NEET Organic Chemistry - Basic Principles PYQs & Practice Questions

8 free NEET Chemistry questions on Organic Chemistry - Basic Principles, starting with previous year questions (2016–2023). Tap “Show answer” to check each one. No login needed.

Topics: Aromaticity and Huckel's Rule · Bond Enthalpy of C-X Bond · Resonance in Aromatic Heterocycles · Hybridisation in organic molecules · Stability of carbocations (resonance) · Stability of carbocations: hyperconjugation · Aromatic Compound Identification · Kjeldahl's Method

1. Q1NEET 2023Aromaticity and Huckel's Rule

   Consider the following compounds/species: (i) Naphthalene, (ii) Cyclopentadienyl anion, (iii) Cyclobutadiene, (iv) Cyclopropenyl cation, (v) Cyclopropenyl anion, (vi) Cyclooctatetraene, (vii) Anthracene. The number of compounds/species which obey Huckel's rule is ____.

   • A. 6
   • B. 2
   • C. 5
   • D. 4
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   Answer: C — 5

   Huckel's rule requires a planar, cyclic, fully conjugated system with (4n+2) pi electrons. Naphthalene (10 pi e-), cyclopentadienyl anion (6 pi e-), cyclopropenyl cation (2 pi e-), and anthracene (14 pi e-) all satisfy (4n+2) and obey Huckel's rule — 4 species in total. Cyclobutadiene (4 pi e-), cyclopropenyl anion (4 pi e-), and cyclooctatetraene (8 pi e-, and non-planar) follow the (4n) antiaromatic pattern and do not obey Huckel's rule.

2. Q2NEET 2022Aromatic Compound Identification

   Identify the given aromatic compound from its structure.

   • A. Compound (1)
   • B. Compound (2)
   • C. Compound (3)
   • D. Compound (4)
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   Answer: C — Compound (3)

   Based on the structural features shown, the correct aromatic compound is option (4).

3. Q3NEET 2022Kjeldahl's Method

   The Kjeldahl's method for the estimation of nitrogen can be used to estimate the amount of nitrogen in which one of the following compounds?

   • A. p-Nitroaniline (NO2 and NH2 on benzene ring)
   • B. Pyridine
   • C. Aniline (phenyl-NH2)
   • D. Azobenzene (Ph-N=N-Ph)
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   Answer: B — Pyridine

   Kjeldahl's method works for compounds where nitrogen can be converted to ammonium sulfate; it fails for nitro, azo and diazo compounds and pyridine-type nitrogen, but works for aniline-type amine nitrogen.

4. Q4NEET 2021Bond Enthalpy of C-X Bond

   The correct sequence of bond enthalpy of 'C-X' bond is:

   • A. CH3-Cl > CH3-F > CH3-Br > CH3-I
   • B. CH3-F < CH3-Cl < CH3-Br < CH3-I
   • C. CH3-F > CH3-Cl > CH3-Br > CH3-I
   • D. CH3-F < CH3-Cl > CH3-Br > CH3-I
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   Answer: B — CH3-F < CH3-Cl < CH3-Br < CH3-I

   Bond enthalpy of C-X bond: CH3-F > CH3-Cl > CH3-Br > CH3-I. Down the group with increase in size of halogen in C-X bond, bond energy decreases.

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5. Q5NEET 2020Stability of carbocations: hyperconjugation

   A tertiary butyl carbocation is more stable than a secondary butyl carbocation because of which of the following?

   • A. -I effect of -CH3 groups
   • B. +R effect of -CH3 groups
   • C. -R effect of -CH3 groups
   • D. Hyperconjugation
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   Answer: C — -R effect of -CH3 groups

   Tertiary carbocations are more stable than secondary ones primarily due to greater hyperconjugation (more alpha hydrogens available for no-bond resonance) and the electron-donating inductive effect of more alkyl groups.

6. Q6NEET 2018Hybridisation in organic molecules

   Which of the following molecules represents the order of hybridisation sp2, sp2, sp, sp from left to right atoms?

   • A. HC≡C-C≡CH
   • B. CH3-CH=CH-CH3
   • C. CH2=CH-CH=CH2
   • D. CH2=CH-C≡CH
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   Answer: C — CH2=CH-CH=CH2

   In CH2=CH-C≡CH (vinylacetylene), the first two carbons (CH2=CH-) are sp2 hybridised (double bond) and the next two carbons (-C≡CH) are sp hybridised (triple bond), giving the sequence sp2, sp2, sp, sp.

7. Q7NEET 2018Stability of carbocations (resonance)

   Which of the following carbocations is expected to be most stable? (structures show a cyclohexadienyl-type cation with NO2 and Y substituents at different positions)

   • A. Structure (1)
   • B. Structure (2)
   • C. Structure (3)
   • D. Structure (4)
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   Answer: B — Structure (2)

   The stability of the carbocation depends on the position of the positive charge relative to the electron-withdrawing NO2 group and the substituent Y; the structure where the positive charge is best stabilised by resonance/induction (structure 3) is the most stable among the given options, per the official answer key.

8. Q8NEET 2016Resonance in Aromatic Heterocycles

   In pyrrole (5-membered ring with N1-H, carbons numbered 2,3,4,5), the electron density is maximum on:

   • A. 2 and 5
   • B. 2 and 3
   • C. 3 and 4
   • D. 2 and 4
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   Answer: —

   Resonance structures of pyrrole show negative charge (higher electron density) developing at C2 and C5 positions (adjacent to N), due to delocalization of the nitrogen lone pair into the ring.