Alkane Questions in English

(D) $n$-pentane and isopentane are both alkanes.
Isopentane $(CH_3-CH(CH_3)-CH_2-CH_3)$ contains a tertiary hydrogen atom attached to the $C-2$ carbon.
Alkanes with a tertiary hydrogen atom can be oxidized by alkaline $KMnO_4$ to the corresponding tertiary alcohol.
$n$-pentane $(CH_3-CH_2-CH_2-CH_2-CH_3)$ contains only primary and secondary hydrogen atoms and is resistant to oxidation by $KMnO_4$ under these conditions.
Therefore,$KMnO_4$ can be used to distinguish between them.

(A) In diamond,each carbon atom is $sp^3$ hybridized and is bonded to four other carbon atoms in a regular tetrahedral structure.

(C) $Tetraethyl$ lead,$(C_2H_5)_4Pb$,is used as an antiknock agent in internal combustion engines.
It helps in reducing knocking by increasing the octane rating of gasoline.

(B) The reaction of an alkane with a halogen in the presence of ultraviolet $(UV)$ light or heat is a free radical substitution reaction.
In this process,a hydrogen atom of the alkane is replaced by a halogen atom.
Since the oxidation state of the carbon atom increases (or the number of bonds to more electronegative atoms increases),this is classified as an oxidative process.

(B) The chlorination of $n$-butane $(CH_3-CH_2-CH_2-CH_3)$ involves free radical substitution.
This process replaces a hydrogen atom with a chlorine atom.
For $n$-butane,the terminal carbon atoms are equivalent,and the internal carbon atoms are equivalent.
Substitution at the terminal carbon yields $1$-chlorobutane.
Substitution at the internal carbon yields $2$-chlorobutane.
Therefore,$n$-butane gives a mixture of $1$-chlorobutane and $2$-chlorobutane.

(A) The reaction $C_3H_8 + Cl_2 \xrightarrow{\text{Light}} C_3H_7Cl + HCl$ is an example of a free-radical substitution reaction.
In this process,a hydrogen atom of the alkane $(C_3H_8)$ is replaced by a halogen atom $(Cl)$ in the presence of light.

(C) The reaction of an alkyl halide with sodium metal in the presence of dry ether is known as the Wurtz reaction.
$2C_2H_5Cl + 2Na \xrightarrow{\text{Dry Ether}} C_4H_{10} + 2NaCl$
Here,$C_2H_5Cl$ (ethyl chloride) reacts to form $C_4H_{10}$ ($n-$butane).

(C) The reaction is a Wurtz reaction involving two different alkyl halides to form an alkane.
To obtain isobutane $(CH_3-CH(CH_3)-CH_3)$,the alkyl groups involved must be an isopropyl group and a methyl group.
The reaction is:
$CH_3-CH(CH_3)-Cl + 2Na + Cl-CH_3 \xrightarrow{\text{Dry Ether}} CH_3-CH(CH_3)-CH_3 + 2NaCl$
Thus,the two chlorine compounds are isopropyl chloride and methyl chloride.

(C) The given reaction involves the coupling of two alkyl halides with sodium metal to form a higher alkane. This is a characteristic reaction known as the Wurtz reaction.
In the Wurtz reaction,two molecules of alkyl halides react with sodium in the presence of dry ether to form alkanes containing double the number of carbon atoms or a mixture of alkanes.

(D) The structure of $2, 6-\text{dimethylheptane}$ is $CH_3-CH(CH_3)-CH_2-CH_2-CH_2-CH(CH_3)-CH_3$.
To determine the number of monochlorinated derivatives,we identify the non-equivalent hydrogen atoms in the molecule.
$1$. The two terminal $CH_3$ groups at the $2$-position are equivalent.
$2$. The $CH$ group at the $2$-position is unique.
$3$. The $CH_2$ group at the $3$-position is unique.
$4$. The $CH_2$ group at the $4$-position is unique.
Due to the symmetry of the molecule,the $CH_2$ at $5$-position is equivalent to the $CH_2$ at $3$-position,the $CH$ at $6$-position is equivalent to the $CH$ at $2$-position,and the $CH_3$ at $7$-position is equivalent to the $CH_3$ at $1$-position.
Thus,there are $4$ distinct types of hydrogen atoms available for substitution,leading to $4$ monochlorinated derivatives.

(B) The reaction of $methyl \ bromide$ $(CH_3Br)$ with $zinc$ $(Zn)$ is a Wurtz-type coupling reaction.
When $2$ moles of $methyl \ bromide$ are heated with $zinc$ dust,they undergo a coupling reaction to form $ethane$ $(C_2H_6)$ and $zinc \ bromide$ $(ZnBr_2)$.
The chemical equation is: $2CH_3Br + Zn \xrightarrow{\Delta} C_2H_6 + ZnBr_2$.

(A) The reaction described is the Wurtz reaction,where alkyl halides react with $Na$ metal in dry ether to form alkanes.
To obtain $2-$methylpropane $(CH_3-CH(CH_3)-CH_3)$,the alkyl groups involved must be an isopropyl group $(CH_3-CH(CH_3)-)$ and a methyl group $(-CH_3)$.
Therefore,the reactants are $2-$chloropropane $(CH_3-CH(Cl)-CH_3)$ and chloromethane $(CH_3Cl)$.
The reaction is: $CH_3-CH(Cl)-CH_3 + CH_3Cl + 2Na \xrightarrow{\text{dry ether}} CH_3-CH(CH_3)-CH_3 + 2NaCl$.

(B) In the Wurtz reaction,two molecules of alkyl halide react with sodium metal in the presence of dry ether to form a higher alkane.
The general reaction is:
$2R-X + 2Na \xrightarrow{\text{dry ether}} R-R + 2NaX$
Therefore,the correct reagent is sodium in dry ether.

(C) The reaction of $1,2-$dibromoethane $(CH_2Br-CH_2Br)$ with metallic sodium $(Na)$ in the presence of dry ether is a Wurtz-type coupling reaction. However,the provided image shows the reaction of $1,1-$dibromoethane $(CH_3-CHBr_2)$ with sodium,which leads to the formation of $2-$butene $(CH_3-CH=CH-CH_3)$.
If the question refers to $1,2-$dibromoethane $(BrCH_2-CH_2Br)$,the reaction with $Na$ would yield ethene $(CH_2=CH_2)$ via elimination. Given the options and the provided solution image,the question likely intended to refer to $1,1-$dibromoethane.
Based on the provided solution image:
$2CH_3CHBr_2 + 4Na \xrightarrow{\text{ether}, \Delta} CH_3-CH=CH-CH_3 + 4NaBr$
Thus,the product is $2-$butene.

(C) The reaction $2CH_3Br + 2Na \to C_2H_6 + 2NaBr$ is known as the Wurtz reaction.
It is a common method used to prepare higher alkanes from alkyl halides.
Methane cannot be prepared by the Wurtz reaction because it requires at least two carbon atoms to form an alkane.

(D) The reaction of an alkyl bromide with $Na$ in dry ether is the Wurtz reaction,which doubles the alkyl chain.
$4, 5-$diethyloctane has the structure: $CH_3-CH_2-CH_2-CH(C_2H_5)-CH(C_2H_5)-CH_2-CH_2-CH_3$.
This molecule is symmetric. Splitting it at the bond between $C_4$ and $C_5$ gives two identical fragments: $CH_3-CH_2-CH_2-CH(C_2H_5)-$.
This fragment corresponds to $1-$bromopentane derivative,specifically $3-$bromopentane is not correct,but $CH_3-CH_2-CH_2-CH(Br)-CH_2-CH_3$ is $3-$bromohexane. Let us re-evaluate the structure.
$4, 5-$diethyloctane is $CH_3CH_2CH_2CH(CH_2CH_3)CH(CH_2CH_3)CH_2CH_2CH_3$.
Splitting at the center gives $CH_3CH_2CH_2CH(Br)CH_2CH_3$,which is $3-$bromohexane.
Comparing this with the options,option $(D)$ is $CH_3(CH_2)_2CH(Br)CH_2CH_3$,which is $3-$bromohexane.

(C) The correct option is $(C)$.
In the $Wurtz$ reaction,two molecules of an alkyl halide react with sodium metal in the presence of dry ether to form a higher alkane.
The general reaction is: $2R-X + 2Na \xrightarrow{\text{Dry Ether}} R-R + 2NaX$.
For example: $CH_3Br + 2Na + BrCH_3 \xrightarrow{\text{Dry Ether}} CH_3-CH_3 + 2NaBr$.

(D) The reaction involves the bromination of isobutane $(CH_3-CH(CH_3)-CH_3)$ using $CH_3OBr$ as a brominating agent,which proceeds via a free radical mechanism.
The tertiary hydrogen atom is the most reactive,leading to the formation of tert-butyl bromide $(CH_3-C(CH_3)_2-Br)$ as the intermediate.
Subsequently,the tert-butyl bromide undergoes $S_N1$ solvolysis in the presence of the solvent methanol $(CH_3OH)$ to form the major product,$2$-methoxy-$2$-methylpropane $(CH_3-C(OCH_3)(CH_3)-CH_3)$.

(B) The Wurtz reaction of methyl iodide $(CH_3I)$ produces ethane $(C_2H_6)$ as compound $X$:
$2CH_3I + 2Na \xrightarrow{\text{dry ether}} C_2H_6 + 2NaI$
Now,let us evaluate the options to see which one yields ethane $(C_2H_6)$:
Option $(A)$: $C_2H_5Cl + Mg \xrightarrow{\text{dry ether}} C_2H_5MgCl$ (Grignard reagent)
Option $(B)$: $C_2H_5Cl + 2[H] \xrightarrow{LiAlH_4} C_2H_6 + HCl$ (Reduction of haloalkane to alkane)
Option $(C)$: $C_2H_5Cl + C_2H_5ONa \xrightarrow{} C_2H_5OC_2H_5 + NaCl$ (Williamson ether synthesis)
Option $(D)$: $2CHCl_3 + 6Ag \xrightarrow{\Delta} C_2H_2 + 6AgCl$ (Formation of ethyne)
Thus,the reaction that yields $X$ (ethane) is option $(B)$.

(C) $CH_3OH$ is an acidic compound due to the presence of the hydroxyl group. $CH_3MgX$ is a Grignard reagent,which acts as a strong base.
The reaction is an acid-base reaction:
$CH_3OH + CH_3MgX \to CH_4 + CH_3OMgX$
Thus,$CH_4$ (Methane) is formed.

(B) Ethylene glycol $(HOCH_2CH_2OH)$ is a dihydric alcohol. Due to the presence of two hydroxyl groups,it forms extensive intermolecular hydrogen bonding. At room temperature $(25,^oC)$,it exists as a viscous,colorless liquid.

(D) When a mixture of methane and oxygen is passed through heated molybdenum oxide,the controlled oxidation of methane occurs to form methanal.
The chemical reaction is:
$CH_4 + O_2 \xrightarrow{\text{Mo-oxide}} HCHO + H_2O$
This method is industrially used for the manufacture of methanal (formaldehyde) by the controlled oxidation of methane in the presence of metallic oxide catalysts.

(B) The heat of combustion is directly proportional to the number of carbon and hydrogen atoms in the molecule.
For the given gases:
$CH_4$ (Methane): $1C, 4H$
$C_2H_6$ (Ethane): $2C, 6H$
$C_2H_4$ (Ethylene): $2C, 4H$
$C_2H_2$ (Acetylene): $2C, 2H$
Comparing the total number of atoms,Ethane $(C_2H_6)$ has the highest number of carbon and hydrogen atoms; therefore,it releases the maximum heat upon complete combustion.

(B) The general combustion reaction for an alkane $C_nH_{2n+2}$ is:
$C_nH_{2n+2} + (\frac{3n+1}{2}) O_2 \to n CO_2 + (n+1) H_2O$
According to the stoichiometry,the volume ratio of $O_2$ to $CO_2$ is given by:
$\frac{Volume \text{ of } O_2}{Volume \text{ of } CO_2} = \frac{\frac{3n+1}{2}}{n} = \frac{7}{4}$
Multiplying both sides:
$4(3n+1) = 14n$
$12n + 4 = 14n$
$2n = 4$
$n = 2$
Substituting $n=2$ into the general formula $C_nH_{2n+2}$,we get $C_2H_6$.

(D) Diazomethane $(CH_2N_2)$ acts as a methylating agent or a carbene source.
$1$. With $CH_3OH$ (alcohol),it forms an ether $(CH_3OCH_3)$ via insertion.
$2$. With $CH_3CH_2NH_2$ (amine),it can undergo alkylation.
$3$. With $CH_2 = CH - CH_3$ (alkene),it undergoes a $[1+2]$ cycloaddition to form a cyclopropane derivative.
$4$. With $CH_3 - CH_2 - CH_2 - CH_3$ (alkane),it does not react under standard conditions because alkanes are chemically inert towards diazomethane.

(B) The reaction of alkanes with nitric acid in the vapor phase at high temperatures $(675 \ K)$ is known as vapor phase nitration.
This process involves the cleavage of $C-C$ and $C-H$ bonds,leading to a mixture of nitroalkanes.
For ethane $(CH_3CH_3)$,the reaction is:
$CH_3CH_3 + HNO_3 \xrightarrow{675 \ K} CH_3CH_2NO_2 + CH_3NO_2 + H_2O$
Thus,the products are nitroethane and nitromethane.

(B) Alkyl magnesium halides (Grignard reagents,$R'-MgX$) are strong bases.
When they react with compounds containing active hydrogen atoms,such as primary amines $(R-NH_2)$,they abstract the proton to form the corresponding alkane $(R'-H)$.
In this specific case,the alkyl group from the Grignard reagent picks up a hydrogen atom from the amine,resulting in the formation of an alkane $(R-H)$.

(C) Isobutylmagnesium bromide is a Grignard reagent with the formula $(CH_3)_2CHCH_2MgBr$.
Grignard reagents react with alcohols $(R'OH)$ to form alkanes and magnesium alkoxy halides.
The reaction is: $(CH_3)_2CHCH_2MgBr + C_2H_5OH \rightarrow (CH_3)_2CHCH_3 + Mg(OC_2H_5)Br$.
Here,$(CH_3)_2CHCH_3$ is isobutane and $Mg(OC_2H_5)Br$ is the magnesium ethoxy bromide byproduct.

(A) chiral compound is one that contains at least one chiral center (a carbon atom bonded to four different groups).
In $2, 3, 4$-trimethylhexane,the carbon atom at position $3$ is bonded to a hydrogen atom,a methyl group $(-CH_3)$,an ethyl group $(-CH_2CH_3)$,and an isopropyl group $(-CH(CH_3)_2)$.
Since all four groups attached to the $C-3$ carbon are different,it is a chiral center.
Therefore,$2, 3, 4$-trimethylhexane is a chiral compound.

(B) The Wurtz reaction of methyl iodide $(CH_3I)$ produces ethane $(C_2H_6)$ as compound $X$.
$2CH_3I + 2Na \xrightarrow{\text{dry ether}} C_2H_6 + 2NaI$
Now,let us analyze the options:
$(A)$ $C_2H_5Cl + Mg \xrightarrow{\text{dry ether}} C_2H_5MgCl$ (Grignard reagent).
$(B)$ $C_2H_5Cl + LiAlH_4 \rightarrow C_2H_6 + HCl$ (Reduction of ethyl chloride produces ethane).
$(C)$ $C_2H_5Cl + C_2H_5ONa \rightarrow C_2H_5OC_2H_5 + NaCl$ (Williamson ether synthesis).
$(D)$ $CHCl_3 \xrightarrow{\text{Ag powder}, \Delta} C_2H_2 + 6AgCl$ (Acetylene is formed).
Therefore,option $B$ also produces ethane $(X)$.

(B) The free radical chlorination of ethane $(C_2H_6)$ is a chain reaction that proceeds through multiple steps,often leading to poly-chlorinated products like $C_2H_4Cl_2$,$C_2H_3Cl_3$,etc.
To maximize the yield of the mono-substituted product $(C_2H_5Cl)$,it is necessary to use an excess of the alkane $(C_2H_6)$.
By using an excess of $C_2H_6$,the probability of a chlorine radical colliding with an ethane molecule is significantly higher than colliding with an already formed $C_2H_5Cl$ molecule.
Therefore,option $B$ provides the correct condition for the maximum yield of $C_2H_5Cl$.

(A) Grignard reagents $(R-MgX)$ react with compounds containing active hydrogen atoms (like $H_2O$,$D_2O$,$ROH$,$RNH_2$) to form alkanes or their deuterated derivatives.
In this reaction,the alkyl group $(R)$ of the Grignard reagent abstracts the deuterium atom from $D_2O$.
The reaction is: $(CH_3)_3CMgCl + D_2O \rightarrow (CH_3)_3CD + Mg(OD)Cl$.
Thus,the product formed is $(CH_3)_3CD$.

(D) The electrolysis of a mixture of sodium ethanoate $(CH_3COONa)$ and sodium propanoate $(CH_3CH_2COONa)$ follows the Kolbe's electrolysis mechanism.
During this process,two types of alkyl radicals are formed: methyl radical $(CH_3^{\bullet})$ and ethyl radical $(CH_3CH_2^{\bullet})$.
These radicals combine in all possible ways to form alkanes:
$1$. Combination of two methyl radicals: $CH_3^{\bullet} + CH_3^{\bullet} \rightarrow CH_3CH_3$ (Ethane)
$2$. Combination of two ethyl radicals: $CH_3CH_2^{\bullet} + CH_3CH_2^{\bullet} \rightarrow CH_3CH_2CH_2CH_3$ (Butane)
$3$. Combination of one methyl and one ethyl radical: $CH_3^{\bullet} + CH_3CH_2^{\bullet} \rightarrow CH_3CH_2CH_3$ (Propane)
Therefore,all three products are obtained.

(A) For isomeric alkanes,the boiling point depends on the surface area of the molecule.
As the branching increases (moving from primary to tertiary structure),the molecule becomes more spherical in shape.
This reduction in surface area leads to weaker van der Waals forces of attraction between the molecules.
Therefore,the boiling point decreases as we move from a primary to a tertiary alkane.

(B) The chlorination of $n$-butane $(CH_3-CH_2-CH_2-CH_3)$ produces $2$-chlorobutane $(CH_3-CH_2-CHCl-CH_3)$ as one of the products.
In $2$-chlorobutane,the carbon atom at position $2$ is a chiral center because it is bonded to four different groups: $-H$,$-Cl$,$-CH_3$,and $-CH_2CH_3$.
Since the chlorination process is non-stereoselective,it produces both $d$ and $l$ enantiomers in equal amounts.
$A$ mixture containing equal amounts of $d$ and $l$ enantiomers is known as a racemic mixture.

(D) homologous series is a group of organic compounds having the same functional group and similar chemical properties,where each successive member differs by a $-CH_2-$ group.
The molecular mass difference between two consecutive members is $14 \ u$ (since $C=12$ and $H=1$,$CH_2 = 12 + 2(1) = 14 \ u$).
Option $A$ is a general formula,but in the context of defining the characteristics of a homologous series,the difference in mass is a fundamental property.
However,since the question asks for a correct statement,and $A, C,$ and $D$ are all technically correct properties of alkanes,in standard chemistry curriculum,the definition of a homologous series is primarily based on the $-CH_2-$ difference.

(B) The reaction of sodium propionate $(CH_3CH_2COONa)$ with sodalime $(NaOH + CaO)$ is a decarboxylation reaction.
In this process,the sodium salt of the carboxylic acid loses a molecule of $Na_2CO_3$ to form an alkane with one carbon atom less than the original salt.
$CH_3CH_2COONa + NaOH \xrightarrow{CaO, \Delta} CH_3CH_3 + Na_2CO_3$.
Thus,sodium propionate ($3$ carbons) yields ethane ($2$ carbons).

(D) The formation of only one monochloroalkane indicates that all hydrogen atoms in the alkane are equivalent.
In $Neopentane$ $(2,2-dimethylpropane)$,all $12$ hydrogen atoms are equivalent.
When $Neopentane$ reacts with $Cl_2$ in the presence of ultraviolet light,it replaces any one of the equivalent hydrogen atoms to form $1-chloro-2,2-dimethylpropane$ as the only product.
Therefore,the correct answer is $Neopentane$.

(A) The given reaction is $2CH_3Br + 2Na \rightarrow CH_3-CH_3 + 2NaBr$.
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.

(A) In $Kolbe's$ electrolysis,the electrolysis of concentrated aqueous solutions of sodium or potassium salts of carboxylic acids yields alkanes at the anode.
However,if the reaction conditions or the nature of the alkyl group allow for disproportionation or secondary reactions,alkenes can also be formed as side products.
Specifically,the reaction involves the formation of free radicals which can undergo coupling to form alkanes or disproportionation to form alkenes and alkanes.
Therefore,$Kolbe's$ electrolysis is the reaction among the given options where both alkanes and alkenes can be obtained.

(A) $C.N.G$ stands for Compressed Natural Gas.
It primarily consists of methane $(CH_4)$ which typically makes up about $80-90\%$ of its composition.
The remaining components include small amounts of ethane,propane,butane,and other higher alkanes.
Therefore,the correct composition is $CH_4$ + propane + butane + higher alkanes $(84\%)$.

(A) The reduction of alkyl halides with zinc and hydrochloric acid $(Zn/HCl)$ is a standard method for the preparation of alkanes.
The reaction is: ${C_2H_5Cl + 2[H] \xrightarrow{Zn/HCl} C_2H_6 + HCl}$.
Thus,$x$ is ethane $(C_2H_6)$.

(A) The boiling point of alkanes increases with an increase in the number of carbon atoms in the chain.
This is because as the molecular size increases,the surface area of the molecule increases,which leads to stronger van der Waals forces of attraction between the molecules.
Consequently,more energy is required to overcome these forces,resulting in a higher boiling point.

(D) The photobromination of alkanes involves the formation of free radicals. The stability of free radicals follows the order: $3^\circ > 2^\circ > 1^\circ$.
In $2$-methylbutane $(CH_3-CH(CH_3)-CH_2-CH_3)$,there are four types of hydrogen atoms.
Abstraction of the hydrogen atom from the $C-2$ position leads to a $3^\circ$ free radical,which is the most stable.
Therefore,the reaction proceeds via the $3^\circ$ free radical intermediate to form $2$-bromo-$2$-methylbutane as the major product.

(B) Alkanes do not undergo addition reactions because they are saturated hydrocarbons and do not contain $C=C$ or $C\equiv C$ bonds.

(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 stability order of alkyl radicals is $3^\circ > 2^\circ > 1^\circ$.
In the reactant $H_3C-CH(D)-CH(CH_3)-CH_3$,the hydrogen at the $C_2$ position (the carbon attached to the methyl group) is a tertiary hydrogen $(3^\circ)$.
Removing this hydrogen creates a $3^\circ$ radical,which is the most stable.
Thus,the major product '$X$' is $H_3C-CH(D)-C^\bullet(CH_3)-CH_3$.

(D) Methane $(CH_4)$ can be prepared by the following methods:
$1$. Decarboxylation: Heating sodium acetate $(CH_3COONa)$ with soda lime $(NaOH + CaO)$ yields methane $(CH_3COONa + NaOH \xrightarrow{\Delta, CaO} CH_4 + Na_2CO_3)$.
$2$. Hydrogenation: Catalytic hydrogenation of carbon monoxide $(CO + 3H_2 \xrightarrow{Ni, 573K} CH_4 + H_2O)$.
$3$. Wurtz reaction: While Wurtz reaction is typically used for higher alkanes,it can produce methane if methyl halides are reacted with sodium in the presence of a suitable catalyst or specific conditions,though it is less common for methane specifically. However,since decarboxylation and hydrogenation are standard methods for methane,and the question asks for preparation methods,'All of the above' is the most appropriate choice.

(B) Marsh gas is primarily composed of methane $(CH_4)$.
It is produced by the anaerobic decomposition of organic matter in wetlands,marshes,and swamps.
Therefore,the correct option is $B$.

(B) In alkanes, the carbon atom is $sp^3$ hybridized.
Due to $sp^3$ hybridization, the geometry is tetrahedral.
The $H - C - H$ bond angle in a perfect tetrahedron is $109^\circ 28'$.
The $C - H$ bond length in alkanes is approximately $112 \, pm$ (or $1.12 \, $ $\mathring{A}$).
Therefore, the correct values are $112 \, pm$ and $109^\circ 28'$.

(A) $1$. Hydroboration-oxidation of propene $(CH_3-CH=CH_2)$ yields propan$-1-$ol,not propane.
$2$. Reduction of $CH_3CH_2CH_2I$ with $HI/P$ yields propane $(CH_3CH_2CH_3)$.
$3$. Wurtz reaction of $CH_3CH_2CH_2Cl$ with $Na$ yields hexane $(CH_3CH_2CH_2CH_2CH_2CH_3)$.
Since both options $A$ and $C$ do not produce propane,the question implies identifying which reaction fails to produce it. However,based on standard chemistry,hydroboration-oxidation $(A)$ produces alcohol,and Wurtz reaction $(C)$ produces a higher alkane. Thus,both $A$ and $C$ do not produce propane.