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101. What are the products of the complete combustion of any alkane in the presence of excess oxygen?
ⓐ. Carbon Dioxide (${CO}_2$) and Water (${H}_2{O}$)
ⓑ. Carbon Monoxide (${CO}$) and Water (${H}_2{O}$)
ⓒ. Carbon (${C}$) and Water (${H}_2{O}$)
ⓓ. Carbon Dioxide (${CO}_2$) and Hydrogen Gas (${H}_2$)
Correct Answer: Carbon Dioxide (${CO}_2$) and Water (${H}_2{O}$)
Explanation: Complete combustion occurs when an alkane (or any hydrocarbon) reacts with a sufficient supply of oxygen. The carbon atoms are fully oxidized to Carbon Dioxide (${CO}_2$), and the hydrogen atoms are oxidized to Water (${H}_2{O}$), releasing a large amount of energy (heat). The general equation is: $C_nH_{2n+2} + \left(\frac{3n+1}{2}\right){O}_2 \rightarrow n{CO}_2 + (n+1){H}_2{O}$.
102. The halogenation of alkanes (e.g., Chlorination of Methane) is an example of which type of organic reaction mechanism?
ⓐ. Electrophilic addition
ⓑ. Nucleophilic substitution
ⓒ. Free-radical substitution
ⓓ. Elimination reaction
Correct Answer: Free-radical substitution
Explanation: Alkanes are saturated and unreactive towards typical polar reagents. When reacting with halogens (${Cl}_2$ or ${Br}_2$) in the presence of heat or ultraviolet light, the reaction proceeds through three steps: initiation, propagation, and termination, involving highly reactive neutral species. This mechanism is classified as free-radical substitution, where a hydrogen atom on the alkane is sequentially replaced by a halogen atom.
103. Which halogen reacts the fastest with methane in a free-radical substitution reaction?
ⓐ. Bromine (${Br}_2$)
ⓑ. Iodine (${I}_2$)
ⓒ. Chlorine (${Cl}_2$)
ⓓ. Fluorine (${F}_2$)
Correct Answer: Fluorine (${F}_2$)
Explanation: The reactivity of halogens in free-radical substitution reactions follows the order: ${F}_2 > {Cl}_2 > {Br}_2 > {I}_2$. Fluorine (${F}_2$) is the most reactive, often reacting explosively, while ${I}_2$ is largely unreactive. Chlorine is fast but manageable, and Bromine is slower and more selective.
104. Incomplete combustion of alkanes, which occurs due to a limited supply of oxygen, typically leads to the formation of:
ⓐ. Carbon Dioxide (${CO}_2$) and Water (${H}_2{O}$)
ⓑ. Carbon Monoxide (${CO}$) or Soot (${C}$) and Water (${H}_2{O}$)
ⓒ. Ethane and Methane
ⓓ. Alkenes and Alkynes
Correct Answer: Carbon Monoxide (${CO}$) or Soot (${C}$) and Water (${H}_2{O}$)
Explanation: When the oxygen supply is insufficient, the carbon atoms in the alkane are not fully oxidized to ${CO}_2$. Instead, the incomplete combustion results in the formation of toxic Carbon Monoxide (${CO}$) or finely divided carbon particles known as Soot (${C}$), alongside water vapor.
105. What is the product formed after the first stage of chlorination of Methane (${CH}_4$) in the presence of UV light?
ⓐ. Methyl chloride (${CH}_3{Cl}$)
ⓑ. Methylene chloride (${CH}_2{Cl}_2$)
ⓒ. Carbon tetrachloride (${CCl}_4$)
ⓓ. Chloroform (${CHCl}_3$)
Correct Answer: Methyl chloride (${CH}_3{Cl}$)
Explanation: Halogenation of methane is a multiple-step substitution reaction. The reaction proceeds sequentially. The first hydrogen atom is replaced by chlorine, producing Methyl chloride (${CH}_3{Cl}$). If the reaction is allowed to continue, ${CH}_3{Cl}$ will be further chlorinated to ${CH}_2{Cl}_2$, ${CHCl}_3$, and finally ${CCl}_4$.
106. When Methane is heated with oxygen at high pressure ($\approx 100 { atm}$) in the presence of Copper (${Cu}$) catalyst, the primary product is:
ⓐ. Methanol (${CH}_3{OH}$)
ⓑ. Formaldehyde (${HCHO}$)
ⓒ. Acetic acid (${CH}_3{COOH}$)
ⓓ. Ethanol (${CH}_3{CH}_2{OH}$)
Correct Answer: Methanol (${CH}_3{OH}$)
Explanation: This is a controlled, low-conversion catalytic oxidation. The reaction of methane with oxygen under specific conditions (high pressure, high temperature, and a ${Cu}$ catalyst) allows the oxidation to be halted after the first step (the substitution of ${-OH}$ for ${-H}$), yielding Methanol (${CH}_3{OH}$). Without precise conditions, the reaction typically continues to ${CO}_2$.
107. The preference of alkanes to undergo substitution reactions rather than addition reactions is due to:
ⓐ. The presence of double or triple bonds.
ⓑ. The strength of the ${C}-{C}$ bonds.
ⓒ. The saturated nature of the carbon atoms.
ⓓ. The ease of forming a carbocation intermediate.
Correct Answer: The saturated nature of the carbon atoms.
Explanation: Alkanes are saturated molecules, meaning all carbon valencies are already occupied by single bonds ($\sigma$ bonds). There are no available $\pi$ bonds to break and no room for additional atoms or groups without displacing a substituent. Therefore, to react, they must substitute an existing atom (${H}$) with a new atom (${X}$), unlike unsaturated compounds (alkenes/alkynes) which undergo addition by breaking their $\pi$ bond.
108. The process of heating higher alkanes in the absence of air and often in the presence of a catalyst to break them into smaller, lower molecular mass hydrocarbons (alkanes and alkenes) is called:
ⓐ. Hydrogenation
ⓑ. Polymerization
ⓒ. Isomerization
ⓓ. Pyrolysis or Cracking
Correct Answer: Pyrolysis or Cracking
Explanation: Pyrolysis, or Cracking, is the thermal decomposition of large alkane molecules into smaller, more volatile hydrocarbons, often including an alkane and an alkene. This process is crucial in the petroleum industry to convert less valuable heavy oils into more valuable gasoline and other light products. Hydrogenation and polymerization are types of addition reactions, and isomerization only rearranges the chain.
109. Which type of hydrogen atom in an alkane is most easily substituted by a halogen atom (${X}$) in a free-radical halogenation reaction?
ⓐ. Primary $\left(1^\circ\right)$ hydrogen
ⓑ. Secondary $\left(2^\circ\right)$ hydrogen
ⓒ. Tertiary $\left(3^\circ\right)$ hydrogen
ⓓ. All are equally reactive
Correct Answer: Tertiary $\left(3^\circ\right)$ hydrogen
Explanation: The ease of substitution is related to the stability of the free radical intermediate formed. Free radical stability increases in the order: Primary $\left(1^\circ\right) < {Secondary} \left(2^\circ\right) < {Tertiary} \left(3^\circ\right)$. Since the reaction proceeds through the most stable intermediate, a tertiary $\left(3^\circ\right)$ hydrogen atom is the most readily substituted, being the most reactive.
110. If an excess amount of Chlorine gas (${Cl}_2$) is reacted with Methane (${CH}_4$) under high temperature or ${UV}$ light, the final product of complete substitution is:
ⓐ. Methyl chloride (${CH}_3{Cl}$)
ⓑ. Methylene chloride (${CH}_2{Cl}_2$)
ⓒ. Chloroform (${CHCl}_3$)
ⓓ. Carbon tetrachloride (${CCl}_4$)
Correct Answer: Carbon tetrachloride (${CCl}_4$)
Explanation: When the halogenation of Methane is carried out with an excess of the halogen and sufficient energy, the substitution continues until all four hydrogen atoms have been replaced by chlorine atoms. The final, fully substituted product is Carbon tetrachloride (${CCl}_4$).
111. What is the general molecular formula for acyclic alkenes containing one carbon-carbon double bond?
ⓐ. $C_nH_{2n}$
ⓑ. $C_nH_{2n+2}$
ⓒ. $C_nH_{2n-2}$
ⓓ. $C_nH_{n}$
Correct Answer: $C_nH_{2n}$
Explanation: Alkenes are unsaturated hydrocarbons defined by the presence of one carbon-carbon double bond (${C}={C}$). The presence of this double bond corresponds to one degree of unsaturation, meaning they have two fewer hydrogen atoms than the corresponding saturated alkane ($C_nH_{2n+2}$). Thus, the general formula for a simple, acyclic alkene is $C_nH_{2n}$. This formula is also shared by cycloalkanes.
112. The two carbon atoms involved in the ${C}={C}$ double bond of an alkene are linked by which combination of bonds?
ⓐ. Two $\pi$ bonds
ⓑ. One $\sigma$ bond and one $\pi$ bond
ⓒ. Two $\sigma$ bonds
ⓓ. Two $\sigma$ bonds and one $\pi$ bond
Correct Answer: One $\sigma$ bond and one $\pi$ bond
Explanation: A carbon-carbon double bond consists of two distinct components. The sigma ($\sigma$) bond is formed by the axial overlap of ${sp}^2$ hybrid orbitals. The pi ($\pi$) bond is formed by the lateral (sideways) overlap of the unhybridized ${p}$ orbitals, which lie perpendicular to the plane of the $\sigma$ bond framework. The combination of these two types of bonds constitutes the double bond.
113. What is the minimum number of carbon atoms required for an alkene to exhibit positional isomerism?
ⓐ. 2
ⓑ. 3
ⓒ. 4
ⓓ. 5
Correct Answer: 4
Explanation: Positional isomerism requires that a functional group (in this case, the double bond) can be located at different positions on the same-length carbon chain.
Ethylene ($C_2$): Only one position (${C}1$).
Propene ($C_3$): Only one position (${C}1$).
Butene ($C_4$): Can have the double bond at ${C}1$ (But-1-ene) or ${C}2$ (But-2-ene).
Therefore, four carbon atoms is the minimum number required to have two different structural isomers based on the double bond’s position.
114. The bond angle around the carbon atoms of the ${C}={C}$ double bond in an alkene is approximately:
ⓐ. $180^\circ$
ⓑ. $109.5^\circ$
ⓒ. $120^\circ$
ⓓ. $90^\circ$
Correct Answer: $120^\circ$
Explanation: The carbon atoms in a double bond are ${sp}^2$ hybridized. The three ${sp}^2$ hybrid orbitals lie in a plane and orient themselves to maximize the distance between electron pairs, which results in a trigonal planar geometry and an ideal bond angle of $120^\circ$. $109.5^\circ$ is for ${sp}^3$ (alkanes), and $180^\circ$ is for ${sp}$ (alkynes).
115. What is the molecular formula for Cyclopentene?
ⓐ. $C_5H_{10}$
ⓑ. $C_5H_{8}$
ⓒ. $C_5H_{12}$
ⓓ. $C_6H_{12}$
Correct Answer: $C_5H_{8}$
Explanation: Cyclopentene is a cyclic hydrocarbon with one double bond. It has two degrees of unsaturation (one ring + one $\pi$ bond). Using the general formula $C_nH_{2n-2}$ where $n=5$, the number of Hydrogens is $2(5)-2 = 8$. Thus, the formula is $C_5H_8$.
116. Which type of orbital overlap forms the $\pi$ bond component of an alkene’s double bond?
ⓐ. Axial overlap of ${sp}^2$ hybrid orbitals
ⓑ. Lateral overlap of ${sp}^2$ hybrid orbitals
ⓒ. Axial overlap of unhybridized ${p}$ orbitals
ⓓ. Lateral overlap of unhybridized ${p}$ orbitals
Correct Answer: Lateral overlap of unhybridized ${p}$ orbitals
Explanation: The $\sigma$ bond is formed by the axial (${head-on}$) overlap of ${sp}^2$ hybrid orbitals. However, the $\pi$ bond is formed by the lateral (${sideways}$) overlap of the single unhybridized ${p}$ orbital present on each of the ${sp}^2$ hybridized carbon atoms. This lateral overlap creates an electron cloud above and below the plane of the $\sigma$ bond framework.
117. Due to the $\pi$ bond, rotation about the carbon-carbon double bond in alkenes is:
ⓐ. Entirely free, leading to conformers.
ⓑ. Slightly restricted, leading to position isomers.
ⓒ. Restricted, leading to geometrical isomerism.
ⓓ. Completely impossible at any temperature.
Correct Answer: Restricted, leading to geometrical isomerism.
Explanation: The energy barrier required to break the $\pi$ bond (which must occur for rotation to happen) is quite high (around $264 { kJ/mol}$). This high energy barrier means that rotation about the ${C}={C}$ bond is highly restricted at room temperature. This restriction is the fundamental condition that gives rise to the existence of non-interconvertible geometrical isomers (cis-trans isomers).
118. Ethene (${C}_2{H}_4$) is the simplest alkene. Which of the following is true about its structure?
ⓐ. The molecule is non-planar (tetrahedral geometry).
ⓑ. All six atoms lie in the same plane.
ⓒ. The ${C}-{H}$ bond angle is $109.5^\circ$.
ⓓ. The carbon atoms are ${sp}$ hybridized.
Correct Answer: All six atoms lie in the same plane.
Explanation: In Ethene (${C}_2{H}_4$), both carbon atoms are ${sp}^2$ hybridized, resulting in a trigonal planar geometry around each carbon. Because the three $\sigma$ bonds around each carbon are in a single plane, and the two ${C}$ atoms are also linked in that plane, all six atoms (two ${C}$ and four ${H}$) of the Ethene molecule lie in the same planar geometry.
119. A compound with the molecular formula $C_6H_{12}$ can be either an alkene or a cycloalkane. This is an example of which type of isomerism?
ⓐ. Chain isomerism
ⓑ. Functional group isomerism
ⓒ. Position isomerism
ⓓ. Metamerism
Correct Answer: Functional group isomerism
Explanation: Functional group isomerism occurs when two or more compounds have the same molecular formula but belong to different homologous series (i.e., contain different functional groups). Both the cycloalkane family (e.g., Cyclohexane) and the simple alkene family (e.g., Hex-1-ene) have the general formula $C_nH_{2n}$ for $n=6$. Since one is saturated (cycloalkane) and one is unsaturated (alkene), they possess different functional groups and are functional group isomers.
120. How many total $\sigma$ bonds and $\pi$ bonds are present in a molecule of Propene (${CH}_3{CH}={CH}_2$)?
ⓐ. $5\sigma$ and $1\pi$
ⓑ. $7\sigma$ and $1\pi$
ⓒ. $8\sigma$ and $1\pi$
ⓓ. $9\sigma$ and $2\pi$
Correct Answer: $8\sigma$ and $1\pi$
Explanation: Propene (${CH}_3{CH}={CH}_2$) has the structure ${H}_3{C}-{CH}={CH}_2$.
The $\sigma$ bonds are: 3 in the ${CH}_3$ group, 1 ${C}-{C}$ single bond, 1 ${C}-{H}$ on the ${C}$ of the double bond, 2 ${C}-{H}$ on the ${CH}_2$, and 1 $\sigma$ bond in the ${C}={C}$ double bond. Total $\sigma$ bonds $= 3+1+1+2+1 = \mathbf{8}$.
The $\pi$ bonds are: The single $\pi$ bond contained within the ${C}={C}$ double bond. Total $\pi$ bonds $= \mathbf{1}$.
121. Which type of isomerism is characterized by having the same molecular formula and the same carbon skeleton, but differing in the position of the functional group or substituent?
ⓐ. Chain isomerism
ⓑ. Position isomerism
ⓒ. Tautomerism
ⓓ. Metamerism
Correct Answer: Position isomerism
Explanation: Position isomerism is a type of structural isomerism where the functional group, multiple bond (double or triple), or a substituent atom occupies a different position on the same parent carbon chain. For example, But-1-ene (${CH}_2={CHCH}_2{CH}_3$) and But-2-ene (${CH}_3{CH}={CHCH}_3$) are position isomers because the double bond is in a different location.
122. Which of the following compounds exhibits geometrical (cis-trans) isomerism?
ⓐ. Propene (${CH}_3{CH}={CH}_2$)
ⓑ. 2-Methylbut-2-ene
ⓒ. 1-Chloropropene (${CH}_2={CHCH}_2{Cl}$)
ⓓ. 1,2-Dichloroethene (${CHCl}={CHCl}$)
Correct Answer: 1,2-Dichloroethene (${CHCl}={CHCl}$)
Explanation: Geometrical isomerism (cis-trans or E/Z) requires that each carbon atom of the double bond must be attached to two different groups.
A. Propene has ${CH}_2$ on one end, meaning two identical ${H}$ atoms on one carbon, so no.
B. 2-Methylbut-2-ene has two identical methyl groups on one carbon, so no.
C. 1-Chloropropene has two identical ${H}$ atoms on the first carbon, so no.
D. 1,2-Dichloroethene (${CHCl}={CHCl}$) has one ${H}$ and one ${Cl}$ on each carbon, satisfying the requirement for geometrical isomerism.
123. The lack of free rotation about a $\sigma$ bond, as opposed to a $\pi$ bond, leads to which specific type of isomerism?
ⓐ. Geometrical isomerism
ⓑ. Structural isomerism
ⓒ. Conformational isomerism
ⓓ. Optical isomerism
Correct Answer: Conformational isomerism
Explanation: The $\sigma$ bond in single-bonded systems (like alkanes) allows for relatively free rotation, leading to different spatial arrangements called conformers or rotamers (e.g., staggered and eclipsed forms of ethane). Conformational isomers are not true isomers because they are rapidly interconvertible without breaking bonds. Geometrical isomerism arises from restricted rotation about a $\pi$ bond, not a $\sigma$ bond.
124. Which pair of compounds represents functional group isomerism?
ⓐ. Ethanol (${CH}_3{CH}_2{OH}$) and Dimethyl ether (${CH}_3{OCH}_3$)
ⓑ. Butan-1-ol and Butan-2-ol
ⓒ. Propanone and Propanal
ⓓ. $n$-Butane and Isobutane
Correct Answer: Ethanol (${CH}_3{CH}_2{OH}$) and Dimethyl ether (${CH}_3{OCH}_3$)
Explanation: Functional group isomers have the same molecular formula (${C}_2{H}_6{O}$) but belong to different homologous series, meaning they possess different functional groups. Ethanol is an alcohol (${-OH}$), while Dimethyl ether is an ether (${-O-}$). Butan-1-ol and Butan-2-ol are position isomers, Propanone and Propanal are functional group isomers (${C}_3{H}_6{O}$), and $n$-Butane and Isobutane are chain isomers.
125. The compound 2,3-Dimethylbut-2-ene does not show geometrical isomerism because:
ⓐ. The chain is too long to be planar.
ⓑ. There is restricted rotation around the double bond.
ⓒ. Each carbon of the double bond is attached to two identical groups (${CH}_3$).
ⓓ. The double bond is at the second position.
Correct Answer: Each carbon of the double bond is attached to two identical groups (${CH}_3$).
Explanation: The essential requirement for geometrical isomerism is that the two groups attached to each carbon atom of the double bond must be different. The structure of 2,3-Dimethylbut-2-ene is ${CH}_3({C})({CH}_3)={C}({CH}_3){CH}_3$. Both carbons in the double bond are attached to two identical methyl (${CH}_3$) groups, thus failing the requirement for cis-trans isomerism.
126. How many total structural isomers (including chain and position) are possible for the molecular formula $C_4H_8$? (Exclude geometrical isomers).
ⓐ. 4
ⓑ. 5
ⓒ. 6
ⓓ. 7
Correct Answer: 5
Explanation: The molecular formula $C_4H_8$ represents alkenes and cycloalkanes. The structural isomers are:
1. But-1-ene
2. But-2-ene
3. 2-Methylpropene
4. Cyclobutane
5. Methylcyclopropane Therefore,
there are 5 distinct structural isomers.
127. Which of the following is not a structural isomer of $n$-Pentane ($C_5H_{12}$)?
ⓐ. 2-Methylbutane
ⓑ. Isopentane
ⓒ. Pent-1-ene
ⓓ. 2,2-Dimethylpropane
Correct Answer: Pent-1-ene
Explanation: Structural isomers must have the same molecular formula. $n$-Pentane has the formula $C_5H_{12}$ (an alkane). Pent-1-ene is an alkene with the formula $C_5H_{10}$. Since the molecular formulas are different, Pent-1-ene is not a structural isomer of $n$-Pentane. 2-Methylbutane (Isopentane) and 2,2-Dimethylpropane are both chain isomers of $n$-Pentane.
128. What distinguishes a cis-isomer from a trans-isomer in a disubstituted alkene?
ⓐ. The number of carbon atoms in the chain.
ⓑ. The position of the double bond.
ⓒ. The relative positions of the identical groups with respect to the double bond.
ⓓ. The hybridization of the carbon atoms.
Correct Answer: The relative positions of the identical groups with respect to the double bond.
Explanation: Geometrical isomers (cis/trans) have identical groups. A cis-isomer has the two identical groups located on the same side of the double bond, which is fixed in space. A trans-isomer has the two identical groups located on opposite sides of the double bond. This difference in spatial arrangement leads to distinct physical properties.
129. Which of the following is the condition for a pair of compounds to be metameric isomers?
ⓐ. They must differ in the position of the multiple bond.
ⓑ. They must be non-interconvertible at room temperature.
ⓒ. They must have different functional groups.
ⓓ. They must have the same functional group but differ in the alkyl groups attached to the functional group’s heteroatom.
Correct Answer: They must have the same functional group but differ in the alkyl groups attached to the functional group’s heteroatom.
Explanation: Metamerism is a type of structural isomerism where the compounds have the same molecular formula and the same functional group (usually an ether, ketone, or secondary amine), but they differ in the size or nature of the alkyl chains attached to the functional group’s heteroatom (${O}$, ${N}$, ${S}$). For example, Diethyl ether (${C}_2{H}_5{OC}_2{H}_5$) and Methyl propyl ether (${CH}_3{OC}_3{H}_7$) are metamers (${C}_4{H}_{10}{O}$).
130. The conversion of a compound from the enol form to the keto form is an example of which specialized type of isomerism?
ⓐ. Tautomerism
ⓑ. Optical isomerism
ⓒ. Positional isomerism
ⓓ. Chain isomerism
Correct Answer: Tautomerism
Explanation: Tautomerism is a specialized type of structural isomerism where two isomers (tautomers), usually the keto form and the enol form, exist in dynamic equilibrium and are readily interconvertible by the migration of a hydrogen atom (proton) and a simultaneous shift of a double bond. This reversible interconversion process is called tautomerization.
131. The dehydration of an alcohol to form an alkene is typically carried out using which type of reagent and condition?
ⓐ. Concentrated ${H}_2{SO}_4$ or ${H}_3{PO}_4$ and heat
ⓑ. Aqueous ${KOH}$ solution at room temperature
ⓒ. ${Na}$ metal in liquid ${NH}_3$
ⓓ. ${H}_2/{Pd}$ catalyst
Correct Answer: Concentrated ${H}_2{SO}_4$ or ${H}_3{PO}_4$ and heat
Explanation: Dehydration is an elimination reaction that removes a water molecule (${H}_2{O}$) from an alcohol (${R}-{OH}$) to form an alkene (${C}={C}$). This reaction requires a strong acid catalyst (like concentrated ${H}_2{SO}_4$ or ${H}_3{PO}_4$) and elevated heat to protonate the hydroxyl group and facilitate its elimination.
132. Which rule governs the major product formed during the dehydrohalogenation (removal of ${H}{X}$) of a secondary alkyl halide?
ⓐ. Markownikov’s rule
ⓑ. Hund’s rule
ⓒ. Saytzeff’s rule (Zaitsev’s rule)
ⓓ. Anti-Markownikov’s rule
Correct Answer: Saytzeff’s rule (Zaitsev’s rule)
Explanation: Dehydrohalogenation is an ${E}2$ elimination reaction that removes a halogen (${X}$) and a hydrogen (${H}$) from adjacent carbons to form a double bond. When a secondary alkyl halide can form two different alkenes, Saytzeff’s rule predicts that the most substituted (most stable) alkene will be the major product. Markownikov’s rule governs addition reactions, not elimination.
133. When 2-Bromobutane is treated with a strong base (like alcoholic ${KOH}$), which alkene is the major product according to Saytzeff’s rule?
ⓐ. But-1-ene
ⓑ. But-2-ene
ⓒ. But-2-yne
ⓓ. 2-Methylpropene
Correct Answer: But-2-ene
Explanation: 2-Bromobutane can lose a hydrogen from ${C}1$ (forming But-1-ene) or from ${C}3$ (forming But-2-ene). But-2-ene is a disubstituted alkene, while But-1-ene is monosubstituted. According to Saytzeff’s rule, the more substituted But-2-ene is the major product formed during dehydrohalogenation because it is thermodynamically more stable.
134. Which of the following conditions is most suitable for the dehydrohalogenation of an alkyl halide (${R}-{X}$)?
ⓐ. Aqueous ${NaOH}$ and heat
ⓑ. ${Na}$ metal in dry ether
ⓒ. Concentrated ${H}_2{SO}_4$ at $170^\circ {C}$
ⓓ. Alcoholic ${KOH}$ and heat
Correct Answer: Alcoholic ${KOH}$ and heat
Explanation: Dehydrohalogenation is an elimination reaction that requires a strong, bulky base. While aqueous ${NaOH}$ or ${KOH}$ favor substitution, the use of alcoholic (ethanolic) ${KOH}$ or ${NaOH}$ ensures that the hydroxide ion acts as a strong base rather than a nucleophile. This base, combined with heat, promotes the ${E}2$ elimination pathway to form the alkene.
135. In the preparation of ethene (${C}_2{H}_4$) via Kolbe’s electrolytic method, the starting material is the aqueous solution of the sodium or potassium salt of which carboxylic acid?
ⓐ. Propanoic acid
ⓑ. Ethanoic acid
ⓒ. Succinic acid (Butanedioic acid)
ⓓ. Acetic acid
Correct Answer: Succinic acid (Butanedioic acid)
Explanation: Kolbe’s synthesis is primarily used for alkanes via radical coupling (${R}-{R}$). However, alkenes can be prepared by the electrolysis of the salts of dicarboxylic acids where the carboxyl groups (${-COO}^-$) are on adjacent carbons. The salt of Succinic acid (potassium succinate) undergoes double decarboxylation and radical coupling at the anode to form Ethene (${C}_2{H}_4$) and two molecules of ${CO}_2$.
136. When an unsymmetrical alcohol like Pentan-2-ol is dehydrated, which type of alkene is predominantly formed?
ⓐ. The least substituted alkene (Hofmann product)
ⓑ. The most substituted alkene (Saytzeff product)
ⓒ. A ketone
ⓓ. An alkane
Correct Answer: The most substituted alkene (Saytzeff product)
Explanation: Dehydration of unsymmetrical secondary or tertiary alcohols typically follows the Saytzeff’s rule (like dehydrohalogenation). This means the hydroxyl (${-OH}$) group and a hydrogen (${-H}$) atom are eliminated in a way that generates the most stable, most substituted alkene (e.g., Pent-2-ene over Pent-1-ene).
137. Which of the following reaction types is classified as a $\beta$-elimination reaction?
ⓐ. Halogenation of Alkanes
ⓑ. Friedel-Crafts reaction
ⓒ. Hydration of Alkenes
ⓓ. Dehydration of Alcohols
Correct Answer: Dehydration of Alcohols
Explanation: A $\beta$-elimination reaction is one in which atoms or groups are removed from adjacent carbon atoms, specifically the $\alpha$-carbon (bearing the functional group) and the $\beta$-carbon (adjacent to the $\alpha$-carbon). Both dehydration of alcohols (loss of ${-OH}$ and ${-H}$) and dehydrohalogenation (loss of ${-X}$ and ${-H}$) are prime examples of $\beta$-elimination reactions used to synthesize alkenes.
138. The conversion of a vicinal dihalide (${R}-{CHX}-{CHX}-{R}$) to an alkene requires treatment with:
ⓐ. ${H}_2/{Ni}$ catalyst
ⓑ. ${Na}$ in liquid ${NH}_3$
ⓒ. Zinc metal dust in alcohol
ⓓ. Concentrated ${H}_2{SO}_4$
Correct Answer: Zinc metal dust in alcohol
Explanation: Vicinal dihalides (halogens on adjacent carbons) can be converted to alkenes by dehalogenation. This is achieved by treating the dihalide with Zinc metal dust in the presence of alcohol and heat. The Zinc removes the two halogen atoms, forming a zinc halide salt (${ZnX}_2$) and introducing a double bond into the hydrocarbon chain.
139. If 1-Propanol (${CH}_3{CH}_2{CH}_2{OH}$) is dehydrated with concentrated ${H}_2{SO}_4$ at $170^\circ {C}$, what is the product?
ⓐ. Propene
ⓑ. Propane
ⓒ. Propyne
ⓓ. Diethyl ether
Correct Answer: Propene
Explanation: The dehydration of 1-Propanol is a straight-forward $\beta$-elimination reaction. The water molecule is eliminated, resulting in the formation of the simple alkene with three carbons, Propene (${CH}_3{CH}={CH}_2$). The temperature of $170^\circ {C}$ strongly favors elimination to form an alkene; a lower temperature (${e.g.}, 140^\circ {C}$) would favor the formation of an ether.
140. Which factor is most important in determining whether the dehydration of an alcohol will follow the ${E}1$ or ${E}2$ mechanism?
ⓐ. The structure $\left(1^\circ, 2^\circ, 3^\circ\right)$ of the alcohol
ⓑ. The type of acid used
ⓒ. The size of the hydroxyl group
ⓓ. The concentration of the water formed
Correct Answer: The structure $\left(1^\circ, 2^\circ, 3^\circ\right)$ of the alcohol
Explanation: The ${E}1$ mechanism proceeds through a carbocation intermediate, which is more stable for $3^\circ$ and $2^\circ$ alcohols. Therefore, tertiary $\left(3^\circ\right)$ and secondary $\left(2^\circ\right)$ alcohols typically undergo dehydration via the ${E}1$ mechanism. Primary $\left(1^\circ\right)$ alcohols, which form less stable carbocations, typically require high temperatures and proceed via the ${E}2$ mechanism.
141. The reaction of an alkene with hydrogen gas (${H}_2$) in the presence of a metal catalyst (${Ni}$, ${Pd}$, or ${Pt}$) is known as:
ⓐ. Hydrogenation
ⓑ. Halogenation
ⓒ. Hydrohalogenation
ⓓ. Hydration
Correct Answer: Hydrogenation
Explanation: Hydrogenation is the process of adding two hydrogen atoms across a carbon-carbon double bond, converting the unsaturated alkene into the corresponding saturated alkane. This reaction typically requires heat and a metal catalyst (Nickel, Platinum, or Palladium) to lower the activation energy, and it is a classic example of an addition reaction.
142. When 1-Butene reacts with ${HBr}$ in the absence of peroxides, the major product is 2-Bromobutane. This is an example of an addition reaction that follows:
ⓐ. Anti-Markownikov’s rule
ⓑ. Saytzeff’s rule
ⓒ. Hund’s rule
ⓓ. Markownikov’s rule
Correct Answer: Markownikov’s rule
Explanation: The addition of an unsymmetrical reagent (${HBr}$) to an unsymmetrical alkene (1-Butene) in the absence of peroxides follows Markownikov’s rule. This rule states that the negative part of the attacking reagent (${Br}^-$) attaches to the carbon atom of the double bond that has the fewer number of hydrogen atoms, which leads to the formation of the more stable carbocation intermediate (secondary carbocation at ${C}2$).
143. The addition of ${Br}_2$ to Ethene is a test for unsaturation. The mechanism for this reaction is an example of:
ⓐ. Free-radical addition
ⓑ. Electrophilic addition
ⓒ. Nucleophilic addition
ⓓ. Electrophilic substitution
Correct Answer: Electrophilic addition
Explanation: The electron-rich $\pi$ bond of the alkene acts as a source of electrons, making it susceptible to attack by an electron-deficient species (an electrophile). In this reaction, the ${Br}_2$ molecule is polarized by the $\pi$ cloud and acts as the electrophile, forming a cyclic bromonium ion intermediate. Therefore, the mechanism is Electrophilic Addition.
144. Which product is formed when Propene (${CH}_3{CH}={CH}_2$) reacts with water in the presence of an acid catalyst (${H}^+$)?
ⓐ. Propan-1-ol
ⓑ. Propan-2-ol
ⓒ. Propane
ⓓ. Propanone
Correct Answer: Propan-2-ol
Explanation: This is a hydration reaction where water is added across the double bond, converting the alkene into an alcohol, following Markownikov’s rule. The ${-OH}$ group adds to the carbon atom of the double bond with the fewer hydrogen atoms (${C}2$), and the ${H}$ adds to the ${C}$ with more hydrogens (${C}1$). The major product is Propan-2-ol (${CH}_3{CH}({OH}){CH}_3$).
145. When ${HBr}$ is added to an alkene in the presence of an organic peroxide (like benzoyl peroxide), the reaction is said to follow:
ⓐ. Markownikov’s rule
ⓑ. Saytzeff’s rule
ⓒ. Anti-Markownikov’s rule
ⓓ. Hofmann’s rule
Correct Answer: Anti-Markownikov’s rule
Explanation: The presence of organic peroxides causes the ${HBr}$ addition to proceed via a free-radical mechanism instead of an ionic one. This mechanism is known as the peroxide effect and causes the orientation of addition to be Anti-Markownikov, meaning the ${Br}$ atom adds to the carbon of the double bond with the greater number of hydrogen atoms, which is often counterintuitive to the standard ionic reaction.
146. Halogenation of an alkene with ${Cl}_2$ or ${Br}_2$ results in the formation of which type of product?
ⓐ. Vicinal dihalide
ⓑ. Geminal dihalide
ⓒ. Alkyne
ⓓ. Alkyl halide
Correct Answer: Vicinal dihalide
Explanation: The addition of a halogen molecule (${X}_2$) across the double bond of an alkene results in the formation of a dihalogenated compound where the two halogen atoms are attached to adjacent carbon atoms. Such a compound is called a vicinal dihalide (${R}-{CHX}-{CHX}-{R}$). A geminal dihalide has both halogens on the same carbon atom.
147. Which product is formed when ${H}_2{O}$ is added to a terminal alkyne like Prop-1-yne in the presence of ${HgSO}_4$ and ${H}_2{SO}_4$?
ⓐ. Ketone (Propanone)
ⓑ. Alcohol (Propan-1-ol)
ⓒ. Alkane (Propane)
ⓓ. Aldehyde (Propanal)
Correct Answer: Ketone (Propanone)
Explanation: The hydration of alkynes is an addition reaction that first produces an enol intermediate, following Markownikov’s rule. The resulting enol is unstable and immediately undergoes tautomerization (an intramolecular proton transfer) to yield a more stable product. For Prop-1-yne, the final product is the ketone, Propanone (Acetone).
148. The addition of ${HBr}$ to 2-Methylpropene (${(CH}_3)_2{C}={CH}_2$) follows Markownikov’s rule. What is the structure of the carbocation intermediate?
ⓐ. A primary carbocation (${CH}_3{CH}^+{CH}_2{CH}_3$)
ⓑ. A secondary carbocation (${CH}_3{CH}^+{CH}_2{CH}_3$)
ⓒ. A tertiary carbocation (${(CH}_3)_2{C}^+{CH}_3$)
ⓓ. A primary carbocation (${(CH}_3)_2{CHCH}_2^+$)
Correct Answer: A tertiary carbocation (${(CH}_3)_2{C}^+{CH}_3$)
Explanation: In the first step of hydrohalogenation, the proton (${H}^+$) adds to the carbon that produces the most stable carbocation. For 2-Methylpropene, adding ${H}^+$ to ${C}2$ would form a primary carbocation (${(CH}_3)_2{CHCH}_2^+$), while adding ${H}^+$ to ${C}1$ forms a tertiary carbocation (${(CH}_3)_2{C}^+{CH}_3$). Since tertiary carbocations are the most stable, this is the favored intermediate.
149. The decolorization of an aqueous solution of cold, dilute, alkaline potassium permanganate (${Baeyer}'{s}$ reagent) is a characteristic test for the presence of:
ⓐ. Alkanes
ⓑ. Aromatics
ⓒ. Unsaturated hydrocarbons (Alkenes/Alkynes)
ⓓ. Saturated hydrocarbons
Correct Answer: Unsaturated hydrocarbons (Alkenes/Alkynes)
Explanation: ${Baeyer}'{s}$ test is used to detect unsaturation. The purple color of the permanganate ion (${MnO}_4^-$) is discharged as it reacts with the double or triple bond (${C}={C}$ or ${C}\equiv {C}$) to form a colorless diol (glycol) and a brown precipitate of manganese dioxide (${MnO}_2$). Alkanes and saturated compounds do not react under these mild conditions.
150. Which type of addition is characteristic of the reaction of ${Br}_2$ with an alkene in an inert solvent?
ⓐ. ${Syn-addition}$ (groups added to the same side)
ⓑ. ${Anti-addition}$ (groups added to opposite sides)
ⓒ. ${Syn}$ or ${Anti}$ depending on the temperature
ⓓ. ${Syn}$ or ${Anti}$ depending on the solvent
Correct Answer: ${Anti-addition}$ (groups added to opposite sides)
Explanation: The mechanism of halogenation involves the formation of a cyclic bromonium ion intermediate. The second bromide ion (${Br}^-$) must then attack this ring from the side opposite to the existing bromine atom. This stereochemical requirement forces the two new ${Br}$ atoms to add to the double bond on opposite sides of the molecule, resulting in ${anti-addition}$.
151. The primary purpose of ozonolysis (${O}_3$ followed by ${Zn}/{H}_2{O}$) in alkene chemistry is to:
ⓐ. Cleave the carbon-carbon double bond to form carbonyl compounds.
ⓑ. Add two hydroxyl (${-OH}$) groups across the double bond (glycol formation).
ⓒ. Convert the alkene into an alkane (reduction).
ⓓ. Form a cyclic ether (epoxide).
Correct Answer: Cleave the carbon-carbon double bond to form carbonyl compounds.
Explanation: Ozonolysis is a powerful analytical and synthetic reaction designed to specifically and cleanly break the ${C}={C}$ double bond and replace it with two carbonyl groups (${C}={O}$), yielding aldehydes or ketones.
152. Ozonolysis of Propene (${CH}_3{CH}={CH}_2$) followed by reductive workup yields which two main products?
ⓐ. Formaldehyde and Acetone
ⓑ. Ethanal and Formaldehyde
ⓒ. Methanol and Ethanol
ⓓ. Ethanal and Acetone
Correct Answer: Ethanal and Formaldehyde
Explanation: Propene is cleaved into two fragments at the double bond. The ${CH}_3{CH}$ part forms ${Ethanal}$ (${CH}_3{CHO}$), and the ${CH}_2$ part forms ${Formaldehyde}$ (${HCHO}$).
153. When But-2-ene (${CH}_3{CH}={CHCH}_3$) undergoes ozonolysis, the single organic product obtained after reductive workup is:
ⓐ. Formaldehyde
ⓑ. Propionaldehyde
ⓒ. Acetone
ⓓ. Ethanal (Acetaldehyde)
Correct Answer: Ethanal (Acetaldehyde)
Explanation: But-2-ene is a symmetrical alkene that cleaves into two identical fragments (${CH}_3{CH}$). Each fragment forms ${Ethanal}$ (${CH}_3{CHO}$), resulting in two moles of the same product.
154. What are the products of the ozonolysis of 2-Methylbut-2-ene (${CH}_3{C}({CH}_3)={CHCH}_3$)?
ⓐ. Ethanal and Propanone (Acetone)
ⓑ. Acetone and Propanal
ⓒ. Only Acetone
ⓓ. Ethanal and Ethanal
Correct Answer: Ethanal and Propanone (Acetone)
Explanation: The double bond cleaves between ${C}2$ and ${C}3$. The left side (${CH}_3{C}({CH}_3)$) forms a ketone, ${Propanone}$ (Acetone), while the right side (${CHCH}_3$) forms an aldehyde, ${Ethanal}$.
155. Ozonolysis of Cyclohexene followed by reductive workup yields a single product, which is a:
ⓐ. Cyclic alkane
ⓑ. Diol
ⓒ. ${Hexane}$-1,6-dial (Adipaldehyde)
ⓓ. ${Hexan}$-2-one
Correct Answer: ${Hexane}$-1,6-dial (Adipaldehyde)
Explanation: Ozonolysis of a cyclic alkene cleaves the ring at the double bond but does not break the carbon chain. Cyclohexene (${C}_6$ ring) breaks to form a straight-chain six-carbon dialdehyde, ${Hexane}$-1,6-dial (${CHO}({CH}_2)_4{CHO}$).
156. The alkene that yields only ${Propanone}$ (Acetone) upon ozonolysis is:
ⓐ. 2-Methylpropene
ⓑ. 2,3-Dimethylbut-2-ene
ⓒ. Pent-2-ene
ⓓ. 3-Methylbut-1-ene
Correct Answer: 2,3-Dimethylbut-2-ene
Explanation: An alkene that yields only one product must be symmetrical, and the product must be a ketone (since ${HCHO}$ is not produced). 2,3-Dimethylbut-2-ene (${(CH}_3)_2{C}={C}({CH}_3)_2$) cleaves symmetrically to yield two moles of ${Propanone}$ (${CH}_3{COCH}_3$).
157. An alkene yields one mole of ${Formaldehyde}$ and one mole of ${Butanone}$ (Methyl Ethyl Ketone) upon ozonolysis. What was the starting alkene?
ⓐ. 2-Methylbut-1-ene
ⓑ. Pent-2-ene
ⓒ. 2-Methylpent-1-ene
ⓓ. 3-Methylpent-2-ene
Correct Answer: 2-Methylbut-1-ene
Explanation: ${Formaldehyde}$ (${HCHO}$) comes from a ${CH}_2$ group, and ${Butanone}$ (${CH}_3{CH}_2{COCH}_3$) comes from a ${CH}_3{CH}_2{C}({CH}_3)$ group. Linking these two fragments at the former carbonyl carbons (${C}={O}$ to ${C}={C}$) gives ${CH}_3{CH}_2{C}({CH}_3)={CH}_2$, which is 2-Methylbut-1-ene.
158. What is the function of the zinc dust (${Zn}$) and water workup (${Zn}/{H}_2{O}$) step after the initial reaction of the alkene with ozone (${O}_3$)?
ⓐ. It performs a reductive cleavage of the unstable ozonide intermediate.
ⓑ. It oxidizes the alkene to form a carboxylic acid.
ⓒ. It promotes the addition of water across the double bond.
ⓓ. It isomerizes the alkene before cleavage.
Correct Answer: It performs a reductive cleavage of the unstable ozonide intermediate.
Explanation: The ${Zn}/{H}_2{O}$ or ${Me}_2{S}$ workup is a reductive workup. Its essential role is to cleave the unstable ozonide and, crucially, to destroy the oxidizing by-product, hydrogen peroxide (${H}_2{O}_2$), which would otherwise further oxidize any resulting aldehydes to carboxylic acids.
159. The intermediate compound formed immediately after the reaction of an alkene with ozone (${O}_3$) but before the workup is called a:
ⓐ. Glycol
ⓑ. Epoxide
ⓒ. Carbocation
ⓓ. Ozonide
Correct Answer: Ozonide
Explanation: The alkene first reacts with ozone to form an unstable intermediate called a molozonide, which rapidly rearranges to the cyclic, explosive intermediate known as an ozonide. This ozonide is then treated with a workup reagent (like ${Zn}/{H}_2{O}$) to yield the final products.
160. Ozonolysis of an unknown alkene yields ${Ethanal}$ (${CH}_3{CHO}$) and ${Propanal}$ (${CH}_3{CH}_2{CHO}$). Identify the structure of the starting alkene.
ⓐ. Hex-3-ene
ⓑ. Pent-2-ene
ⓒ. 2-Methylbut-2-ene
ⓓ. 3-Methylpent-2-ene
Correct Answer: Pent-2-ene
Explanation: To find the starting alkene, combine the two carbonyl products by removing the oxygen atoms and forming a double bond between the two carbon atoms that held the oxygen. ${Ethanal}$ has two carbons, and ${Propanal}$ has three carbons. Joining them gives a five-carbon straight-chain alkene with the double bond at ${C}2$: Pent-2-ene (${CH}_3{CH}={CHCH}_2{CH}_3$).
161. Polymerization is generally defined as a chemical process where:
ⓐ. Small organic molecules are broken down into simpler compounds.
ⓑ. Two different molecules combine to form a single, larger molecule with the elimination of a small molecule.
ⓒ. Many small molecules (monomers) combine to form a single large chain molecule (polymer).
ⓓ. An atom or group is substituted by another atom or group in an existing molecule.
Correct Answer: Many small molecules (monomers) combine to form a single large chain molecule (polymer).
Explanation: Polymerization is the process of linking a large number of simple, repeating units called monomers (such as ethene or chloroethene) together through covalent bonds to form a long-chain macromolecule known as a polymer. Alkenes undergo addition polymerization, where no small molecules are eliminated.
162. Which of the following is the monomer used to produce Polyvinyl Chloride (PVC)?
ⓐ. Ethene
ⓑ. Propene
ⓒ. Chloroethene (Vinyl Chloride)
ⓓ. Tetrafluoroethene
Correct Answer: Chloroethene (Vinyl Chloride)
Explanation: Polyvinyl Chloride is a polymer abbreviated as PVC. The repeating unit, or monomer, of this polymer is Chloroethene, which is commonly known by its trivial name, Vinyl Chloride (${CH}_2={CHCl}$).
163. Polyethylene (PE) is formed from the monomer Ethene. What is the repeating structural unit in the Polyethylene chain?
ⓐ. ${-(CH}_2-{CH}_2{-)}_n$
ⓑ. ${-(CH}_2-{CH}{-)}_n$
ⓒ. ${-(CH}_2-{CHCl}{-)}_n$
ⓓ. ${-(CH}={CH}{-)}_n$
Correct Answer: ${-(CH}_2-{CH}_2{-)}_n$
Explanation: The monomer ${Ethene}$ (${CH}_2={CH}_2$) breaks its $\pi$ bond and links with other monomers. The repeating unit in the resulting polymer, ${Polyethylene}$, is the original molecule with the double bond replaced by two single bonds, written as the dimethylene unit: ${-(CH}_2-{CH}_2{-)}_n$.
164. The formation of Polyethylene from Ethene is an example of which type of polymerization reaction?
ⓐ. Condensation polymerization
ⓑ. Step-growth polymerization
ⓒ. Addition polymerization
ⓓ. Cationic polymerization
Correct Answer: Addition polymerization
Explanation: In Addition Polymerization, monomers with a double bond (like alkenes) simply add to each other without the elimination of any small molecules (such as water or ammonia). This is the key difference from condensation polymerization, which does eliminate small molecules.
165. The two main types of Polyethylene, Low-Density Polyethylene (LDPE) and High-Density Polyethylene (HDPE), primarily differ in their:
ⓐ. Monomer source
ⓑ. Color and smell
ⓒ. Molecular weight alone
ⓓ. Degree of branching and resulting density
Correct Answer: Degree of branching and resulting density
Explanation: ${LDPE}$ is produced via high pressure and high temperature, resulting in significant short and long-chain branching, which prevents close packing and gives it low density. ${HDPE}$ is produced at lower pressure with Ziegler-Natta catalysts, resulting in a more linear chain that packs tightly, leading to high density.
166. Polyvinyl Chloride (PVC) is considered a thermoplastic material, meaning it:
ⓐ. Decomposes irreversibly upon heating.
ⓑ. Can be repeatedly softened by heating and solidified by cooling.
ⓒ. Does not melt or soften when heated.
ⓓ. Is primarily used for electrical insulation only.
Correct Answer: Can be repeatedly softened by heating and solidified by cooling.
Explanation: Thermoplastics are polymers whose chains are held together by relatively weak intermolecular forces. When heated, these forces are overcome, allowing the polymer to soften or melt. This allows the material to be reshaped, and upon cooling, it solidifies again. ${PVC}$ is a common thermoplastic.
167. Which catalyst system is typically used in the industrial production of ${HDPE}$ to create a highly linear polymer chain with minimal branching?
ⓐ. Peroxides (Free Radical Initiator)
ⓑ. ${AlCl}_3$ (Lewis Acid)
ⓒ. Concentrated ${H}_2{SO}_4$ (Strong Acid)
ⓓ. Ziegler-Natta catalysts (${TiCl}_4/{Al}({C}_2{H}_5)_3$)
Correct Answer: Ziegler-Natta catalysts (${TiCl}_4/{Al}({C}_2{H}_5)_3$)
Explanation: High-Density Polyethylene (${HDPE}$) is produced using specialized Ziegler-Natta catalysts (typically a mixture of titanium tetrachloride and an organoaluminum compound). These catalysts allow for polymerization at low temperatures and pressures, leading to highly linear, unbranched polymer chains.
168. The primary difference between a homopolymer (like polyethylene) and a copolymer is:
ⓐ. The number of carbon atoms in the polymer chain.
ⓑ. The use of a catalyst during polymerization.
ⓒ. The number of different types of monomers used to form the chain.
ⓓ. Whether the polymer is formed via addition or condensation.
Correct Answer: The number of different types of monomers used to form the chain.
Explanation: A homopolymer (e.g., Polyethylene from only ${Ethene}$) is formed from a single type of monomer. A copolymer is formed from the polymerization of two or more different types of monomers (e.g., Styrene and Butadiene).
169. What is the approximate range of the degree of polymerization (number of monomer units, $n$) in commercial polyethylene?
ⓐ. $n = 10 { to } 100$
ⓑ. $n = 1000 { to } 50,000$ or more
ⓒ. $n = 1 { to } 5$
ⓓ. $n = 100 { to } 1000$
Correct Answer: $n = 1000 { to } 50,000$ or more
Explanation: Commercial polymers are macromolecules with very high molecular weights. The degree of polymerization for common polyethylene products typically ranges from $1,000$ up to $50,000$ or even hundreds of thousands of monomer units, resulting in a very long chain molecule.
170. What is the hybridization of the carbon atoms in the Polyethylene polymer chain?
ⓐ. ${sp}^2$
ⓑ. ${sp}$
ⓒ. ${sp}^3$
ⓓ. ${d}^2{sp}^3$
Correct Answer: ${sp}^3$
Explanation: The monomer ${Ethene}$ has ${sp}^2$ hybridized carbons. During addition polymerization, the $\pi$ bond is broken, and each carbon atom forms two new $\sigma$ bonds to adjacent monomer units. In the final, saturated polymer chain, all carbon atoms are linked by single bonds, making them ${sp}^3$ hybridized and resulting in a tetrahedral geometry around each carbon.
171. What is the general molecular formula for an acyclic alkyne containing one carbon-carbon triple bond?
ⓐ. $C_nH_{2n-2}$
ⓑ. $C_nH_{2n+2}$
ⓒ. $C_nH_{2n}$
ⓓ. $C_nH_{n}$
Correct Answer: $C_nH_{2n-2}$
Explanation: Alkynes are unsaturated hydrocarbons containing one carbon-carbon triple bond (${C}\equiv {C}$). The triple bond represents two degrees of unsaturation, meaning they have four fewer hydrogen atoms than the corresponding saturated alkane ($C_nH_{2n+2}$). Thus, the general formula for a simple, acyclic alkyne is $C_nH_{2n-2}$. This formula is also shared by dienes (compounds with two double bonds).
172. The carbon atoms involved in the ${C}\equiv {C}$ triple bond of an alkyne are linked by which combination of bonds?
ⓐ. Two $\sigma$ bonds and one $\pi$ bond
ⓑ. One $\sigma$ bond and two $\pi$ bonds
ⓒ. Three $\sigma$ bonds
ⓓ. Three $\pi$ bonds
Correct Answer: One $\sigma$ bond and two $\pi$ bonds
Explanation: A triple bond consists of one strong sigma ($\sigma$) bond formed by the axial overlap of ${sp}$ hybrid orbitals, and two weaker pi ($\pi$) bonds formed by the lateral (sideways) overlap of the two unhybridized ${p}$ orbitals on each carbon atom. These two $\pi$ bonds lie in planes perpendicular to each other and to the $\sigma$ bond.
173. What is the minimum number of carbon atoms required for a compound to be classified as an alkyne?
ⓐ. 1
ⓑ. 2
ⓒ. 3
ⓓ. 4
Correct Answer: 2
Explanation: An alkyne requires the presence of a carbon-carbon triple bond (${C}\equiv {C}$). Therefore, the smallest possible alkyne must contain two carbon atoms. The simplest alkyne is Ethyne (${C}_2{H}_2$), commonly known as Acetylene.
174. The hybridization of the carbon atoms involved in the triple bond of an alkyne is:
ⓐ. ${sp}^3$
ⓑ. ${sp}^2$
ⓒ. ${sp}$
ⓓ. ${d}^2{sp}^3$
Correct Answer: ${sp}$
Explanation: In an alkyne, each carbon atom involved in the triple bond needs two hybrid orbitals for the $\sigma$ bonds (one to the other carbon and one to an attached hydrogen or alkyl group) and two unhybridized ${p}$ orbitals for the $\pi$ bonds. This structure requires the mixing of one ${s}$ and one ${p}$ orbital, resulting in ${sp}$ hybridization.
175. What is the molecular formula for Hex-1-yne?
ⓐ. $C_6H_{10}$
ⓑ. $C_6H_{12}$
ⓒ. $C_6H_{14}$
ⓓ. $C_5H_{8}$
Correct Answer: $C_6H_{10}$
Explanation: Hex-1-yne is an alkyne with $n=6$ carbon atoms. Using the general formula $C_nH_{2n-2}$:
Hydrogen atoms $= 2(6) – 2 = 12 – 2 = 10$.
Therefore, the molecular formula for Hex-1-yne (and any acyclic alkyne with six carbons) is $C_6H_{10}$.
176. What is the geometry and bond angle around the carbon atoms of the ${C}\equiv {C}$ triple bond in a simple alkyne?
ⓐ. Trigonal planar, $120^\circ$
ⓑ. Tetrahedral, $109.5^\circ$
ⓒ. Linear, $180^\circ$
ⓓ. Bent, $104.5^\circ$
Correct Answer: Linear, $180^\circ$
Explanation: The ${sp}$ hybridized carbon atoms have two ${sp}$ orbitals oriented $180^\circ$ apart to minimize electron repulsion. This orientation dictates a linear geometry for the ${C}-{C}\equiv {C}$ fragment with a bond angle of $180^\circ$.
177. Alkynes are classified as terminal or internal based on:
ⓐ. The total number of carbon atoms in the chain.
ⓑ. The hybridization of the carbon atoms.
ⓒ. Whether they are acyclic or cyclic.
ⓓ. Whether the triple bond is located at the end or in the middle of the carbon chain.
Correct Answer: Whether the triple bond is located at the end or in the middle of the carbon chain.
Explanation: A terminal alkyne has the triple bond located at the end of the carbon chain (i.e., between ${C}1$ and ${C}2$), meaning the ${C}1$ carbon is bonded to one hydrogen atom (${R}-{C}\equiv {CH}$). An internal alkyne has the triple bond located elsewhere in the chain (${R}-{C}\equiv {C}-{R}’$), with no hydrogen atoms directly bonded to the ${sp}$ carbons.
178. The carbon-carbon triple bond length in Ethyne ($\approx 1.20 { Å}$) compared to the double bond length in Ethene ($\approx 1.34 { Å}$) and the single bond length in Ethane ($\approx 1.54 { Å}$) is:
ⓐ. The longest, due to weak $\pi$ bonds.
ⓑ. Shorter than the single bond but longer than the double bond.
ⓒ. Equivalent to the single bond length.
ⓓ. The shortest, due to a high degree of ${s}$-character and three shared electron pairs.
Correct Answer: The shortest, due to a high degree of ${s}$-character and three shared electron pairs.
Explanation: The triple bond is the shortest of the ${C}-{C}$ bonds. This is due to two main factors: the presence of three shared electron pairs pulling the nuclei closer, and the high $50\%$ s-character in the ${sp}$ hybrid orbitals, which results in tighter orbitals and shorter, stronger $\sigma$ bonds.
179. A compound with the molecular formula $C_5H_8$ can represent either a simple alkyne or:
ⓐ. A simple alkene
ⓑ. A cycloalkane
ⓒ. A diene (two double bonds)
ⓓ. An aromatic compound
Correct Answer: A diene (two double bonds)
Explanation: The formula $C_5H_8$ has two degrees of unsaturation (${D.U.}=2$). A ${D.U.}$ of 2 can be satisfied by:
1. One triple bond (simple alkyne, e.g., Pent-1-yne).
2. Two double bonds (a diene, e.g., Penta-1,3-diene).
3. One ring and one double bond (a cycloalkene, e.g., Cyclopentene).
Since option C provides the simplest alternative structure for this formula, it is the correct choice.
180. How many total $\sigma$ bonds and $\pi$ bonds are present in a molecule of But-2-yne (${CH}_3{C}\equiv {CCH}_3$)?
ⓐ. $6\sigma$ and $2\pi$
ⓑ. $9\sigma$ and $2\pi$
ⓒ. $10\sigma$ and $1\pi$
ⓓ. $8\sigma$ and $2\pi$
Correct Answer: $9\sigma$ and $2\pi$
Explanation: But-2-yne (${C}_4{H}_6$).
Total $\pi$ bonds: The triple bond contains 2 $\pi$ bonds.
Total $\sigma$ bonds: The total number of atoms is $4 ({C}) + 6 ({H}) = 10$. The total number of $\sigma$ bonds in a non-cyclic molecule is always (Number of atoms – 1) $= 10 – 1 = 9$.
Alternatively: $3 { C}-{H}$ (${CH}_3$) $+ 1 { C}-{C}$ (${CH}_3$ to ${sp}$ carbon) $+ 1 { C}\equiv {C}$ ($\sigma$ part) $+ 1 { C}-{C}$ (${sp}$ carbon to ${CH}_3$) $+ 3 { C}-{H}$ (${CH}_3$) $= 9 \sigma$ bonds.
Therefore, the molecule has $9\sigma$ and $2\pi$ bonds.
181. The preparation of an alkyne from a vicinal dihalide involves the sequential elimination of two molecules of ${HX}$. Which type of reaction mechanism is involved in these elimination steps?
ⓐ. Nucleophilic substitution (${S}_{{N}}2$)
ⓑ. Free-radical addition
ⓒ. Beta-elimination ($\beta$-elimination)
ⓓ. Alpha-elimination ($\alpha$-elimination)
Correct Answer: Beta-elimination ($\beta$-elimination)
Explanation: The removal of a halogen (${X}$) and a hydrogen (${H}$) from adjacent carbon atoms (the $\alpha$ and $\beta$ carbons) to form a new $\pi$ bond is the defining characteristic of a $\beta$-elimination reaction, which is typically an ${E}2$ mechanism.
182. Which reagent is typically used for the first dehydrohalogenation step when converting a vicinal dihalide (e.g., 1,2-Dibromopropane) into a haloalkene?
ⓐ. Alcoholic ${KOH}$ and heat
ⓑ. Concentrated ${H}_2{SO}_4$
ⓒ. Aqueous ${KOH}$
ⓓ. ${Na}$ metal in liquid ${NH}_3$
Correct Answer: Alcoholic ${KOH}$ and heat
Explanation: Alcoholic (${ethanolic}$) ${KOH}$ is a strong base that, when heated, promotes the first ${E}2$ elimination of ${HX}$ over the substitution (${S}_{{N}}$) pathway that aqueous ${KOH}$ would favor.
183. The second elimination step, converting a haloalkene into an alkyne, is often difficult because it removes a hydrogen from a carbon that is part of a double bond. What extremely strong base is required to complete this step, especially for terminal alkynes?
ⓐ. Lithium aluminum hydride (${LiAlH}_4$)
ⓑ. Sodium hydroxide (${NaOH}$)
ⓒ. Sodium amide (${NaNH}_2$) in liquid ammonia
ⓓ. Zinc metal dust (${Zn}$)
Correct Answer: Sodium amide (${NaNH}_2$) in liquid ammonia
Explanation: Due to the high basicity needed to abstract the second, less acidic proton, and to prevent unwanted substitution reactions, a powerful base like Sodium amide (${NaNH}_2$) in an inert solvent like liquid ammonia is necessary to complete the formation of the triple bond.
184. The dehydrohalogenation of a terminal geminal dihalide (1,1-dihalide) primarily yields which type of alkyne?”
ⓐ. Internal alkyne
ⓑ. Diene
ⓒ. Cyclic alkyne
ⓓ. Terminal alkyne
Correct Answer: Terminal alkyne
Explanation: The structure of a geminal dihalide (${R}-{CH}_2-{CHX}_2$) ensures that the final ${C}\equiv {C}$ triple bond forms at the end of the chain, resulting in a terminal alkyne (${R}-{C}\equiv {CH}$) after the double elimination.
185. Starting with 1,1-Dichloropropane and performing double dehydrohalogenation, what is the final alkyne product?
ⓐ. But-1-yne
ⓑ. Prop-1-yne
ⓒ. Prop-2-yne
ⓓ. Ethyne
Correct Answer: Prop-1-yne
Explanation: 1,1-Dichloropropane has three carbon atoms. Double dehydrohalogenation only removes the halogens and adjacent hydrogens, but does not alter the carbon chain length. Therefore, the product must be the three-carbon alkyne, Prop-1-yne (${CH}_3{C}\equiv {CH}$).
186. Kolbe’s electrolytic method for the preparation of alkynes involves the electrolysis of the potassium or sodium salt of which specific class of organic compounds?
ⓐ. Unsaturated dicarboxylic acids
ⓑ. Saturated monocarboxylic acids
ⓒ. Saturated dicarboxylic acids
ⓓ. $\beta$-Keto acids
Correct Answer: Unsaturated dicarboxylic acids
Explanation: To synthesize an alkyne, the electrolysis must lead to the formation of a triple bond. This is achieved through the double decarboxylation and radical coupling of salts of unsaturated dicarboxylic acids, such as maleic or fumaric acid.
187. In Kolbe’s synthesis of Ethyne (${C}_2{H}_2$), the starting material is the potassium or sodium salt of which parent acid?
ⓐ. Propanoic acid
ⓑ. Ethanoic acid
ⓒ. Oxalic acid
ⓓ. Maleic acid (or Fumaric acid)
Correct Answer: Maleic acid (or Fumaric acid)
Explanation: The ${Ethyne}$ product has two carbons linked by a triple bond. This is formed by the elimination of the two carboxylate groups from the two carbons of an unsaturated ${C}_4$ acid, specifically ${Maleic Acid}$ (cis-But-2-endioic acid) or ${Fumaric Acid}$ (trans-But-2-endioic acid).
188. Which of the following intermediates is formed at the anode during the Kolbe’s electrolytic synthesis of alkynes?
ⓐ. Carbocation
ⓑ. Carbonyl ion
ⓒ. Free radical
ⓓ. Carbanion
Correct Answer: Free radical
Explanation: Similar to alkane synthesis, the process involves the oxidation of the carboxylate ion at the anode, which leads to the formation of a carboxyl free radical (${RCOO}\bullet$). This radical then quickly loses ${CO}_2$ to generate another radical that ultimately couples or eliminates to form the alkyne.
189. Dehydrohalogenation of an unsymmetrical vicinal dihalide (e.g., 1,2-Dibromobutane) will preferentially form the internal alkyne (But-2-yne) over the terminal alkyne (But-1-yne). This is because the internal alkyne is:
ⓐ. Less acidic than the terminal alkyne.
ⓑ. Less prone to nucleophilic attack.
ⓒ. Easier to purify.
ⓓ. More substituted and therefore more stable.
Correct Answer: More substituted and therefore more stable.
Explanation: The formation of the more stable intermediate and final product is favored. Internal alkynes (where ${R}$ groups replace ${H}$ on the ${sp}$ carbons) are generally more substituted and thus more thermodynamically stable than terminal alkynes, making them the major product of elimination reactions.
190. After an alkyne salt (${R}-{C}\equiv {C}^-{Na}^+$) is formed using ${NaNH}_2$, how is the neutral terminal alkyne (${R}-{C}\equiv {CH}$) typically regenerated?
ⓐ. By treatment with ${H}_2$ gas and a catalyst.
ⓑ. By heating it vigorously.
ⓒ. By neutralization with water or a dilute mineral acid.
ⓓ. By reaction with an alkyl halide.
Correct Answer: By neutralization with water or a dilute mineral acid.
Explanation: Alkyne salts are strong bases. They readily undergo a simple acid-base reaction with a weak acid like water or a dilute mineral acid to become protonated (${R}-{C}\equiv {C}^- + {H}_2{O} \rightarrow {R}-{C}\equiv {CH} + {OH}^-$), thus regenerating the neutral alkyne.
191. The terminal hydrogen atom of ethyne (${HC}\equiv {CH}$) is acidic because the carbon atom involved is:
ⓐ. ${sp}$ hybridized, which has high s-character, making the ${C}-{H}$ bond polar.
ⓑ. ${sp}^2$ hybridized, which stabilizes the resulting carbanion.
ⓒ. Less electronegative than the carbon atoms in alkanes.
ⓓ. Unable to participate in resonance stabilization of the conjugate base.
Correct Answer: ${sp}$ hybridized, which has high s-character, making the ${C}-{H}$ bond polar.
Explanation: The ${sp}$ orbital has $\mathbf{50\%}$ ${s}$-character. This high ${s}$-character makes the ${sp}$ hybridized carbon highly electronegative compared to ${sp}^2$ or ${sp}^3$ carbons. This increased electronegativity strongly pulls the electron density from the ${C}-{H}$ bond, making the hydrogen atom sufficiently acidic to be removed by a strong base.
192. Which reagent can be used to distinguish between a terminal alkyne (like Prop-1-yne) and an internal alkyne (like But-2-yne)?
ⓐ. Bromine water (${Br}_2$ in ${CCl}_4$)
ⓑ. Tollen’s reagent (Ammoniacal Silver Nitrate)
ⓒ. Concentrated Sulfuric Acid (${H}_2{SO}_4$)
ⓓ. Alcoholic Potassium Hydroxide (${KOH}$)
Correct Answer: Tollen’s reagent (Ammoniacal Silver Nitrate)
Explanation: Only terminal alkynes have the acidic ${C}-{H}$ proton. This proton reacts with the weak base in ${Tollen}'{s}$ reagent to form a solid, white precipitate of a metal acetylide (e.g., silver acetylide). Internal alkynes, lacking the acidic proton, do not react, allowing the two types of alkynes to be distinguished.
193. The first molecule of ${HBr}$ is added to an alkyne following which rule?
ⓐ. Anti-Markownikov’s rule, always
ⓑ. Markownikov’s rule, always
ⓒ. Saytzeff’s rule
ⓓ. Hofmann’s rule
Correct Answer: Markownikov’s rule, always
Explanation: The ionic addition of the first molecule of an unsymmetrical reagent (${HX}$) to an alkyne (hydrohalogenation) follows Markownikov’s rule. The hydrogen atom adds to the carbon of the triple bond that results in the formation of the more stable intermediate (often a secondary vinyl carbocation).
194. The reaction of Ethyne with hydrogen gas (${H}_2$) in the presence of Nickel (${Ni}$) catalyst yields which final product?
ⓐ. Ethene (${CH}_2={CH}_2$)
ⓑ. But-1-ene (${CH}_2={CHCH}_2{CH}_3$)
ⓒ. Vinyl chloride (${CH}_2={CHCl}$)
ⓓ. Ethane (${CH}_3{CH}_3$)
Correct Answer: Ethane (${CH}_3{CH}_3$)
Explanation: Using a highly active catalyst like Nickel (${Ni}$), Platinum (${Pt}$), or Palladium (${Pd}$) under standard conditions causes the complete hydrogenation of the alkyne, adding two moles of ${H}_2$. Ethyne is converted first to ethene and then immediately to the fully saturated alkane, Ethane.
195. Which catalyst is used to selectively stop the hydrogenation of an alkyne at the alkene stage (i.e., synthesize an alkene from an alkyne)?
ⓐ. Concentrated ${H}_2{SO}_4$
ⓑ. Nickel (${Ni}$) powder
ⓒ. Lindlar’s catalyst (${Pd}/{CaCO}_3$ poisoned with lead salts or sulfur)
ⓓ. Aluminum Chloride (${AlCl}_3$)
Correct Answer: Lindlar’s catalyst (${Pd}/{CaCO}_3$ poisoned with lead salts or sulfur)
Explanation: To halt the reduction at the alkene stage, a deactivated or “poisoned” catalyst is required. Lindlar’s catalyst is a specific palladium catalyst (often deposited on calcium carbonate and treated with a lead compound or quinoline) that is partially deactivated, allowing it to reduce alkynes but preventing the further reduction of the resulting alkene.
196. The reaction of Prop-1-yne (${CH}_3{C}\equiv {CH}$) with Tollen’s reagent produces:
ⓐ. Propan-2-one
ⓑ. Silver Propan-1-ylide (${CH}_3{C}\equiv {CAg}$)
ⓒ. Propane
ⓓ. Prop-1-ene
Correct Answer: Silver Propan-1-ylide (${CH}_3{C}\equiv {CAg}$)
Explanation: Prop-1-yne is a terminal alkyne with an acidic hydrogen. ${Tollen}'{s}$ reagent (Ammoniacal Silver Nitrate) reacts via an acid-base mechanism to form the insoluble salt, Silver Propan-1-ylide (or Silver Propylide), identified as a white precipitate.
197. The addition of water (${H}_2{O}$) to Ethyne (${C}_2{H}_2$) in the presence of ${HgSO}_4$ and ${H}_2{SO}_4$ yields an intermediate enol, which quickly tautomerizes to form the final product:
ⓐ. Ethanal (Acetaldehyde)
ⓑ. Ethanol
ⓒ. Acetone (Propanone)
ⓓ. Vinyl alcohol
Correct Answer: Ethanal (Acetaldehyde)
Explanation: The hydration of Ethyne follows Markownikov’s rule (though symmetrical here) and yields an unstable enol, Vinyl alcohol (${CH}_2={CHOH}$). This enol rapidly tautomerizes (a proton migrates) to form the final, stable product, the aldehyde, Ethanal (${CH}_3{CHO}$).
198. The reaction of an internal alkyne with ${Br}_2$ in an inert solvent results in the formation of a product with which stereochemistry?
ⓐ. ${Syn}$-addition (halogens added to the same side)
ⓑ. ${Anti}$-addition (halogens added to the opposite sides)
ⓒ. A racemic mixture of addition and substitution products
ⓓ. No reaction, as internal alkynes are inert to halogens
Correct Answer: ${Anti}$-addition (halogens added to the opposite sides)
Explanation: Similar to alkenes, the halogenation of alkynes proceeds via an electrophilic addition mechanism involving a cyclic halonium ion intermediate. The second halogen ion attacks from the opposite face of the triple bond, resulting in ${anti}$-addition to form a trans-dihaloalkene.
199. Treatment of a terminal alkyne with a strong base like ${NaNH}_2$ followed by reaction with an alkyl halide (${R}'{X}$) is an effective method for:
ⓐ. Reducing the alkyne to an alkane.
ⓑ. Converting the terminal alkyne into a ketone.
ⓒ. Cleaving the alkyne into two smaller fragments.
ⓓ. Increasing the carbon chain length of the alkyne (alkylation).
Correct Answer: Increasing the carbon chain length of the alkyne (alkylation).
Explanation: The initial reaction forms the nucleophilic acetylide ion (${R}-{C}\equiv {C}^-$). This ion is a strong nucleophile and can attack an alkyl halide in an ${S}_{{N}}2$ reaction, resulting in a carbon-carbon bond formation and the synthesis of a new, larger alkyne, thereby increasing the carbon chain length (alkylation).
200. When Prop-1-yne is hydrated (${H}_2{O}, {HgSO}_4, {H}_2{SO}_4$), the final product is ${Propanone}$ (Acetone). This is because the initial ${OH}$ addition follows Markownikov’s rule, placing the ${OH}$ group on:
ⓐ. The terminal ${C}$ atom (${C}1$), leading to an aldehyde.
ⓑ. The internal ${C}$ atom (${C}2$), leading to an aldehyde.
ⓒ. The internal ${C}$ atom (${C}2$), leading to an unstable enol that forms a ketone.
ⓓ. The terminal ${C}$ atom (${C}1$), leading to an unstable enol that forms an acid.
Correct Answer: The internal ${C}$ atom (${C}2$), leading to an unstable enol that forms a ketone.
Explanation: Markownikov’s addition places the ${OH}$ group on the internal, more substituted carbon (${C}2$), resulting in the enol ${CH}_3{C}({OH})={CH}_2$. Tautomerization of this enol leads to the final, stable ketone product, Propanone (${CH}_3{COCH}_3$).
You are now on Class 11 Chemistry MCQs – Chapter 13: Hydrocarbons (Part 2). This part continues your journey into the world of hydrocarbons. The focus here is on reactions and properties of alkenes and alkynes, including hydrogenation, halogenation, and addition reactions.
Topics covered in Part 2 (100 MCQs): Reactions of alkenes and alkynes, polymerization, uses of alkenes in industry, and the significance of functional groups like alcohols and ethers derived from hydrocarbons.
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