Organic Chemistry: Some Basic Principles and Techniques NEET PYQReal NEET & AIPMT 2005–2021 questions, year-tagged, with solutions
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Organic Chemistry: Some Basic Principles and Techniques previous year questions
29 questions. Tap an option to check your answer.
Q1NEET 2021
The compound which shows metamerism is :
Not quite — the answer is D.
Metamerism arises when different alkyl groups are attached to the same polyvalent functional group, e.g. ethers. C₄H₁₀O gives CH₃–O–C₃H₇ and C₂H₅–O–C₂H₅, which are metamers. C₃H₈O has only one ether (CH₃OC₂H₅), so it cannot show metamerism.
Q2NEET 2020 · Q108
Paper chromatography is an example of :
Not quite — the answer is B.
In paper chromatography water held in the cellulose fibres is the stationary phase, and components separate by partition between it and the moving solvent, so it is partition chromatography. Thin layer and column chromatography are adsorption methods using a solid adsorbent.
Q3NEET 2020 · Q112
A tertiary butyl carbocation is more stable than a secondary butyl carbocation because of which of the following ?
Not quite — the answer is D.
The tert-butyl cation has nine α C–H bonds that can donate electron density into the empty p orbital, against fewer in the sec-butyl cation, so it gets more hyperconjugative stabilisation. Methyl groups are +I (electron-releasing), not −I, and they have no resonance effect here.
Q4NEET 2019 · Q46
The number of sigma (σ) and pi (π) bonds in pent-2-en-4-yne is:
Not quite — the answer is A.
Pent-2-en-4-yne is CH₃–CH=CH–C≡CH: 4 C–C σ bonds + 6 C–H σ bonds = 10 σ; the double bond adds 1 π and the triple bond 2 π, total 3 π. Counting every bond line of the multiple bonds as π would overcount.
Q5NEET 2018 · Q63
Which of the following molecules represents the order of hybridisation sp², sp², sp, sp from left to right atoms?
Not quite — the answer is B.
In CH₂=CH–C≡CH the two doubly bonded carbons are sp² and the two triply bonded carbons are sp. Buta-1,3-diene is all sp², and but-2-ene has sp³ end carbons.
Q6NEET 2017 · Q3
The correct statement regarding electrophile is:
Not quite — the answer is D.
An electrophile is an electron-deficient species: it may be positively charged (H⁺, NO₂⁺) or neutral (BF₃, AlCl₃, carbonyl carbon), and it bonds by accepting an electron pair from a nucleophile. Saying electrophiles are 'generally neutral' ignores the many cationic electrophiles, and negatively charged species are nucleophiles.
Q7NEET 2016 Phase 1 · Q90
The pair of electron in the given carbanion, CH₃C≡C⁻, is present in which of the following orbitals?
Not quite — the answer is D.
The negatively charged carbon is triple-bonded, so it is sp hybridised; its lone pair occupies the remaining sp orbital, pointing away from the triple bond. The two unhybridised 2p orbitals are used for the π bonds, so the lone pair is not in a p orbital.
Q8AIPMT 2015
The total number of π-bond electrons in the following structure is: (CH₃)₂C=CH–C(=CH₂)–CH=CH–CH=C(CH₃)₂
Not quite — the answer is C.
The structure contains four C=C double bonds (including the exocyclic =CH₂), i.e. four π bonds. Each π bond has 2 electrons, so there are 8 π electrons; 4 is the number of π bonds, not electrons.
Q9AIPMT 2015
In Dumas' method for estimation of nitrogen, 0.25 g of an organic compound gave 40 mL of nitrogen collected at 300 K temperature and 725 mm pressure. If the aqueous tension at 300 K is 25 mm, the percentage of nitrogen in the compound is:
Not quite — the answer is D.
Dry gas pressure = 725 − 25 = 700 mm. V(STP) = 40 × 700 × 273/(300 × 760) = 33.5 mL; mass of N₂ = 28 × 33.5/22400 = 0.0419 g. %N = 0.0419/0.25 × 100 = 16.76. Not subtracting aqueous tension gives a higher value.
Q10AIPMT 2014
In the Kjeldahl's method for estimation of nitrogen present in a soil sample, ammonia evolved from 0.75 g of sample neutralized 10 mL of 1 M H₂SO₄. The percentage of nitrogen in the soil is:
Not quite — the answer is D.
10 mL of 1 M H₂SO₄ = 10 mmol, which neutralises 20 mmol NH₃, containing 20 mmol N = 0.28 g. %N = 0.28/0.75 × 100 = 37.33.
Q11NEET 2013
The structure of isobutyl group in an organic compound is
Not quite — the answer is A.
Isobutyl is (CH₃)₂CH–CH₂–, attached through a CH₂ next to the branched carbon. CH₃–CH(–)–CH₂–CH₃ is sec-butyl, (CH₃)₃C– is tert-butyl and the straight chain is n-butyl.
Q12NEET 2013
The order of stability of the following tautomeric compounds is (I) CH₂=C(OH)–CH₂–CO–CH₃ (II) CH₃–CO–CH₂–CO–CH₃ (III) CH₃–C(OH)=CH–CO–CH₃
Not quite — the answer is B.
III is the enol of acetylacetone; its C=C is conjugated with the C=O and it forms a six-membered intramolecular hydrogen bond, so it is the most stable. The keto form II comes next, and I, whose C=C is not conjugated with the carbonyl, is least stable: III > II > I.
Q13AIPMT 2012 Prelims
Which nomenclature is not according to IUPAC system?
Not quite — the answer is B.
Br–CH₂–CH=CH₂ must be numbered to give the double bond the lowest locant, so its IUPAC name is 3-bromoprop-1-ene, not 1-bromoprop-2-ene. The other names are correct: 5-oxohexanoic acid, 4-bromo-2,4-dimethylhexane (locants 2,4,4 beat 3,3,5) and (2-methylpentan-3-yl)benzene.
Q14AIPMT 2011 Mains
The IUPAC name of the compound ClC(CH₃)=C(I)CH₂CH₃, in which Cl and I lie on opposite sides of the double bond, is
Not quite — the answer is C.
The longest chain containing the double bond has five carbons, numbered from the end nearer the double bond: C2 carries Cl and C3 carries I, giving 2-chloro-3-iodopent-2-ene. Cl and I, the heavier substituents on each alkene carbon, are on opposite sides, so it is trans. Names like 3-iodo-4-chloro-3-pentene number from the wrong end.
Q15AIPMT 2011 Prelims
In Dumas' method of estimation of nitrogen 0.35 g of an organic compound gave 55 mL of nitrogen collected at 300 K temperature and 715 mm pressure. The percentage composition of nitrogen in the compound would be (Aqueous tension at 300 K = 15 mm)
Not quite — the answer is C.
Pressure of dry N₂ = 715 − 15 = 700 mm. Volume at STP = 55 × (700/760) × (273/300) = 46.1 mL. Mass of N₂ = 28 × 46.1/22400 = 0.0576 g, so %N = 0.0576/0.35 × 100 ≈ 16.5 (16.45). Forgetting the aqueous tension correction gives a higher value.
Q16AIPMT 2011 Prelims
The Lassaigne's extract is boiled with conc. HNO₃ while testing for halogens. By doing so it
Not quite — the answer is B.
If N or S is present, Na₂S and NaCN are in the Lassaigne extract and would give black Ag₂S or white AgCN with AgNO₃, interfering with the halide test. Boiling with HNO₃ decomposes them (releasing H₂S and HCN). It does not change the solubility product of AgCl.
Q17AIPMT 2010 Mains · Q99
Which of the following species is not electrophilic in nature?
Not quite — the answer is C.
Electrophiles are electron-deficient: Cl⁺ and NO₂⁺ are positive and electron-poor, and BH₃ has an empty orbital on boron. In H₃O⁺ the oxygen has a complete octet and still a lone pair, so it acts as an acid (proton donor) but not as an electrophile.
Q18AIPMT 2010 Mains · Q113
The IUPAC name of the compound CH₃CH=CHC≡CH is
Not quite — the answer is B.
Number the chain so the multiple bonds get the lowest locants: from the C≡CH end the triple bond is at C-1 and the double bond at C-3, giving pent-3-en-1-yne. Numbering from the CH₃ end gives the higher set {2,4}, so pent-2-en-4-yne is wrong.
Q19AIPMT 2009 Prelims · Q151
The state of hybridization of C2, C3, C5 and C6 of the hydrocarbon CH₃–C(CH₃)₂–CH=CH–CH(CH₃)–C≡CH (numbered from the right: C≡CH carbons are C1 and C2, the CH(CH₃) carbon is C3, CH=CH are C4 and C5, the C(CH₃)₂ carbon is C6, end CH₃ is C7) is in the following sequence:
Not quite — the answer is D.
C2 is a triple-bonded carbon (sp). C3 is the CH(CH₃) carbon with only single bonds (sp³). C5 is a double-bonded carbon (sp²). C6 is the quaternary C(CH₃)₂ carbon (sp³). So the sequence is sp, sp³, sp², sp³.
Q20AIPMT 2009 Prelims · Q183
The IUPAC name of the compound having the formula CH≡C–CH=CH₂ is:
Not quite — the answer is C.
CH₂=CH–C≡CH: numbering from the double-bond end gives the double bond at C1 and the triple bond at C3, and when locants tie the double bond gets the lower number. The name is but-1-en-3-yne, written here as 1-butene-3-yne. Names with 'yne' cited first are wrong since 'ene' precedes 'yne'.
Q21AIPMT 2008 Prelims · Q85
In the hydrocarbon CH₃(6)–CH(5)=CH(4)–CH₂(3)–C(2)≡CH(1) The state of hybridization of carbons 1, 3 and 5 are in the following sequence
Not quite — the answer is C.
C-1 is part of the C≡C triple bond (sp), C-3 is a CH₂ with only single bonds (sp³), and C-5 is part of the C=C double bond (sp²). The sp, sp², sp³ sequence wrongly gives C-3 a double bond.
Q22AIPMT 2008 Prelims · Q88
How many stereoisomers does the molecule have? CH₃CH=CHCH₂CHBrCH₃
Not quite — the answer is B.
The C=C bond has two different groups on each carbon, so it shows cis–trans isomerism (2 forms). The CHBr carbon carries H, Br, CH₃ and CH₂CH=CHCH₃, so it is a stereocentre (2 forms). These are independent, giving 2 × 2 = 4 stereoisomers. Counting only one source of isomerism would give 2.
Q23AIPMT 2008 Prelims · Q91
Base strength of (a) H₃C–CH₂⁻ (b) H₂C=CH⁻ (c) H–C≡C⁻ is in the order of
Not quite — the answer is A.
The conjugate acids get more acidic from ethane (sp³) to ethene (sp²) to ethyne (sp), because more s-character holds the lone pair more tightly. The stronger the conjugate acid, the weaker the base, so the basicity order is ethyl > vinyl > acetylide anion. The reverse order (acetylide strongest) is the stability order, not the base strength.
Q24AIPMT 2006 Prelims · Q78
Which of the following is not chiral ?
Not quite — the answer is C.
In 3-bromopentane, C-3 carries H, Br and two identical ethyl groups, so there is no stereocentre and the molecule is achiral. 2-Butanol and lactic acid each have a carbon with four different groups, and 2,3-dibromopentane has two such carbons with no internal symmetry.
Q25AIPMT 2006 Prelims · Q92
The correct order regarding the electronegativity of hybrid orbitals of carbon is :
Not quite — the answer is B.
The more s-character in a hybrid orbital, the closer its electrons are held to the nucleus and the more electronegative it is. s-character is 50% in sp, 33% in sp² and 25% in sp³, so electronegativity is sp > sp² > sp³. This is why ethyne's C–H is the most acidic.
Q26AIPMT 2006 Prelims · Q100
The IUPAC name of CH₃CH₂–CH(CH₃)–CH(CH₃)–COCl is :
Not quite — the answer is D.
The principal chain includes the –COCl carbon and has five carbons; numbering starts from that carbon (C-1). Methyl groups sit on C-2 and C-3, so the name is 2,3-dimethylpentanoyl chloride. Numbering from the other end, or choosing a four-carbon chain with an ethyl branch, breaks the IUPAC rules.
Q27AIPMT 2005 Prelims · Q51
Which amongst the following is the most stable carbocation ?
Not quite — the answer is B.
Carbocation stability is 3° > 2° > 1° > methyl because of hyperconjugation and the +I effect of alkyl groups. (CH₃)₃C⁺ is tertiary and most stable; the methyl cation is least stable.
Q28AIPMT 2005 Prelims · Q54
Which one of the following pairs represents stereoisomerism ?
Not quite — the answer is D.
Stereoisomers have the same connectivity but different spatial arrangement; geometrical and optical isomerism are the two types. Chain, structural and linkage isomerism change connectivity.
Q29AIPMT 2005 Prelims · Q95
Names of some compounds are given. Which one is not correct in IUPAC system ?
Not quite — the answer is D.
For CH₃CH₂CH₂–CH(C₂H₅)–CH(CH₃)–CH₂CH₃ the substituents must be cited alphabetically: 4-ethyl-3-methylheptane, so '3-methyl-4-ethyl heptane' is incorrect. The other three names correctly identify the parent chain and lowest locants.
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