NEET Chemistry: Organic Basics and Hydrocarbons · Chapter 2

Organic Chemistry: Some Basic Principles

What to remember

Carbon is tetravalent and forms sp³ (tetrahedral, 109.5°), sp² (trigonal planar, 120°) and sp (linear, 180°) hybrid orbitals; the s-character is 25 %, 33.3 % and 50 %, and bond length falls in the order C–C > C=C > C≡C.
Stability orders to know: carbocations and free radicals 3° > 2° > 1° > methyl; carbanions methyl > 1° > 2° > 3°. Carbocations are stabilised by +I, hyperconjugation and resonance.
Electronic effects explain properties: the inductive effect falls rapidly with distance, resonance (+R or −R) delocalises π electrons or lone pairs, and hyperconjugation uses the σ bond of an α C–H. Acid strength of Cl₃CCOOH > Cl₂CHCOOH > ClCH₂COOH > CH₃COOH follows from the −I effect of chlorine.
IUPAC names are built from the longest chain containing the principal functional group, with the lowest locant to that group. The priority order is –COOH > –SO₃H > –COOR > –COCl > –CONH₂ > –CN > –CHO > C=O > –OH > –NH₂ > C=C > C≡C.

Tetravalency, hybridisation and shape

Carbon has the ground-state configuration 1s² 2s² 2p². One 2s electron is promoted to 2p, giving 2s¹ 2p³ with four unpaired electrons. Carbon therefore forms four covalent bonds (tetravalency). It can bond to other carbon atoms in chains and rings (catenation) and can form single, double and triple bonds.

Hybridisation	Orbitals mixed	Shape	Bond angle	s-character	Example	Carbon–carbon bond length
sp³	one s + three p	Tetrahedral	109.5°	25 %	CH₄, C₂H₆	C–C about 154 pm
sp²	one s + two p	Trigonal planar	120°	33.3 %	CH₂=CH₂	C=C about 134 pm
sp	one s + one p	Linear	180°	50 %	HC≡CH	C≡C about 120 pm

A sigma (σ) bond is formed by head-on overlap of orbitals. A pi (π) bond is formed by sideways overlap of unhybridised p orbitals. A π bond is weaker than a σ bond and is the reactive part of a multiple bond. A double bond is one σ + one π; a triple bond is one σ + two π. Ethene has five σ and one π bond; ethyne has three σ and two π bonds. With greater s-character the carbon is more electronegative and holds the bond pair more tightly, so a bond to an sp carbon is shorter.

Structural representations and classification

Complete structural formula: every atom and every bond is shown.
Condensed formula: groups are written together, such as CH₃CH₂CH₂CH₃ or CH₃(CH₂)₂CH₃.
Bond-line formula: carbon and hydrogen atoms on carbon are not written; each bend or end of a line is a carbon atom, and hydrogen atoms are assumed to complete four bonds.

Class	Meaning	Example
Acyclic (open chain, aliphatic)	No ring	CH₃CH₂CH₃
Alicyclic	Closed ring of carbon atoms with aliphatic properties	Cyclohexane
Aromatic	Benzene and related compounds (benzenoid, or non-benzenoid such as tropolone)	Benzene, naphthalene
Heterocyclic	Ring contains an atom other than carbon	Pyridine, furan

A functional group is an atom or group that decides the chemical properties of a compound. A homologous series is a family of compounds with the same functional group and a similar general formula, in which successive members differ by a –CH₂– unit (14 u). Members show a regular gradation in physical properties and similar chemical properties. Alkanes are CₙH₂ₙ₊₂, alkenes CₙH₂ₙ, alkynes CₙH₂ₙ₋₂.

IUPAC nomenclature

Steps for an aliphatic compound:

1. Choose the longest chain containing the principal functional group (and, if possible, the maximum number of multiple bonds).
2. Number the chain to give the lowest locant to the principal functional group, then to multiple bonds, then to substituents. If still equal, give the lowest number to the substituent that comes first alphabetically.
3. Write substituents as prefixes in alphabetical order. Prefixes di, tri and tetra are ignored in alphabetical order. Name = prefixes + root + (en or yne) + suffix.

Root names: C₁ meth, C₂ eth, C₃ prop, C₄ but, C₅ pent, C₆ hex, C₇ hept, C₈ oct. Halo (fluoro, chloro, bromo, iodo), nitro, alkoxy and alkyl groups are named only as prefixes.

Functional group	Suffix	Prefix form	Example
Carboxylic acid –COOH	-oic acid	carboxy	CH₃COOH ethanoic acid
Sulphonic acid –SO₃H	-sulphonic acid	sulpho	C₆H₅SO₃H benzenesulphonic acid
Ester –COOR	-oate	alkoxycarbonyl	CH₃COOCH₃ methyl ethanoate
Acid chloride –COCl	-oyl chloride	chloroformyl	CH₃COCl ethanoyl chloride
Amide –CONH₂	-amide	carbamoyl	CH₃CONH₂ ethanamide
Nitrile –CN	-nitrile	cyano	CH₃CN ethanenitrile
Aldehyde –CHO	-al	oxo or formyl	CH₃CHO ethanal
Ketone >C=O	-one	oxo	CH₃COCH₃ propanone
Alcohol –OH	-ol	hydroxy	CH₃CH₂OH ethanol
Amine –NH₂	-amine	amino	CH₃NH₂ methanamine
Alkene C=C	-ene	–	CH₂=CHCH₃ propene
Alkyne C≡C	-yne	–	HC≡CCH₃ propyne

Examples of polyfunctional and substituted names: HOCH₂CH₂COOH is 3-hydroxypropanoic acid; CH₃CH(OH)CH₂CHO is 3-hydroxybutanal; CH₂=CHCH₂OH is prop-2-en-1-ol; CH₃CH(CH₃)CH₂CH₃ is 2-methylbutane; CH₃CH(Cl)CH₃ is 2-chloropropane; CH₃CH₂CH(CH₃)CH₂CH₃ is 3-methylpentane.

Substituted benzenes. Benzene is the parent. Two substituents are at 1,2 (ortho, o-), 1,3 (meta, m-) or 1,4 (para, p-). Common names kept by IUPAC include toluene (methylbenzene), phenol, aniline, benzaldehyde, benzoic acid and nitrobenzene. Examples: 1,2-dimethylbenzene (o-xylene); 1,3-dichlorobenzene (m-dichlorobenzene); 4-nitrophenol (p-nitrophenol); 2,4,6-trinitrophenol (picric acid). In a ring with phenol, the –OH carbon is number 1.

Isomerism

Isomers have the same molecular formula but different structures or arrangements.

Type	Difference	Example (formula)
Chain	Different carbon skeleton	Butane and 2-methylpropane (C₄H₁₀)
Position	Same skeleton, different place of the functional group or multiple bond	Propan-1-ol and propan-2-ol (C₃H₈O); but-1-ene and but-2-ene (C₄H₈)
Functional	Different functional groups	Ethanol and methoxymethane (C₂H₆O); propanal and propanone (C₃H₆O)
Metamerism	Different alkyl groups on the same polyvalent atom (O, S, N)	Diethyl ether and methyl propyl ether (C₄H₁₀O)
Tautomerism (in brief)	Two forms in rapid equilibrium by shift of a hydrogen atom	Keto form CH₃COCH₃ ⇌ enol form CH₃C(OH)=CH₂

Stereoisomers have the same structural formula but different arrangement in space.

Geometrical isomerism arises from restricted rotation about a double bond when each doubly bonded carbon has two different groups. The cis isomer has similar groups on the same side; the trans isomer has them on opposite sides (cis-but-2-ene and trans-but-2-ene). But-1-ene and propene have no cis-trans isomers, because one carbon of the double bond carries two identical hydrogen atoms.
Optical isomerism (in brief): a carbon bonded to four different groups is a chiral carbon (asymmetric carbon). Such a molecule has a non-superimposable mirror image. The two forms are enantiomers. They rotate the plane of plane-polarised light by equal amounts in opposite directions. An equimolar mixture of both is a racemic mixture and is optically inactive. Examples: lactic acid CH₃CH(OH)COOH and butan-2-ol CH₃CH(OH)CH₂CH₃ each have one chiral carbon.

Fundamental concepts in reaction mechanism

A covalent bond can break in two ways.

Homolytic fission: each atom takes one electron of the shared pair. It gives free radicals (species with an unpaired electron). It is favoured in non-polar solvents, by heat, light or peroxides. A half-headed (fish-hook) arrow shows it.
Heterolytic fission: the shared pair goes to one atom. It gives a cation and an anion, for example a carbocation and a halide ion. It is favoured by polar solvents and by a polar bond. A full-headed arrow shows it.

Species	Hybridisation and shape	Stability order	Reason
Carbocation (R₃C⁺)	sp², planar, empty p orbital	3° > 2° > 1° > CH₃⁺	+I effect and hyperconjugation by alkyl groups spread the positive charge
Carbanion (R₃C⁻)	sp³, pyramidal, lone pair	CH₃⁻ > 1° > 2° > 3°	Alkyl groups (+I) push electrons to carbon and increase the negative charge, which is unfavourable
Free radical (R₃C•)	Nearly sp², planar	3° > 2° > 1° > CH₃•	Hyperconjugation by alkyl groups

Allyl and benzyl carbocations are more stable than simple alkyl carbocations because of resonance.

An electrophile is an electron-deficient species that seeks a pair of electrons (examples: H⁺, NO₂⁺, carbocations, BF₃, AlCl₃, the carbon of C=O). A nucleophile is an electron-rich species that donates a pair of electrons (examples: OH⁻, CN⁻, Cl⁻, NH₃, H₂O, carbanions). Neutral molecules with a lone pair, such as NH₃ and H₂O, are nucleophiles.

Electron displacement effects

Effect	Nature	Cause	Key points
Inductive effect	Permanent	Polarisation of a σ bond by an electronegative atom or group	Decreases rapidly along the chain and is almost negligible after three carbon atoms. Hydrogen is the reference
Resonance (mesomeric) effect	Permanent	Delocalisation of π electrons or lone pairs through a conjugated system	+R: groups donating electrons (–OH, –OR, –NH₂, –Cl, –Br). −R: groups withdrawing electrons (–NO₂, –CN, –CHO, –COOH, >C=O)
Electromeric effect	Temporary	Complete transfer of the π electron pair of a multiple bond to one atom at the demand of an attacking reagent	Disappears when the reagent is removed
Hyperconjugation	Permanent	Delocalisation of the σ electrons of an α C–H bond into an adjacent p orbital or π bond	"No-bond resonance"; more α hydrogen atoms mean more stability

Inductive effect. Groups that push electrons (+I): alkyl groups, and the order (CH₃)₃C– > (CH₃)₂CH– > CH₃CH₂– > CH₃–. Groups that pull electrons (−I): –NO₂ > –CN > –F > –Cl > –Br > –I.

Resonance. A molecule that cannot be shown by one structure is described by two or more contributing (canonical) structures. The real molecule is the resonance hybrid, which is more stable than any one contributor. The difference in energy is the resonance energy. Rules: only electrons move, not atoms; all contributors have the same number of unpaired electrons; the more equivalent the contributors and the lower their energy, the greater the stability. Benzene is the hybrid of two Kekulé structures.

Hyperconjugation and stability.

Carbocation	Number of α C–H bonds	Relative stability
CH₃CH₂⁺ (ethyl)	3	Lowest of the three
(CH₃)₂CH⁺ (isopropyl)	6	Intermediate
(CH₃)₃C⁺ (tert-butyl)	9	Highest

In alkenes, more alkyl groups on the double-bond carbons give more hyperconjugative structures, so the alkene is more stable. Thus but-2-ene (six α hydrogen atoms) is more stable than but-1-ene (two α hydrogen atoms), and ethene, which has no α hydrogen, is less stable than propene.

Types of organic reactions

Type	Meaning	Example
Substitution	One atom or group replaces another	CH₄ + Cl₂ → CH₃Cl + HCl (light)
Addition	Two molecules combine to one by breaking a π bond	CH₂=CH₂ + Br₂ → BrCH₂CH₂Br
Elimination	A small molecule is removed from adjacent atoms and a multiple bond forms	CH₃CH₂Br + alcoholic KOH → CH₂=CH₂ + KBr + H₂O
Rearrangement	The atoms of a molecule reorganise to give an isomer	Ammonium cyanate → urea on heating

Acidity and basicity trends

Acid strength depends on the stability of the conjugate base. Anything that stabilises the anion by spreading or reducing its negative charge makes the acid stronger.

Acid	pKa (about)	Reason
Cl₃CCOOH	0.65	Three −I chlorine atoms stabilise the anion strongly
Cl₂CHCOOH	1.29	Two −I chlorine atoms
ClCH₂COOH	2.86	One −I chlorine atom
HCOOH	3.75	No +I alkyl group
C₆H₅COOH	4.20	Phenyl group is slightly −I
CH₃COOH	4.76	+I methyl group destabilises the anion

Lower pKa means a stronger acid. Among haloacetic acids, FCH₂COOH > ClCH₂COOH > BrCH₂COOH > ICH₂COOH, following the −I order. The nearer the −I group is to the –COOH, the stronger the acid.

Phenol is more acidic than ethanol because the phenoxide ion is stabilised by resonance, whereas the ethoxide ion is destabilised by the +I alkyl group. p-Nitrophenol is more acidic than phenol because the –NO₂ group (−I, −R) stabilises the phenoxide ion.

Basicity depends on the availability of the lone pair on nitrogen. Aliphatic amines are more basic than ammonia because alkyl groups (+I) increase electron density on nitrogen. Aniline is less basic than ammonia because its lone pair is delocalised into the benzene ring by resonance.

NEET traps

Carbocation stability increases 1° < 2° < 3°, but carbanion stability decreases in the same direction. Do not reverse them.
A carbocation is sp² and planar. A carbanion is sp³ and pyramidal.
Hyperconjugation is called no-bond resonance. The number of α C–H bonds (not β) decides it.
The inductive effect is a σ-bond effect and falls with distance. Resonance is a π-electron or lone-pair effect.
The electromeric effect is temporary and needs an attacking reagent; the resonance effect is permanent.
The +R effect of –Cl, –Br is present, but the −I effect of halogens is stronger.
Chloroacetic acid is a stronger acid than acetic acid because of −I of Cl, not +I.
A homologous series differs by CH₂; isomers share the molecular formula. Do not mix them.
Metamers have different alkyl groups on the same polyvalent atom. Diethyl ether and methyl propyl ether are metamers; ethanol and methoxymethane are functional isomers.
A chiral carbon needs four different groups; propan-2-ol has no chiral carbon.

One-liners

Carbon is tetravalent and shows catenation.
sp³, sp² and sp carbon have 25 %, 33.3 % and 50 % s-character.
The C≡C bond (120 pm) is shorter than C=C (134 pm), which is shorter than C–C (154 pm).
A homologous series differs by one CH₂ unit.
The IUPAC suffix for an aldehyde is -al and for a ketone -one.
Halo, nitro and alkoxy groups are only prefixes.
Position isomers differ in the place of the functional group.
Diethyl ether and methyl propyl ether are metamers.
A chiral carbon has four different groups.
A racemic mixture is optically inactive.
Homolytic fission gives free radicals; heterolytic fission gives ions.
Methyl carbocation is the least stable carbocation in this series.
Hyperconjugation is also called no-bond resonance.
The –NO₂ group shows −I and −R effects.
Trichloroacetic acid is the strongest acid among the chloro-substituted acetic acids and acetic acid.

Practice questions

The hybridisation of each carbon atom in ethyne is:
1. sp²
2. sp
3. sp³d
4. sp³
Answer
B. sp
Each carbon of HC≡CH has a triple bond and one other bond, so two σ bonds: sp, linear, 180°.
The number of σ and π bonds in propyne, CH₃C≡CH, is:
1. 6 σ and 3 π
2. 5 σ and 3 π
3. 6 σ and 2 π
4. 7 σ and 1 π
Answer
C. 6 σ and 2 π
Four C–H σ bonds + two C–C σ bonds = 6 σ; the triple bond has two π bonds.
The IUPAC name of (CH₃)₂CHCH₂CH₂Cl is:
1. 1-chloro-3-methylbutane
2. 3-methyl-1-chloropentane
3. 1-chloro-2-methylbutane
4. 4-chloro-2-methylbutane
Answer
A. 1-chloro-3-methylbutane
The longest chain has four carbons; numbering from the carbon bearing Cl gives 1-chloro-3-methylbutane.
The IUPAC name of CH₃CH₂CH(OH)CH₃ is:
1. butan-1-ol
2. 2-methylpropan-1-ol
3. butan-3-ol
4. butan-2-ol
Answer
D. butan-2-ol
Numbering from the end that gives –OH the lower locant gives butan-2-ol.
The IUPAC name of CH₃COCH₂CH₃ is:
1. butan-2-ol
2. butan-3-one
3. butanal
4. butan-2-one
Answer
D. butan-2-one
A four-carbon chain with a ketone group on C-2 is butan-2-one.
The IUPAC name of CH₂=CHCH₂OH is:
1. prop-2-en-1-ol
2. propen-3-ol
3. prop-2-en-3-ol
4. prop-1-en-3-ol
Answer
A. prop-2-en-1-ol
The –OH group (principal group) gets locant 1, so the double bond is at C-2.
The IUPAC name of HOCH₂CH₂CHO is:
1. 1-hydroxypropanal
2. 3-oxopropan-1-ol
3. 3-hydroxypropanal
4. 1-hydroxypropan-3-al
Answer
C. 3-hydroxypropanal
The aldehyde has higher priority than –OH, so it is the suffix -al (C-1) and –OH is the 3-hydroxy prefix.
Which pair represents metamers?
1. CH₃CH₂OH and CH₃OCH₃
2. CH₃OCH₂CH₂CH₃ and CH₃CH₂OCH₂CH₃
3. CH₃CH₂CH₂CH₃ and (CH₃)₃CH
4. CH₃CH₂CH₂OH and CH₃CH(OH)CH₃
Answer
B. CH₃OCH₂CH₂CH₃ and CH₃CH₂OCH₂CH₃
Both are C₄H₁₀O ethers with different alkyl groups on oxygen.
The correct order of stability of carbocations is:
1. (CH₃)₂CH⁺ > (CH₃)₃C⁺ > CH₃CH₂⁺ > CH₃⁺
2. (CH₃)₃C⁺ > (CH₃)₂CH⁺ > CH₃CH₂⁺ > CH₃⁺
3. (CH₃)₃C⁺ > CH₃CH₂⁺ > (CH₃)₂CH⁺ > CH₃⁺
4. CH₃⁺ > CH₃CH₂⁺ > (CH₃)₂CH⁺ > (CH₃)₃C⁺
Answer
B. (CH₃)₃C⁺ > (CH₃)₂CH⁺ > CH₃CH₂⁺ > CH₃⁺
More alkyl groups give more +I and hyperconjugation, so 3° > 2° > 1° > methyl.
Which of the following carbanions is the most stable?
1. CH₃CH₂⁻
2. (CH₃)₂CH⁻
3. (CH₃)₃C⁻
4. CH₃⁻
Answer
D. CH₃⁻
Alkyl groups (+I) increase electron density on the carbanion carbon, so methyl carbanion is most stable.
Which of the following is a nucleophile?
1. BF₃
2. AlCl₃
3. NH₃
4. NO₂⁺
Answer
C. NH₃
NH₃ has a lone pair to donate; BF₃, NO₂⁺ and AlCl₃ are electron-deficient electrophiles.
Which of the following compounds shows optical activity?
1. CH₃CH(OH)CH₂CH₃
2. (CH₃)₂CHOH
3. (CH₃)₃COH
4. CH₃CH₂CH₂OH
Answer
A. CH₃CH(OH)CH₂CH₃
Only butan-2-ol has a carbon (C-2) bonded to four different groups: H, OH, CH₃ and C₂H₅.
Homolytic fission of a covalent bond produces:
1. two carbanions
2. a carbocation and a carbanion
3. two carbocations
4. free radicals
Answer
D. free radicals
In homolytic fission each atom takes one electron of the shared pair, giving free radicals.
Which of the following shows geometrical (cis-trans) isomerism?
1. CH₃CH=CH₂
2. CH₃CH=CHCH₃
3. (CH₃)₂C=CH₂
4. CH₃CH₂CH=CH₂
Answer
B. CH₃CH=CHCH₃
Each double-bonded carbon of but-2-ene has two different groups (H and CH₃).
The correct increasing order of acidic strength is:
1. CH₃COOH < ClCH₂COOH < Cl₂CHCOOH < Cl₃CCOOH
2. CH₃COOH < Cl₃CCOOH < Cl₂CHCOOH < ClCH₂COOH
3. ClCH₂COOH < CH₃COOH < Cl₂CHCOOH < Cl₃CCOOH
4. Cl₃CCOOH < Cl₂CHCOOH < ClCH₂COOH < CH₃COOH
Answer
A. CH₃COOH < ClCH₂COOH < Cl₂CHCOOH < Cl₃CCOOH
More chlorine atoms mean stronger −I effect and a more stable anion.
The number of α C–H bonds (hyperconjugative hydrogens) in the tert-butyl carbocation (CH₃)₃C⁺ is:
1. 3
2. 6
3. 9
4. 12
Answer
C. 9
Each of the three CH₃ groups attached to C⁺ has three hydrogen atoms: 3 × 3 = 9.
Given below are two statements: Statement I: An sp-hybridised carbon is more electronegative than an sp³-hybridised carbon. Statement II: The C≡C bond is shorter than the C–C bond.
1. Both Statement I and Statement II are correct
2. Both Statement I and Statement II are incorrect
3. Statement I is correct but Statement II is incorrect
4. Statement I is incorrect but Statement II is correct
Answer
A. Both Statement I and Statement II are correct
Greater s-character makes carbon more electronegative, and the triple bond is about 120 pm against 154 pm for C–C.
Given below are two statements: Statement I: Carbocations are sp² hybridised and planar. Statement II: Carbanions are sp² hybridised and planar.
1. Both Statement I and Statement II are correct
2. Statement I is correct but Statement II is incorrect
3. Both Statement I and Statement II are incorrect
4. Statement I is incorrect but Statement II is correct
Answer
B. Statement I is correct but Statement II is incorrect
Carbanions are sp³ hybridised and pyramidal.
Given below are two statements: Statement I: A resonance hybrid is more stable than any of its contributing structures. Statement II: In resonance, atoms change their positions in the contributing structures.
1. Both Statement I and Statement II are correct
2. Both Statement I and Statement II are incorrect
3. Statement I is correct but Statement II is incorrect
4. Statement I is incorrect but Statement II is correct
Answer
C. Statement I is correct but Statement II is incorrect
Only electrons are shifted; the positions of atoms stay the same.
Given below are two statements: Statement I: The electromeric effect is a permanent effect. Statement II: Hyperconjugation is also called no-bond resonance.
1. Both Statement I and Statement II are correct
2. Both Statement I and Statement II are incorrect
3. Statement I is correct but Statement II is incorrect
4. Statement I is incorrect but Statement II is correct
Answer
D. Statement I is incorrect but Statement II is correct
The electromeric effect is temporary and appears only in the presence of an attacking reagent.
Given below are two statements: Statement I: The stability order of carbanions is 3° > 2° > 1° > CH₃⁻. Statement II: Homolytic fission of a bond gives ions.
1. Both Statement I and Statement II are correct
2. Both Statement I and Statement II are incorrect
3. Statement I is correct but Statement II is incorrect
4. Statement I is incorrect but Statement II is correct
Answer
B. Both Statement I and Statement II are incorrect
Carbanion stability is CH₃⁻ > 1° > 2° > 3°, and homolytic fission gives free radicals.
Given below are two statements: Statement I: Chloroacetic acid is a stronger acid than acetic acid. Statement II: Trichloroacetic acid is a stronger acid than dichloroacetic acid.
1. Both Statement I and Statement II are incorrect
2. Statement I is correct but Statement II is incorrect
3. Both Statement I and Statement II are correct
4. Statement I is incorrect but Statement II is correct
Answer
C. Both Statement I and Statement II are correct
The −I effect of chlorine stabilises the anion, and more chlorine atoms increase the effect.
Given below are two statements: Statement I: Propan-1-ol and propan-2-ol are chain isomers. Statement II: Ethanol and methoxymethane are functional isomers.
1. Both Statement I and Statement II are correct
2. Both Statement I and Statement II are incorrect
3. Statement I is correct but Statement II is incorrect
4. Statement I is incorrect but Statement II is correct
Answer
D. Statement I is incorrect but Statement II is correct
Propan-1-ol and propan-2-ol are position isomers.
Which of the following statements are correct? 1. Phenol is more acidic than ethanol. 2. Aniline is a stronger base than ammonia. 3. p-Nitrophenol is more acidic than phenol.
1. 1 and 3 only
2. 1 and 2 only
3. 2 and 3 only
4. 1, 2 and 3
Answer
A. 1 and 3 only
The lone pair of aniline is delocalised into the ring, so it is a weaker base than ammonia.
Which of the following statements are correct? 1. Ethene has five σ bonds and one π bond. 2. Ethyne has three σ bonds and two π bonds. 3. Both carbon atoms in ethyne are sp² hybridised.
1. 1 and 3 only
2. 2 and 3 only
3. 1 and 2 only
4. 1, 2 and 3
Answer
C. 1 and 2 only
The carbons of ethyne are sp hybridised.
Given below are two statements: Statement I: Butan-2-ol has a chiral carbon atom. Statement II: Propan-2-ol has a chiral carbon atom.
1. Both Statement I and Statement II are correct
2. Both Statement I and Statement II are incorrect
3. Statement I is incorrect but Statement II is correct
4. Statement I is correct but Statement II is incorrect
Answer
D. Statement I is correct but Statement II is incorrect
Propan-2-ol has two identical CH₃ groups on C-2, so it is achiral.
Assertion (A): Trichloroacetic acid is a stronger acid than acetic acid. Reason (R): The −I effect of the chlorine atoms stabilises the trichloroacetate ion.
1. Both A and R are true and R is the correct explanation of A
2. Both A and R are true but R is not the correct explanation of A
3. A is true but R is false
4. A is false but R is true
Answer
A. Both A and R are true and R is the correct explanation of A
Electron withdrawal by Cl disperses the negative charge of the anion.
Assertion (A): A tertiary carbocation is more stable than a primary carbocation. Reason (R): More alkyl groups give more +I effect and more hyperconjugation.
1. Both A and R are true but R is not the correct explanation of A
2. Both A and R are true and R is the correct explanation of A
3. A is true but R is false
4. A is false but R is true
Answer
B. Both A and R are true and R is the correct explanation of A
The alkyl groups spread the positive charge.
Assertion (A): But-2-ene is more stable than but-1-ene. Reason (R): But-2-ene has fewer α hydrogen atoms than but-1-ene.
1. Both A and R are true and R is the correct explanation of A
2. Both A and R are true but R is not the correct explanation of A
3. A is false but R is true
4. A is true but R is false
Answer
D. A is true but R is false
But-2-ene has six α hydrogens against two in but-1-ene, so R is false.
Assertion (A): Phenol is less acidic than ethanol. Reason (R): The phenoxide ion is stabilised by resonance.
1. A is false but R is true
2. Both A and R are true and R is the correct explanation of A
3. Both A and R are true but R is not the correct explanation of A
4. A is true but R is false
Answer
A. A is false but R is true
Phenol is more acidic than ethanol, so A is false; R is true.
Assertion (A): Ethyne is a linear molecule. Reason (R): A C≡C bond is shorter than a C=C bond.
1. Both A and R are true and R is the correct explanation of A
2. Both A and R are true but R is not the correct explanation of A
3. A is true but R is false
4. A is false but R is true
Answer
B. Both A and R are true but R is not the correct explanation of A
Linear shape comes from sp hybridisation, not from the bond length.
Assertion (A): The inductive effect decreases as the distance from the substituent increases. Reason (R): The inductive effect is a temporary effect.
1. Both A and R are true and R is the correct explanation of A
2. Both A and R are true but R is not the correct explanation of A
3. A is true but R is false
4. A is false but R is true
Answer
C. A is true but R is false
The inductive effect is a permanent effect, so R is false.
Match List I with List II. List I: a. CH₃CHO b. CH₃COCH₃ c. CH₃COOH d. CH₃CN List II: i. Propanone ii. Ethanenitrile iii. Ethanal iv. Ethanoic acid
1. a-i, b-iii, c-iv, d-ii
2. a-iii, b-iv, c-i, d-ii
3. a-iii, b-i, c-ii, d-iv
4. a-iii, b-i, c-iv, d-ii
Answer
D. a-iii, b-i, c-iv, d-ii
CH₃CHO is ethanal, CH₃COCH₃ propanone, CH₃COOH ethanoic acid, CH₃CN ethanenitrile.
Match List I with List II. List I: a. Chain isomerism b. Position isomerism c. Functional isomerism d. Metamerism List II: i. Propanal and propanone ii. Butane and 2-methylpropane iii. Diethyl ether and methyl propyl ether iv. But-1-ene and but-2-ene
1. a-ii, b-iv, c-i, d-iii
2. a-ii, b-iv, c-iii, d-i
3. a-ii, b-i, c-iv, d-iii
4. a-iv, b-ii, c-i, d-iii
Answer
A. a-ii, b-iv, c-i, d-iii
Match each pair with its type of isomerism.
Match List I with List II. List I: a. Carbon in CH₄ b. Carbon in CH₂=CH₂ c. Carbon in HC≡CH d. Carbon in a carbocation R₃C⁺ List II: i. Linear, 50 % s-character ii. Tetrahedral, 25 % s-character iii. Trigonal planar, 120°, with one π bond on it iv. Planar with a vacant p orbital
1. a-iii, b-ii, c-i, d-iv
2. a-ii, b-iii, c-i, d-iv
3. a-ii, b-i, c-iii, d-iv
4. a-ii, b-iii, c-iv, d-i
Answer
B. a-ii, b-iii, c-i, d-iv
CH₄ carbon is sp³, ethene carbon sp² with a π bond, ethyne carbon sp, and a carbocation carbon sp² with an empty p orbital.
Match List I with List II. List I: a. Inductive effect b. Electromeric effect c. Hyperconjugation d. Resonance effect List II: i. Delocalisation of σ electrons of an α C–H bond ii. Polarisation of a σ bond iii. Temporary shift of the π pair in the presence of an attacking reagent iv. Delocalisation of π electrons through a conjugated system
1. a-ii, b-iii, c-iv, d-i
2. a-iii, b-ii, c-i, d-iv
3. a-ii, b-iii, c-i, d-iv
4. a-ii, b-i, c-iii, d-iv
Answer
C. a-ii, b-iii, c-i, d-iv
Match each effect with its description.
Match List I with List II. List I: a. NO₂⁺ b. OH⁻ c. CH₃• d. CH₃⁻ List II: i. Nucleophile ii. Electrophile iii. Free radical iv. Carbanion with an sp³ carbon
1. a-ii, b-i, c-iii, d-iv
2. a-i, b-ii, c-iii, d-iv
3. a-iii, b-i, c-ii, d-iv
4. a-ii, b-i, c-iv, d-iii
Answer
A. a-ii, b-i, c-iii, d-iv
NO₂⁺ is an electrophile, OH⁻ a nucleophile, CH₃• a free radical and CH₃⁻ a carbanion.
Match List I with List II. List I: a. Substitution b. Addition c. Elimination d. Rearrangement List II: i. NH₄CNO → urea on heating ii. CH₂=CH₂ + Br₂ → BrCH₂CH₂Br iii. CH₃CH₂Br + alcoholic KOH → CH₂=CH₂ iv. CH₄ + Cl₂ → CH₃Cl + HCl in light
1. a-iv, b-iii, c-ii, d-i
2. a-iv, b-ii, c-iii, d-i
3. a-iv, b-ii, c-i, d-iii
4. a-ii, b-iv, c-iii, d-i
Answer
B. a-iv, b-ii, c-iii, d-i
Chlorination of methane is substitution; bromine on ethene is addition; dehydrohalogenation is elimination; ammonium cyanate to urea is rearrangement.
The IUPAC name of CH₃CH(CH₃)CH(OH)CH₃ is:
1. 2-methylbutan-2-ol
2. 2-methylbutan-3-ol
3. 3-methylbutan-2-ol
4. 3-methylbutan-1-ol
Answer
C. 3-methylbutan-2-ol
Numbering from the end nearer –OH gives –OH at C-2 and the methyl group at C-3.
The IUPAC name of CH₃CH₂CH(Cl)CH(CH₃)CH₃ is:
1. 3-chloro-4-methylpentane
2. 2-methyl-3-chloropentane
3. 2-chloro-3-methylpentane
4. 3-chloro-2-methylpentane
Answer
D. 3-chloro-2-methylpentane
The lowest set of locants is {2,3}; prefixes are in alphabetical order, so chloro comes before methyl.
In CH₂=CH–C≡CH the hybridisation of the carbon atoms C-1 to C-4 respectively is:
1. sp², sp², sp, sp
2. sp², sp², sp², sp
3. sp³, sp², sp, sp
4. sp², sp³, sp, sp
Answer
A. sp², sp², sp, sp
The double-bond carbons are sp²; the triple-bond carbons are sp.
Which of the following is the strongest acid?
1. ClCH₂COOH
2. CH₃COOH
3. FCH₂COOH
4. BrCH₂COOH
Answer
C. FCH₂COOH
Fluorine has the strongest −I effect among the halogens.
Which of the following alkenes is the most stable, based on the number of hyperconjugative structures?
1. CH₃CH=CH₂
2. (CH₃)₂C=CHCH₃
3. CH₃CH=CHCH₃
4. (CH₃)₂C=CH₂
Answer
B. (CH₃)₂C=CHCH₃
(CH₃)₂C=CHCH₃ has nine α C–H bonds; the others have 3, 6 and 6.
Which of the following alkyl chlorides is optically active?
1. CH₃CH₂CH₂CH₂Cl
2. CH₃CH(Cl)CH₃
3. CH₃CH₂Cl
4. CH₃CH₂CH(Cl)CH₃
Answer
D. CH₃CH₂CH(Cl)CH₃
Only 2-chlorobutane has a carbon bonded to four different groups: H, Cl, CH₃ and C₂H₅.
Heterolytic fission of the C–Br bond in (CH₃)₃C–Br gives (CH₃)₃C⁺. This carbocation is:
1. sp² hybridised and planar
2. sp hybridised and linear
3. sp³ hybridised and tetrahedral
4. sp³ hybridised and pyramidal
Answer
A. sp² hybridised and planar
A carbocation has three bonds and a vacant p orbital, so it is sp² and planar.

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1.Sunil made a paste of rice flour and added a drop of iodine solution to the paste. What will he see?
2.The process used for removing gangue from the ore works on the basis of difference of their chemical or physical properties and is called ______ of ores.
3.Which of the following is an example of a third-class lever?
4.Which of the following methods is NOT used for preventing iron from rusting?
5.Which of the following methods is most commonly used nowadays to determine nitrogen and protein content in foods?
6.Which of the following is the reverse of electroplating?
7.Which process is used to get petrol, petroleum gas, diesel, kerosene and fuel oil from petroleum?
8.Which technique is useful for the separation of colloidal particles?
9.Which of the following procedures would you use to separate ammonium chloride from sodium chloride?
10.Which of the following parts are cleaned using ultrasound?

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