25 questions in 25 minutes on the NTA computer-based test interface, all from Organic Chemistry: Some Basic Principles and Techniques. +4 / -1 marking, instant score, full solutions. Free, no login.
A chapter test is the fastest honest check on whether a chapter has actually landed. This one draws 25 questions from Organic Chemistry: Some Basic Principles and Techniques, of which 17 are real NEET previous-year questions and 6 sit at the harder end of the bank. Sit it in one 25-minute block, the way you would in the hall.
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Duration: 180 minutes · 180 questions · 720 marks maximum
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How many questions are in this Organic Chemistry: Some Basic Principles and Techniques mock test?
25 questions in 25 minutes, marked +4 for a correct answer and -1 for a wrong one, exactly like the real paper. 17 of them are NEET previous-year questions.
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Both, and that is the point. Take it cold to find out what you actually do not know, revise from the chapter notes, then retake it to confirm the gap closed.
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Chemistry
- Q1How many structural isomers (including chain and position) are possible for C₄H₁₀?
- 4
- 1
- 2
- 3
Answer: (C) 2
C₄H₁₀ has only two isomers: n-butane (CH₃CH₂CH₂CH₃) and isobutane/2-methylpropane. No position or functional isomerism is possible for an alkane this small.
- Q2Which technique among the following, is most appropriate in separation of a mixture of 100 mg of p-nitrophenol and picric acid ?
- Steam distillation
- 2-5 ft long column of silica gel
- Sublimation
- Preparative TLC (Thin Layer Chromatography)
Answer: (D) Preparative TLC (Thin Layer Chromatography)
D. Tests: matching a separation technique to the sample scale and to the physical properties of the components. Why D: The sample is only 100 mg, a milligram scale mixture of two closely related, non volatile phenolic solids. Preparative TLC is built for exactly this: the mixture is applied as a band on a thick silica plate, the plate is developed, and the separated bands are scraped off and extracted. The two compounds adsorb on silica quite differently, since picric acid carries three nitro groups against one in p-nitrophenol, so their R_f values are far apart and the bands resolve cleanly. Why not A: Steam distillation requires one component to be steam volatile and sparingly water soluble; both p-nitrophenol and picric acid are non volatile hydrogen bonded solids, so neither is carried over by steam and nothing separates [misconception A: “steam distillation separates any pair of aromatic compounds”]. Why not B: A 2 to 5 ft silica column is designed for gram scale work; a 100 mg sample would be spread through an enormous bed, diluted through litres of eluent and largely lost, so it is not appropriate at this scale [misconception B: “a longer column always gives a better separation”]. Why not C: Sublimation needs a solid that vaporises directly on heating while the other component does not; neither of these sublimes cleanly and picric acid is thermally unstable, so the technique does not apply [misconception C: “nitro aromatics sublime the way naphthalene does”]. Remember: milligrams of a non volatile mixture means preparative TLC, grams means a column.
- Q3The number of alpha (sp³) C-H bonds available for hyperconjugation in the tert-butyl carbocation (CH₃)₃C⁺ is:
- 9
- 3
- 12
- 6
Answer: (A) 9
The cationic carbon is flanked by three CH₃ groups. Each methyl supplies 3 C-H bonds adjacent (alpha) to the empty p-orbital: 3 × 3 = 9 hyperconjugative C-H bonds, which is why the tert-butyl cation is so stable.
- Q4Given below are two statements: One is labelled as Assertion A and the other is labelled as Reason R Assertion A : Thin layer chromatography is an adsorption chromatography. Reason R : A thin layer of silica gel is spread over a glass plate of suitable size in thin layer chromatography which acts as an adsorbent. In the light of the above statements, choose the correct answer from the options given below
- A is false but R is true
- Both A and R are true but R is NOT the correct explanation of A
- Both A and R are true and R is the correct explanation of A
- A is true but R is false
Answer: (C) Both A and R are true and R is the correct explanation of A
A. Tests: whether the stated reason genuinely explains why TLC counts as adsorption chromatography. Why A: The assertion is true. In TLC the stationary phase is a solid, silica gel or alumina, and components separate because they stick to that solid surface with different strengths, which is the definition of adsorption chromatography. The reason is true as well, and it is the cause: the thin layer of silica gel spread on the glass plate is the adsorbent, and it is exactly the presence of that solid adsorbent as the stationary phase that classifies the technique as adsorption chromatography. So R correctly explains A. Why not B: R is not an unrelated true fact; the silica adsorbent it describes is the very thing that makes the technique adsorption based, so it does explain A [misconception B: “a description of the apparatus can never count as an explanation of the principle”]. Why not C: R is not false; a thin layer of silica gel or alumina really is spread over a glass plate and really does act as the adsorbent [misconception C: “the glass plate itself is the adsorbent”]. Why not D: A is not false; TLC is standard adsorption chromatography, unlike paper chromatography which is partition chromatography [misconception D: “all chromatography run on a flat surface is partition chromatography”]. Remember: TLC uses a solid adsorbent (silica or alumina) so it is adsorption; paper uses trapped water so it is partition.
- Q5Which carbocation is the most stable overall?
- Isopropyl carbocation
- tert-butyl carbocation
- Ethyl carbocation (CH₃CH₂⁺)
- Benzyl carbocation (C₆H₅CH₂⁺)
Answer: (D) Benzyl carbocation (C₆H₅CH₂⁺)
The benzyl carbocation is resonance-stabilised by the aromatic ring (charge delocalised over multiple positions), making it more stable than even the tertiary tert-butyl cation.
- Q6Which one of the following is NOT chiral?
- 2-hydroxypropanoic acid
- 2-butanol
- 2,3-dibromopentane
- 3-bromopentane
Answer: (D) 3-bromopentane
3-bromopentane is CH₃CH₂-CHBr-CH₂CH₃: the carbon bearing Br has two identical ethyl groups, so it is not asymmetric and the molecule is achiral. The other three each contain a carbon with four different groups.
- Q7Which one of the following can exhibit cis-trans isomerism?
- Cl-CH=CH-Cl
- ClCH₂-CH₂Cl
- CH₃-CHCl-COOH
- H-C≡C-Cl
Answer: (A) Cl-CH=CH-Cl
Cis-trans isomerism requires a C=C with two different groups on each doubly-bonded carbon. Only 1,2-dichloroethene (Cl-CH=CH-Cl) qualifies: each carbon bears H and Cl, giving distinct cis and trans forms. The triple-bonded and single-bonded options cannot.
- Q8Which of the following organic compounds has the same hybridisation as its combustion product CO₂?
- Ethane
- Ethanol
- Ethene
- Ethyne
Answer: (D) Ethyne
In CO₂ (O=C=O) the carbon forms two sigma bonds and is sp hybridised. Ethyne (HC≡CH) also has sp carbons. Ethane is sp³, ethene is sp², ethanol is sp³ – so only ethyne matches CO₂.
- Q9During Lassaigne’s test, the elements present in an organic compound are converted from:
- Ionic form to ionic form
- Covalent form to covalent form
- Ionic form to covalent form
- Covalent form to ionic form
Answer: (D) Covalent form to ionic form
B. Tests: why an organic compound must be fused with sodium before nitrogen, sulphur or halogen can be detected. Why B: In an organic compound N, S and X are held by COVALENT bonds to carbon, and covalently bound atoms give none of the ordinary precipitation tests. Fusing the compound with hot sodium metal breaks those bonds and converts the elements into IONIC sodium salts: NaCN, Na₂S, NaX and NaSCN. The fused mass is extracted with water to give Lassaigne’s extract, in which CN⁻, S²⁻ and X⁻ are free ions and can be tested with FeSO₄, sodium nitroprusside or AgNO₃. Why not A: the elements do not start as ions at all, they start covalently bonded inside the organic molecule, so there is nothing ionic to begin with. Why not C: if they stayed covalent the whole point of the fusion would be lost, since a covalently held nitrogen gives no prussian blue with ferrous sulphate. Why not D: this is the reverse of what sodium fusion does, and it would make the elements harder, not easier, to detect [misconception: “the fusion locks the elements into the compound”]. Remember: Lassaigne’s fusion exists for one reason, to turn covalent N, S and X into ions that ordinary inorganic tests can see.
- Q10Given below are four compounds: (a) n-propyl chloride (b) iso-propyl chloride (c) sec-butyl chloride (d) neo-pentyl chloride Percentage of carbon in the one which exhibits optical isomerism is:
- 46
- 56
- 40
- 52
Answer: (D) 52
C. Tests: identifying the chiral compound first, then converting its formula into a mass percentage. Why C: Step 1, find the stereocentre. n-propyl chloride CH₃CH₂CH₂Cl has its Cl on a CH₂, iso-propyl chloride (CH₃)₂CHCl has two identical methyls on the C-Cl carbon, and neo-pentyl chloride (CH₃)₃CCH₂Cl again has Cl on a CH₂, so none of these three is chiral. sec-butyl chloride is 2-chlorobutane, CH₃-CHCl-CH₂CH₃, and its C2 carries four different groups, H, Cl, CH₃ and C₂H₅, so it is optically active. Step 2, compute the carbon percentage of C₄H₉Cl: molar mass = 4(12) + 9(1) + 35.5 = 48 + 9 + 35.5 = 92.5 g/mol, so %C = (48/92.5) x 100 = 51.9, which rounds to 52. Why not A: 46 is the carbon percentage of a propyl chloride, C₃H₇Cl, molar mass 36 + 7 + 35.5 = 78.5, giving (36/78.5) x 100 = 45.9, so it is what you get by picking (a) or (b) as the chiral one. Why not B: no consistent substitution gives this, none of the four chlorides has a carbon percentage near 40. Why not D: 56 is the carbon percentage of neo-pentyl chloride, C₅H₁₁Cl, molar mass 60 + 11 + 35.5 = 106.5, giving (60/106.5) x 100 = 56.3, but its C-Cl carbon is a CH₂ with two identical hydrogens, so it has no stereocentre [misconception: “a bulky branched halide must be chiral”]. Remember: optical activity needs four DIFFERENT groups on one carbon, and sec-butyl is the smallest simple alkyl chloride that manages it.
- Q11Tautomerism will be exhibited by:
- RCH₂NO₂
- R₂C=NO₂H
- (CH₃)₂NH
- (CH₃)₃C-NO₂
Answer: (A) RCH₂NO₂
Tautomerism (here the nitro-aci/nitro-azo type) needs an alpha-hydrogen adjacent to the -NO₂ group. RCH₂NO₂ has alpha C-H atoms and can tautomerise to its aci-nitro form; (CH₃)₃C-NO₂ has no alpha-hydrogen.
- Q12Identify correct statement/s: (A) -OCH₃ and -NHCOCH₃ are activating group. (B) -CN and -OH are meta directing group. (C) -CN and -SO₃H are meta directing group. (D) Activating groups act as ortho and para directing groups. (E) Halides are activating groups. Choose the correct answer from the options given below:
- (A), (C) and (D) only
- (A) and (C) only
- (A), (B) and (E) only
- (A) only
Answer: (A) (A), (C) and (D) only
A. Tests: sorting substituents into activating or deactivating and into o/p or meta directing. Why A: take the statements one by one. (A) is true, since -OCH₃ and -NHCOCH₃ both hold a lone pair on the atom attached to the ring and donate it by resonance, raising ring electron density. (C) is true, since -CN and -SO₃H both pull electron density out through a multiply bonded electronegative centre and have no lone pair to give back, so both are meta directors. (D) is true, because an activating group works by pushing electron density into the ring, and resonance puts that extra density specifically at ortho and para, so activators direct o/p. (B) is false, because -OH is a lone pair donor and directs o/p, not meta. (E) is false, because halides deactivate the ring by the dominant -I effect even though they direct o/p. The correct set is (A), (C) and (D). Why not B: it accepts (B) and (E); -OH is o/p directing because oxygen feeds its lone pair into the ring, and halides are the classic deactivating yet o/p directing exception, so both fail [misconception B: “an electronegative atom on the ring always means meta directing”]. Why not C: (A) and (C) are indeed correct, but stopping there discards (D), which is also correct, and the question asks for every correct statement [misconception C: “activating power and o/p direction are unrelated properties”]. Why not D: taking (A) alone drops both (C) and (D), and (C) correctly labels -CN and -SO₃H as meta directors [misconception D: “only lone pair donors can be classified with confidence”]. Remember: activating means o/p directing, deactivating means meta directing, and the halogens are the single exception to that pairing.
- Q13In the hydrocarbon CH₃-CH=CH-CH₂-C≡CH (carbons numbered 6,5,4,3,2,1 from left), the state of hybridisation of carbons 1, 3 and 5 respectively is:
- sp, sp³, sp²
- sp², sp³, sp
- sp³, sp², sp
- sp, sp², sp³
Answer: (A) sp, sp³, sp²
C-1 is the terminal ≡CH (triple bond) → sp. C-3 is the -CH₂- (only single bonds) → sp³. C-5 is the =CH- (double bond) → sp². Hence C1, C3, C5 = sp, sp³, sp².
- Q14The correct order of increasing bond length of C-H, C-O, C-C and C=C is:
- C-H < C=C < C-C < C-O
- C-C < C=C < C-O < C-H
- C-H < C-O < C-C < C=C
- C-O < C-H < C-C < C=C
Answer: (A) C-H < C=C < C-C < C-O
Typical bond lengths: C-H ≈ 0.109 nm, C=C ≈ 0.134 nm, C-C ≈ 0.154 nm, C-O ≈ 0.143 nm. Ordering them: C-H (0.109) < C=C (0.134) < C-O (0.143) < C-C (0.154). Hence C-H < C=C < C-O < C-C.
- Q15How many structural (constitutional) isomers are possible for the molecular formula C₅H₁₂?
- 5
- 3
- 2
- 4
Answer: (B) 3
C₅H₁₂ (pentane) has three chain isomers: n-pentane, 2-methylbutane (isopentane) and 2,2-dimethylpropane (neopentane). DoU = (2×5+2−12)/2 = 0, so all are saturated open chains.
- Q16Which of the following reactions is an example of nucleophilic substitution?
- 2RX + 2Na → R-R + 2NaX
- RX + KOH → ROH + KX
- RX + Mg → RMgX
- RX + H₂ → RH + HX
Answer: (B) RX + KOH → ROH + KX
In RX + KOH → ROH + KX, the nucleophile OH⁻ replaces the halide X⁻ on the saturated carbon – a nucleophilic substitution. The Wurtz reaction (b), reduction (c) and Grignard formation (d) are not nucleophilic substitutions.
- Q17Which one of the following series contains only electrophiles?
- AlCl₃, R-NH₂, PCl₃
- BF₃, SO₃, NO⁺
- H₂O, R₂N⁻, H₃O⁺
- R-OH, Cl⁺, NH₃
Answer: (B) BF₃, SO₃, NO⁺
C. Tests: sorting reagents into electrophiles and nucleophiles by counting electrons at the reacting atom. Why C: An electrophile is electron deficient and accepts a pair of electrons, either because it is positively charged or because its central atom has an incomplete octet or an empty acceptor orbital. BF₃ has only six electrons around boron and an empty p orbital, SO₃ has a strongly electron poor sulphur that alkenes and arenes attack, and NO⁺ is a cation with a positive nitrogen. All three accept electron pairs, so this series is purely electrophilic. Why not A: only Cl⁺ is an electrophile here, since R-OH and NH₃ both carry lone pairs on an electron rich atom and act as nucleophiles. Why not B: H₃O⁺ is an electrophile, but H₂O and especially the amide anion R₂N⁻ are electron rich lone pair donors, so the series is a mixture. Why not D: AlCl₃ is a genuine electron deficient electrophile, but R-NH₂ and PCl₃ both donate their lone pairs, which is exactly why PCl₃ acts as a ligand and an amine acts as a base [misconception: “anything with chlorine in it must be an electrophile”]. Remember: look at the reacting atom, an empty orbital or a positive charge means electrophile, an available lone pair or a negative charge means nucleophile.
- Q18Which of the following acts as a nucleophile?
- CN⁻
- NO₂⁺
- AlCl₃
- BF₃
Answer: (A) CN⁻
A nucleophile is electron-rich and donates an electron pair. CN⁻ has lone pairs and a negative charge. NO₂⁺, BF₃ and AlCl₃ are electron-deficient electrophiles (Lewis acids).
- Q19The number of σ bonds, π bonds and lone pair of electrons in pyridine, respectively are:
- 11, 3, 1
- 12, 3, 0
- 12, 2, 1
- 11, 2, 0
Answer: (A) 11, 3, 1
B. Tests: counting sigma bonds, pi bonds and lone pairs from a heterocyclic structure. Why B: pyridine is C₅H₅N, a six membered aromatic ring in which one CH unit of benzene is replaced by N. Count the sigma bonds: the ring itself has 6 sigma bonds, and only the five carbons carry hydrogen, giving 5 C-H sigma bonds, so total σ = 6 + 5 = 11. Count the pi bonds: the ring is aromatic with three alternating double bonds, so π = 3. Count the lone pairs: nitrogen is sp² hybridised, uses two of its sp² orbitals for the two ring sigma bonds and one p orbital for the ring pi system, which leaves one sp² orbital in the ring plane holding a lone pair, so lone pairs = 1. The answer is 11, 3, 1. Why not A: 12, 3, 0 uses benzene’s C-H count of 6 and then denies nitrogen its lone pair; pyridine’s nitrogen carries no hydrogen, so there are only 5 C-H sigma bonds and the vacated position holds the lone pair instead [misconception A: “pyridine is benzene with a nitrogen simply inserted, hydrogen and all”]. Why not C: 12, 2, 1 repeats that same benzene-like sigma count of 12 and additionally drops one of the three ring pi bonds; an aromatic six membered ring needs three pi bonds to hold six delocalised electrons [misconception C: “the heteroatom uses up one of the ring pi bonds”]. Why not D: 11, 2, 0 gets the sigma count right but loses one pi bond and the nitrogen lone pair, the very lone pair that makes pyridine a base [misconception D: “the nitrogen lone pair is part of the aromatic sextet, so it is not a lone pair”]. Remember: pyridine keeps benzene’s three pi bonds, has one fewer C-H bond, and parks the nitrogen lone pair in the ring plane.
- Q20The stability order of carbanions (3°, 2°, 1°, methyl) is the reverse of carbocations because:
- Carbanions are sp hybridised
- Carbanions are stabilised by electron-donating alkyl groups
- Alkyl groups withdraw electrons from carbanions
- Carbanions carry an extra electron pair, so electron-donating alkyl groups destabilise them
Answer: (D) Carbanions carry an extra electron pair, so electron-donating alkyl groups destabilise them
A carbanion already has a lone pair and negative charge. Electron-donating alkyl groups (+I) increase electron density and destabilise it, so methyl/1° carbanions are more stable than 3°: CH₃⁻ > 1° > 2° > 3°.
- Q21During hearing of a court case, the judge suspected that some changes in the documents had been carried out. He asked the forensic department to check the ink used at two different places. According to you which technique can give the best results?
- Column chromatography
- Thin layer chromatography
- Distillation
- Solvent extraction
Answer: (B) Thin layer chromatography
D. Tests: choosing the analytical technique that suits a trace sample and a side by side comparison. Why D: Ink is a mixture of dyes, and the question is whether two inks are the SAME mixture, not how to obtain them in bulk. Thin layer chromatography answers exactly that. A tiny spot of each ink is placed on the same silica plate, one solvent front is run, and the two lanes develop side by side. Identical inks give identical R_f values and identical spot patterns, different inks do not. The sample needed is minute, which matters when all you have is ink lifted from a document, and the result is a direct visual comparison. Why not A: column chromatography does separate the dyes, but it is a preparative method needing a much larger sample and it delivers fractions one after another, which makes a direct comparison of two inks slow and wasteful of scarce evidence. Why not B: solvent extraction only moves compounds between two immiscible solvents by solubility, so it can pull the dye off the paper but cannot resolve the mixture into components or compare two mixtures. Why not C: distillation separates volatile liquids by boiling point, and the coloured dyes in ink are non-volatile solids, so there is nothing to distil [misconception: “any separation problem can be handled by distillation”]. Remember: tiny sample plus a same plate comparison of R_f values means TLC, column chromatography is for isolating material in quantity.
- Q22With respect to the conformers of ethane, which statement is true?
- Bond angle remains same but bond length changes
- Both bond angle and bond length remain same
- Both bond angle and bond length change
- Bond angle changes but bond length remains same
Answer: (B) Both bond angle and bond length remain same
Conformers of ethane interconvert by rotation about the C-C single bond only. No bonds are broken, so both bond lengths and bond angles stay the same; only the dihedral (torsion) angle and the energy change.
- Q23Match List – I with List – II. List – I (Pair of Compounds) A. 2-Methylpropene and but-1-ene B. Cis-but-2-ene and trans-but-2-ene C. 2-Butanol and diethyl ether D. But-1-ene and but-2-ene List – II (Type of Isomers) I. Stereoisomers II. Position isomers III. Chain isomers IV. Functional group isomers Choose the correct answer from the options given below:
- A-III, B-I, C-II, D-IV
- A-II, B-I, C-IV, D-III
- A-III, B-I, C-IV, D-II
- A-I, B-IV, C-III, D-II
Answer: (C) A-III, B-I, C-IV, D-II
A. Tests: naming the exact isomer relationship within each pair rather than just spotting that they are isomers. Why A: Pair A, 2-methylpropene (CH₃)₂C=CH₂ and but-1-ene CH₂=CH-CH₂CH₃, are both C₄H₈ and differ in the carbon SKELETON, one branched and one straight, so they are chain isomers (III). Pair B, cis and trans but-2-ene, have identical connectivity and differ only in the spatial arrangement across the double bond, so they are stereoisomers (I). Pair C, 2-butanol and diethyl ether, are both C₄H₁₀O but carry different functional groups, alcohol versus ether, so they are functional group isomers (IV). Pair D, but-1-ene and but-2-ene, share the same skeleton and the same functional group and differ only in where the double bond sits, so they are position isomers (II). That gives A-III, B-I, C-IV, D-II. Why not B: it labels pair A as position isomers, but moving from a branched to a straight four carbon skeleton is a change of chain, not just of a locant, and it then leaves pair D as chain isomers when both butenes have the same straight chain. Why not C: it swaps the first two, calling the chain isomers stereoisomers and the cis-trans pair functional isomers, even though cis and trans but-2-ene contain exactly the same functional group. Why not D: pairs A and B are right, but C and D are exchanged, which calls an alcohol and an ether position isomers although their functional groups are different [misconception: “different oxygen placement means position isomerism”]. Remember: same skeleton and same group with a moved locant is POSITION, different skeleton is CHAIN, different group is FUNCTIONAL, same connectivity in a different arrangement is STEREO.
- Q24The correct order of decreasing acidic strength of trichloroacetic acid (A), trifluoroacetic acid (B), acetic acid (C) and formic acid (D) is:
- A > B > C > D
- B > D > A > C
- B > A > D > C
- A > C > B > D
Answer: (C) B > A > D > C
Electron-withdrawing groups stabilise the carboxylate and increase acidity. -CF₃ withdraws more than -CCl₃ (F more electronegative), so trifluoroacetic (B) > trichloroacetic (A). Both beat the alpha-H-only acids: formic (D) > acetic (C) (the +I methyl in acetic acid reduces acidity). Order: B > A > D > C.
- Q25Which purification technique is used to separate two miscible liquids whose boiling points are very close?
- Steam distillation
- Simple distillation
- Fractional distillation
- Crystallisation
Answer: (C) Fractional distillation
Fractional distillation, using a fractionating column for repeated vaporisation-condensation cycles, separates liquids with close boiling points (e.g. crude oil fractions).