JEE Advanced (One or More Correct Option): Some Basic Principles of Organic Chemistry | Chapter-wise Tests for JEE Main & Advanced PDF Download

Q.1. Which of the following species is/are aromatic
(a)
(b)
(c)
(d)

Correct Answer is option (a, b and c)

There are 4 conditions a molecule must meet in order for it to be aromatic. If any of these conditions are violated, no aromaticity is possible.

• First, it must be cyclic.
• Second, every atom in the ring must be conjugated.
• Third, the molecule must have [4n+2] pi electrons.
• Fourth, the molecule must be flat (usually true if conditions 1-3 are met, but there are rare exceptions) 

In the given case, 

(a) Benzyne is the aromatic compound as it is cyclic, planar in geometry. It has a conjugated 6π electron system.  
(b) Cyclopropenyl Cation has 2 π-electrons exhibiting cyclic delocalisation, and the 4n+2 (n=0) rule predicts aromaticity.
(c) Cyclobutadienyl Dication has 2 electrons, the delocalization of 2 electrons is possible through empty p orbitals. Hence, it is Aromatic.
(d) Cyclopentadiene is non-aromatic because it does not meet the criteria for aromaticity. Cyclopentadiene has 5 pi electrons, which does not meet the Hückel's rule, and thus it is not considered to be non aromatic.

Q.2. In which of the following cases the first compound has higher dipole moment than the second?
(a) CH₂ = CH—Cl & CH₃—CH₂—Cl
(b)
(c)
(d) CH₃—CH=CH—CHO and CH₃CH₂CH₂CHO

Correct Answer is option (b and d)

(a) In this case, second compound is having more dipole moment than the first.

(b) If you see the resonance structure of nitrobenzene, the two oxygen atoms pull electrons have negative charges hence resulting in much higher dipole moment.

(c) In this case, second compound is having more dipole moment than the first, due to more delocalization 

(d) In this case, first compound is having more dipole moment as larger charge separation leads to higher dipole moment. 

Q.3. The rearrangements in which  the alkyl isocyanate (RNCO) is formed as an intermediate is  are
(a) Hoffmann-bromamide reaction
(b) Schmidt reaction
(c) Curtius reaction
(d) Lossen reaction

Correct Answer is option (a, b, c and d)

In each of these reactions, an acyl azide or its precursor is involved, and they can lead to the formation of an alkyl isocyanate as an intermediate, which can then react further to produce different products. So, the correct answer is indeed (a, b, c and d)
Here is the summary of Hoffmann and Curtius Reaction 

Q.4. Stability of alkyl carbocation can be explained by
(a) Inductive effect
(b) Hyperconjugation
(c) Electromeric effect
(d) None of these

Correct Answer is option (a and b)

The stability of carbocations increases from primary to secondary to tertiary carbons. The stability is attributed to two main effects: inductive effect and hyperconjugation effect.

• Alkyl groups have a +I (positive inductive) effect, donating electron density to the positively charged carbon in the carbocation. Tertiary carbocations (three alkyl groups) are most stable, followed by secondary (two alkyl groups), and then primary (one alkyl group) carbocations. The methyl carbocation (no alkyl groups) is the least stable among carbocations.

• Hyperconjugation involves the interaction of adjacent C-H sigma bonds with the carbocation's empty p orbital. More adjacent alkyl groups lead to more hyperconjugation interactions and increased stability.
  

Q.5. Choose the correct reaction as per the given major product (s)
(a)
(b)
(c)
(d)

Correct Answer is option (a and c)

(A)