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101. The electromeric effect occurs when:
ⓐ. A pi bond is shifted to an adjacent atom in the presence of an attacking reagent
ⓑ. Electrons are delocalized in conjugated systems
ⓒ. Electrons are donated from a sigma bond to an adjacent empty orbital
ⓓ. A molecule forms a permanent dipole moment
Correct Answer: A pi bond is shifted to an adjacent atom in the presence of an attacking reagent
Explanation: The electromeric effect refers to the shift of electron density in a pi bond to an adjacent atom when an attacking reagent, such as an electrophile or nucleophile, interacts with the molecule. This effect is temporary and reversible, unlike resonance, which involves the delocalization of electrons over a longer period.
102. Which of the following is an example of the electromeric effect?
ⓐ. The movement of electron density in a benzene ring
ⓑ. The shift of electrons in the double bond of ethene when an electrophile attacks
ⓒ. The donation of electron density from a lone pair in amines
ⓓ. The electron withdrawal by a nitro group in an aromatic compound
Correct Answer: The shift of electrons in the double bond of ethene when an electrophile attacks
Explanation: The electromeric effect is observed when a pi bond in a molecule, such as ethene, shifts electrons to one of the atoms involved in the bond in response to the attack by an electrophile. This causes the molecule to temporarily lose its double bond character as the electron density is pushed to one atom.
103. The electromeric effect is:
ⓐ. A permanent shift of electron density in a molecule
ⓑ. A type of hybridization in molecules
ⓒ. A type of resonance involving sigma bonds
ⓓ. A temporary shift of electron density in response to an external force
Correct Answer: A temporary shift of electron density in response to an external force
Explanation: The electromeric effect is temporary and occurs when a molecule responds to an external reagent, typically an electrophile or nucleophile, by shifting electron density from a pi bond to an adjacent atom. Unlike resonance, which is a stable distribution of electrons, the electromeric effect is a temporary phenomenon.
104. Which of the following functional groups can exhibit the electromeric effect?
ⓐ. Alcohols
ⓑ. Alkenes
ⓒ. Alkynes
ⓓ. Both B and C
Correct Answer: Both B and C
Explanation: Both alkenes and alkynes can exhibit the electromeric effect because they contain pi bonds that can shift electron density when attacked by electrophiles or nucleophiles. Alcohols, on the other hand, do not have the required pi bonds for the electromeric effect to occur.
105. The electromeric effect is typically observed in:
ⓐ. Saturated compounds with no pi bonds
ⓑ. Compounds with conjugated pi bonds
ⓒ. Compounds with only sigma bonds
ⓓ. Aromatic compounds with resonance structures
Correct Answer: Compounds with conjugated pi bonds
Explanation: The electromeric effect is most commonly observed in compounds with conjugated pi bonds, such as alkenes or alkynes. These conjugated systems allow for the temporary shift of electron density in response to an external force, such as the attack of an electrophile.
106. Which of the following is the main characteristic of the electromeric effect?
ⓐ. Permanent delocalization of electron density
ⓑ. Transfer of electrons through a sigma bond
ⓒ. Temporary shift of electrons to a particular atom in the molecule
ⓓ. Movement of electrons in a conjugated system without any external influence
Correct Answer: Temporary shift of electrons to a particular atom in the molecule
Explanation: The electromeric effect is characterized by the temporary shift of electron density from a pi bond to an adjacent atom when a reagent, such as an electrophile, interacts with the molecule. This effect is reversible and is different from resonance, which involves a more stable delocalization of electrons.
107. Which of the following compounds will most likely exhibit the electromeric effect?
ⓐ. Ethene
ⓑ. Butane
ⓒ. Methanol
ⓓ. Benzene
Correct Answer: Ethene
Explanation: Ethene (C₂H₄) contains a double bond (pi bond) between the two carbon atoms. In the presence of an electrophile, the pi electrons of the double bond can shift temporarily to one of the carbon atoms, demonstrating the electromeric effect. Butane, methanol, and benzene do not exhibit this effect due to their lack of conjugated pi bonds or reactivity.
108. Which of the following is true about the electromeric effect in conjugated systems?
ⓐ. It is permanent and leads to the formation of a stable resonance structure
ⓑ. It occurs only in the presence of an external reagent, such as an electrophile
ⓒ. It involves the donation of lone pairs from heteroatoms
ⓓ. It stabilizes the molecule by distributing electron density evenly
Correct Answer: It occurs only in the presence of an external reagent, such as an electrophile
Explanation: The electromeric effect occurs in conjugated systems when an external reagent, such as an electrophile, interacts with the molecule, causing a temporary shift of electron density from the pi bond to an adjacent atom. This effect is not permanent and is reversed once the reagent is no longer interacting with the molecule.
109. In which of the following reactions is the electromeric effect most likely to be observed?
ⓐ. Electrophilic addition to an alkene
ⓑ. Nucleophilic substitution of an alkyl halide
ⓒ. Free radical substitution of an alkane
ⓓ. Dehydration of an alcohol
Correct Answer: Electrophilic addition to an alkene
Explanation: The electromeric effect is most likely to be observed in electrophilic addition reactions, such as the addition of HX (e.g., HCl, HBr) to an alkene. The pi bond of the alkene shifts electron density to one of the carbon atoms, making it more susceptible to attack by the electrophile.
110. Which of the following statements best describes the effect of the electromeric shift on the stability of a molecule?
ⓐ. It increases the stability of the molecule by delocalizing electron density
ⓑ. It has no effect on the stability of the molecule
ⓒ. It destabilizes the molecule by concentrating electron density on one atom
ⓓ. It stabilizes the molecule by donating electrons from a lone pair
Correct Answer: It destabilizes the molecule by concentrating electron density on one atom
Explanation: The electromeric shift causes a temporary concentration of electron density on one atom, which can destabilize the molecule, especially if this leads to an electron-deficient center. The effect is temporary and usually occurs to facilitate the reaction with an external reagent, such as an electrophile.
111. Which of the following is an example of an electrophilic substitution reaction?
ⓐ. Nucleophilic substitution in alkyl halides
ⓑ. Substitution of a hydrogen atom with a halogen in benzene
ⓒ. Addition of a nucleophile to an alkene
ⓓ. Free radical halogenation of alkanes
Correct Answer: Substitution of a hydrogen atom with a halogen in benzene
Explanation: Electrophilic substitution occurs when an electrophile (such as a halogen) replaces a hydrogen atom on an aromatic ring like benzene. The reaction mechanism involves the attack of the electrophile on the electron-rich benzene ring.
112. Which of the following best describes a nucleophilic substitution reaction?
ⓐ. A halogen replaces a hydrogen atom in an alkane
ⓑ. An electron-rich nucleophile attacks a carbon center, replacing a leaving group
ⓒ. A hydrogen atom is added to a double bond in alkenes
ⓓ. A free radical abstracts a hydrogen atom from a molecule
Correct Answer: An electron-rich nucleophile attacks a carbon center, replacing a leaving group
Explanation: In nucleophilic substitution, a nucleophile (an electron-rich species) attacks a carbon atom, displacing the leaving group. This type of substitution occurs in compounds such as alkyl halides and can follow either an SN1 or SN2 mechanism.
113. What is the characteristic feature of a free radical substitution reaction?
ⓐ. The reaction proceeds via a free radical intermediate
ⓑ. The substitution involves an electron pair donation from the nucleophile
ⓒ. The leaving group is replaced by an electrophile
ⓓ. The reaction involves the addition of a halogen to an alkene
Correct Answer: The reaction proceeds via a free radical intermediate
Explanation: Free radical substitution reactions involve the formation of free radicals during the reaction mechanism. The most common example is the halogenation of alkanes (like the chlorination of methane), where chlorine radicals replace hydrogen atoms in a stepwise process.
114. Which of the following is an example of an electrophilic aromatic substitution reaction?
ⓐ. The reaction between an alkyl halide and a nucleophile
ⓑ. The addition of HBr to an alkene
ⓒ. The nitration of benzene
ⓓ. The chlorination of methane
Correct Answer: The nitration of benzene
Explanation: Electrophilic aromatic substitution reactions occur when an electrophile (such as a nitronium ion, $NO_2^+$) attacks the electron-rich aromatic ring (benzene), replacing a hydrogen atom. Nitration of benzene with nitric acid is a classic example of this reaction.
115. In a nucleophilic substitution reaction, the rate-determining step in an SN1 mechanism involves:
ⓐ. The nucleophile attacking the carbon atom
ⓑ. The rearrangement of the carbocation
ⓒ. The departure of the leaving group
ⓓ. The formation of a carbocation intermediate
Correct Answer: The formation of a carbocation intermediate
Explanation: In an SN1 mechanism, the rate-determining step involves the departure of the leaving group, resulting in the formation of a carbocation intermediate. This intermediate is then attacked by the nucleophile. The reaction proceeds in two steps.
116. Which of the following is the key characteristic of an SN2 mechanism?
ⓐ. A stable carbocation is formed
ⓑ. The nucleophile attacks the substrate before the leaving group departs
ⓒ. The reaction occurs in two steps, with the formation of an intermediate
ⓓ. The rate depends only on the concentration of the leaving group
Correct Answer: The nucleophile attacks the substrate before the leaving group departs
Explanation: In an SN2 mechanism, the nucleophile attacks the carbon atom simultaneously as the leaving group departs, forming a transition state. The reaction occurs in a single step and is characterized by a backside attack, resulting in inversion of configuration.
117. Which of the following is an example of a free radical substitution reaction?
ⓐ. The formation of an alkyl halide from an alkene and HBr
ⓑ. The halogenation of an alkane in the presence of UV light
ⓒ. The addition of a nucleophile to a carbonyl group
ⓓ. The substitution of a hydrogen atom with a methyl group in a benzene ring
Correct Answer: The halogenation of an alkane in the presence of UV light
Explanation: Free radical substitution reactions often involve the halogenation of alkanes, such as the chlorination of methane, in the presence of UV light. This reaction proceeds via a chain mechanism, where chlorine radicals initiate the process and abstract hydrogen atoms from the alkane.
118. In an electrophilic substitution reaction, the leaving group:
ⓐ. Is always an alkyl group
ⓑ. Is replaced by a nucleophile
ⓒ. Is usually a halide ion
ⓓ. Departs as a neutral molecule, leaving behind a positively charged intermediate
Correct Answer: Departs as a neutral molecule, leaving behind a positively charged intermediate
Explanation: In electrophilic substitution reactions, such as the nitration of benzene, the leaving group (typically a hydrogen atom or a halide ion) departs as a neutral molecule, and the intermediate carbocation or complex formed is positively charged. The electrophile then attacks the molecule.
119. What is the result of a free radical halogenation reaction of an alkane?
ⓐ. The alkane undergoes polymerization
ⓑ. The alkane forms an alkene
ⓒ. One hydrogen atom is replaced by a halogen atom
ⓓ. The alkane reacts with a nucleophile to form a new functional group
Correct Answer: One hydrogen atom is replaced by a halogen atom
Explanation: Free radical halogenation of an alkane involves the substitution of one hydrogen atom with a halogen atom, such as in the chlorination of methane. The reaction proceeds via a chain mechanism, with the halogen radical initiating the process.
120. Which of the following factors increases the rate of an SN1 nucleophilic substitution reaction?
ⓐ. A polar solvent that stabilizes the carbocation
ⓑ. A bulky nucleophile
ⓒ. A strong base as a nucleophile
ⓓ. A highly substituted alkyl group
Correct Answer: A polar solvent that stabilizes the carbocation
Explanation: In an SN1 mechanism, the rate is determined by the formation of a carbocation intermediate. A polar solvent, especially a polar protic solvent, stabilizes the carbocation, thereby increasing the reaction rate. A strong nucleophile is less important in SN1 reactions because the rate-determining step involves the carbocation formation.
121. Which of the following is an example of an electrophilic addition reaction?
ⓐ. Addition of HBr to an alkene
ⓑ. Hydrolysis of an alkyl halide
ⓒ. Nucleophilic substitution in alkyl halides
ⓓ. Addition of Br₂ to an alkyl group
Correct Answer: Addition of HBr to an alkene
Explanation: The addition of HBr to an alkene is a classic example of an electrophilic addition reaction. In this reaction, the electrophile (H⁺) adds to the double bond, creating a carbocation intermediate, which is then attacked by the bromide ion (Br⁻).
122. Which of the following is an example of a free radical addition reaction?
ⓐ. Addition of HCl to an alkene
ⓑ. Hydrogenation of an alkene
ⓒ. Addition of HBr in the presence of light
ⓓ. Addition of water to an alkene
Correct Answer: Addition of HBr in the presence of light
Explanation: The addition of HBr to an alkene in the presence of light follows a free radical mechanism. The reaction starts with the homolytic cleavage of the H-Br bond, generating free radicals, which then add to the alkene.
123. In an electrophilic addition reaction to an alkene, the intermediate species formed is:
ⓐ. A free radical
ⓑ. A carbocation
ⓒ. A nucleophile
ⓓ. A transition state
Correct Answer: A carbocation
Explanation: In an electrophilic addition reaction, such as the addition of HCl to an alkene, the intermediate species formed is a carbocation. The electrophile attacks the double bond, leading to the formation of a positively charged carbocation, which is then attacked by the nucleophile (Cl⁻).
124. Which of the following compounds undergoes nucleophilic addition reactions?
ⓐ. Alkynes
ⓑ. Alcohols
ⓒ. Alkenes
ⓓ. Aldehydes and ketones
Correct Answer: Aldehydes and ketones
Explanation: Aldehydes and ketones undergo nucleophilic addition reactions because they have a polar carbonyl group ($C=O$), where the carbon is electrophilic and can be attacked by nucleophiles. This is typical in reactions like the addition of hydride ions or alcohols to carbonyl compounds.
125. In a free radical addition reaction, the first step is:
ⓐ. Formation of a carbocation
ⓑ. Homolytic cleavage of a bond to form free radicals
ⓒ. The nucleophile attacking the electrophile
ⓓ. Addition of an electrophile to a double bond
Correct Answer: Homolytic cleavage of a bond to form free radicals
Explanation: The first step in a free radical addition reaction is the homolytic cleavage of a bond (usually a halogen-halogen bond in the presence of light) to form free radicals. These radicals then react with the alkene, initiating the chain reaction.
126. Which of the following best describes a nucleophilic addition reaction?
ⓐ. A nucleophile attacks a carbon atom, replacing a leaving group
ⓑ. A nucleophile attacks a carbocation intermediate
ⓒ. A nucleophile adds to a carbonyl carbon, forming a tetrahedral intermediate
ⓓ. A nucleophile adds to an alkene in the presence of a catalyst
Correct Answer: A nucleophile adds to a carbonyl carbon, forming a tetrahedral intermediate
Explanation: In nucleophilic addition reactions, the nucleophile attacks the electrophilic carbonyl carbon of aldehydes or ketones, forming a tetrahedral intermediate. This is a key step in reactions like the addition of hydride in reduction reactions or alcohols in condensation reactions.
127. Which of the following is an example of an electrophilic addition to an alkene?
ⓐ. Addition of water to an alkene in the presence of an acid
ⓑ. Addition of NaOH to an alkene
ⓒ. Nucleophilic substitution in alkyl halides
ⓓ. Addition of a halogen to a benzene ring
Correct Answer: Addition of water to an alkene in the presence of an acid
Explanation: The addition of water (hydration) to an alkene in the presence of an acid is an electrophilic addition reaction. The alkene reacts with the proton (H⁺) to form a carbocation intermediate, which is then attacked by the water molecule.
128. Which of the following is true about free radical addition reactions?
ⓐ. The reaction proceeds through a carbocation intermediate
ⓑ. The reaction involves an electrophilic attack
ⓒ. The reaction involves the addition of a nucleophile to a carbonyl group
ⓓ. The reaction proceeds through a free radical intermediate
Correct Answer: The reaction proceeds through a free radical intermediate
Explanation: Free radical addition reactions proceed through the formation of free radical intermediates. These radicals react with the double bond of alkenes or other substrates, leading to the addition of atoms like halogens or hydrogen across the double bond.
129. What is the rate-determining step in an electrophilic addition reaction to an alkene?
ⓐ. The departure of the leaving group
ⓑ. The formation of a free radical intermediate
ⓒ. The attack of the electrophile on the double bond
ⓓ. The attack of the nucleophile on the carbocation
Correct Answer: The attack of the electrophile on the double bond
Explanation: In an electrophilic addition reaction, such as the addition of HCl to an alkene, the rate-determining step is the attack of the electrophile (H⁺) on the electron-rich alkene, which forms a carbocation intermediate.
130. Which of the following is an example of an electrophilic addition reaction in organic chemistry?
ⓐ. Addition of bromine to an alkene
ⓑ. Addition of sodium hydroxide to an alkyl halide
ⓒ. Addition of an alcohol to an aldehyde
ⓓ. Nucleophilic substitution in alkyl halides
Correct Answer: Addition of bromine to an alkene
Explanation: The addition of bromine (Br₂) to an alkene is an electrophilic addition reaction. The alkene’s double bond attacks the bromine molecule, forming a bromonium ion intermediate, which is then attacked by a bromide ion, leading to the addition of bromine across the double bond.
131. In the E1 elimination mechanism, the rate-determining step involves:
ⓐ. The departure of the leaving group
ⓑ. The formation of a carbocation intermediate
ⓒ. The attack of a nucleophile
ⓓ. The abstraction of a proton by a base
Correct Answer: The formation of a carbocation intermediate
Explanation: In the E1 (unimolecular elimination) mechanism, the rate-determining step is the formation of a carbocation intermediate after the leaving group departs. Once the carbocation is formed, a base abstracts a proton from an adjacent carbon, leading to the formation of the double bond.
132. Which of the following is true about the E2 elimination mechanism?
ⓐ. The base abstracts a proton and the leaving group departs simultaneously in one step.
ⓑ. The reaction occurs in two steps, with the formation of a carbocation intermediate.
ⓒ. The reaction rate depends only on the concentration of the leaving group.
ⓓ. The E2 mechanism is favored by a polar, protic solvent.
Correct Answer: The base abstracts a proton and the leaving group departs simultaneously in one step.
Explanation: The E2 (bimolecular elimination) mechanism involves the concerted removal of a proton by a base and the departure of the leaving group in a single step. The rate of the reaction depends on the concentration of both the substrate and the base.
133. In which of the following conditions is the E1 elimination mechanism most likely to occur?
ⓐ. A strong base and a high concentration of substrate
ⓑ. A weak nucleophile and a low temperature
ⓒ. A strong base and a polar aprotic solvent
ⓓ. A weak base and a polar, protic solvent
Correct Answer: A weak base and a polar, protic solvent
Explanation: The E1 elimination mechanism is favored by a weak base and a polar, protic solvent, which helps stabilize the carbocation intermediate formed during the reaction. These conditions slow down the reaction but allow for the formation of the carbocation necessary for elimination.
134. Which of the following statements is correct about the E2 mechanism?
ⓐ. It forms a carbocation intermediate.
ⓑ. It occurs via a single-step, concerted mechanism.
ⓒ. It is favored by a weak base.
ⓓ. It requires the formation of a three-membered ring intermediate.
Correct Answer: It occurs via a single-step, concerted mechanism.
Explanation: The E2 mechanism occurs in a single, concerted step, where the base abstracts a proton from the beta-carbon while the leaving group departs simultaneously. This mechanism does not involve the formation of a carbocation intermediate, unlike E1.
135. The E1 mechanism is most favored when the substrate is:
ⓐ. A tertiary alkyl halide
ⓑ. A secondary alkyl halide
ⓒ. A primary alkyl halide
ⓓ. An alkene
Correct Answer: A tertiary alkyl halide
Explanation: The E1 mechanism is favored by tertiary alkyl halides because the formation of a tertiary carbocation intermediate is highly stabilized. The more substituted the carbocation, the more stable it is, making E1 the preferred mechanism in these cases.
136. Which of the following is a key feature of the E2 mechanism?
ⓐ. The base and the leaving group act simultaneously in one step.
ⓑ. The leaving group departs before the base abstracts a proton.
ⓒ. The reaction proceeds through a carbocation intermediate.
ⓓ. The reaction is unimolecular.
Correct Answer: The base and the leaving group act simultaneously in one step.
Explanation: The E2 mechanism is a concerted process where the base abstracts a proton, and the leaving group departs at the same time, in a single step. This mechanism does not involve the formation of a carbocation, unlike E1.
137. Which of the following conditions would favor an E1 reaction over an E2 reaction?
ⓐ. Strong base and high temperature
ⓑ. Weak base and low temperature
ⓒ. Polar protic solvent and weak base
ⓓ. Strong base and polar aprotic solvent
Correct Answer: Polar protic solvent and weak base
Explanation: E1 reactions are favored by polar protic solvents, which stabilize the carbocation intermediate, and weak bases that can assist in the formation of the carbocation. These conditions are not conducive to E2, which requires a strong base and concerted elimination.
138. Which of the following is true for the E2 elimination mechanism?
ⓐ. The reaction is favored by a weak base.
ⓑ. The reaction mechanism involves the formation of a carbocation intermediate.
ⓒ. The reaction proceeds in a single step.
ⓓ. The leaving group departs after proton abstraction.
Correct Answer: The reaction proceeds in a single step.
Explanation: The E2 mechanism is a one-step, concerted reaction where the base abstracts a proton while the leaving group departs simultaneously. There is no carbocation intermediate involved in E2 reactions, unlike E1.
139. Which of the following is a factor that can influence whether an elimination reaction follows the E1 or E2 mechanism?
ⓐ. The solvent polarity and strength of the base
ⓑ. The presence of a halogen atom in the substrate
ⓒ. The temperature of the reaction
ⓓ. The number of electrons on the leaving group
Correct Answer: The solvent polarity and strength of the base
Explanation: The mechanism of elimination (E1 or E2) is influenced by the strength of the base and the polarity of the solvent. E1 reactions are favored by weak bases and polar protic solvents, whereas E2 reactions are favored by strong bases and polar aprotic solvents.
140. What is the product of an E2 reaction in which an alkyl halide undergoes elimination?
ⓐ. A ketone
ⓑ. An alkene
ⓒ. An alcohol
ⓓ. A carboxylic acid
Correct Answer: An alkene
Explanation: In an E2 elimination reaction, the product is an alkene, as the base abstracts a proton from a β-carbon, and the leaving group (usually a halide) departs simultaneously. This leads to the formation of a double bond between the alpha and beta carbons, resulting in an alkene product.
141. Which of the following best describes a rearrangement reaction?
ⓐ. A reaction in which a molecule undergoes a change in structure, but not in molecular formula
ⓑ. A reaction where two molecules combine to form a new product
ⓒ. A reaction where a molecule is broken down into smaller products
ⓓ. A reaction where the position of electrons in a molecule is shifted without changing the atom positions
Correct Answer: A reaction in which a molecule undergoes a change in structure, but not in molecular formula
Explanation: A rearrangement reaction involves a change in the structure of a molecule without altering its molecular formula. This often involves the migration of atoms or groups within the molecule, resulting in different structural isomers.
142. Which of the following is an example of a rearrangement reaction?
ⓐ. Addition of HX to an alkene
ⓑ. Hydration of an alkene to form an alcohol
ⓒ. The Wagner-Meerwein rearrangement
ⓓ. Electrophilic aromatic substitution
Correct Answer: The Wagner-Meerwein rearrangement
Explanation: The Wagner-Meerwein rearrangement is an example of a rearrangement reaction, where a carbocation undergoes a shift in position to form a more stable carbocation, leading to the formation of a new product. This rearrangement typically occurs in the context of the formation of alkyl groups in organic synthesis.
143. Which of the following reactions is commonly associated with a hydride shift during the rearrangement?
ⓐ. E1 elimination
ⓑ. E2 elimination
ⓒ. SN1 substitution
ⓓ. The formation of a more stable carbocation in the E1 mechanism
Correct Answer: The formation of a more stable carbocation in the E1 mechanism
Explanation: In the E1 mechanism, a carbocation intermediate forms after the departure of the leaving group, and a hydride shift (the movement of a hydrogen atom with its bonding electrons) can occur to stabilize the carbocation. This results in the formation of a more stable carbocation and helps guide the elimination reaction.
144. What type of rearrangement occurs in the formation of tert-butyl carbocation from neopentyl carbocation?
ⓐ. Methyl shift
ⓑ. Hydride shift
ⓒ. Ring closure
ⓓ. Halide ion displacement
Correct Answer: Hydride shift
Explanation: he neopentyl carbocation is $(CH_3)_3C-CH_2^+$. This is a primary carbocation. The carbon adjacent to the positive charge has no hydrogen atoms to donate. Therefore, a hydride shift is impossible36. The rearrangement occurs via a 1,2-methyl shift to form a more stable tertiary carbocation.
145. Which of the following factors can favor a rearrangement reaction?
ⓐ. A highly stable leaving group
ⓑ. Low temperature and weak solvents
ⓒ. The presence of a strong nucleophile
ⓓ. The formation of a more stable intermediate or product
Correct Answer: The formation of a more stable intermediate or product
Explanation: Rearrangement reactions are favored when they result in the formation of a more stable intermediate or product. This typically happens when a carbocation or radical can shift to a position that stabilizes its charge, such as in the case of alkyl shifts leading to more substituted carbocations.
146. In a ring closure rearrangement, the reaction leads to:
ⓐ. The formation of a ring structure
ⓑ. The loss of a leaving group
ⓒ. The attack of a nucleophile on a carbonyl group
ⓓ. The formation of a free radical intermediate
Correct Answer: The formation of a ring structure
Explanation: A ring closure rearrangement occurs when a molecule undergoes an intramolecular shift, leading to the formation of a ring structure. This type of rearrangement can be observed in various organic reactions, such as the formation of cyclic ethers or cyclic esters.
147. Which of the following reactions is an example of a molecular rearrangement involving an alkyl group shift?
ⓐ. The Beckmann rearrangement
ⓑ. The formation of a Grignard reagent
ⓒ. The addition of water to an alkene
ⓓ. The nucleophilic substitution in alkyl halides
Correct Answer: The Beckmann rearrangement
Explanation: The Beckmann rearrangement is a molecular rearrangement involving the migration of an alkyl group. This reaction typically starts with the formation of an oxime, which undergoes rearrangement to form a cyclic ketone. The alkyl group shifts in the process, leading to the new product.
148. Which of the following is true regarding the rearrangement in the formation of the more stable carbocation?
ⓐ. It always involves a hydride shift
ⓑ. It involves the migration of a group to a more substituted carbon atom
ⓒ. It occurs only in SN2 reactions
ⓓ. It only occurs in the presence of a nucleophile
Correct Answer: It involves the migration of a group to a more substituted carbon atom
Explanation: The formation of a more stable carbocation often involves the migration of a group (such as a hydride or alkyl group) to a more substituted carbon atom, resulting in a more stable carbocation. This rearrangement is commonly seen in the E1 and SN1 mechanisms, where carbocation stability is key.
149. In the Hofmann rearrangement, the key step involves:
ⓐ. A nucleophilic attack on an amide
ⓑ. The loss of a halide ion
ⓒ. A shift in the carbonyl group of the amide
ⓓ. The formation of an isocyanate intermediate
Correct Answer: The formation of an isocyanate intermediate
Explanation: The Hofmann rearrangement converts an amide to a primary amine. The key step is not a “shift in the carbonyl group”39. The key step is the migration of the alkyl group (R) from the carbonyl carbon to the nitrogen atom, which results in the formation of an isocyanate intermediate (R-N=C=O). The carbonyl group is eventually lost as carbonate ($CO_3^{2-}$) during aqueous workup40.
150. In a reaction involving a Wagner-Meerwein rearrangement, which of the following intermediates is involved?
ⓐ. Primary carbocation
ⓑ. Secondary carbocation
ⓒ. Free radical intermediate
ⓓ. Tertiary carbocation
Correct Answer: Tertiary carbocation
Explanation: The Wagner-Meerwein rearrangement involves the migration of a group (such as a hydride or alkyl group) to form a more stable tertiary carbocation. This type of rearrangement is often observed in the formation of alkyl carbocations in reactions such as hydride shifts or alkyl shifts.
151. Which of the following statements correctly describes a carbocation?
ⓐ. A species with a negative charge on carbon
ⓑ. A species with an incomplete octet and a positive charge on carbon
ⓒ. A species containing a lone pair on carbon
ⓓ. A neutral carbon atom in an sp³ hybridized state
Correct Answer: A species with an incomplete octet and a positive charge on carbon
Explanation: A carbocation is positively charged and has only six electrons around carbon (incomplete octet). It is highly reactive and electron-deficient. The carbon is usually sp² hybridized, giving a trigonal planar geometry. Due to electron deficiency, carbocations often undergo rearrangements to become more stable.
152. Which carbocation is the most stable?
ⓐ. Primary carbocation
ⓑ. Secondary carbocation
ⓒ. Tertiary carbocation
ⓓ. Methyl carbocation
Correct Answer: Tertiary carbocation
Explanation: Tertiary carbocations are the most stable because they are stabilized by +I (inductive) effect and hyperconjugation from three alkyl groups. More alkyl groups donate electron density to the positively charged carbon, reducing electron deficiency. Methyl and primary carbocations are least stable due to minimum electron donation.
153. Which of the following is the least stable carbocation?
ⓐ. Secondary carbocation
ⓑ. Tertiary carbocation
ⓒ. Allylic carbocation
ⓓ. Methyl carbocation
Correct Answer: Methyl carbocation
Explanation: Methyl carbocation ($CH_3^+$) is the least stable because there are no alkyl groups to donate electron density via +I or hyperconjugation. Allylic carbocations are resonance-stabilized, and tertiary carbocations are stabilized by hyperconjugation and inductive effects.
154. The stability order due to hyperconjugation is:
ⓐ. Tertiary < Secondary < Primary < Methyl
ⓑ. Methyl < Primary < Secondary < Tertiary
ⓒ. Primary < Methyl < Secondary < Tertiary
ⓓ. Secondary < Tertiary < Primary < Methyl
Correct Answer: Methyl < Primary < Secondary < Tertiary
Explanation: Hyperconjugation increases with the number of adjacent C–H bonds available to donate electron density. Tertiary has the maximum hyperconjugative structures, followed by secondary, primary, and least in methyl. This stabilizes the carbocation by delocalizing the positive charge.
155. What type of rearrangement commonly occurs to form a more stable carbocation?
ⓐ. Electrophilic substitution
ⓑ. Hydride or alkyl shift
ⓒ. Free radical substitution
ⓓ. Nucleophilic addition
Correct Answer: Hydride or alkyl shift
Explanation: Carbocations rearrange through hydride shifts (movement of H with its bonding electrons) or alkyl shifts (movement of CH₃ or larger groups) to form more stable carbocations. For example, a secondary carbocation may shift to become a tertiary carbocation.
156. Which of the following carbocations is aromatic and therefore highly stable?
ⓐ. Cyclopropyl carbocation
ⓑ. Cyclohexyl carbocation
ⓒ. Tropylium carbocation
ⓓ. Benzyl carbocation
Correct Answer: Tropylium carbocation
Explanation: The tropylium ion ($C_7H_7^+$) is aromatic due to 6 π-electrons (follows Hückel’s rule, 4n+2). This delocalization greatly stabilizes the carbocation. Although benzyl is resonance-stabilized, tropylium is even more stable due to complete cyclic conjugation.
157. Which reaction mechanism typically forms a carbocation intermediate?
ⓐ. SN1
ⓑ. E2
ⓒ. SN2
ⓓ. Free radical substitution
Correct Answer: SN1
Explanation: SN1 reactions proceed in two steps: first, the leaving group departs forming a carbocation; second, the nucleophile attacks. The intermediate carbocation’s stability determines the rate and feasibility of SN1 reactions. Tertiary carbocations undergo SN1 easily, but primary almost never.
158. What is the hybridization state of carbon in a carbocation?
ⓐ. sp
ⓑ. dsp²
ⓒ. sp³
ⓓ. sp²
Correct Answer: sp²
Explanation: Carbocations are sp² hybridized because the positively charged carbon has three sigma bonds and an empty p-orbital. This gives it a trigonal planar shape. The empty p-orbital is responsible for resonance stabilization in allylic and benzylic carbocations.
159. Which carbocation benefits from resonance stabilization?
ⓐ. Ethyl carbocation
ⓑ. Isopropyl carbocation
ⓒ. Benzyl carbocation
ⓓ. t-Butyl carbocation
Correct Answer: Benzyl carbocation
Explanation: The benzyl carbocation has a positive charge adjacent to an aromatic ring, allowing resonance delocalization across the ring. This extensive delocalization makes it highly stable—often more stable than many alkyl (non-resonance stabilized) carbocations.
160. In which situation does a carbocation rearrangement not occur?
ⓐ. When a more stable carbocation can form
ⓑ. When a hydride shift increases carbocation stability
ⓒ. When resonance stabilization is already present
ⓓ. When the carbocation is secondary
Correct Answer: When resonance stabilization is already present
Explanation: Carbocations that are resonance-stabilized (allylic or benzylic) are already very stable, so they typically do not rearrange. Rearrangement happens only when significant stabilization is possible — for example, secondary → tertiary. Resonance carbocations, however, keep their structure due to extensive charge delocalization.
161. Which of the following statements correctly describes a carbanion?
ⓐ. A species with a positive charge on carbon
ⓑ. A species with a lone pair and a negative charge on carbon
ⓒ. A species that is always sp³ hybridized
ⓓ. A species that has an incomplete octet
Correct Answer: A species with a lone pair and a negative charge on carbon
Explanation: A carbanion has three bonds and one lone pair on carbon, giving it a negative charge. It is electron-rich and highly reactive toward electrophiles. Unlike carbocations, carbanions have a complete octet and often prefer hybridization forms that increase s-character (sp > sp² > sp³) to stabilize the negative charge.
162. Which carbanion is the most stable?
ⓐ. Benzyl carbanion
ⓑ. Tertiary carbanion
ⓒ. Primary carbanion
ⓓ. Methyl carbanion
Correct Answer: Benzyl carbanion
Explanation: Benzyl carbanion is stabilized by resonance delocalization into the benzene ring. While tertiary carbocations are stable, tertiary carbanions are highly unstable due to +I effects from alkyl groups. Resonance stabilization is the strongest stabilizing factor for carbanions, making benzyl the most stable among them.
163. Which of the following destabilizes a carbanion?
ⓐ. Presence of an electron-withdrawing group
ⓑ. Presence of resonance
ⓒ. Increase in s-character
ⓓ. Presence of electron-donating alkyl groups
Correct Answer: Presence of electron-donating alkyl groups
Explanation: Electron-donating alkyl groups destabilize carbanions by increasing electron density on an already negatively charged carbon. Carbanions prefer electron-withdrawing groups (–NO₂, –CN), resonance, and high s-character (sp) because these stabilize the negative charge.
164. Which hybridization state gives maximum stability to a carbanion?
ⓐ. sp³
ⓑ. sp²
ⓒ. sp
ⓓ. dsp²
Correct Answer: sp
Explanation: Stability of carbanions increases as s-character increases because the electron pair is held closer to the nucleus. The order is:
sp (50% s) > sp² (33% s) > sp³ (25% s).
Thus, carbanions such as acetylide anions (sp) are significantly more stable compared to sp² or sp³ carbanions.
165. Which of the following carbanions is stabilized by resonance?
ⓐ. Ethyl carbanion
ⓑ. Propyl carbanion
ⓒ. tert-Butyl carbanion
ⓓ. Allyl carbanion
Correct Answer: Allyl carbanion
Explanation: Allyl carbanion is stabilized by resonance, where the negative charge is delocalized over two carbon atoms. This extensive charge spreading lowers the energy and increases stability. Simple alkyl carbanions (ethyl, propyl, tert-butyl) lack resonance, making them less stable.
166. Which group increases the stability of a carbanion through the –I effect?
ⓐ. –CH₃
ⓑ. –NO₂
ⓒ. –OH
ⓓ. –NH₂
Correct Answer: –NO₂
Explanation: –NO₂ is a strong electron-withdrawing group, pulling electron density away from the negatively charged carbon through the –I effect. This stabilizes the carbanion. Electron-donating groups like –CH₃, –OH, –NH₂ destabilize it by increasing electron density.
167. Which of the following statements about carbanion structure is correct?
ⓐ. Carbanions are typically pyramidal with an sp³-like structure
ⓑ. Carbanions always exist in a planar geometry
ⓒ. Carbanions always rearrange to become more stable
ⓓ. Carbanions contain an empty p-orbital
Correct Answer: Carbanions are typically pyramidal with an sp³-like structure
Explanation: Most carbanions are pyramidal (like NH₃) because the lone pair occupies one hybrid orbital. They can flatten toward sp² if resonance stabilization is possible (as in benzyl or allyl carbanions). They do not contain empty p-orbitals — that is a feature of carbocations.
168. Which of the following is the correct stability order for simple alkyl carbanions?
ⓐ. Tertiary > Secondary > Primary > Methyl
ⓑ. Secondary > Primary > Methyl > Tertiary
ⓒ. Primary > Methyl > Secondary > Tertiary
ⓓ. Methyl > Primary > Secondary > Tertiary
Correct Answer: Methyl > Primary > Secondary > Tertiary
Explanation: Alkyl groups destabilize carbanions through electron-donation (+I effect). Thus, fewer alkyl groups means greater stability:
Methyl > Primary > Secondary > Tertiary.
This is the exact opposite of carbocation stability trends.
169. Which factor is the MOST important for stabilizing carbanions?
ⓐ. Hyperconjugation
ⓑ. Resonance
ⓒ. Steric hindrance
ⓓ. Backside attack
Correct Answer: Resonance
Explanation: Resonance is the strongest stabilizing effect for carbanions because it delocalizes the negative charge over multiple atoms. This reduces charge density and lowers the energy of the species. Hyperconjugation stabilizes carbocations and radicals more than carbanions.
170. Which carbanion is stabilized by both resonance and inductive effects?
ⓐ. Cyanide ion ($⁻C≡N$)
ⓑ. Ethyl carbanion
ⓒ. Isopropyl carbanion
ⓓ. Methyl carbanion
Correct Answer: Cyanide ion ($⁻C≡N$)
Explanation: The cyanide ion (⁻C≡N) is highly stabilized because:
• It has sp hybridization (50% s-character)
• It has a strong –I effect from the electronegative nitrogen
• The triple bond allows some resonance form contribution
This combination makes cyanide one of the most stable carbanions known.
171. Which of the following correctly describes a free radical?
ⓐ. A species with a positive charge on carbon
ⓑ. A species with a negative charge and a lone pair
ⓒ. A species containing an unpaired electron
ⓓ. A species with a complete octet and no charge
Correct Answer: A species containing an unpaired electron
Explanation: A free radical contains one unpaired electron, making it highly reactive. Free radicals are neutral species, neither positively nor negatively charged. They are typically formed by homolytic bond cleavage and show behaviors similar to both carbocations and carbanions but with unique stability factors like hyperconjugation and resonance.
172. Which free radical is the most stable?
ⓐ. Primary free radical
ⓑ. Secondary free radical
ⓒ. Tertiary free radical
ⓓ. Methyl free radical
Correct Answer: Tertiary free radical
Explanation: Stability of free radicals follows: 3° > 2° > 1° > methyl, due to hyperconjugation and +I effect from alkyl groups. More alkyl groups provide more C–H sigma bonds for hyperconjugation, spreading the electron density of the unpaired electron and stabilizing the radical.
173. Which factor stabilizes a free radical the most?
ⓐ. Steric hindrance
ⓑ. Hyperconjugation
ⓒ. Negative inductive effect
ⓓ. Backside attack
Correct Answer: Hyperconjugation
Explanation: Free radicals are greatly stabilized by hyperconjugation, where adjacent C–H bonds donate electron density to the radical center. The greater the number of hyperconjugative structures, the more stable the radical. Resonance also stabilizes radicals but hyperconjugation is the dominant effect for simple alkyl radicals.
174. Which of the following radicals is stabilized by resonance?
ⓐ. Ethyl radical
ⓑ. Methyl radical
ⓒ. Allyl radical
ⓓ. tert-Butyl radical
Correct Answer: Allyl radical
Explanation: The allyl radical ($CH_2–CH–CH_2•$) is resonance stabilized because the unpaired electron is delocalized over two carbon atoms. This delocalization significantly lowers the energy of the radical, making allyl radicals highly stable compared to simple alkyl radicals.
175. Which hybridization is most stable for a carbon-centered free radical?
ⓐ. sp²
ⓑ. sp³
ⓒ. sp
ⓓ. dsp²
Correct Answer: sp²
Explanation: Simple alkyl free radicals are most stable with sp² hybridization. This gives a trigonal planar geometry, which allows the unpaired electron to reside in a p-orbital. This planar structure maximizes the stabilizing effect of hyperconjugation with adjacent C-H bonds.
176. Which order represents the stability of simple alkyl free radicals?
ⓐ. Methyl > Primary > Secondary > Tertiary
ⓑ. Tertiary > Secondary > Primary > Methyl
ⓒ. Primary > Methyl > Secondary > Tertiary
ⓓ. Secondary > Tertiary > Primary > Methyl
Correct Answer: Tertiary > Secondary > Primary > Methyl
Explanation: Stability increases with the number of alkyl groups attached to the radical carbon. Alkyl groups stabilize the unpaired electron through hyperconjugation and the +I effect. Tertiary radicals have maximum hyperconjugation, making them most stable.
177. Which of the following radicals is aromatic and highly stable?
ⓐ. Benzyl radical
ⓑ. Ethyl radical
ⓒ. n-Propyl radical
ⓓ. Isobutyl radical
Correct Answer: Benzyl radical
Explanation: The benzyl radical ($C_6H_5–CH_2•$) is highly stabilized because the unpaired electron is delocalized into the aromatic ring through resonance. Aromatic stabilization makes benzyl radicals even more stable than tertiary alkyl radicals.
178. Free radicals are generally formed by:
ⓐ. Heterolytic cleavage
ⓑ. Proton abstraction
ⓒ. Hydride shift
ⓓ. Homolytic cleavage
Correct Answer: Homolytic cleavage
Explanation: Free radicals form when a bond breaks homolytically, meaning each atom takes one electron from the bond. This can be induced by heat, UV light, or peroxides. Heterolytic cleavage forms ions, not radicals.
179. Which of the following statements is TRUE about the reactivity of free radicals?
ⓐ. More stable radicals react faster
ⓑ. Radical stability does not affect reaction rate
ⓒ. All radicals have equal reactivity
ⓓ. Less stable radicals react faster
Correct Answer: Less stable radicals react faster
Explanation: Free radicals follow the general principle: Less stable = more reactive. A radical with little hyperconjugation or no resonance (like methyl or primary radicals) reacts much faster to achieve stability. More stable radicals (benzyl, tertiary) are less reactive due to electron delocalization.
180. Why is the tertiary butyl radical more stable than a secondary radical?
ⓐ. Because it is aromatic
ⓑ. Because it has more resonance structures
ⓒ. Because it has greater hyperconjugation and inductive effects
ⓓ. Because it forms a planar structure
Correct Answer: Because it has greater hyperconjugation and inductive effects
Explanation: The tertiary butyl radical has nine C–H bonds available for hyperconjugation, while secondary radicals have only six. More alkyl groups also donate electron density via +I effect, stabilizing the unpaired electron. This combination makes tertiary radicals significantly more stable.
181. Which of the following correctly describes a carbene?
ⓐ. A species with a positive charge on carbon
ⓑ. A neutral species with a divalent carbon and two nonbonding electrons
ⓒ. A negatively charged carbon species
ⓓ. A species containing an unpaired electron only
Correct Answer: A neutral species with a divalent carbon and two nonbonding electrons
Explanation: A carbene is a neutral divalent carbon species with two bonds and two nonbonding electrons. It can exist as a singlet carbene (paired electrons) or triplet carbene (unpaired electrons). Carbenes are highly reactive intermediates used in cyclopropanation and insertion reactions.
182. Carbenes commonly exist in two forms. What are they?
ⓐ. Cationic and anionic
ⓑ. Heterolytic and homolytic
ⓒ. Singlet and triplet
ⓓ. Electrophilic and nucleophilic
Correct Answer: Singlet and triplet
Explanation: Carbenes exist in singlet (paired electrons, sp²-like) and triplet (two unpaired electrons, sp-hybrid-like) forms. Singlet carbenes are more reactive in insertion reactions, while triplet carbenes behave like free radicals due to their two unpaired electrons.
183. Which of the following increases carbene stability?
ⓐ. Electron-withdrawing groups
ⓑ. Highly acidic hydrogens
ⓒ. Aromatic rings
ⓓ. Alkyl groups by hyperconjugation
Correct Answer: Alkyl groups by hyperconjugation
Explanation: Alkyl groups stabilize carbenes through hyperconjugation and +I effect, distributing electron density to the electron-deficient carbon. This stabilizes singlet carbenes more than triplet carbenes. Electron-withdrawing groups decrease carbene stability.
184. What is the electronic configuration of a singlet carbene?
ⓐ. Two unpaired electrons in separate orbitals
ⓑ. One unpaired electron in a p-orbital
ⓒ. Two paired electrons in one orbital
ⓓ. A fully filled octet
Correct Answer: Two paired electrons in one orbital
Explanation: Singlet carbenes have paired electrons in the nonbonding orbital (sp²-like). They are electrophilic because the carbon has an empty p-orbital. Triplet carbenes, in contrast, have unpaired electrons.
185. Which type of carbene is more reactive toward insertion into C–H bonds?
ⓐ. Singlet carbene
ⓑ. Triplet carbene
ⓒ. Benzyl carbene
ⓓ. Vinylic carbene
Correct Answer: Singlet carbene
Explanation: Singlet carbenes are more electrophilic and insert readily into C–H and X–H bonds due to their paired nonbonding electrons and empty p-orbital. Triplet carbenes instead tend to undergo radical-type reactions.
186. Which of the following correctly describes a nitrene?
ⓐ. A species with a positive charge on nitrogen
ⓑ. A neutral species with a divalent nitrogen and one lone pair
ⓒ. A species containing a nitrogen atom with six valence electrons and two nonbonding electrons
ⓓ. A nitrogen ion with resonance stabilization
Correct Answer: A species containing a nitrogen atom with six valence electrons and two nonbonding electrons
Explanation: A nitrene (general formula R-N) is the nitrogen analog of a carbene. The nitrogen is bonded to only one group (R), making it monovalent, not divalent. It is a neutral species with 6 valence electrons (2 in the bond, and 4 non-bonding electrons).
187. Nitrenes commonly form from the decomposition of:
ⓐ. Aldehydes
ⓑ. Alkyl halides
ⓒ. Azides
ⓓ. Ketones
Correct Answer: Azides
Explanation: Nitrenes are commonly generated by thermal or photochemical decomposition of azides (RN₃). This produces a highly reactive nitrene intermediate used in rearrangements and insertions.
188. Which is the correct stability order of nitrenes?
ⓐ. Alkylnitrene > Aryl nitrene > Acyl nitrene
ⓑ. Alkyl nitrene > Acyl nitrene > Aryl nitrene
ⓒ. Aryl nitrene > Alkylnitrene > Acyl nitrene
ⓓ. Acyl nitrene > Aryl nitrene > Alkylnitrene
Correct Answer: Acyl nitrene > Aryl nitrene > Alkylnitrene
Explanation: Acyl nitrenes are most stable due to resonance involving the carbonyl group. Aryl nitrenes gain partial stabilization from the aromatic ring. Alkylnitrenes lack resonance or stabilization and are the least stable.
189. What is the electronic configuration of a triplet nitrene?
ⓐ. Two electrons paired in one orbital
ⓑ. Two unpaired electrons in separate orbitals
ⓒ. Fully filled octet with one lone pair
ⓓ. Carbon-like sp hybridization
Correct Answer: Two unpaired electrons in separate orbitals
Explanation: Triplet nitrenes have two unpaired electrons, similar to triplet carbenes. This makes them behave like radicals. Triplet nitrenes are more stable than singlet nitrenes due to reduced electron–electron repulsion.
190. Nitrenes can undergo rearrangements similar to:
ⓐ. Carbenes
ⓑ. Carbocations
ⓒ. Free radicals
ⓓ. Carbanions
Correct Answer: Carbenes
Explanation: Both carbenes and nitrenes are neutral, electron-deficient species capable of insertion, addition, and rearrangement. Nitrenes often rearrange (e.g., Curtius, Hofmann reactions) just like carbenes rearrange via hydride/alkyl shifts.
191. Crystallization is primarily used for the purification of:
ⓐ. Liquids with very low boiling points
ⓑ. Solids that contain small amounts of impurities
ⓒ. Solutions containing insoluble salts
ⓓ. Gases dissolved in liquids
Correct Answer: Solids that contain small amounts of impurities
Explanation: Crystallization is a purification technique used mainly for solid organic compounds. It works because the pure compound forms well-defined crystals while impurities remain dissolved in the solvent. This method is effective when the desired solid is significantly less soluble in cold solvent than in hot solvent.
192. Which of the following is the MOST important property of a good crystallization solvent?
ⓐ. The solute must be equally soluble in hot and cold solvent
ⓑ. The solute must be insoluble in both hot and cold solvent
ⓒ. The solute must be highly soluble in hot solvent but sparingly soluble in cold solvent
ⓓ. The solvent must react with the solute
Correct Answer: The solute must be highly soluble in hot solvent but sparingly soluble in cold solvent
Explanation: A suitable crystallization solvent dissolves the impure solid at high temperature, but the pure compound crystallizes out on cooling. If the solute is too soluble even in cold conditions, crystals won’t form; and if insoluble in hot solvent, it won’t dissolve at all.
193. What is the purpose of using activated charcoal during crystallization?
ⓐ. To increase the solubility of the compound
ⓑ. To change the color of the solvent
ⓒ. To remove colored impurities from the solution
ⓓ. To speed up the cooling process
Correct Answer: To remove colored impurities from the solution
Explanation: Activated charcoal (animal charcoal) adsorbs colored organic impurities without reacting with the solute. During hot filtration, charcoal is used to decolorize the solution, giving purer crystals upon cooling.
194. Why is hot filtration used during crystallization?
ⓐ. To slow down the cooling process
ⓑ. To remove insoluble impurities before cooling
ⓒ. To form crystals at higher temperatures
ⓓ. To filter out the pure product
Correct Answer: To remove insoluble impurities before cooling
Explanation: Hot filtration ensures impurities like dust, insoluble salts, or charcoal are removed while the solute is completely dissolved. If filtration is done cold, premature crystallization may clog the filter, reducing efficiency.
195. Why is slow cooling preferred in crystallization?
ⓐ. It forms larger, purer crystals
ⓑ. It forms very fine crystals that trap impurities
ⓒ. It speeds up the crystallization process
ⓓ. It avoids the need for filtration
Correct Answer: It forms larger, purer crystals
Explanation: Slow cooling allows molecules to arrange neatly into a solid lattice, producing larger and purer crystals. Rapid cooling traps impurities and gives small, less pure crystals. Slow cooling is ideal for high-purity separation.
196. Which technique is used to check the purity of crystals obtained after crystallization?
ⓐ. Measuring boiling point
ⓑ. Measuring density
ⓒ. Measuring refractive index
ⓓ. Measuring melting point
Correct Answer: Measuring melting point
Explanation: A pure organic compound has a sharp melting point, while impurities lower and broaden the melting point range. Thus, melting point measurement is commonly used to test purity after crystallization.
197. Which solvent is generally avoided in crystallization due to its strong hydrogen bonding?
ⓐ. Ether
ⓑ. Ethanol
ⓒ. Water
ⓓ. Benzene
Correct Answer: Water
Explanation: Water has strong hydrogen bonding which sometimes prevents proper dissolution and crystallization. However, it is used when compounds are sufficiently soluble. In many cases, organic solvents like ethanol or acetone are preferred due to less excessive hydrogen bonding.
198. What is the purpose of scratching the container during crystallization?
ⓐ. To provide surfaces for crystal nucleation
ⓑ. To remove impurities
ⓒ. To increase solubility
ⓓ. To break large crystals
Correct Answer: To provide surfaces for crystal nucleation
Explanation: Scratching creates small rough spots on the container, allowing solute molecules to start forming crystals. This is useful when crystallization does not begin on its own due to lack of nucleation sites.
199. Why is crystallization NOT suitable for removing volatile impurities?
ⓐ. Because crystals dissolve volatile compounds
ⓑ. Because crystallization requires high temperature
ⓒ. Because crystallization only separates based on boiling point
ⓓ. Because volatile impurities evaporate easily and cannot be trapped
Correct Answer: Because volatile impurities evaporate easily and cannot be trapped
Explanation: Crystallization removes non-volatile impurities. Volatile impurities escape during heating or evaporation and cannot be separated through crystallization. Techniques like distillation are better suited for volatile impurities.
200. Which of the following statements about crystallization is TRUE?
ⓐ. Faster cooling gives better purity
ⓑ. Impurities crystallize first before the solute
ⓒ. Pure compounds form well-defined crystals
ⓓ. Crystallization requires the solute to react chemically with the solvent
Correct Answer: Pure compounds form well-defined crystals
Explanation: Pure compounds have structured, repeating molecular arrangements that form well-defined crystals. Impurities disrupt this order, so they remain dissolved. Faster cooling or reactive solvents can lead to impure or incomplete crystallization.
You’re on Class 11 Chemistry MCQs – Chapter 12: Organic Chemistry (Part 2). This section continues with the second set of 100 MCQs focused on reaction mechanisms, electronic effects, and intermediate stability. Topics include carbocations, carbanions, and free radicals, along with their formation and rearrangements in organic reactions. Perfect for Boards, JEE Main, and NEET preparation where concept-based questions dominate.
- 👉 Total in chapter: 300 MCQs (100 + 100 + 100)
- 👉 This page: Second set of 100 solved MCQs with explanations
- 👉 Next: Continue with Part 3 for the final set of Organic Chemistry MCQs