane.
$2$. $Alkyl \text{ group}$: An alkyl group is formed by the removal of one hydrogen atom from an alkane molecule. Its general formula is $C_nH_{2n+1}$. It is a reactive radical or substituent. For example,removing one $H$ atom from methane $(CH_4)$ gives the methyl group $(-CH_3)$.

(N/A) Carbon hydrides of the type $(C_nH_{2n+2})$ are known as electron-precise hydrides.
They possess the exact number of electrons required to form the necessary covalent bonds in their structure.
Since they have no lone pairs to donate and no vacant orbitals to accept electrons,they do not exhibit any tendency to gain or lose electrons.
Therefore,they act neither as Lewis acids nor as Lewis bases.

(N/A) Catenation is the unique property of an element,particularly carbon,to form stable covalent bonds with other atoms of the same element,leading to the formation of long chains,branched structures,or rings.

(N/A) The important methods for the preparation of alkanes are as follows:
$(i)$ From unsaturated hydrocarbons (Hydrogenation):
$R-CH=CH_2 + H_2 \xrightarrow{Pt/Pd/Ni} R-CH_2-CH_3$
$R-C \equiv CH + 2H_2 \xrightarrow{Pt/Pd/Ni} R-CH_2-CH_3$
$(ii)$ From alkyl halides (Reduction):
$R-X + H_2 \xrightarrow{Zn + HCl \text{ (dil.)}} R-H + HX$
$(iii)$ Wurtz reaction:
$2R-X + 2Na \xrightarrow{\text{dry ether}} R-R + 2NaX$
$(iv)$ From carboxylic acids (Decarboxylation):
$R-COONa + NaOH \xrightarrow{CaO, \Delta} R-H + Na_2CO_3$
Soda lime is a mixture of $NaOH$ and $CaO$.

(N/A) The process of adding dihydrogen gas to alkenes and alkynes in the presence of finely divided catalysts like $Pt$,$Pd$,or $Ni$ to form alkanes is known as hydrogenation.
Catalytic Mechanism: These metals adsorb dihydrogen gas on their surfaces,which activates the $H-H$ bond. $Pt$ and $Pd$ catalyze the reaction at room temperature,whereas $Ni$ requires higher temperature and pressure.
Examples:
$(i)$ $CH_2=CH_2 + H_2 \xrightarrow{Pt/Pd \text{ or } Ni, \Delta} CH_3-CH_3$ (Ethene to Ethane)
$(ii)$ $CH_3-CH=CH_2 + H_2 \xrightarrow{Pt/Pd \text{ or } Ni, \Delta} CH_3-CH_2-CH_3$ (Propene to Propane)
$(iii)$ $CH_3-C \equiv CH + 2H_2 \xrightarrow{Pt/Pd \text{ or } Ni, \Delta} CH_3-CH_2-CH_3$ (Propyne to Propane)

Alkyl halides (except fluorides) on reduction with zinc and dilute hydrochloric acid give alkanes.
$R-X + H_2 \xrightarrow{Zn + H^+} R-H + HX$
Example of reaction:
$(i)$ $CH_3Cl + H_2 \xrightarrow{Zn + H^+} CH_4 + HCl$
Chloromethane $\rightarrow$ Methane
$(ii)$ $CH_3CH_2Cl + H_2 \xrightarrow{Zn + H^+} CH_3CH_3 + HCl$
Chloroethane $\rightarrow$ Ethane
$(iii)$ $CH_3CH_2CH_2Cl + H_2 \xrightarrow{Zn + H^+} CH_3CH_2CH_3 + HCl$
$1$-Chloropropane $\rightarrow$ Propane
$(iv)$ $CH_3CH_2Br + H_2 \xrightarrow{Zn + H^+} CH_3CH_3 + HBr$
Bromoethane $\rightarrow$ Ethane
$(v)$ $CH_3CH(Br)CH_2CH_3 + H_2 \xrightarrow{Zn + H^+} CH_3CH_2CH_2CH_3 + HBr$
$2$-Bromobutane $\rightarrow$ Butane

(N/A) Wurtz reaction: Alkyl halides,when treated with sodium metal in a dry ethereal (moisture-free) solution,yield higher alkanes. This reaction is known as the Wurtz reaction and is primarily used for the preparation of higher alkanes containing an even number of carbon atoms.
Examples:
$i$. Formation of ethane from bromomethane $(CH_3Br)$:
$2CH_3Br + 2Na \xrightarrow{\text{dry ether}} CH_3-CH_3 + 2NaBr$
$ii$. Preparation of butane from bromoethane $(C_2H_5Br)$:
$2C_2H_5Br + 2Na \xrightarrow{\text{dry ether}} C_4H_{10} (n-\text{butane}) + 2NaBr$
Limitations of the Wurtz reaction:
$i$. It is generally limited to the synthesis of symmetric alkanes.
$ii$. If a mixture of two different alkyl halides is used,a mixture of alkanes is formed,which is difficult to separate.
$iii$. It is not suitable for the preparation of alkanes with an odd number of carbon atoms in high yield due to the formation of side products.

(N/A) In the Wurtz reaction,an alkyl halide $(R-X)$ reacts with sodium in the presence of dry ether to form a symmetric alkane with double the number of carbon atoms present in the alkyl group.
If we attempt to prepare an alkane with an odd number of carbon atoms by using a mixture of two different alkyl halides ($R-X$ and $R'-X$),a mixture of three different alkanes is formed: $R-R$,$R'-R'$,and $R-R'$.
For example,to prepare propane $(C_3H_8)$,a mixture of methyl bromide $(CH_3Br)$ and ethyl bromide $(C_2H_5Br)$ is used:
$2CH_3Br + 2Na \xrightarrow{\text{Dry ether}} CH_3-CH_3 + 2NaBr$
$2C_2H_5Br + 2Na \xrightarrow{\text{Dry ether}} C_2H_5-C_2H_5 + 2NaBr$
$CH_3Br + C_2H_5Br + 2Na \xrightarrow{\text{Dry ether}} CH_3-C_2H_5 + 2NaBr$
As seen,the product is a mixture of ethane,$n$-butane,and propane. Since the boiling points of these alkanes are very close,their separation is extremely difficult and costly. Therefore,the Wurtz reaction is not preferred for the synthesis of alkanes with an odd number of carbon atoms.