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Test: Heterocyclic Level - 1 - Chemistry MCQ


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20 Questions MCQ Test Organic Chemistry - Test: Heterocyclic Level - 1

Test: Heterocyclic Level - 1 for Chemistry 2024 is part of Organic Chemistry preparation. The Test: Heterocyclic Level - 1 questions and answers have been prepared according to the Chemistry exam syllabus.The Test: Heterocyclic Level - 1 MCQs are made for Chemistry 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Heterocyclic Level - 1 below.
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Test: Heterocyclic Level - 1 - Question 1

Pyrrole does not show basic properties, because:

Detailed Solution for Test: Heterocyclic Level - 1 - Question 1

Option 2 : lone pair on nitrogen is delocalized with p-orbital electrons of carbon

Test: Heterocyclic Level - 1 - Question 2

The heterocyclic diene employed in cyclo – addition reactions is:

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Test: Heterocyclic Level - 1 - Question 3

Furan on prolonged heating with dimethyl acetylenedicarboxylate yields:

Test: Heterocyclic Level - 1 - Question 4

The structures of the products A and B formed in the above reaction scheme respectively are:

Test: Heterocyclic Level - 1 - Question 5

The decreasing order of the reactivity of the following compounds towards electrophiles is:



Detailed Solution for Test: Heterocyclic Level - 1 - Question 5

Pyrole is most reactive towards the electrophiles because in this case it is carrying one lone pair of electron .and it is attached with one hydrogen atom , resulting less resonance stabilize than that of pyridine one the another hand thiophene is more reactive because it's carry 2 lone pair of electrons at last pyrole is more reactive than thiophene because nitrogen is smaller in size than that of sulphur so the availability of electrons cloud is maximum than sulphur ... consequently it is highly reactive towards electrophiles

Test: Heterocyclic Level - 1 - Question 6

Pyridine undergoes electrophilic nitration at elevated temperature to given the following as a major product:

Detailed Solution for Test: Heterocyclic Level - 1 - Question 6

There is a relatively less partial positive charge on meta positions of Pyridine. Hence, when an electrophile attacks it attacks on meta position.
Hence D is correct.

Test: Heterocyclic Level - 1 - Question 7

A pyridine derivative (P) reacts with (Y). (Y) can be a free radical, cation or anion. The structure of intermediate (Q) formed in the reacts is given below. (P) and (Y) respectively are:

Test: Heterocyclic Level - 1 - Question 8

Pyrrole + PhMgBr → E + F
E + MeCl → G + H
F + MeCl → no reaction without a catalyst. 







The structure of products E–H, respectively are:

Test: Heterocyclic Level - 1 - Question 9

Chose the correct answer from the following four choices.
Statement: Pyridine is more basic than pyrrole.
Reason: The nitrogen in pyrrole carries a proton while the nitrogen in pyridine does not.
Assertion: Nitrogen in trigonal geometry are generally more basic than nitrogens in a tetrahedral geometry. 

Detailed Solution for Test: Heterocyclic Level - 1 - Question 9
  • Pyridine is more basic than pyrrole because lone pair of electrons on N in pyridine and pyrrole are different in nature. These form a part of aromatic sextet in pyrrole, while not in pyridine.
  • Pyrrole, C4H4NH (in which N contributes a lone pair) has a pKa−3.8 but pyridine (where N is part of the ring's double bond) has a pKa 5.14. Electron pair availability indicates the strength of basicity. In this case, pyridine is the stronger base.
  • In pyrrole, the lone pair electrons of the nitrogen atom gets involved with the 2 carbon-carbon double bonds in the 5-member ring to form a conjugated system of pi electrons, leading to greater stability of the molecule.
  • Pyridine, on the other hand, already has a stable conjugated system of 3 double bonds in the aromatic hexagonal ring, like benzene. Hence, the lone pair electrons on the N atom in pyridine can be easily donated to a H+ ion or a Lewis acid.
Test: Heterocyclic Level - 1 - Question 10

In the following sequence of reactions, the major product Q is:

Test: Heterocyclic Level - 1 - Question 11

 What will be the reagent used for the completion of the following reaction?

Detailed Solution for Test: Heterocyclic Level - 1 - Question 11
  • In the case of pyrrole coupling reactions, dilute acids such as dilute HCl or dilute sulfuric acid (H₂SO₄) are often used.
  • Pyrrole is sensitive to strong acids (like concentrated acids), which can lead to protonation of the nitrogen atom and polymerization or degradation of the molecule.
  • Dilute acid provides mild reaction conditions to facilitate coupling without breaking the pyrrole structure.

Thus, dilute acid is used as the reagent.

Test: Heterocyclic Level - 1 - Question 12

The correct order of the basicity of the following compound is:

Test: Heterocyclic Level - 1 - Question 13

Identify the major Product P in the following two–step reaction:

Test: Heterocyclic Level - 1 - Question 14

Match the structures in List–I with their correct names in List–II.



Test: Heterocyclic Level - 1 - Question 15

The most acidic species is:

Test: Heterocyclic Level - 1 - Question 16

The compound that is NOT oxidized by KMnO4 is:

Test: Heterocyclic Level - 1 - Question 17

Thiophene reacts with HCHO in presence of aqueous HCl to give:

Detailed Solution for Test: Heterocyclic Level - 1 - Question 17

The CH2O/HCl reagent is suitable for chloromethylation of various polycyclic aromatic hydrocarbons as well as heterocyclic aromatic compounds. Thiophene gives 2-(chloromethyl)thiophene in 40-41% isolated yield, and 4-methylimidazole gives the 5-chloromethyl derivative (51-68%).

Test: Heterocyclic Level - 1 - Question 18

The reaction of 2-methylfuran with DMF-POCl3 would give:

Test: Heterocyclic Level - 1 - Question 19

In the following reaction the major product (X) is:

Test: Heterocyclic Level - 1 - Question 20

Two regions of cimetidine are susceptible to metabolism. Which regions?

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