Heterocyclic Compounds MCQ Quiz - Objective Question with Answer for Heterocyclic Compounds - Download Free PDF

Concept:-

Lewis concept of acids and bases:

• Species that can accept a pair of electrons is a lewis acid.
• Lewis acid has a generally positive charge.
• Lewis acids have an empty orbital.
• Electron pairs from donor atoms are accepted in these orbitals.
• Species that can donate a pair of electrons are called lewis bases.

• It is the ability of a compound to donate its pair of electrons.

​Bronsted Lowry's theory of acids and bases:

According to this theory,

• ​A base is a substance that is capable of accepting a proton from another compound in a reaction.
• The new protonated compound thus formed is called its conjugate acid.
• An acid is a substance that is capable of donating a proton in a solution or to other compounds.

• Molecules, where the electron pair is delocalized by resonance, will be less basic.

Explanation:-

• Basicity is proportional to the Ease of donation of lone pair of electron

• Basicity is directly proportional to the ease of donation of lone pair of electrons as

Basicity \(\propto\frac{1}{{{\rm{\% \;}}of\;s - \;character\;of\;N}}\)

\({\rm{\;\;}}\propto\;\frac{1}{{Electronegativity\;of\;N}}\)

• Electronegativity is a chemical property that describes the tendency of an atom to attract a shared pair of electrons or electron density toward itself. 
• An atom's electronegativity is affected by both its atomic number and the distance at which its valence electrons reside from the charged nucleus.
• The higher the associated electronegativity number, the more an atom or a substituent group attracts electrons toward itself.

I.​

• For the above compound, the lone pair of electrons of the N atom is involved in the aromaticity of the ring.  Thus, the  lone pair of electrons of the N atom is involved in delocalization with the other pi electrons of the ring 
• Thus, the lone pair of electrons on the N atom is not available for donation, and compound I is less basic.

II. 

• Similarly for the above compound,  the lone pair of electrons of the N atom is involved in the aromaticity of the ring and hence, compound II is also less basic.
• In compound II, the lone pair of electrons of the N atom is involved in delocalization with the two aromatic ring in compare to compound I, where it is involved in delocalization with only one ring.
• Hence, compound II is the least basic.

III and IV.

• Compounds III and IV both will be more basic than compound I and II, as in compounds III and IV the lone pair of electrons of the N atom is not involved in the aromaticity.
• In between compounds III and IV, compound IV is more basic in comparison to IIII as the lone pair of electrons of the N atom in compound IV is situated in an sp3 hybridized orbital whereas, in compound III the lone pair of electrons of the N atom in compound III is situated in an sp2 hybridized orbital.
• As an sp2 hybridized orbital is more electronegative than an sp3 hybridized orbital, thus ease of donation of lone pair of electrons for compound IV is easier in comparison to IIII.
• Thus, compound IV will be the most basic amine among the four amines.
• Hence, the correct order of basicity will be

​IV>III>I>II

Conclusion:-

• Hence, option 3 is the correct answer.

​

Additional InformationIndole is more basic than pyrrole because in N-protonated indolium cation aromaticity of benzene is retain while aromaticity of pyrrolium ion is loss after the protonation of indole & pyrrole respectively. Among pyridine & piperidine; piperidine is more basic (localized lone pair on sp³ hybridized atom is more basic than the localized lone pair on sp² hybridized atom. Order of the basicity:-

Concept:

Aromaticity and Resonance in Five-Membered Heterocyclic Compounds

• Five-membered heterocyclic compounds such as furan, thiophene, and pyrrole exhibit aromaticity due to delocalized π-electrons.
• Furan is the least aromatic because oxygen is highly electronegative and withdraws electron density, reducing resonance stabilization.
• Thiophene is more aromatic than furan due to sulfur’s lower electronegativity and ability to participate in resonance with 3d-orbitals.
• Pyrrole is highly aromatic due to strong delocalization of the lone pair on nitrogen into the π-system.
• The Paal-Knorr synthesis is a method for synthesizing furan, thiophene, and pyrrole from 1,4-dicarbonyl compounds.

Explanation:

• Statement 1: "Furan is least aromatic in nature among furan, thiophene, and pyrrole."
  • Correct, because furan's oxygen withdraws electron density, reducing its aromatic stability.
• Statement 2: "Thiophene can have more number of canonical structures using 3d-orbitals."
  • Correct, because sulfur can utilize its 3d-orbitals, allowing additional resonance structures.
• Statement 3: "Pyrrole has maximum double bond character among furan, thiophene, and pyrrole."
  • Incorrect, because in pyrrole, the nitrogen lone pair is highly delocalized into the ring.
  • This reduces the double bond character compared to thiophene, which has stronger resonance stabilization.
• Statement 4: "Furan, thiophene, and pyrrole can be synthesized using Paal-Knorr synthesis."
  • Correct, as the Paal-Knorr synthesis efficiently produces these heterocycles from 1,4-diketones.

Therefore, the incorrect statement is "Pyrrole has maximum double bond character among furan, thiophene, and pyrrole."

Concept:

Basicity in Nonpolar Aprotic Solvent

• In a nonpolar aprotic solvent, the strength of a base is determined by the availability of the lone pair of electrons on the nitrogen atom. The less the nitrogen is involved in resonance or electron-withdrawing effects, the stronger the base is, as the lone pair is more readily available for protonation.
• In nonpolar aprotic solvents, the solvent does not solvate anions well, allowing the basicity of the compound to be more influenced by the structure of the base itself.
• The compounds to be compared are:
  • In option 1 (aniline, C₆H₅NH₂), the nitrogen's lone pair is somewhat involved in resonance with the aromatic ring, reducing its availability for protonation.
  • In option 2 (pyrrole, C₄H₄N), the nitrogen's lone pair is involved in the aromatic system, making it less available for protonation, thus a weaker base.
  • In option 3 (pyrrolidine, C₄H₉N), the nitrogen is not involved in resonance, and its lone pair is readily available, making it the strongest base among the options.
  • In option 4 (pyridine, C₅H₅N), the nitrogen's lone pair is not involved in an aromatic system, although less basic than non-aromatic amines due to the sp2 hybridization of the nitrogen atom. 

Hence, the correct answer is Pyrrolidine (option 3).

Concept:

Perkin Reaction

• The Perkin reaction is a classic organic reaction used to form α,β-unsaturated carboxylic acids (also known as cinnamic acids) by the condensation of an aromatic aldehyde with an acid anhydride.

Explanation:

Mechanism

Therefore, the correct option is 2.

Concept:

Pyridine and its Properties

• Pyridine is an aromatic compound similar to benzene but with one of its carbons replaced by a nitrogen atom, making it a basic heterocyclic compound.
• It is less basic than pyrrole because the nitrogen atom in pyridine has a lone pair of electrons that are not involved in the aromaticity of the ring, making them available for protonation, but the nitrogen in pyrrole is part of the ring’s aromatic system and thus less available for protonation.
• The carbon-carbon bonds in pyridine are of equal length, and the atoms in pyridine are sp² hybridized, meaning the carbon atoms in the ring form bonds with 120° angles between them.
• Pyridine has a 117 kJ/mol (close to 125.5) resonance energy. The value can vary depending on the source.

Explanation:

• The statement "Pyridine is less basic than pyrrole" is incorrect because pyridine is generally more basic than pyrrole due to the availability of the nitrogen's lone pair for protonation, unlike pyrrole where the lone pair is involved in the aromatic system and is less available for protonation.
• The other statements are correct:
  • Carbon-carbon bonds in pyridine are of equal length due to resonance in the ring.
  • Pyridine’s atoms are sp² hybridized, and the lone pair of electrons on nitrogen is in an sp² orbital.

Therefore, the incorrect statement is: Pyridine is less basic than pyrrole.

Concept: Triflic anhydride is useful for converting ketones into enol triflates. In a representative application, is used to convert an imine into a NTf group. It will convert phenols into a triflic ester, which enables cleavage of the C-O bond.

Hydrazine is a powerful, endothermic reducing agent.

Explanation: Triflic anhydride first get attached to the ring and then lone pair of hydrazine will attack to triflate carbon followed by proton transfer and then ring cyclization.

It is Wolf Kishner clemmensen reaction.

Conclusion: option A is correct.

Concept:

→ The reaction you described is a bromination reaction, where 3-methylthiophene is reacted with N-bromosuccinimide (NBS) in the presence of a radical initiator such as (PhCOO)₂ and benzene (C₆H₆) under heat.

→ The reaction mechanism involves the formation of a bromine radical from NBS, which reacts with the 3-methylthiophene molecule to form a thiophene radical intermediate.

→ This radical intermediate then reacts with molecular bromine to form a more stable intermediate. This intermediate then undergoes rearrangement to form 3-(bromomethyl)thiophene as the major product.

Explanation:

The reaction can be summarized as follows:

→Step 1: Formation of a bromine radical from NBS:

NBS + heat → Br• + NsH

→ Step 2: Reaction of the bromine radical with 3-methylthiophene to form a thiophene radical intermediate:

Br• + 3-methylthiophene → 3-methylthiophene radical

→ Step 3: Reaction of the thiophene radical intermediate with molecular bromine to form a more stable intermediate:

3-methylthiophene radical + Br₂ →

→ Step 4: Rearrangement of the intermediate to form 3-(bromomethyl)thiophene as the major product:

→ 

Conclusion: Therefore, the major product formed in the reaction of 3-methylthiophene in the presence of NBS, (PhCOO)₂, C₆H₆, heat is 3-(bromomethyl)thiophene. Therefore, the correct answer is option 1 .

Concept:

Decarboxylation:

• Decarboxylation is the process of removal of a molecule of carbon dioxide from a compound.
• The rate of decarboxylation may be different for different molecules depending upon the nature of the substituent present in the molecule.
• The decarboxylation reaction is a kind of reverse Friedel craft reaction, in which a proton (provided by the carboxylic acid) itself acts as an electrophile. The protonation may occur anywhere, but it leads to decarboxylation only if it occurs where there is a -CO2H group.

Explanation:

• The electrophilic aromatic substitution reaction works best for indole in 3-position. The -CO2H group is present at the 3-position in indole.
• The electrophilic aromatic substitution reaction works best for pyridine in the 3-position. The -CO2H group is present at the 2-position in pyridine. This concludes the decarboxylation reaction will be fast for indol than for pyridine.
• The N atom present at the beta position facilitates the decarboxylation reaction.
• Due to the absence of an N atom at the beta position, benzoic acid will undergo decarboxylation at the slowest rate.
• The mechanism of the decarboxylation reaction is

Conclusion:

• Hence, the ease of decarboxylation is in the order 

C > B > A.

Concept:

• Methyl amine (MeNH₂) can act as a base as well as the nucleophile.  
• if substrate has acidic proton, it will preferably act as the base and abstracts the most acidic proton.
• the H is considered acidic if the negative charge formed after  its abstraction can be stabilized by conjugation or presence of some electronegative atom.

Explanation:

• In the first step, Methyl amine will abstract the most acidic proton attached to N in 5-membered ring.

          The generated negative charge is resonance stablized.

• In the next step, negative charge will move to C and will substitute Br^- to form 3 membered ring (SN²).

• Next step will follow up with  nucleophillic attack of another methylamine molecule at electrophillic carbon centre of Carbonyl bond.

• Finally, the back conjugation of negative change on O, will facilitate the breaking of  3-membered ring (which is unstable) in such a way that the aromaticity of 5-membered ring is regained.

Conclusion:

The final product of the reaction is :

Concept:

• Organolithium reagents contain carbon–lithium (C-Li) bonds. 
• The large difference in electronegativity between the carbon atom and the lithium atom results in highly ionic Carbon−Lithium bonds.
• Because of the polar nature of the C−Li bond, organolithium reagents can act as good nucleophiles as well as strong bases.

Explanation:

• The reaction pathway is shown below:

• From the above reaction, we see that in the first step, the organolithium reagent (^tBuLi) will act as a strong base and will abstract an acidic -H from the pyridine derivative.
• In the last step of the reaction, the resulting carbanion will react with the generated I2 in the reaction medium to give the final product.
• In the reaction medium ICH₂CH₂I will form I₂.

Conclusion:

• Hence, the major product formed in the following reaction sequence is

.

Concept:

→ The rate of thermal decarboxylation of organic carboxylic acids depends on several factors, including the stability of the carboxylate ion, the strength of the carbon-oxygen bond, and the electronic properties of the substituents on the aromatic ring.

Explanation:

→ Benzoic acid does not have an electron-withdrawing group attached to the aromatic ring, and the carbon-oxygen bond in the carboxylate group is relatively strong, making it more difficult to break. Therefore, the decarboxylation of A is expected to be the slowest among the three compounds.

→ Picolinic acid also has an electron-withdrawing group (-COOH) attached to the aromatic ring, but it is less stable than the carboxylate ion in C because the nitrogen atom in the pyridine ring can donate electron density to the carboxylate group.

However, the carbon-oxygen bond in the carboxylate group is still relatively weak, making it easier to break than in A. Therefore, the decarboxylation of B is expected to be faster than A but slower than C.

→ (2-(pyridin-2-yl)acetic acid) has an electron-withdrawing group (-COOH) attached to the aromatic ring, which makes the carboxylate ion more stable.

Additionally, the carbon-oxygen bond in the carboxylate group is relatively weak, making it easier to break. Therefore, the decarboxylation of C is expected to be faster than the other two compounds

Conclusion:
In summary, the correct order for the rate of thermal decarboxylation of the given compounds is C > B > A, with C being the fastest and A being the slowest.

Concept:

The given reaction involves sulfonation of pyridine, followed by bromination in the presence of heat. Pyridine, being an electron-deficient aromatic system, reacts via its nitrogen to form an intermediate that stabilizes the substitution process. This leads to a sulfonated intermediate before further bromination occurs.

• Sulfonation Reaction: Pyridine undergoes electrophilic substitution in the presence of sulfur trioxide (SO₃).

• Bromination Reaction: Bromine (Br₂) then substitutes at the position adjacent to the sulfonate group upon heating, forming a brominated pyridine intermediate.

• Aqueous Treatment: The reaction is neutralized with sodium carbonate (Na₂CO₃), which helps to stabilize the product by removing any excess acid.

Explanation: 

• Mechanism:

  • ​Step 1: Nitrogen donates its lone pair to Sulphur of SO₃ and form a adduct.

  • Step 2: Bromination takes place.

    • 

Conclusion:

• The correct intermediate in this reaction is a sulfonated pyridinium ion with bromination at the 2-position, as shown in option 2.

Concept:

• A nucleophilic substitution reaction is a class of chemical reactions in which an electron-rich chemical species or nucleophile reacts with another electron-deficient molecule and replaces a functional group within. 
• The molecule that contains the electrophile and the leaving group is known as substrate.
• Lithium diisopropylamide or LDA is a chemical compound with the molecular formula LiN(CH(CH₃)₂)₂. It is a strong base and has good solubility in non-polar organic solvents and
• LDA is non-nucleophilic.

Explanation:

• The reaction pathway is shown below:

• From the above reaction, we can see that LDA act as a strong base and abstract a proton from the substrate in the first step of the recation.
• The intermediate carbanion will act as a nucleophile and react with  PhCh₂Br (Electrophile) via a nucleophilic substitution reaction. The reaction can lead to two products depending upon the streic effect experienced by the incoming group with the substituent presents in intermediate carbanion.
• The phenyl group (-Ph) being a bulky substituent will cause more steric hindrance than -Me group. Also, the incoming electrophile is a much bulkier substituent than the methyl group.
• The bulky phenyl group (-Ph) is present above the plane in the intermediate carbanion.
• This indicates that the incoming group (-CH₂Ph) will face more steric effect when it will be placed in above the plane.
• Thus, in the major product incoming electrophile will be placed below the plane. 

 ​Conclusion:

• Hence, the major products formed in the following reaction is 

​ 

Explanation:- 

• Ac₂O  or (CH₃CO)₂O is an isolated anhydride of carboxylic acid and is widely used as a reagent for organic reactions.
• It is colourless with a strong smell of acetic acid in it.

Here, it attacks the electronegative oxygen and the complex so formed is rearranged in order to get neutralized and become stable.

Reaction:- 

Mechanism :

Concept:

• Pyrrole is a heterocyclic compound and a  much weaker base than secondary amines.
• It has a weak acidic character with pKa value of 17.5.
• The structure of Pyrrole is: 

• The proton bound to the nitrogen atom is acidic in nature and can be abstracted by strong bases such as sodium amide, ammonia, and n-butyl lithium in n-hexane.
• Pyrrole can react with Grignard reagents. Pyrrole is an aromatic system and undergoes electrophilic substitution reactions easily.
• Pyrrole does not undergo nucleophilic substitution reactions.

Explanation:

• Pyrrole and furan undergo electrophilic substitution reactions.
• They are more reactive towards electrophilic substitution reaction than benzene.
• Electrophiles majorly attack the 2 positions rather than the third position in these compounds.
• This occurs because when the electrophile attacks on the 2nd position, the intermediate is more resonance stabilized.
• When the electrophile attacks on the 3rd position, only two resonating structures are possible.

• The carbocationic intermediate has to be resonance stabilized, so when the electrophilic substitution takes place in Furan, it is not stabilized, as much as pyrrole.
• This is because, oxygen being more electronegative than nitrogen, likes to keep its electron density towards itself.
• Pyrrole is a weaker base than Pyridine because, in pyridine, the lone pair of electrons are not involved in delocalization, whereas it is involved in resonance in the case of pyrrole.

Hence, the false statement is: Pyridine is less basic than pyrrole.