465 questions on Some Basic Principles of Organic Chemistry for NEET 2027, 1 of them from previous NEET papers. Below are 15 to try; tap a question to see its answer and explanation.
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Answer: (A) 3°>2°>1°>CH₃⁺
More alkyl groups → more hyperconjugation → more stable.
Quick tip: Carbocation stability: 3°>2°>1°>methyl. Carbanion: reverse. Radical: 3°>2°>1°.
NCERT Class 11, Chapter 13, page 361
Answer: (A) Hexa-1,3,5-triene
Hexa-1,3,5-triene (CH₂=CH-CH=CH-CH=CH₂) consists solely of alternating double and single bonds where every single carbon is bonded to three other atoms (two carbons and one hydrogen, or one carbon and two hydrogens), making all six carbons sp² hybridized. In buta-1,2-diene C-2 is sp, in propenal oxygen is present though all carbons are sp², and in acrylonitrile the nitrile carbon is sp. Among pure hydrocarbon polyenes listed, hexa-1,3,5-triene has every carbon in the sp² state.
Answer: (C) 3,6-dioxohexanoic acid
The carboxylic acid group (-COOH) takes highest priority and is assigned C1. The chain has 6 carbons. When both ketone (=O) and aldehyde (=O at terminal position) are treated as prefix substituents in the presence of -COOH, both can be designated as 'oxo' groups at positions 3 and 6. Therefore, the name is 3,6-dioxohexanoic acid.
Answer: (B) Hepta-1,3-dien-6-yne
The principal chain must be numbered from the end that gives the lowest set of locants for all unsaturations. Numbering from right to left gives locants 1, 3 for double bonds and 6 for the triple bond (set: 1, 3, 6), whereas numbering from left to right gives 1 for triple bond and 4, 6 for double bonds (set: 1, 4, 6). The set (1, 3, 6) is lower at the second point of difference (3 < 4), so the name is hepta-1,3-dien-6-yne.
Answer: (A) 7
The structure of the 2,3-dimethylbut-2-yl cation is (CH₃)₂C⁺-CH(CH₃)₂. The carbocation center is bonded to two methyl groups (3 + 3 = 6 α-hydrogens) and one isopropyl methine group containing 1 α-hydrogen (C-H). Total α-hydrogens = 3 + 3 + 1 = 7.
Answer: (A) Both (A) and (R) are true and (R) is the correct explanation of (A)
A racemic mixture consists of equimolar proportions of two enantiomers (+) and (-). The net optical rotation is zero because the rotation caused by one enantiomer is externally compensated by the equal and opposite rotation of the other. Thus, both (A) and (R) are true, and (R) correctly explains (A).
Answer: (A) pent-1-en-4-yn-3-ol
The principal functional group is the -OH group, which is located at C-3 whether numbered from left or right. To decide between numbering directions, we compare the unsaturations: numbering from left gives double bond at 1 and triple bond at 4 (set 1, 4); numbering from right gives triple bond at 1 and double bond at 4 (set 1, 4). When the sets are identical, the double bond receives preference over the triple bond for the lower locant. Hence, the correct name is pent-1-en-4-yn-3-ol.
Answer: (B) 4-hydroxypentan-2-one
Between a ketone (>C=O) and an alcohol (-OH), the ketone has higher priority according to IUPAC rules and acts as the principal functional group. Numbering starts from the right to give the carbonyl carbon the lowest possible locant (C-2), making the -OH group a substituent at C-4. Therefore, the IUPAC name is 4-hydroxypentan-2-one.
Answer: (D) Attack of an electrophile (H⁺) to form a more stable carbocation
Hydration of an alkene in acidic medium is an electrophilic addition. The initial step is the electrophilic attack of H⁺ on the π-electron cloud to form the more stable secondary carbocation.
Answer: (A) 3
DBE = C − (H + X)/2 + N/2 + 1 = 7 − (10 + 0)/2 + 0/2 + 1 = 3. Oxygen does not change the DBE.
Answer: (C) Inversion of configuration occurs without the formation of an intermediate carbocation
An SN2 reaction takes place in a single concerted step through a pentacoordinate transition state (not an intermediate), resulting in complete inversion of configuration (Walden inversion).
Answer: (A) Both (A) and (R) are true and (R) is the correct explanation of (A)
Tropylium bromide ionizes to produce a bromide anion and a cycloheptatrienyl (tropylium) cation. The cation is planar, completely conjugated with 6 π electrons, satisfying Huckel's rule (4n+2 with n=1) for aromaticity, giving it extreme ionic character and water solubility.
Answer: (A) 0
DBE = C − (H + X)/2 + N/2 + 1 = 6 − (15 + 0)/2 + 1/2 + 1 = 0. Oxygen does not change the DBE.
Answer: (A) 9 σ and 9 π
Tetracyanoethylene consists of a central C=C bond (1 σ, 1 π), four C-CN single bonds (4 σ), and four C≡N triple bonds (each has 1 σ and 2 π, contributing 4 σ and 8 π). In total, there are 1 + 4 + 4 = 9 σ bonds and 1 + 8 = 9 π bonds.
Answer: (C) 6
Each of the 6 carbon atoms in benzene is sp² hybridized and contributes one unhybridized 2p orbital perpendicular to the molecular plane. These 6 p-orbitals overlap laterally to form a delocalized π molecular orbital system encompassing all 6 π electrons.