Haloalkanes MCQ Quiz - Objective Question with Answer for Haloalkanes - Download Free PDF

CONCEPT:

SN1 Reaction (Unimolecular Nucleophilic Substitution)

• The SN1 reaction is a two-step process where the first step is the rate-determining step.
• The first step involves the formation of a carbocation intermediate by the loss of the leaving group.
• The rate of the SN1 reaction depends on the stability of the carbocation intermediate formed.
• More stable carbocations form faster, leading to a faster SN1 reaction.

EXPLANATION:

• Comparing the given compounds:
  1. Chloromethane (CH3Cl) - Forms a methyl carbocation, which is highly unstable.
     
  2. 2-bromo-3-methylbutane - Forms a secondary carbocation, which is relatively more stable than a methyl carbocation.
     
  3. 2-chloro-3-methylbutane - Also forms a secondary carbocation, but chlorine is a poorer leaving group compared to bromine.
  4. 
  5. 2-bromo-2-methylpropane - Forms a tertiary carbocation, which is highly stable.
     
• Among the given options, the 2-bromo-2-methylpropane forms a tertiary carbocation which is the most stable.
• Because the stability of the carbocation intermediate is the key factor in the SN1 reaction, the compound that forms the most stable carbocation will undergo the SN1 reaction the fastest.

Therefore, the compound that would undergo the SN1 reaction faster is 2-bromo-2-methylpropane (option 4).

CONCEPT:

SN1 and SN2 Reactions

• The SN1 reaction is a unimolecular nucleophilic substitution where the rate-determining step involves the formation of a carbocation intermediate.
• The SN2 reaction is a bimolecular nucleophilic substitution where the nucleophile attacks the electrophilic carbon in a single, concerted step.

EXPLANATION:

• Statement I:

  CH3-CH2-CH2-O-CH3 will undergo SN2 reaction at a faster rate than SN1 reaction, though it is an ether.

  • The structure is propyl methyl ether.
  • In SN2 reactions, the nucleophile attacks the less hindered carbon, proceeding through a single-step mechanism.
  • SN1 reactions are less likely because they require the formation of a stable carbocation intermediate, which is difficult in ethers without good leaving groups forming stable carbocations.
  • Therefore, SN2 is more favorable for this ether due to the steric accessibility and concerted mechanism, avoiding carbocation formation.
• Statement II:

  CH3-CH2-CH2-CH3 will undergo an SN1 reaction faster than an SN2 reaction, despite being an alkane, due to the structure and stability considerations.

  • The structure is n-butane, a simple straight-chain alkane.
  • Alkanes do not typically participate in SN1 or SN2 reactions since they lack functional groups (e.g., halides) that could readily leave and form intermediates.
  • SN1 reactions are very unlikely because forming a carbocation from n-butane would be highly unstable and energetically unfavorable.
  • SN2 reactions require a good leaving group, which n-butane does not have under normal conditions.
  • Thus, the statement is incorrect as n-butane does not favor either SN1 or SN2 mechanisms without functionalization.

Additional Information

•  SN1 Mechanism:
  • Favored in tertiary and secondary carbocations.
  • Carbocation stability is key.
  • Occurs in polar protic solvents (e.g., water, alcohols).
• SN2 Mechanism:
  • Favored in primary and methyl halides.
  • Requires a good leaving group and a strong nucleophile.
  • Occurs in polar aprotic solvents (e.g., acetone, DMSO).

Therefore, Statement I is correct, while Statement II is incorrect.

CONCEPT:

Ortho- and Para-Directing Groups in Electrophilic Aromatic Substitution

• In electrophilic aromatic substitution reactions, substituents already present on the benzene ring influence the position where the new electrophile will attack.
• Substituents can either be activating or deactivating. Activating groups typically direct electrophiles to the ortho and para positions, while deactivating groups direct to the meta position.
• Chlorine is unique because it is an electron withdrawing group through inductive effect, yet it is ortho- and para-directing due to its resonance effect.

EXPLANATION:

• Chlorine withdraws electrons through its inductive effect due to its high electronegativity:
  • Chlorine is more electronegative than carbon, leading to a withdrawal of electron density from the benzene ring through the sigma bonds (inductive effect).
• Despite this, chlorine releases electrons through resonance, which stabilizes the intermediate carbocation formed during electrophilic substitution:
  • Chlorine has lone pairs of electrons that can participate in resonance with the benzene ring. This delocalization of electrons can stabilize the positively charged intermediate formed during the substitution.
  • Resonance donation of electron density to the ortho and para positions makes these positions more reactive towards electrophilic attack.

Therefore, the correct answer is (A), (B), and (D) only.

CONCEPT:

SN1 Reaction (Unimolecular Nucleophilic Substitution)

• The SN1 reaction is a two-step process where the first step is the rate-determining step.
• The first step involves the formation of a carbocation intermediate by the loss of the leaving group.
• The rate of the SN1 reaction depends on the stability of the carbocation intermediate formed.
• More stable carbocations form faster, leading to a faster SN1 reaction.

EXPLANATION:

• Comparing the given compounds:
  1. Chloromethane (CH3Cl) - Forms a methyl carbocation, which is highly unstable.
     
  2. 2-bromo-3-methylbutane - Forms a secondary carbocation, which is relatively more stable than a methyl carbocation.
     
  3. 2-chloro-3-methylbutane - Also forms a secondary carbocation, but chlorine is a poorer leaving group compared to bromine.
  4. 
  5. 2-bromo-2-methylpropane - Forms a tertiary carbocation, which is highly stable.
     
• Among the given options, the 2-bromo-2-methylpropane forms a tertiary carbocation which is the most stable.
• Because the stability of the carbocation intermediate is the key factor in the SN1 reaction, the compound that forms the most stable carbocation will undergo the SN1 reaction the fastest.

Therefore, the compound that would undergo the SN1 reaction faster is 2-bromo-2-methylpropane (option 4).

CONCEPT:

• The given alcohol X has the formula C₄H₁₀O. Possible isomers include butan-2-ol and tert-butanol.
• When alcohol reacts with conc. HCl, a nucleophilic substitution occurs, replacing the –OH with –Cl to form an alkyl halide.
• Y is given as C₄H₉Cl. Among possible isomers, tert-butyl chloride forms from tert-butanol, due to the stability of the tertiary carbocation.
• When tert-butanol is passed over heated copper at 573 K, it undergoes elimination (dehydration) to form an alkene, specifically isobutene (2-methylpropene).

EXPLANATION:

• X (C₄H₁₀O): tert-butanol (2-methylpropan-2-ol)
• Y (C₄H₉Cl): tert-butyl chloride via SN1 reaction with conc. HCl.
• Reaction with copper at 573 K: Dehydration occurs, forming an alkene → isobutene (2-methylpropene)

Therefore, the correct answer is: Option 2 (Alkene - isobutene)

Explanation:-

sp³ hybridized carbon atom show fast nucleophilic substitution reaction than sp² hybridized carbon atom. 

The ccorrect answer is option 1 that is Benzyl chloride.

Concept:

• Compounds that can produce a stable carbocation intermediate are more likely to undergo a nucleophilic substitution reaction that exclusively proceeds via an SN1 mechanism.

Explanation:

• The chemical that, in the list of choices, can produce a stable benzylic carbocation when it loses the chloride ion is benzyl chloride (C₆H₅CH₂Cl). This is because of its resonance stabilization with the phenyl group.
• Ethyl chloride (CH₃CH₂Cl) often undergoes an SN₂ mechanism because it creates a less stable primary carbocation, which makes it less appropriate for an SN₁ reaction.
• Due to the aromatic ring's resonance stabilization, chlorobenzene (C₆H₅Cl) typically does not undergo nucleophilic substitution.
• In order to increase the benzene ring's resistance to both the SN1 and SN2 reactions, the reaction would necessitate breaking its aromaticity, which is energetically unfavorable.

CONCEPT:

Saytzeff's Rule in Elimination Reactions

• Saytzeff's (Zaitsev's) rule states that in elimination reactions, the more substituted alkene (one with more alkyl groups attached to the double bond) is generally the major product.
• This rule is applicable to reactions like dehydrohalogenation and dehydration where the base or reagent promotes the formation of an alkene.
• However, some bulky bases or specific reaction conditions may lead to exceptions, producing the less substituted alkene as the major product (anti-Saytzeff or Hoffman product).

EXPLANATION:

• Reaction a: This involves dehydration using concentrated sulfuric acid, which follows Saytzeff's rule, leading to the formation of the more substituted alkene.
• Reaction b: Alcoholic KOH, in a dehydrohalogenation reaction, also typically follows Saytzeff's rule, forming the more substituted alkene.
• Reaction c: In the presence of a bulky base like tert-butoxide (CH₃)₃CO^-, the less substituted alkene (anti-Saytzeff or Hoffman product) is formed.
• Reaction d: This reaction involves heat, leading to elimination that produces an unsaturated aldehyde. This does not follow the standard Saytzeff mechanism.

Conclusion:-

The reactions that do not produce a Saytzeff product are reaction c (due to bulky base). Therefore, the correct answer is: 4) c only

Explanation:-

sp³ hybridized carbon atom show fast nucleophilic substitution reaction than sp² hybridized carbon atom. 

Concept:

Favorskii Rearrangement: 

• The transformation of alpha halo ketones to esters with rearranged carbon skeleton by the treatment with alkoxide ions is known as Favorskii Rearrangement.
• In the case of cyclic α-halo ketones, the Favorskii rearrangement constitutes a ring contraction.

Explanation:-

• The reaction pathway is shown below:

• In the above reaction, it is shown that in the first step of the reaction the base alkoxide (methoxy) first abstracts alpha hydrogen to produce a carbanion.
• Intramolecular nucleophilic attack on the carbon bearing the bromine displaces the bromine atom with the formation of a transient symmetrical cyclopropane ring.
• The alkoxide then attacks the carbonyl carbon to open up the ring with equal ease on either side of the carbonyl carbon to give carbanions which then take up a proton to give the corresponding ester.

​Conclusion:-

• Hence, option 2 is the correct answer.

Concept - 

SN1 reactions -

• These are substitution nucleophilic unimolecular reactions.
• These are called unimolecular because the rate of these reactions depends upon the concentration of one reactant species only.
• These reactions complete in two steps.
• First step is the slowest step and the rate-determining step.

Reaction mechanism - Reaction steps -

1. Formation of carbocation
2. Attack of nucleophile

Explanation:

The order of reactivity towards SN1 reaction depends on the stability of intermediate carbocation formed in step 1 of the reaction.

The reaction which involves the most stable carbonation formation will be the more reactive towards SN1 reaction.

Carbocation stability order - The stability order of carbocation is tertiary butyl > benzylic carbocation > allylic > 3°> 2°> 1°. 

Check the given options, first step of the given reactants will be- 

1) 

2)\(\xrightarrow{-Cl}\)  

3) \(\xrightarrow{-Cl}\)

4) \(\xrightarrow{-Cl}\) 

Thus, as per the stability order of carbonation,  is the most stable carbonation among these and hence will undergo SN1 reaction predominantly.

Conclusion:

will be most reactive for Sn1 reaction.

Concept:

SN2 reactions - 

• These nucleophilic substitution reactions follow second-order kinetics and are hence called nucleophilic substitution bimolecular reactions.
• The rate of reaction depends upon both; the reactants and participating nucleophile.
• It is one step reaction.
• Both, formation of carbocation and the removal of leaving group take place in a single step.
• Inversion of configuration occurs in SN2 reactions.

In SN2 reactions, the attack of nucleophile and removal of leaving group occurs in a single step.

Therefore, the bulky substrate has an inhibiting effect due to high stearic hindrance.

SN2 reactions depend on the following factors -

1. Strong nucleophile
2. Good leaving group
3. Non-bulky substrate
4. Polar-aprotic solvent

CONCEPT:

• The given alcohol X has the formula C₄H₁₀O. Possible isomers include butan-2-ol and tert-butanol.
• When alcohol reacts with conc. HCl, a nucleophilic substitution occurs, replacing the –OH with –Cl to form an alkyl halide.
• Y is given as C₄H₉Cl. Among possible isomers, tert-butyl chloride forms from tert-butanol, due to the stability of the tertiary carbocation.
• When tert-butanol is passed over heated copper at 573 K, it undergoes elimination (dehydration) to form an alkene, specifically isobutene (2-methylpropene).

EXPLANATION:

• X (C₄H₁₀O): tert-butanol (2-methylpropan-2-ol)
• Y (C₄H₉Cl): tert-butyl chloride via SN1 reaction with conc. HCl.
• Reaction with copper at 573 K: Dehydration occurs, forming an alkene → isobutene (2-methylpropene)

Therefore, the correct answer is: Option 2 (Alkene - isobutene)

CONCEPT:

SN1 and SN2 Reactions

• The SN1 reaction is a unimolecular nucleophilic substitution where the rate-determining step involves the formation of a carbocation intermediate.
• The SN2 reaction is a bimolecular nucleophilic substitution where the nucleophile attacks the electrophilic carbon in a single, concerted step.

EXPLANATION:

• Statement I:

  CH3-CH2-CH2-O-CH3 will undergo SN2 reaction at a faster rate than SN1 reaction, though it is an ether.

  • The structure is propyl methyl ether.
  • In SN2 reactions, the nucleophile attacks the less hindered carbon, proceeding through a single-step mechanism.
  • SN1 reactions are less likely because they require the formation of a stable carbocation intermediate, which is difficult in ethers without good leaving groups forming stable carbocations.
  • Therefore, SN2 is more favorable for this ether due to the steric accessibility and concerted mechanism, avoiding carbocation formation.
• Statement II:

  CH3-CH2-CH2-CH3 will undergo an SN1 reaction faster than an SN2 reaction, despite being an alkane, due to the structure and stability considerations.

  • The structure is n-butane, a simple straight-chain alkane.
  • Alkanes do not typically participate in SN1 or SN2 reactions since they lack functional groups (e.g., halides) that could readily leave and form intermediates.
  • SN1 reactions are very unlikely because forming a carbocation from n-butane would be highly unstable and energetically unfavorable.
  • SN2 reactions require a good leaving group, which n-butane does not have under normal conditions.
  • Thus, the statement is incorrect as n-butane does not favor either SN1 or SN2 mechanisms without functionalization.

Additional Information

•  SN1 Mechanism:
  • Favored in tertiary and secondary carbocations.
  • Carbocation stability is key.
  • Occurs in polar protic solvents (e.g., water, alcohols).
• SN2 Mechanism:
  • Favored in primary and methyl halides.
  • Requires a good leaving group and a strong nucleophile.
  • Occurs in polar aprotic solvents (e.g., acetone, DMSO).

Therefore, Statement I is correct, while Statement II is incorrect.

CONCEPT:

Picric Acid and its Synthesis

• Picric acid, also known as 2,4,6-trinitrophenol, is a yellow crystalline solid commonly used in explosives.
• Statement I mistakenly identifies picric acid as 2,4,6-trinitrotoluene (TNT), which is an entirely different compound. TNT is used as an explosive, but it is not picric acid.
• Statement II describes a synthesis method for picric acid. Phenol-2,4-disulphuric acid can be treated with concentrated nitric acid (HNO₃) to produce picric acid.

Explanation:-

• Statement I: Incorrect, as picric acid is 2,4,6-trinitrophenol, not 2,4,6-trinitrotoluene.
• Statement II: Correct, as it correctly describes the synthesis of picric acid from phenol-2,4-disulphuric acid and conc. HNO₃.

picric acid

(2, 4, 6 – trinitrophenol)

The most appropriate answer is tatement I is incorrect but Statement II is correct.