61 questions on this topic of Some Basic Principles of Organic Chemistry (NEET 2027 syllabus). Here are 12; tap one to see its answer and explanation.
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Answer: (B) Free radicals
In homolysis each atom takes one electron of the shared pair, forming neutral species with unpaired electrons (free radicals).
NCERT Class 11
Answer: (B) HC≡C⁻
The stability of carbanions depends strongly on the percentage of s-character in the hybridised orbital holding the unshared electron pair. In HC≡C⁻, the carbon is sp-hybridised (50% s-character), which is more electronegative and holds the negative charge closest to the nucleus, rendering it the most stable among the given options.
Answer: (A) Both Statement I and Statement II are correct.
Singlet carbene has an sp² hybridised carbon with a paired set of non-bonding electrons in one sp² orbital (diamagnetic). Triplet carbene has an sp hybridised carbon with two unpaired electrons in two unhybridised p-orbitals possessing parallel spins according to Hund's rule (paramagnetic).
Answer: (C) (C₆H₅)₃C⁺
Triphenylmethyl carbocation, (C₆H₅)₃C⁺, is exceptionally stable because the positive charge is delocalised over three benzene rings via resonance, producing a total of ten contributing canonical structures.
Answer: (A) 8
In an alkyl carbanion like CH₃-CH₂⁻, the negatively charged carbon atom forms three single covalent bonds (sharing 6 electrons) and carries one unshared electron pair (2 electrons), making a total of 8 valence electrons (an octet).
Answer: (B) Cleavage (2)
In cleavage (2), the resulting triphenylmethyl cation ((C₆H₅)₃C⁺) is exceptionally stable due to resonance over three phenyl rings, and the cyanide anion (CN⁻) is also exceptionally stable due to high electronegativity of the sp-hybridised carbon and nitrogen. In contrast, methyl carbanion (1), methyl cation (3), and OH⁺ (4) are extremely unstable species.
Answer: (D) I > III > IV > II
Electron-donating groups stabilize carbocations, whereas electron-withdrawing groups destabilize them. The -OCH₃ group provides strong resonance donation (+R), -CH₃ stabilizes through hyperconjugation and +I, while -NO₂ exerts a strong electron-withdrawing effect (-R, -I). Thus, the stability order is I > III > IV > II.
Answer: (C) (CH₃)₃C⁺ and Br⁻
Due to the higher electronegativity of bromine compared to carbon, heterolytic bond cleavage causes the shared pair of electrons to depart with bromine, yielding a bromide anion (Br⁻) and a stable tertiary carbocation ((CH₃)₃C⁺).
Answer: (A) Both Statement I and Statement II are correct
Carbenes (:CR₂) are neutral divalent carbon intermediates with 6 valence electrons (sextet). In a triplet carbene, carbon is sp hybridised with two degenerate p-orbitals containing one electron each with parallel spins, rendering it a diradical.
Answer: (B) Statement I is correct but Statement II is incorrect
Carbanions carry eight valence electrons, including an unshared pair, making them electron-rich nucleophiles (Statement I is correct). An alkyl carbanion typically undergoes sp³ hybridisation with three bond pairs and one lone pair, assuming a pyramidal geometry rather than a planar trigonal geometry (Statement II is incorrect).
Answer: (C) Extensive delocalisation of positive charge across three benzene rings
In the triphenylmethyl cation [(C₆H₅)₃C⁺], the positive charge is delocalized over 3 phenyl rings through resonance, giving a large number of contributing canonical structures that strongly stabilize the carbocation.
Answer: (B) Trigonal planar, +1
Heterolytic cleavage of 2-bromopropane releases a bromide ion (Br⁻) and an isopropyl carbocation [(CH₃)₂CH⁺]. The central carbon atom is sp² hybridized, possesses a formal charge of +1, and has a trigonal planar geometry.