Class 11 Chemistry MCQs | Chapter 13: Hydrocarbons – Part 1

Explanation: The name Hydrocarbons itself is derived from the two constituent elements: Hydrogen and Carbon. By definition, hydrocarbons are organic compounds consisting exclusively of hydrogen and carbon atoms. All other organic compounds can be considered derivatives of hydrocarbons where one or more hydrogen atoms are replaced by functional groups containing elements like oxygen, nitrogen, sulfur, or halogens. Therefore, they form the parent class for virtually all organic compounds.

Explanation: Alkanes are saturated open-chain hydrocarbons, meaning they contain only carbon-carbon single bonds and have the maximum possible number of hydrogen atoms. The general formula, $C_nH_{2n+2}$, where $n$ is the number of carbon atoms, correctly accounts for this saturation. For example, for $n=2$ (Ethane), the formula is $C_2H_{2(2)+2} = C_2H_6$. The formulas $C_nH_{2n}$ (Alkenes) and $C_nH_{2n-2}$ (Alkynes) represent unsaturated hydrocarbons.

Explanation: Ethyne ($H-Cequiv C-H$) contains a carbon-carbon triple bond. A carbon atom involved in a triple bond forms two sigma ($sigma$) bonds (one with the other carbon and one with a hydrogen) and two pi ($pi$) bonds. To accommodate the two $sigma$ bonds, the carbon atom uses $sp$ hybridization, which involves mixing one $s$ and one $p$ orbital to form two $sp$ hybrid orbitals. The remaining two unhybridized $p$ orbitals form the two $pi$ bonds. This results in a linear geometry.

Explanation: The structure of Propene is ${C}^{(1)}{H}_3-{C}^{(2)}{H}={C}^{(3)}{H}_2$.

1. ${C}^{(1)}{H}_3$ group has $3$ ${C}-{H}$ $sigma$ bonds.

2. The ${C}^{(1)}-{C}^{(2)}$ single bond is $1$ $sigma$ bond.

3. The ${C}^{(2)}={C}^{(3)}$ double bond consists of $1$ $sigma$ bond and $1$ $pi$ bond.

4. ${C}^{(2)}{H}$ group has $1$ ${C}-{H}$ $sigma$ bond.

5. ${C}^{(3)}{H}_2$ group has $2$ ${C}-{H}$ $sigma$ bonds.

Total $sigma$ bonds: $3 + 1 + 1 + 1 + 2 = 8$. Total $pi$ bonds: $1$.

Explanation: Alkanes are characterized by having only single covalent bonds and are generally quite unreactive towards most common reagents like acids, bases, or oxidizing agents under ordinary conditions. The older term Paraffins comes from the Latin words ‘parum’ (meaning little) and ‘affinis’ (meaning affinity), reflecting this low chemical reactivity. Alkenes are called Olefins, and Alkynes do not have a separate common group name like this.

Explanation: Chain isomers are compounds that share the same molecular formula but possess a different arrangement of the carbon skeleton (straight chain vs. branched chain). Both $n$-Butane and Isobutane (2-Methylpropane) have the molecular formula $C_4H_{10}$. $n$-Butane has a straight chain of four carbons, while Isobutane has a branched chain with a methyl group on the second carbon. Propane and the $C_4H_8$ compounds (But-1-ene, Cyclobutane) have different molecular formulae, so they cannot be isomers of Butane.

Explanation: Toluene is the common, historically accepted name for the simplest alkyl-substituted aromatic hydrocarbon. In $IUPAC$ nomenclature, the name is derived by considering the methyl group (${CH}_3$) as a substituent attached to the parent benzene ring. Therefore, its correct systematic name is Methylbenzene. 1,2-Dimethylbenzene is $o$-Xylene, and 1,4-Dimethylbenzene is $p$-Xylene, both of which are different compounds with the formula $C_6H_4({CH}_3)_2$.

Explanation: Hydrocarbons are classified into Saturated (only single bonds, ${C}_n{H}_{2n+2}$) and Unsaturated. Unsaturated hydrocarbons are those that contain at least one ${C}-{C}$ multiple bond (double or triple bond). The presence of a double bond (Alkenes) or triple bond (Alkynes) means the molecule contains fewer than the maximum number of hydrogen atoms possible, and can thus undergo addition reactions to become saturated. Aromatic hydrocarbons are a special subclass of unsaturated cyclic compounds.

Explanation: A homologous series is a group of compounds where consecutive members differ by a fixed structural unit. For the Alkane series ($CH_4, C_2H_6, C_3H_8, dots$), the difference between Methane ($CH_4$) and Ethane ($C_2H_6$) is $CH_2$. Similarly, the difference between Ethane ($C_2H_6$) and Propane ($C_3H_8$) is also $CH_2$. This fixed difference, known as the methylene group ($CH_2$), results in a constant increase in molecular mass of $14 { u}$ (${atomic mass units}$) for each successive member.

Explanation: This is a calculation-based question using the general formula for a straight-chain alkane. The general formula is $C_nH_{2n+2}$, where ‘n’ is the number of carbon atoms.

Here, $n=8$.

Number of Hydrogen atoms $= 2n + 2 = 2(8) + 2$

Number of Hydrogen atoms $= 16 + 2 = 18$.

The resulting compound is ${Octane}$ with the molecular formula $C_8H_{18}$. The options 14 and 16 would correspond to hydrocarbons with degrees of unsaturation, such as Alkynes or Alkenes/Cycloalkanes, respectively.

Explanation: In alkanes, every carbon atom is ${sp}^3$ hybridized, which gives the molecule a tetrahedral geometry. The ideal bond angle in a perfect tetrahedron is $109^circ 28’$ (or $109.5^circ$). This angle minimizes the repulsive forces between the four bond pairs of electrons surrounding the central carbon atom. Bond angles of $120^circ$ are found in ${sp}^2$ hybridized atoms (alkenes), and $180^circ$ are found in ${sp}$ hybridized atoms (alkynes).

Explanation: The molecular formula $C_6H_{14}$ corresponds to Hexane. The five possible structural isomers (chain isomers) are:

$n$-Hexane (straight chain)

2-Methylpentane

3-Methylpentane

2,2-Dimethylbutane

2,3-Dimethylbutane

These isomers have the same molecular formula but differ in the arrangement of the carbon skeleton, making the total number of non-stereoisomeric structures five.

Explanation: Decarboxylation is a method for preparing alkanes where the carbon chain is shortened by one carbon atom. The reaction involves heating a sodium salt of a carboxylic acid (${RCOONa}$) with a mixture of solid sodium hydroxide (${NaOH}$) and calcium oxide (${CaO}$), commonly known as soda lime. The ${CaO}$ is used to keep the ${NaOH}$ dry and prevent it from melting or absorbing atmospheric moisture. The general reaction is ${RCOONa} + {NaOH} xrightarrow{{CaO, heat}} {R}-{H} + {Na}_2{CO}_3$.

Explanation: Alkanes are saturated and contain strong ${C}-{H}$ bonds, making them largely unreactive towards ionic reagents. The halogenation of alkanes requires high temperature or UV light (${h}nu$) to break the halogen molecule (${X}_2$) via homolytic fission, creating highly reactive free radicals (${X}bullet$). These radicals then initiate a chain reaction, substituting a hydrogen atom of the alkane with a halogen atom. The entire process follows a free-radical substitution mechanism, which proceeds through three steps: chain initiation, chain propagation, and chain termination.

Explanation: Isomeric alkanes have the same molecular formula and molar mass, but different structures. Increased branching causes the molecule to become more spherical or compact. A more compact structure leads to a smaller surface area, which in turn reduces the extent of the weak, attractive van der Waals forces (dispersion forces) between molecules. Since less energy is required to overcome these reduced forces, the boiling point decreases as the degree of branching increases.

Explanation: The complete combustion of an alkane follows the general reaction: $C_n H_{2n+2} + left(dfrac{3n+1}{2}right)O_2 rightarrow nCO_2 + (n+1)H_2O$.

For Propane ($n=3$):

$C_3H_8 + left(dfrac{3(3)+1}{2}right)O_2 rightarrow 3CO_2 + (3+1)H_2O$

$C_3H_8 + 5O_2 rightarrow 3CO_2 + 4H_2O$

The stoichiometric coefficient of $O_2$ is $5$. The stoichiometric coefficient of $CO_2$ is $3$.

The sum of the coefficients of $O_2$ and $CO_2$ is $5 + 3 = 8$.

Explanation: The general equation for the complete combustion of an alkane is:

$C_n H_{2n+2} + left(dfrac{3n+1}{2}right)O_2 rightarrow nCO_2 + (n+1)H_2O$

For Pentane ($n=5$):

$C_5H_{12} + left(dfrac{3(5)+1}{2}right)O_2 rightarrow 5CO_2 + (5+1)H_2O$

The balanced equation is: $C_5H_{12} + 8O_2 rightarrow 5CO_2 + 6H_2O$.

The stoichiometric coefficient of $O_2$ is $8$, and the coefficient of $CO_2$ is $5$.

Therefore, the sum is $8 + 5 = 13$.

Explanation: The staggered conformation of ethane is the most stable form. In this conformation, the hydrogen atoms on the front carbon are positioned exactly in the spaces between the hydrogen atoms on the back carbon when viewed along the ${C}-{C}$ axis. This arrangement maximizes the distance between the electron clouds of the ${C}-{H}$ bonds on adjacent carbons, thereby minimizing the repulsive interactions known as torsional strain. The eclipsed conformation, where the hydrogen atoms are directly aligned, has the highest energy (maximum strain).

Explanation: The Wurtz reaction involves reacting two molecules of an alkyl halide (${R}-{X}$) with sodium metal in dry ether to form a symmetrical alkane (${R}-{R}$), for example: $2{CH}_3{Cl} + 2{Na} rightarrow {CH}_3{CH}_3 + 2{NaCl}$. However, if a mixture of two different alkyl halides (${R}-{X}$ and ${R}’-{X}$) is used, three possible products are formed (${R}-{R}$, ${R}’-{R}’$, and ${R}-{R}’$). This resulting mixture of alkanes is difficult to separate due to their close boiling points, making the reaction impractical for preparing unsymmetrical alkanes.

Explanation: Cracking (or Pyrolysis) is the thermal decomposition of higher molecular mass alkanes into smaller molecular mass hydrocarbons. This process is usually carried out by heating the alkanes to high temperatures (about $700-800 { K}$) in the absence of air. This process is highly significant in the petroleum industry as it converts heavy oil fractions (higher alkanes) into valuable volatile fuels like gasoline (which consists of lower alkanes and alkenes). The ${C}-{C}$ bonds are broken, following a free-radical mechanism.

Explanation: The two carbon atoms forming a carbon-carbon double bond (${C}={C}$) in an alkene are both ${sp}^2$ hybridized. This hybridization involves the mixing of one ${s}$ and two ${p}$ orbitals, resulting in three ${sp}^2$ hybrid orbitals that lie in one plane with a bond angle close to $120^circ$. The remaining unhybridized ${p}$ orbital on each carbon overlaps laterally to form the $pi$ bond, which is essential for the double bond structure.

Explanation: The ${C}={C}$ double bond in alkenes consists of one $sigma$ bond and one $pi$ bond. Due to the presence of the $pi$ bond and the higher ${s}$-character (33.3%) of the ${sp}^2$ hybrid orbitals compared to ${sp}^3$ (25%), the bond is stronger and shorter than a ${C}-{C}$ single bond. The standard ${C}={C}$ bond length is approximately $1.34 { Angstroms}$ ($134 { pm}$). $1.54 { Å}$ is the ${C}-{C}$ single bond length, and $1.20 { Å}$ is the ${C}equiv {C}$ triple bond length.

Explanation: The $pi$ component of the ${C}={C}$ double bond restricts the rotation of the carbon atoms, holding the substituents in fixed positions relative to the double bond. If each carbon atom of the double bond is attached to two different groups, the molecule can exist as two distinct spatial isomers: the cis isomer (identical groups on the same side) and the trans isomer (identical groups on opposite sides). This phenomenon is known as geometrical or cis-trans isomerism.

Explanation: Geometrical isomerism requires that each carbon atom of the double bond must be attached to two different groups. In 1-Butene (${CH}_2={CHCH}_2{CH}_3$), the first carbon (${C}1$) is attached to two identical hydrogen atoms. Since the two groups on ${C}1$ are the same, cis-trans isomerism is impossible. All other options, 2-Butene (${CH}_3{CH}={CHCH}_3$), 1,2-Dichloroethene (${CHCl}={CHCl}$), and 2-Pentene (${CH}_3{CH}={CHCH}_2{CH}_3$), can exist as cis and trans isomers.

Explanation: The addition of an unsymmetrical reagent like Hydrogen Bromide (${HBr}$) to an unsymmetrical alkene like Propene is an electrophilic addition reaction. This reaction strictly follows Markownikov’s rule, which states that the negative part of the adding molecule (the ${Br}^-$ ion in this case) attaches to the carbon atom of the double bond that has the least number of hydrogen atoms. Thus, the major product is 2-Bromopropane (${CH}_3{CH}({Br}){CH}_3$).

Explanation: Ozonolysis cleaves the ${C}={C}$ double bond and forms carbonyl compounds. The reaction can be written as:

$${CH}_3{CH}={CHCH}_3 xrightarrow[{Ozonide}]{{O}_3} {CH}_3{CHO} + {CH}_3{CHO}$$

Since the double bond is between ${C}2$ and ${C}3$, the molecule splits exactly in the middle. The products are two identical molecules of ${Ethanal}$ (${CH}_3{CHO}$), also known as Acetaldehyde. The ${Zn}/{H}_2{O}$ step is a reductive workup that prevents the further oxidation of the aldehyde product to a carboxylic acid.

Explanation: Bayer’s test uses cold, dilute, and alkaline Potassium Permanganate (${KMnO}_4$) as a mild oxidizing agent. When reacted with an alkene, it results in the ${cis}$-dihydroxylation of the double bond, meaning two hydroxyl (${-OH}$) groups are added across the double bond. The resulting product is a vicinal diol, commonly called a glycol. The purple color of ${KMnO}_4$ disappears, and a brown precipitate of ${MnO}_2$ forms, which is used as a test for unsaturation.

Explanation: Addition polymerization is a key reaction of alkenes where many small monomer molecules containing a double bond, such as Ethene (${CH}_2={CH}_2$), add up to form a large, single chain molecule without the elimination of any byproducts. The reaction for Ethene is:

$$n({CH}_2={CH}_2) xrightarrow{{catalyst, heat, pressure}} (-{CH}_2-{CH}_2-)_n$$

This process forms the polymer Polyethylene (Polythene), which is a common plastic. The other options are substitution or simple addition reactions.

Explanation: The conversion of an alkene to an alkane is called catalytic hydrogenation. It involves the addition of a hydrogen molecule (${H}_2$) across the ${C}={C}$ double bond, saturating the compound. The reaction is typically carried out by heating the alkene with hydrogen gas at high pressure over a finely divided catalyst like Nickel (${Ni}$), Platinum (${Pt}$), or Palladium (${Pd}$). The general reaction is: ${Alkene} + {H}_2 xrightarrow{{Ni}/{Pt}/{Pd}} {Alkane}$.

Explanation: 2-Methylpropene is an unsymmetrical alkene. The addition of ${HCl}$ (where ${H}^+$ is the electrophile and ${Cl}^-$ is the nucleophile) follows Markownikov’s rule. The ${H}^+$ adds to the carbon of the double bond with more hydrogen atoms (${C}2$), leading to the formation of the more stable tertiary carbocation (${C}^+$ on ${C}1$). The ${Cl}^-$ then attacks the tertiary carbocation at ${C}1$.

The double bond is between ${C}1$ (${CH}_2$) and ${C}2$ (${C}$). ${C}1$ has two ${H}$ atoms; ${C}2$ has zero ${H}$ atoms. ${Cl}$ adds to ${C}2$ (least ${H}$’s), and ${H}$ adds to ${C}1$ (most ${H}$’s). The product is 2-Chloro-2-methylpropane.

$${(CH}_3)_2{C}={CH}_2 + {HCl} rightarrow {(CH}_3)_2{C}({Cl}){CH}_3$$

Explanation: Alkynes, due to the presence of a triple bond, have four fewer hydrogen atoms than the corresponding saturated alkane ($C_nH_{2n+2}$) with the same number of carbon atoms. The formula $C_nH_{2n-2}$ indicates this unsaturation (two degrees of unsaturation). For instance, Ethyne ($n=2$) is $C_2H_{2(2)-2} = C_2H_2$. The formula $C_nH_{2n}$ is for alkenes, and $C_nH_{2n+2}$ is for alkanes.

Explanation: The carbon-carbon triple bond in alkynes is formed by one $sigma$ bond and two $pi$ bonds, involving ${sp}$ hybridized carbon atoms. The ${sp}$ orbital has the highest ${s}$-character (50%) among ${sp}^3$, ${sp}^2$, and ${sp}$. Increased ${s}$-character results in stronger and shorter bonds. Consequently, the ${C}equiv {C}$ bond is the shortest ${C}-{C}$ bond, measuring approximately $1.20 { Angstroms}$ ($120 { pm}$).

Explanation: The two carbon atoms forming the triple bond in an alkyne are ${sp}$ hybridized. The ${sp}$ hybrid orbitals are directed away from each other at an angle of $180^circ$ to minimize electron repulsion. This $180^circ$ bond angle forces the substituents attached to the triple bond carbons to lie along a straight line, resulting in a linear geometry for the atoms directly involved in the triple bond.

Explanation: The acidity of terminal alkynes (${R}-{C}equiv {C}-{H}$) arises because the carbon atom is ${sp}$ hybridized, possessing 50% ${s}$-character. Orbitals with higher ${s}$-character are held closer to the nucleus, making the ${sp}$ hybridized carbon highly electronegative. This strong electronegativity pulls electron density away from the ${C}-{H}$ bond, making the hydrogen atom slightly positive and easier to remove as a proton (${H}^+$), thus exhibiting weak acidity.

Explanation: Similar to alkenes, alkynes react with cold, dilute, alkaline ${KMnO}_4$ (Bayer’s reagent) via oxidation and ${cis}$-dihydroxylation. The triple bond is only partially oxidized, and the initial addition of two hydroxyl groups leads to the formation of a diol, which rapidly tautomerizes. The overall mild oxidation of Ethyne (${HC}equiv {CH}$) initially forms an unstable dihydroxy alkene, which further reacts to form Ethane-1,2-diol (Glycol), often via a transient diketone intermediate depending on reaction conditions.

Explanation: The complete hydrogenation of an alkyne to an alkane uses ${Pt}$ or ${Ni}$ and is non-selective. To stop the reduction at the alkene stage, a deactivated palladium catalyst known as Lindlar’s catalyst (${Pd}/{BaSO}_4$ poisoned with sulfur or quinoline) is used. This catalyst promotes the ${syn}$-addition of hydrogen, resulting exclusively in the formation of the cis-alkene isomer.

Explanation: Propyne is an unsymmetrical alkyne. According to Markownikov’s rule, the negative part of ${HBr}$ (${Br}^-$) adds to the carbon atom of the triple bond that has the least number of hydrogen atoms, and ${H}^+$ adds to the carbon with more ${H}$ atoms. In propyne, the carbon with zero ${H}$ atoms is ${C}2$ and the one with one ${H}$ is ${C}3$. Thus, the ${Br}$ atom attacks ${C}2$, yielding the vinyl halide 2-Bromopropene (${CH}_3{C}({Br})={CH}_2$) as the major product.

Explanation: This is a key industrial reaction for the synthesis of aromatic compounds. When Ethyne gas is passed through a red-hot iron tube at $873 { K}$, three molecules of Ethyne undergo cyclic polymerization (trimerization). This reaction causes the triple bonds to open up and form three new $sigma$ bonds between the molecules, resulting in the formation of the aromatic compound Benzene (${C}_6{H}_6$).

Explanation: Reduction using ${Na}$ in liquid ${NH}_3$ is a standard method for the selective conversion of an alkyne to an alkene. Unlike Lindlar’s catalyst, this method involves a free-radical mechanism that favors the formation of the more stable trans-alkene isomer. The reduction proceeds through the addition of the two hydrogen atoms from opposite sides of the triple bond (${anti}$-addition), leading to the less sterically hindered trans-product.

Explanation: To count the bonds, we look at the structure ${C}^{(1)}{H}_3-{C}^{(2)}equiv {C}^{(3)}{H}$:

1. ${C}^{(1)}{H}_3$ group has $3$ ${C}-{H}$ $sigma$ bonds.

2. The ${C}^{(1)}-{C}^{(2)}$ single bond is $1$ $sigma$ bond.

3. The ${C}^{(2)}equiv {C}^{(3)}$ triple bond consists of $1$ $sigma$ bond and $2$ $pi$ bonds.

4. The ${C}^{(3)}-{H}$ bond is $1$ $sigma$ bond.

Total $sigma$ bonds: $3 + 1 + 1 + 1 = 6$.

Total $pi$ bonds: $2$.

Explanation: This is a cyclic polymerization reaction. When Ethyne is passed through a red-hot iron tube, three molecules of ethyne polymerize to form the aromatic compound Benzene. This reaction is significant as it links aliphatic hydrocarbons (open chain) to aromatic hydrocarbons (closed chain).

$$3{CH}equiv{CH} xrightarrow{{Red hot Fe tube, 873 K}} {C}_6{H}_6$$

Explanation: But-1-yne is a terminal alkyne (${CH}_3{CH}_2{C}equiv{CH}$), meaning it possesses an acidic hydrogen atom attached to the $sp$-hybridized carbon. This acidic hydrogen reacts with Ammoniacal Silver Nitrate to form a white precipitate of Silver Butynide. But-2-yne is an internal alkyne (${CH}_3{C}equiv{CCH}_3$) and lacks this acidic hydrogen, so it does not react.

Explanation: Both compounds share the same molecular formula, ${C}_4{H}_6$ (General formula ${C}_nH_{2n-2}$). However, they possess different functional groups/unsaturation patterns. But-1-yne contains a triple bond, classifying it as an Alkyne. Buta-1,3-diene contains two double bonds, classifying it as a Diene. Isomers with the same formula but different functional groups are Functional Isomers.

Explanation: The hydration of Propyne follows Markownikov’s rule. The water adds across the triple bond to form an unstable enol intermediate (${CH}_3{C(OH)}={CH}_2$). This enol immediately undergoes tautomerization (rearrangement of protons and electrons) to form the more stable ketone, Propanone (${CH}_3{COCH}_3$). Aldehydes are only formed from Ethyne in this reaction.

Explanation: Alkenes and Alkynes decolorize bromine water (test for unsaturation). Ethyne reacts with two equivalents of ${Br}_2$. The first equivalent forms 1,2-Dibromoethene. The second equivalent adds across the remaining double bond to form the saturated compound 1,1,2,2-Tetrabromoethane.

$${HC}equiv{CH} + 2{Br}_2 rightarrow {CHBr}_2-{CHBr}_2$$

Explanation: Benzene is a planar, cyclic molecule. To maintain the flat hexagonal shape and allow for the delocalization of $pi$-electrons (resonance), every carbon atom in the ring must be $sp^2$ hybridized. This leaves one unhybridized $p$-orbital on each carbon perpendicular to the ring to form the $pi$-cloud.

Explanation: Naphthalene consists of two fused benzene rings and contains a total of 5 double bonds. Each double bond contributes 2 $pi$ electrons, giving a total of $10$ $pi$ electrons. According to Hückel’s rule:

$$4n + 2 = 10$$

$$4n = 8 implies n = 2$$

Therefore, the integer value is 2.

Explanation: Dimethylbenzene contains a benzene ring with two methyl substituents. The position of the second methyl group relative to the first determines the isomer. There are three possible positions:

1. 1,2-Dimethylbenzene (ortho-xylene)

2. 1,3-Dimethylbenzene (meta-xylene)

3. 1,4-Dimethylbenzene (para-xylene)

Explanation: Resonance energy is a measure of the extra stability a molecule gains through electron delocalization. A high resonance energy ($150 { kJ mol}^{-1}$) implies that Benzene is significantly more stable (lower in potential energy) than the hypothetical structure with localized double bonds (cyclohexatriene). This stability makes it resistant to oxidation and addition reactions.

Explanation: Aromatic compounds have a high ratio of Carbon to Hydrogen (e.g., Benzene is ${C}_1{H}_1$, while an alkane like Hexane is approx ${C}_1{H}_{2.3}$). During combustion in air, the oxygen supply is often insufficient to completely oxidize all the carbon. The unburnt carbon particles glow yellow and form black smoke (soot), resulting in a sooty flame.

Explanation: The concept of resonance in Benzene arose from the structures first proposed by ${Friedrich Kekulé}$ in 1865. He suggested that the benzene molecule oscillates between two structures, each having three single and three double bonds in alternating positions. However, the true structure is a resonance hybrid, which is a blend of these two forms, leading to equal bond lengths and a delocalized $pi$ electron cloud.

Explanation: When six atomic ${p}$-orbitals overlap laterally, they form six $pi$ molecular orbitals (MOs). Half of these MOs will be bonding and half will be antibonding. Therefore, six ${p}$-orbitals combine to form three bonding molecular orbitals and three antibonding molecular orbitals. The six delocalized $pi$ electrons of benzene occupy these three lowest energy bonding molecular orbitals, resulting in high stability.

Explanation: Anti-aromatic compounds are cyclic, planar, and fully conjugated, but they possess $4n$ $pi$ electrons (where $n=1, 2, 3, dots$). The $pi$-electron counts 4, 8, 12, etc., fall into this category. The $4pi$ electron system (for $n=1$) is highly unstable and is the count for anti-aromaticity. Aromatic compounds follow the $(4n+2)$ rule, corresponding to $2, 6, 10, dots$ $pi$ electrons.

Explanation: The ${Resonance Energy}$ (or delocalization energy) is the difference between the heat of hydrogenation of Benzene and the theoretical heat of hydrogenation of the hypothetical cyclohexa-1,3,5-triene. The experimental value for the resonance energy of Benzene is approximately $150 { kJ/mol}$ ($36 { kcal/mol}$). This large value demonstrates the immense thermodynamic stability that the delocalized $pi$-electron cloud imparts to the molecule.

Explanation: Cyclobutadiene is a cyclic, planar molecule with continuous conjugation. It possesses 4 $pi$ electrons (two double bonds), which fits the $4n$ rule for $n=1$. Due to its anti-aromatic character, it is extremely unstable and highly reactive, making it difficult to isolate. The cyclopentadienyl anion ($6 pi$ electrons) and cyclopropenyl cation ($2 pi$ electrons) are both aromatic.

Explanation: Each carbon atom in the benzene ring is ${sp}^2$ hybridized, contributing one unhybridized ${p}$-orbital perpendicular to the ring plane. The lateral overlap of all six of these ${p}$-orbitals creates a continuous, circular cloud of $6 pi$ electrons above and below the ring. This continuous delocalization is the source of the ring’s extraordinary stability, which is characteristic of aromaticity.

Explanation: The primary reason Benzene undergoes electrophilic substitution instead of the addition reactions typical of alkenes is the preservation of aromaticity. Addition reactions would convert the ${sp}^2$ carbons to ${sp}^3$ carbons, breaking the cyclic conjugation and losing the high resonance energy. Substitution replaces a ${H}$ atom with an electrophile (${E}^+$) and restores the aromatic system, maintaining the high stability.

Explanation: Polycyclic Aromatic Hydrocarbons (PAHs) are compounds composed of two or more fused benzene rings. While Naphthalene (two rings) is a PAH, it is not highly carcinogenic. ${Benzo(a)pyrene}$ (five fused rings) is a well-known, potent carcinogen found in soot, coal tar, tobacco smoke, and combustion exhaust. Its planar structure allows it to intercalate with DNA, leading to mutations and cancer.

Explanation: The ${Molecular Orbital (MO) theory}$ best describes the Benzene structure. It states that the six ${p}$-orbitals overlap continuously to form a single, large ring of electron density both above and below the plane of the ${C}$ atoms. This represents a ${6-electron}$ cloud that is delocalized over all six carbon atoms, giving every ${C}-{C}$ bond the same character and length.

Explanation: To achieve electrophilic substitution (${C}_6{H}_6 + {Cl}_2 rightarrow {C}_6{H}_5{Cl} + {HCl}$), a Lewis acid catalyst like anhydrous ${FeCl}_3$ or ${AlCl}_3$ is required. The catalyst reacts with the halogen (${Cl}_2$) to generate the highly electrophilic species (${Cl}^+$), which attacks the electron-rich benzene ring. Using ${UV}$ light without a Lewis acid favors free-radical addition, which is undesirable as it destroys the aromaticity.

Explanation: Heptane is an alkane with $n=7$ carbon atoms. The general formula for alkanes is $C_nH_{2n+2}$.

Substituting $n=7$ into the formula:

Hydrogen atoms $= 2(7) + 2 = 14 + 2 = 16$.

Therefore, the molecular formula for Heptane is $C_7H_{16}$. The other options represent an alkene ($C_7H_{14}$), an eight-carbon alkane ($C_8H_{18}$), or a six-carbon alkane ($C_6H_{14}$).

Explanation: Alkanes are defined as saturated hydrocarbons, meaning they contain only carbon-carbon single bonds (${C}-{C}$) and carbon-hydrogen single bonds (${C}-{H}$). All single bonds are formed by the axial overlap of hybrid orbitals and are classified as sigma ($sigma$) bonds. The absence of $pi$ bonds (which form double or triple bonds) makes alkanes relatively unreactive compared to alkenes and alkynes.

Explanation: In ${sp}^3$ hybridization, one ${s}$ orbital and three ${p}$ orbitals mix to form four equivalent ${sp}^3$ hybrid orbitals. These four orbitals arrange themselves in three-dimensional space to minimize electron repulsion, leading to a tetrahedral geometry with a bond angle of approximately $109.5^circ$. The ${s}$-character is $25%$, and $120^circ$ angles are characteristic of ${sp}^2$ hybridization.

Explanation: In an alkane like propane, all carbon atoms are ${sp}^3$ hybridized. The ${C}-{C}$ single bonds that connect the carbon chain are formed by the head-on (axial) overlap of one ${sp}^3$ hybrid orbital from the first carbon atom and one ${sp}^3$ hybrid orbital from the second carbon atom. The ${sp}^3-{s}$ overlap forms the ${C}-{H}$ bonds.

Explanation: The molecular formula of $n$-Butane is $C_4H_{10}$. The formula itself directly tells you the total number of hydrogen atoms is 10. Since every ${C}-{H}$ bond in a saturated alkane is a sigma bond, the number of ${C}-{H}$ sigma bonds is equal to the number of hydrogen atoms, which is 10. The total number of $sigma$ bonds in $n$-Butane is 13 (${C}_4{H}_{10}$ has $n+n+1 = 4+10-1 = 13$ total $sigma$ bonds).

Explanation: Isomerization is the process where a straight-chain alkane is converted into its branched-chain isomer. This reaction occurs when the alkane is heated with a catalyst such as anhydrous ${AlCl}_3$ and ${HCl}$. The resulting branched isomers (like Isobutane from $n$-Butane) have lower boiling points, which is important for refining petroleum products. Pyrolysis and Cracking involve breaking the carbon chain, not just rearranging it.

Explanation: Chain isomers are compounds with the same molecular formula but different carbon skeletons. $n$-Butane has the formula $C_4H_{10}$ and a straight chain. The only other structural isomer with the formula $C_4H_{10}$ is Isobutane (systematic name: 2-Methylpropane), which has a branched-chain structure. The other options are either unsaturated hydrocarbons or alkanes with a different number of carbon atoms.

Explanation: Unlike the restricted rotation in alkenes, the ${C}-{C}$ single bond in alkanes allows for relatively free rotation. This rotation results in different temporary spatial arrangements of atoms, which can be interconverted without breaking any bonds. These different arrangements are called conformers or rotamers. The staggered and eclipsed forms of Ethane are classical examples of conformers.

Explanation: Two consecutive members of a homologous series, such as Methane (${CH}_4$) and Ethane (${C}_2{H}_6$), differ by a methylene group (${CH}_2$).

The difference in molecular mass is calculated as:

Atomic mass of Carbon (${C}$) $approx 12 { g/mol}$

Atomic mass of Hydrogen (${H}$) $approx 1 { g/mol}$

Difference in mass (${CH}_2$) $= 12 + 2(1) = 14 { g/mol}$. This difference is constant across the entire alkane homologous series.

Explanation: A carbon atom is classified based on the number of other carbon atoms directly attached to it. The carbon atoms at the ends of any straight chain are attached to only one other carbon atom in the chain, irrespective of the length of the chain. Therefore, the end carbons are always classified as primary $left(1^circright)$ carbon atoms.

Explanation: To name this compound, we first identify the longest continuous carbon chain. The chain is four carbons long (Butane). Next, we number the chain to give the substituent (the methyl group) the lowest possible number. Numbering from the left gives the methyl group position 2, while numbering from the right gives position 3. We choose 2. Therefore, the correct name is 2-Methylbutane.

Explanation: In IUPAC nomenclature, the suffix is determined by the class of the compound. The suffix ‘-ene’ is specifically used to denote the presence of one or more carbon-carbon double bonds in the longest parent chain, characterizing the compound as an alkene. The suffix ‘-yne’ is used for triple bonds, and ‘-ol’ is used for hydroxyl groups (alcohols).

Explanation: First, identify the longest continuous chain. While a four-carbon chain (${butane}$) might appear obvious, tracing the path from the far left ${CH}_3$ to the far right ${CH}_3$ and including the carbons in the ethyl group reveals a continuous chain of five carbons (Pentane). The chain is numbered to give the methyl substituent the lowest number. The substituent is a methyl group at position 3, so the correct name is 3-Methylpentane.

Explanation: For disubstituted benzene derivatives, specific common prefixes are used in addition to numerical locants. The meta- (${m-}$) prefix is used when the two substituents are at positions 1 and 3 relative to each other (separated by one carbon). The ortho- (${o-}$) prefix is for 1 and 2 positions, and the para- (${p-}$) prefix is for 1 and 4 positions.

Explanation: The parent chain is determined by the triple bond, which dictates the suffix ‘-yne’. The longest chain containing the triple bond has four carbons (Butyne). The chain must be numbered to give the triple bond the lowest possible locant. Since the triple bond starts at ${C}1$, the correct name is But-1-yne. But-2-yne would have the triple bond between ${C}2$ and ${C}3$.

Explanation: When both a double bond and a triple bond are present (an enyne), the parent chain is numbered to give the multiple bond system the overall lowest set of locants. If there is a tie in the numbering (i.e., both are equidistant from their respective ends), the double bond takes priority for the lower number. However, the rule generally followed is that the chain is numbered to give the first site of unsaturation the lowest number, regardless of whether it is a double or triple bond. If a tie occurs, the double bond gets the lower number.

Explanation: When an organic compound has multiple different substituents, they must be listed in the name in alphabetical order. The prefixes indicating the number of identical substituents (di-, tri-, tetra-, etc.) are ignored for alphabetizing purposes. For instance, the ‘e’ in ethyl is compared against the ‘m’ in methyl, so ethyl is listed before methyl, regardless of which group has the lower numbering.

Explanation: The longest chain containing the double bond is three carbons (Propene). The chain is numbered to give the double bond the lowest possible number (position 1). The chlorine substituent is at position 3. The halogen is treated as a substituent, and the double bond defines the parent and suffix. Thus, the name is 3-Chloroprop-1-ene. The common name is Allyl chloride, but the question asks for the IUPAC name.

Explanation: When two identical substituents are attached to a benzene ring, the positions are designated numerically to give the lowest possible numbers. The two nitro groups (${-NO}_2$) are at the opposite sides of the ring, which corresponds to positions 1 and 4 (or $p$-). According to strict IUPAC rules, the numerical designation is preferred, so the name is 1,4-Dinitrobenzene. The common name, $p$-Dinitrobenzene, is also acceptable in some contexts.

Explanation: The name Hexane indicates an alkane with $n=6$ carbon atoms. The general formula for alkanes is $C_nH_{2n+2}$.

Substituting $n=6$ into the formula:

Hydrogen atoms $= 2(6) + 2 = 12 + 2 = 14$.

Therefore, the molecular formula for $n$-Hexane is $C_6H_{14}$. $C_6H_{12}$ is Hexene (alkene or cycloalkane), and $C_5H_{12}$ is Pentane.

Explanation: Chain isomers are compounds that share the same molecular formula but differ in the structure of their carbon skeleton (straight chain vs. branched chain). Both $n$-Butane (${CH}_3{CH}_2{CH}_2{CH}_3$) and Isobutane (${CH}_3{CH}({CH}_3){CH}_3$) have the molecular formula $C_4H_{10}$. They differ because $n$-Butane has a straight chain, and Isobutane has a branched chain. The other options represent different homologous series, position isomerism, or chain length differences.

Explanation: The molecular formula $C_5H_{12}$ belongs to the pentane family of alkanes. There are exactly three structural isomers that can be drawn for this formula, all exhibiting chain isomerism:

1. $n$-Pentane (straight chain)

2. Isopentane (2-Methylbutane)

3. Neopentane (2,2-Dimethylpropane)

All three compounds are saturated and share the same general formula $C_nH_{2n+2}$, but they have distinct carbon backbones.

Explanation: Isomerism, by definition, is the phenomenon where compounds have the same molecular formula but possess different arrangements of atoms. For structural isomers (including chain isomers), the chemical formula must be identical (e.g., $C_4H_{10}$), but the sequence in which the atoms are bonded must be different. Having the same functional group or different properties is characteristic of specific types of isomerism, but the same molecular formula is the fundamental requirement.

Explanation: The structure of 2,2-Dimethylpropane (${C}({CH}_3)_4$) has a central quaternary carbon $left(4^circright)$ atom (attached to four other carbons). Each of the four surrounding methyl groups contains a carbon atom attached to only the central carbon. Therefore, there are four methyl groups, and each of these carbons is a primary $left(1^circright)$ carbon atom. The total count of primary carbons is four.

Explanation: $n$-Hexane is a straight-chain alkane with the formula $C_6H_{14}$. 2-Methylpentane is a branched alkane, also with the formula $C_6H_{14}$. Since they share the same molecular formula but have different arrangements of their carbon skeletons (six-carbon straight chain vs. five-carbon chain with a methyl branch), they are classified as chain isomers. Position isomers would have the same chain length but different substituent positions.

Explanation: To identify the parent chain, we look at the core name of the compound:

A. 2,2-Dimethylpentane has a five-carbon parent chain.

B. 3-Ethylpentane has a five-carbon parent chain.

C. 2,2,3-Trimethylbutane has a four-carbon parent chain (${C}_4$), with three methyl groups attached to it.

D. $n$-Heptane has a seven-carbon parent chain.

Explanation: Chain isomerism requires a minimum number of carbon atoms to allow for branching. Since Propane ($C_3H_8$) has only three carbon atoms, it is impossible to form any arrangement other than the straight chain (${CH}_3{CH}_2{CH}_3$). Therefore, propane has no chain isomers. Butane (4 carbons) and all higher alkanes do exhibit chain isomerism.

Explanation: Chain isomers have the same molecular mass, so their van der Waals forces are the dominant factor affecting physical properties. As the degree of branching increases (moving from a straight chain to a compact, spherical branched structure), the molecular surface area decreases. This reduction in surface area leads to weaker van der Waals forces between molecules, and consequently, the boiling point decreases.

Explanation: Positional isomerism requires that the substituent (not the chain itself) be moved to a different position on the same length parent chain. However, if the parent chain is the straight ${n}$-Hexane chain, there are no substituents to move. Therefore, ${n}$-Hexane only exists as one structure when defined as the six-carbon straight chain, meaning it has only one possible positional arrangement within that definition.

Explanation: $n$-Octane has the formula $C_8H_{18}$ and an eight-carbon straight chain. 2-Methylheptane also has the formula $C_8H_{18}$ (since ${methyl} + {heptane} = 1+7=8$ carbons) but possesses a seven-carbon parent chain with a methyl branch. Since they share the same molecular formula but have different carbon skeletons, they are chain isomers. Options A and C have different molecular formulas. Option D (${2-Ethylhexane}$) is an incorrect IUPAC name for ${3-Methylheptane}$ (${C}_8{H}_{18}$), which is also a chain isomer, but the question asks for the relationship between the two listed.

Explanation: The Wurtz reaction joins two alkyl groups (${R}$) from two alkyl halide molecules (${R}-{X}$) to form a symmetrical alkane (${R}-{R}$) in the presence of sodium metal. To get a single product in good yield, only one type of alkyl halide must be used. Ethane (${C}_2{H}_6$) is the result of coupling two methyl groups (${CH}_3$) from methyl halides (${CH}_3{X}$). All odd-numbered alkanes (like Propane, ${C}_3{H}_8$) or larger even-numbered alkanes require the coupling of different alkyl groups or result in side products, respectively.

Explanation: This is the decarboxylation reaction, which involves the removal of a carbon dioxide group (${-COO}^-$) from the sodium salt of a carboxylic acid, resulting in an alkane with one less carbon atom.

$${CH}_3{CH}_2{COONa} + {NaOH} xrightarrow{{CaO, heat}} {CH}_3{CH}_3 + {Na}_2{CO}_3$$

Since sodium propanoate has three carbons, the resulting alkane will have two carbons, which is Ethane (${C}_2{H}_6$). The carbon chain is always reduced by one unit.

Explanation: Kolbe’s electrolytic method involves the electrolysis of the aqueous solution of a sodium or potassium salt of a carboxylic acid (${RCOONa}$). At the anode, the carboxylate ion (${RCOO}^-$) loses an electron to form an acyloxy radical, which then decomposes by losing ${CO}_2$, generating an alkyl free radical (${R}bullet$). Two such free radicals then combine (couple) to form the symmetrical alkane (${R}-{R}$).

Explanation: To prepare an unsymmetrical alkane like Propane (${C}_3{H}_8$), a mixture of two different alkyl halides (e.g., methyl halide, ${R}_1{X}$, and ethyl halide, ${R}_2{X}$) must be used. The Wurtz reaction will produce three possible alkanes: ${R}_1-{R}_1$, ${R}_2-{R}_2$, and ${R}_1-{R}_2$. For Propane, the products would be Ethane, Butane, and Propane itself. This resulting mixture of three alkanes is difficult to separate due to their close boiling points, rendering the method inefficient for unsymmetrical alkanes.

Explanation: While the primary reactant is sodium hydroxide (${NaOH}$), the Calcium Oxide (${CaO}$) is mixed with it to form soda lime. ${CaO}$ does not participate directly in the decarboxylation of the carboxylic acid salt but plays a vital role in preventing the fusion or melting of solid ${NaOH}$ when heated. It also prevents ${NaOH}$ from absorbing moisture from the atmosphere, thus acting as a drying agent and ensuring the reaction proceeds efficiently.

Explanation: Sodium ethanoate (${CH}_3{COONa}$) has an ${R}$ group equal to a methyl group (${CH}_3$). In Kolbe’s synthesis, the ${R}$ groups couple at the anode.

$$2{CH}_3{COO}^- xrightarrow{{-2e}^-} 2{CH}_3{COO}bullet rightarrow 2{CH}_3bullet + 2{CO}_2$$

$$2{CH}_3bullet rightarrow {CH}_3-{CH}_3$$

The coupling of two methyl radicals (${CH}_3bullet$) forms Ethane (${C}_2{H}_6$).

Explanation: The Wurtz reaction requires the use of highly reactive sodium metal, which reacts violently with water or alcohol. Therefore, the alkyl halide must be dissolved in a completely anhydrous (water-free) solvent. Dry Ether is the standard solvent used because it is non-polar, unreactive toward sodium metal, and capable of dissolving both the sodium and the organic reactants, enabling the reaction to occur safely and effectively.

Explanation: Kolbe’s reaction, like the Wurtz reaction, proceeds via the coupling of identical free radicals (${R}bullet + {R}bullet rightarrow {R}-{R}$). This means it is highly effective for synthesizing symmetrical alkanes (those with an even number of carbon atoms). Trying to prepare unsymmetrical or odd-numbered alkanes using different carboxylic acid salts results in a mixture of three alkanes that are difficult to separate, making the method less useful for these purposes.

Explanation: Decarboxylation removes one carbon atom from the starting material. It is the only practical method for preparing the simplest alkane, Methane (${CH}_4$), from the sodium salt of the next higher acid, ${Ethanoic Acid}$ (Sodium acetate, ${CH}_3{COONa}$).

$${CH}_3{COONa} + {NaOH} xrightarrow{{CaO, heat}} {CH}_4 + {Na}_2{CO}_3$$

Other options are incorrect as it reduces the carbon chain, not increases it (A), and reduces the chain from 4 to 3 (Propane from Butanoic acid, C), or is used for linear, not cyclic alkanes (D).

Explanation: 1-Chloropropane (${R}-{X}$, where ${R}={CH}_3{CH}_2{CH}_2-$) is an alkyl halide with three carbon atoms. The Wurtz reaction couples two identical ${R}$ groups (${R}-{R}$).

$${2}({CH}_3{CH}_2{CH}_2{Cl}) + 2{Na} rightarrow {CH}_3{CH}_2{CH}_2-{CH}_2{CH}_2{CH}_3 + 2{NaCl}$$

The coupling of two propyl groups results in a six-carbon symmetrical alkane, $n$-Hexane, with the structure ${CH}_3{CH}_2{CH}_2{CH}_2{CH}_3$ ($C_6H_{14}$).