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Test: MCQs (One or More Correct Option): Organic Chemistry - Some Basic Principles & Technique | JEE Advanced - JEE MCQ


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18 Questions MCQ Test 35 Years Chapter wise Previous Year Solved Papers for JEE - Test: MCQs (One or More Correct Option): Organic Chemistry - Some Basic Principles & Technique | JEE Advanced

Test: MCQs (One or More Correct Option): Organic Chemistry - Some Basic Principles & Technique | JEE Advanced for JEE 2024 is part of 35 Years Chapter wise Previous Year Solved Papers for JEE preparation. The Test: MCQs (One or More Correct Option): Organic Chemistry - Some Basic Principles & Technique | JEE Advanced questions and answers have been prepared according to the JEE exam syllabus.The Test: MCQs (One or More Correct Option): Organic Chemistry - Some Basic Principles & Technique | JEE Advanced MCQs are made for JEE 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: MCQs (One or More Correct Option): Organic Chemistry - Some Basic Principles & Technique | JEE Advanced below.
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*Multiple options can be correct
Test: MCQs (One or More Correct Option): Organic Chemistry - Some Basic Principles & Technique | JEE Advanced - Question 1

Resonance structures of a molecule should have :

Detailed Solution for Test: MCQs (One or More Correct Option): Organic Chemistry - Some Basic Principles & Technique | JEE Advanced - Question 1

Resonating structures differ in bonding pattern.

*Multiple options can be correct
Test: MCQs (One or More Correct Option): Organic Chemistry - Some Basic Principles & Technique | JEE Advanced - Question 2

Phenol is less acidic than :

Detailed Solution for Test: MCQs (One or More Correct Option): Organic Chemistry - Some Basic Principles & Technique | JEE Advanced - Question 2

Higher the stability of the corresponding anion, more will be the acidic character of the parent compound.

Higher stability of acetate ions than phenoxide ion is due to equivalent resonating structures in the former

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*Multiple options can be correct
Test: MCQs (One or More Correct Option): Organic Chemistry - Some Basic Principles & Technique | JEE Advanced - Question 3

Dipole moment is shown by :

Detailed Solution for Test: MCQs (One or More Correct Option): Organic Chemistry - Some Basic Principles & Technique | JEE Advanced - Question 3

1, 4-Dichlorobenzene (p-dichlorobenzene) and trans1, 2-dichloroethene have zero dipole moment because of their symmetrical structures.

*Multiple options can be correct
Test: MCQs (One or More Correct Option): Organic Chemistry - Some Basic Principles & Technique | JEE Advanced - Question 4

Only two isomeric monochloro derivatives are possible for:

Detailed Solution for Test: MCQs (One or More Correct Option): Organic Chemistry - Some Basic Principles & Technique | JEE Advanced - Question 4

In n-butane, Cl can add at either the first or second carbon giving two isomers.

 will give three isomers with Cl group at either of the CH3 groups, second C-atom and 3rd C-atom.
Benzene forms only one single derivative.

Option (d) :  will again give two isomers with Cl at either one of the CH3 groups or on the central C-atom

*Multiple options can be correct
Test: MCQs (One or More Correct Option): Organic Chemistry - Some Basic Principles & Technique | JEE Advanced - Question 5

Which of the following have asymmetric carbon atom?

Detailed Solution for Test: MCQs (One or More Correct Option): Organic Chemistry - Some Basic Principles & Technique | JEE Advanced - Question 5

An asymmetric carbon atom is one which is attached with 4 different groups. Hence (c) & (d) are correct.

*Multiple options can be correct
Test: MCQs (One or More Correct Option): Organic Chemistry - Some Basic Principles & Technique | JEE Advanced - Question 6

What is the decreasing order of strength of the bases

Detailed Solution for Test: MCQs (One or More Correct Option): Organic Chemistry - Some Basic Principles & Technique | JEE Advanced - Question 6

TIPS/Formulae : Conjugate base of strong acid is weak while conjugate base of a weak acid is stronger.
Acidic strength of acids, HOH > CH ≡ CH > NH3 > CH3.CH3 Hence the order of strength of bases,

*Multiple options can be correct
Test: MCQs (One or More Correct Option): Organic Chemistry - Some Basic Principles & Technique | JEE Advanced - Question 7

Which of the following compounds will show geometrical isomerism?

Detailed Solution for Test: MCQs (One or More Correct Option): Organic Chemistry - Some Basic Principles & Technique | JEE Advanced - Question 7


Only 2-butene and 1-phenylpropene can show geometrical isomerism (cis-and trans-isomers).

*Multiple options can be correct
Test: MCQs (One or More Correct Option): Organic Chemistry - Some Basic Principles & Technique | JEE Advanced - Question 8

Among the following compounds, the strongest acid is

Detailed Solution for Test: MCQs (One or More Correct Option): Organic Chemistry - Some Basic Principles & Technique | JEE Advanced - Question 8

Order of acidic strength CH3OH > CH ≡ CH > C6H6 > C2H6 ; CH3OH is most acidic because O is more electronegative than C and capable of accommodating negative charge in CH3O– Although alcohols are neutral towards the litmus paper.

*Multiple options can be correct
Test: MCQs (One or More Correct Option): Organic Chemistry - Some Basic Principles & Technique | JEE Advanced - Question 9

Tautomerism is exhibited by

Detailed Solution for Test: MCQs (One or More Correct Option): Organic Chemistry - Some Basic Principles & Technique | JEE Advanced - Question 9

TIPS/Formulae :

For a carbonyl compound to show tautomerism, it must have at least one H at the α – carbon atom. (a), (c) and (d) show tautomerism.


 Tautomerism is not possible

*Multiple options can be correct
Test: MCQs (One or More Correct Option): Organic Chemistry - Some Basic Principles & Technique | JEE Advanced - Question 10

An aromatic molecule will

Detailed Solution for Test: MCQs (One or More Correct Option): Organic Chemistry - Some Basic Principles & Technique | JEE Advanced - Question 10

NOTE : An aromatic species will have :
(4n + 2) p electrons     (by Huckel’s Rule)
planar structure           (due to resonance)
cyclic structure             (due to presence of sp2 - hybrid carbon atoms).

*Multiple options can be correct
Test: MCQs (One or More Correct Option): Organic Chemistry - Some Basic Principles & Technique | JEE Advanced - Question 11

The correct statements(s) concerning the structures E,F and G is (are) –

Detailed Solution for Test: MCQs (One or More Correct Option): Organic Chemistry - Some Basic Principles & Technique | JEE Advanced - Question 11

E and F ; and also E and G differ in position of atom (H), so these are tautomers (not resonating structures.
Geometrical isomers are also diastereomers).

*Multiple options can be correct
Test: MCQs (One or More Correct Option): Organic Chemistry - Some Basic Principles & Technique | JEE Advanced - Question 12

The correct statement(s) about the compound given below is (are)

Detailed Solution for Test: MCQs (One or More Correct Option): Organic Chemistry - Some Basic Principles & Technique | JEE Advanced - Question 12

The given molecule although posseses neither centre of symmetry nor a plane of symmetry (hence optically active) but it has an axis of symmetry (Cn).
NOTE : A Cn axis of symmetry is an axis about which the molecule can be rotated by 360°/n to produce a molecule indistinguishable from the original molecule.

*Multiple options can be correct
Test: MCQs (One or More Correct Option): Organic Chemistry - Some Basic Principles & Technique | JEE Advanced - Question 13

The correct statement(s) about the compound

H3C(HO)HC–CH = CH – CH(OH)CH3 (X) is(are)

Detailed Solution for Test: MCQs (One or More Correct Option): Organic Chemistry - Some Basic Principles & Technique | JEE Advanced - Question 13

*Multiple options can be correct
Test: MCQs (One or More Correct Option): Organic Chemistry - Some Basic Principles & Technique | JEE Advanced - Question 14

In the Newman projection for 2,2-dimethylbutane

X and Y can respectively be

Detailed Solution for Test: MCQs (One or More Correct Option): Organic Chemistry - Some Basic Principles & Technique | JEE Advanced - Question 14

Structural formula of 2, 2-dimethylbutane is

(I) Newman projection using C1–C2 bond

(II) Newman projection using C3–C2 bond

*Multiple options can be correct
Test: MCQs (One or More Correct Option): Organic Chemistry - Some Basic Principles & Technique | JEE Advanced - Question 15

Amongst the given options, the compound(s) in which all the atoms are in one plane in all the possible conformations (if any), is (are)

*Multiple options can be correct
Test: MCQs (One or More Correct Option): Organic Chemistry - Some Basic Principles & Technique | JEE Advanced - Question 16

Which of the following molecules, in pure form, is (are) unstable at room temperature ?

Detailed Solution for Test: MCQs (One or More Correct Option): Organic Chemistry - Some Basic Principles & Technique | JEE Advanced - Question 16

b and c, being antiaromatic, are unstable at room temperature.

*Multiple options can be correct
Test: MCQs (One or More Correct Option): Organic Chemistry - Some Basic Principles & Technique | JEE Advanced - Question 17

Which of the given statement(s) about N, O, P and Q with respect to M is (are) correct ?

Detailed Solution for Test: MCQs (One or More Correct Option): Organic Chemistry - Some Basic Principles & Technique | JEE Advanced - Question 17

Converting all the structures in the Fischer projection


M and N are diastereoisomers M and O are identical M and P are enantiomers M and Q are diastereoisomers Hence, the correct options are a, b, c.

*Multiple options can be correct
Test: MCQs (One or More Correct Option): Organic Chemistry - Some Basic Principles & Technique | JEE Advanced - Question 18

The hyperconjugative stabilities of tert-butyl cation and 2-butene, respectively, are due to

Detailed Solution for Test: MCQs (One or More Correct Option): Organic Chemistry - Some Basic Principles & Technique | JEE Advanced - Question 18

In tert butyl cation, carbon bearing positive charge has one vacant p-orbital hence it is σ–p (empty) conjugation or hyperconjugation.

In 2-butene, hyperconjugation is between σ→π* bond.

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