CBSE Class 11 Chemistry Organic Chemistry Some Basic Principles and Techniques MCQs Set 06

Answer: (b) HO(CH₂)₃CH(CH₃)CH(CH₃)₂

Explanation:
1. In a condensed formula, identical groups written one after another are put in brackets with a subscript.
2. HOCH₂CH₂CH₂— has three CH₂ groups, so it is written as HO(CH₂)₃—.
3. This is followed by CH(CH₃) and then CH(CH₃)CH₃, i.e. CH(CH₃)₂.
4. Hence, the correct condensed formula is HO(CH₂)₃CH(CH₃)CH(CH₃)₂.

Teacher's Note:
1. Count the repeating CH₂ units carefully before writing the subscript.
2. Two CH₃ groups on the same carbon are written as CH(CH₃)₂.

Answer: (a) dashed wedge, solid wedge

Explanation:
1. X points to a dashed (hashed) wedge, i.e. a bond going away from the observer (behind the plane of paper).
2. Y points to a solid wedge, i.e. a bond projecting out of the plane of paper towards the observer.
3. So, X and Y are dashed wedge and solid wedge respectively.

Teacher's Note:
1. Solid wedge = towards you; dashed wedge = away from you; plain line = in the plane of paper.
2. This wedge-dash notation is used to show the 3D tetrahedral shape of carbon on paper.

Answer: (c)

Explanation:
1. Cyclohexane is a six-membered ring, drawn in bond-line form as a hexagon.
2. Chlorocyclohexane has one Cl attached to a ring carbon, so it is a hexagon with —Cl on one corner.
3. (a) is chlorocyclopentane, (b) chlorocyclobutane and (d) chlorocyclopropane.

Teacher's Note:
1. In bond-line formulas each corner of the polygon is a carbon carrying enough H atoms to complete four bonds.
2. Count the corners: 3, 4, 5 and 6 corners stand for propane, butane, pentane and hexane rings.

Answer: (b)

Explanation:
1. Alicyclic (aliphatic cyclic) compounds contain carbon atoms joined in the form of a ring but do not have aromatic character.
2. Homocyclic means the ring contains only carbon atoms.
3. Options (a), (c) and (d) contain O or S in the ring, so they are heterocyclic.
4. Only (b), cyclobutane, has a ring made of carbon atoms alone.

Teacher's Note:
1. Homocyclic (carbocyclic) rings contain only C atoms; heterocyclic rings contain atoms like O, N or S.
2. Cyclopentane, cyclohexane, cyclohexene and cyclobutane are examples of homocyclic alicyclic compounds.

Answer: (d) All of these

Explanation:
1. Furan (O in ring), thiophene (S in ring) and pyridine (N in ring) all have a hetero atom in the ring.
2. All three follow the aromatic character, so all are heterocyclic aromatic compounds.

Teacher's Note:
1. Furan, thiophene and pyrrole are five-membered aromatic heterocycles; pyridine is six-membered.
2. The lone pair of O or S in furan or thiophene takes part in the aromatic π-system.

Answer: (b) Benzenoid aromatic compound

Explanation:
1. Naphthalene consists of two fused benzene rings.
2. Since the compound contains the benzene nucleus, naphthalene is a benzenoid aromatic compound.

Teacher's Note:
1. Aromatic compounds containing one or more benzene rings are called benzenoid.
2. Tropolone and azulene are examples of non-benzenoid aromatic compounds.

Answer: (c) 5

Explanation:
1. A 1° carbon is attached directly to only one other carbon atom.
2. In CH₃—CH(CH₃)—C(CH₃)₂—CH₂—CH₃, the two terminal CH₃ groups, the CH₃ on the CH and the two CH₃ groups on the quaternary C are 1° carbons.
3. So, there are 5 primary carbons.

Teacher's Note:
1. Every CH₃ group in an alkane chain is a primary carbon.
2. The C attached to four other carbons here is a quaternary (4°) carbon.

Answer: (b) 1-chloro-4-methyl-2-nitrobenzene

Explanation:
1. For tri or higher substituted benzene derivatives, the positions on the ring are given by the lowest locant rule.
2. Numbering from the C bearing Cl gives Cl = 1, NO₂ = 2, CH₃ = 4, i.e. the lowest set of locants (1, 2, 4).
3. The substituents are then written in alphabetical order: chloro, methyl, nitro.
4. Hence, the name is 1-chloro-4-methyl-2-nitrobenzene.

Teacher's Note:
1. First find the lowest locant set, then arrange prefixes alphabetically.
2. Option (a) has the right numbers but the prefixes are not in alphabetical order.

Answer: (a) 3-keto-2-methylhex-4-enal

Explanation:
1. The —CHO group gets higher priority (acts as principal group) over >C=O and C=C, so the —CHO carbon is C-1.
2. The six-carbon chain is numbered from CHO: CH₃ at C-2, C=O at C-3 and the double bond between C-4 and C-5.
3. Hence, the name is 3-keto-2-methylhex-4-enal (present IUPAC form: 2-methyl-3-oxohex-4-enal).

Teacher's Note:
1. When an aldehyde and a ketone are both present, the ketone is shown as a prefix (keto/oxo).
2. The suffix ‘-al’ always takes locant 1, so it is not written in the name.

Answer: (a) 2-ethyl-3,4-dimethylpentanal

Explanation:
1. The longest chain must include the —CHO carbon, which is C-1.
2. From CHO, the longest chain runs through CH—CH(CH₃)—CH(CH₃)—CH₃, giving a five-carbon chain (pentanal).
3. Substituents: ethyl at C-2, methyl at C-3 and C-4.
4. IUPAC name: 2-ethyl-3,4-dimethylpentanal.

Teacher's Note:
1. The principal chain must contain the principal functional group even if a longer carbon chain exists elsewhere.
2. Prefixes are arranged alphabetically: ethyl comes before methyl.

Answer: (d) 4-oxopentanoic acid

Explanation:
1. When more than one functional group is present in the main chain, the name is based on the group of higher priority.
2. Carboxylic acid (—COOH) has more priority than ketone (>C=O).
3. Numbering from COOH (C-1), the keto group is at C-4 of a five-carbon chain.
4. IUPAC name: 4-oxopentanoic acid.

Teacher's Note:
1. Priority order: —COOH > —SO₃H > ester > acid halide > amide > —CN > —CHO > >C=O > —OH > —NH₂.
2. A lower-priority ketone group is written as the prefix ‘oxo’.

Answer: (b) 7-hydroxyheptan-2-one

Explanation:
1. HOCH₂(CH₂)₃CH₂COCH₃ = HO—CH₂—CH₂—CH₂—CH₂—CH₂—CO—CH₃, a seven-carbon chain.
2. A compound containing both an alcohol and a keto group is named as a hydroxy alkanone, since the keto group is preferred over the hydroxy group.
3. Numbering from the end nearer the C=O: C=O at C-2 and OH at C-7.
4. IUPAC name: 7-hydroxyheptan-2-one.

Teacher's Note:
1. Ketone ranks higher than alcohol, so the suffix is ‘-one’ and OH becomes the prefix ‘hydroxy’.
2. Count all carbons including the CH₃ of the COCH₃ group.

Answer: (c) 2,2-dimethyl propyl

Explanation:
1. Neo-pentyl group is (CH₃)₃C—CH₂—.
2. The longest chain starting from the carbon with the free valency has three carbons (propyl).
3. Two alkyl (methyl) groups are present at C-2.
4. So, it is named 2,2-dimethylpropyl.

Teacher's Note:
1. In a substituent group, numbering always starts from the carbon that carries the free valency.
2. iso and neo are common-name prefixes, not systematic IUPAC names.

Answer: (d) 2-(4-chlorophenyl) propanoic acid

Explanation:
1. The parent chain CH₃—CH—COOH is propanoic acid, with COOH carbon as C-1.
2. The substituted phenyl ring is attached at C-2.
3. In the ring, the carbon attached to the chain is C-1 and Cl is at the para position, C-4: 4-chlorophenyl.
4. IUPAC name: 2-(4-chlorophenyl)propanoic acid.

Teacher's Note:
1. A substituted benzene ring used as a substituent is named as a ‘phenyl’ with its own numbering in brackets.
2. The ring carbon attached to the main chain is always numbered 1 in the substituent.

Answer: (a)

Explanation:
1. ‘iso’ means one methyl group is present at the end (penultimate carbon) of the side chain.
2. Hence, iso-butyl group is (CH₃)₂CH—CH₂—.
3. The suffix ‘-yl’ is used for a group with one H less than the parent hydrocarbon.
4. (b) is butane, (c) n-butyl and (d) tert-butyl.

Teacher's Note:
1. The four butyl groups are n-butyl, sec-butyl, iso-butyl and tert-butyl.
2. iso-butyl is 2-methylpropyl in IUPAC naming.

Answer: (b)

Explanation:
1. In 2-methyl-2-cyclohexen-1-ol (2-methylcyclohex-2-en-1-ol), the —OH group is on C-1.
2. The double bond is between C-2 and C-3, and the CH₃ group is on C-2.
3. In option (b), OH is on C-1, CH₃ on the adjacent C-2 and the double bond starts from C-2, so (b) is correct.
4. (a) has OH on a double-bonded carbon, (c) has CH₃ on C-3 and (d) is 4-methylcyclohex-2-en-1-ol.

Teacher's Note:
1. Number the ring from the carbon carrying the principal group (—OH) and go towards the double bond.
2. The locant of the double bond in a cycloalkene is the lower of the two carbons it joins.

Answer: (c) trans-2-chloro-3-iodo-2-pentene

Explanation:
1. The main chain is CH₃—C(Cl)=C(I)—CH₂CH₃, a five-carbon chain (pentene) numbered so that the double bond gets the lowest locant: C-2.
2. Cl is on C-2 and I is on C-3; prefixes in alphabetical order give 2-chloro-3-iodo-2-pentene.
3. The chain carbons CH₃ (C-1) and CH₂CH₃ (C-4, C-5) lie on opposite sides of the double bond, so it is the trans isomer.
4. Name: trans-2-chloro-3-iodo-2-pentene.

Teacher's Note:
1. cis/trans is decided by the positions of the main-chain parts across the C=C bond.
2. Lowest locant goes to the double bond first; then halogen prefixes are arranged alphabetically.

Answer: (a) 3-ethyl-4,4-dimethylheptane

Explanation:
1. The structure is CH₃CH₂CH₂C(CH₃)₂CH(C₂H₅)CH₂CH₃, whose longest chain has seven carbons (heptane).
2. Numbering from the end nearer the substituents gives ethyl at C-3 and two methyls at C-4 (locant set 3, 4, 4).
3. While writing the IUPAC name, alkyl groups are written in alphabetical order; so ethyl comes before methyl.
4. Note: prefixes di, tri, tetra are not included in alphabetical order. Name: 3-ethyl-4,4-dimethylheptane.

Teacher's Note:
1. Alphabetise by the substituent name (ethyl, methyl), ignoring multiplying prefixes like di- and tri-.
2. ‘bis’ is used only for complex substituents, not for simple methyl groups.

Answer: (a) 3,5-dimethylcyclohexene

Explanation:
1. In a cycloalkene, the double bond carbons are numbered C-1 and C-2, and the locant of the double bond is not written.
2. Numbering in the direction that gives substituents the lowest numbers puts the two CH₃ groups at C-3 and C-5.
3. IUPAC name: 3,5-dimethylcyclohexene.

Teacher's Note:
1. In cyclohexene the double bond is always between C-1 and C-2, so ‘1’ is omitted.
2. After fixing the double bond, choose the direction that gives lower locants to the substituents.

Answer: (c) 1-bromo-3-methylbutane

Explanation:
1. The longest chain is Br—CH₂—CH₂—CH(CH₃)—CH₃, i.e. four carbons (butane).
2. Numbering from the end nearer Br gives Br at C-1 and CH₃ at C-3.
3. Alphabetical order (bromo before methyl) gives 1-bromo-3-methylbutane.

Teacher's Note:
1. Halogens are always written as prefixes (fluoro, chloro, bromo, iodo).
2. This compound is commonly called isopentyl (isoamyl) bromide.

Answer: (a) Br—CH₂—CH=CH₂  (1-bromoprop-2-ene)

Explanation:
1. In IUPAC nomenclature, preference is given to the multiple bond over the halogen substituent while numbering.
2. So, in Br—CH₂—CH=CH₂ the double bond carbon nearest the end gets C-1: Br is on C-3.
3. The correct name is 3-bromoprop-1-ene, so the name in option (a) is not in accordance with IUPAC system.
4. The names in options (b), (c) and (d) are correct.

Teacher's Note:
1. A halogen is only a prefix, so it never decides the direction of numbering over a C=C bond.
2. Allyl bromide (CH₂=CH—CH₂Br) is 3-bromoprop-1-ene.

Answer: (d) one cation and one anion

Explanation:
1. In heterolytic fission, the shared pair of electrons goes with one of the atoms.
2. So, Cl—Cl gives Cl⁺ (cation) and Cl⁻ (anion) due to transfer of the electron pair from the shared bond to one Cl atom.

Teacher's Note:
1. Homolytic fission gives two free radicals; heterolytic fission gives a pair of ions.
2. Carbocation and carbanion are names used only when the charged atom is carbon.

Answer: (c) CH₃—C^⊕H—CH₂—CH₂—CH₃

Explanation:
1. (a), (b) and (c) are 2° carbocations, while (d) is a 1° carbocation.
2. Stability of carbocations increases with the number of α-H atoms, which increases the number of hyperconjugating structures.
3. α-H atoms: (a) 3, (b) 4, (c) 5, (d) 2.
4. So, CH₃—C^⊕H—CH₂CH₂CH₃ has maximum hyperconjugation and is the most stable.

Teacher's Note:
1. Count the H atoms on carbons directly attached to the positive carbon (α-H).
2. A C(CH₃)₃ group attached to C⁺ has no α-H, so it gives no hyperconjugation.

Answer: (d) Tertiary

Explanation:
1. Stability order of free radicals: 3° > 2° > 1° > methyl.
2. The greater the number of alkyl groups attached to the carbon carrying the odd electron, the greater the delocalisation (hyperconjugation and +I effect) of the odd electron.
3. So, a tertiary free radical is the most stable.

Teacher's Note:
1. Free radicals and carbocations both follow 3° > 2° > 1° > CH₃.
2. Benzyl and allyl free radicals are even more stable due to resonance.

Answer: (c) Both (a) and (b)

Explanation:
1. In CH₃Cl, carbon is sp³-hybridised and tetrahedral (bond angle 109°28′).
2. Homolysis gives CH₃^∙ + Cl^∙. The methyl free radical is sp²-hybridised with a singly occupied p-orbital.
3. It is planar with bond angle 120°, so both (a) and (b) are correct.

Teacher's Note:
1. Alkyl free radicals and carbocations are planar (sp²), while carbanions are pyramidal (sp³).
2. The odd electron stays in the unhybridised p-orbital perpendicular to the plane.

Answer: (b) II > III > I > IV

Explanation:
1. The benzyl radical (II) is the most stable due to resonance.
2. In the remaining free radicals, the more alkyl groups bonded to the electron deficient carbon, the more stable is the radical.
3. III is 3°, I is 2° and IV is 1°.
4. Order of stability: II > III > I > IV.

Teacher's Note:
1. Resonance stabilisation (benzyl, allyl) outweighs hyperconjugation.
2. Among alkyl radicals the order is 3° > 2° > 1° > methyl.

Answer: (c)

Explanation:
1. (a) is a 3° alkyl carbocation, (b) a phenyl (vinylic type) carbocation which is least stable, (c) the benzyl carbocation and (d) a 2° alkyl carbocation.
2. The positive charge of the benzyl carbocation is delocalised over the benzene ring (ortho and para positions) through resonance.
3. Order of stability: (b) < (d) < (a) < (c). So, the benzyl carbocation is the most stable.

Teacher's Note:
1. Positive charge on an sp² ring carbon (phenyl cation) cannot be delocalised, so it is very unstable.
2. Resonance with the benzene ring makes the benzyl cation more stable than simple alkyl cations here.

Answer: (b) sp²

Explanation:
1. The shape of CH₃⁺ may be considered as being derived from the overlap of three equivalent C(sp²) hybridised orbitals with 1s-orbital of each of the three hydrogen atoms.
2. Each bond may be represented as C(sp²)—H(1s) sigma bond.
3. The remaining carbon orbital is perpendicular to the molecular plane and contains no electrons.
4. 3 σ-bonds and no lone pair ⇒ sp²-hybridisation.

Teacher's Note:
1. A carbocation is trigonal planar with an empty unhybridised p-orbital.
2. Number of σ bonds + lone pairs on C = 3 means sp².

Answer: (c) CH₃—C^⊕H—CH₂—CH₂—CH₃

Explanation:
1. CH₃—C^⊕H—CH₂CH₂CH₃ is the most stable carbocation among the given carbocations.
2. It is because the number of α-H atoms attached to the carbocation is maximum (5) in CH₃CH^⊕CH₂CH₂CH₃.
3. Thus, it has the maximum number of hyperconjugating structures; hence, it is most stable.

Teacher's Note:
1. More α-H atoms means more hyperconjugation structures and more stability.
2. A (CH₃)₃C— group gives +I effect but no α-H for hyperconjugation.

Answer: (b) I > II > III > IV

Explanation:
1. I is a 3° carbocation stabilised by two phenyl groups (resonance).
2. II is a 2° benzylic carbocation stabilised by one phenyl group.
3. III and IV are 1° carbocations not conjugated with any phenyl group; III has one α-H (hyperconjugation) while IV has none.
4. Order of stability: I > II > III > IV.

Teacher's Note:
1. Benzylic carbocations are stabilised by resonance; more phenyl groups on the C⁺ give more stability.
2. A phenyl on the carbon next to C⁺ cannot delocalise the charge by resonance.

Answer: (a) II > I > III

Explanation:
1. I (CH₃—C^⊕H—CH₃): weak +I effect (and hyperconjugation) of the two methyl groups stabilises the carbocation.
2. II (CH₃—C^⊕H—OCH₃): strong +R effect of the —OCH₃ group (lone pair of O) stabilises the carbocation.
3. III (CH₃—C^⊕H—CH₂—OCH₃): −I effect of the —OCH₃ group destabilises the carbocation.
4. Hence, the stability of carbocations decreases as II > I > III.

Teacher's Note:
1. A lone pair on an atom directly attached to C⁺ stabilises it strongly by resonance.
2. The same group one carbon away can act only through its electron-withdrawing −I effect.

Answer: (c) CH₃⁺

Explanation:
1. CH₃Cl + AlCl₃ → CH₃⁺ + AlCl₄⁻; the CH₃⁺ (methyl carbocation) is the electrophile that attacks benzene.
2. Hence, the CH₃⁺ group acts as an intermediate in the reaction of benzene with CH₃Cl in the presence of anhydrous AlCl₃.
3. This reaction is called Friedel-Crafts alkylation.

Teacher's Note:
1. The Lewis acid AlCl₃ pulls Cl⁻ from CH₃Cl to generate the carbocation.
2. Friedel-Crafts alkylation is an electrophilic substitution reaction of benzene.

Answer: (a) A-1, B-1, C-2

Explanation:
1. Free radicals are formed by homolytic fission, e.g. CH₃^∙; C is sp²-hybridised, so trigonal planar: A → 1.
2. Carbocations are formed by heterolytic fission when carbon is attached to a more electronegative atom, e.g. CH₃⁺; sp², trigonal planar: B → 1.
3. Carbanions are formed by heterolytic fission when carbon is attached to a more electropositive atom, e.g. CH₃⁻; sp³, pyramidal: C → 2.

Teacher's Note:
1. Carbocation and free radical: sp² and planar; carbanion: sp³ and pyramidal (lone pair occupies one hybrid orbital).
2. None of these common intermediates is linear.

Answer: (d) R₃P < R₃N

Explanation:
1. Nucleophilic strength increases down the group (larger, more polarisable atom), so R₃P > R₃N.
2. Hence, the comparison R₃P < R₃N is incorrect.
3. The other comparisons are correct: bulky (CH₃)₃CO^⊖ is a weaker nucleophile than OH^⊖, NH₃ is a better nucleophile than H₂O and RS^⊖ is better than RO^⊖.

Teacher's Note:
1. Within a group, nucleophilicity rises with size: I⁻ > Br⁻ > Cl⁻, RS⁻ > RO⁻, R₃P > R₃N.
2. Steric bulk lowers nucleophilicity even when basicity is high.

Answer: (c) CH₃⁻

Explanation:
1. Across a period, nucleophilicity varies inversely with electronegativity of the attacking atom: electronegativity ∝ \( \frac{1}{\text{nucleophilicity}} \).
2. Electronegativity order: F > O > N > C.
3. ∴ CH₃⁻ has the highest nucleophilicity.

Teacher's Note:
1. The less electronegative the atom, the more readily it donates its lone pair.
2. So in the same period: CH₃⁻ > NH₂⁻ > OH⁻ > F⁻.

Answer: (b) AlCl₃

Explanation:
1. A Lewis base (Cl⁻) reacts only with a Lewis acid (electron pair acceptor).
2. AlCl₃ is electron deficient (Lewis acid): AlCl₃ + Cl⁻ → AlCl₄⁻.
3. CH₄ + Cl⁻, OH⁻ + Cl⁻ and CH₃⁻ + Cl⁻ give no reaction.

Teacher's Note:
1. Al in AlCl₃ has only six electrons in its valence shell, so it accepts a lone pair.
2. Two electron-rich species (two Lewis bases) do not combine with each other.

Answer: (c) NO₂⁺

Explanation:
1. Electrophiles are electron deficient species.
2. Among the given species, the nitronium ion (NO₂⁺) acts as an electrophile due to lack of electrons.
3. NH₃ and H₂O have lone pairs of electrons and act as nucleophiles.

Teacher's Note:
1. NO₂⁺ is the electrophile in the nitration of benzene.
2. Neutral molecules with lone pairs (NH₃, H₂O) are nucleophiles.

Answer: (b) AlCl₃, SO₃, NO₂⁺

Explanation:
1. AlCl₃ and SO₃ are Lewis acids (electron deficient) and NO₂⁺ is a positive ion; all three are electrophiles.
2. In (c), CH₃⁻ is a carbanion, which is a nucleophile.
3. In (d), C₂H₅⁻ is a nucleophile, and in (a) NO₂ is a neutral odd-electron molecule, not the electrophile NO₂⁺.
4. Hence, group (b) contains only electrophiles.

Teacher's Note:
1. Electrophiles include cations (H⁺, NO₂⁺, carbocations) and neutral Lewis acids (BF₃, AlCl₃, SO₃).
2. Any species with a negative charge or available lone pair is a nucleophile.

Answer: (a) A-1, B-4, C-2, D-3

Explanation:
1. A. Carbocation: H₃C⁺; loss of e⁻ makes its p-orbital empty and carbon is sp²-hybridised → 1.
2. B. Nucleophile: nucleus loving, i.e. having negative charge or excess of electrons; it supplies a pair of electrons → 4.
3. C. sp-hybridisation: ethyne, HC≡CH → 2.
4. D. Electrophile: electron loving, i.e. positive charge or lack of electrons; it receives a pair of electrons → 3.

Teacher's Note:
1. Nucleophiles are Lewis bases (pair donors); electrophiles are Lewis acids (pair acceptors).
2. Triple-bonded carbons (as in ethyne) are sp hybridised.

Answer: (b)

Explanation:
1. In the benzoate ion, there is resonance in the benzene nucleus as well as in the carboxylate ion (two equivalent structures).
2. Thus, it is maximum stabilised, i.e. it has the maximum resonance energy.
3. CH₃—CH₂^⊖ shows no resonance.

Teacher's Note:
1. Resonance between equivalent structures (like the two C—O⁻ forms of carboxylate) gives large stabilisation.
2. In phenoxide, the structures with negative charge on ring carbons are less important than those with charge on O.

Answer: (c) Positive electromeric effect

Explanation:
1. In the given reaction, the π-electrons of the multiple bond are transferred to that atom on which the attacking reagent (H⁺) gets attached.
2. So, it is an example of positive electromeric effect (+E-effect).

Teacher's Note:
1. Electromeric effect is temporary and operates only in the presence of an attacking reagent.
2. +E: π-electrons move to the atom where the reagent attaches; −E: they move to the other atom.

Answer: (d)

Explanation:
1. Ethynic H is attached to the —C≡C— system (sp carbon).
2. NO₂ is electron-withdrawing, so the (C—H) bond strength is weakened to a greater extent when NO₂ is attached.
3. Since the benzene nucleus is also resonance stabilised and transmits the electron withdrawal, the acetylenic H is released most easily in (d).

Teacher's Note:
1. Electron-withdrawing groups stabilise the acetylide anion and make the terminal H more acidic.
2. An aromatic ring transmits the −R effect of NO₂, while a cyclohexane ring cannot.

Answer: (a) A → hyperconjugation, B → empty 2p-orbital of carbon, C → C_sp³—H_1s bond

Explanation:
1. In CH₃C⁺H₂ (ethyl cation), the positively charged carbon atom has an empty p-orbital (B).
2. One of the C—H bonds of the methyl group (a C_sp³—H_1s σ bond, C) can align in the plane of this empty p-orbital.
3. The electrons of this C—H bond can then be delocalised into the empty p-orbital; this is hyperconjugation (A).

Teacher's Note:
1. Hyperconjugation is σ–p (or σ–π) delocalisation and is a permanent effect.
2. The methyl carbon in the ethyl cation is sp³, so its C—H bond is C_sp³—H_1s.

Answer: (b) IV < III < II < I

Explanation:
1. The greater the number of alkyl groups attached to a positively charged carbon atom, the greater is the hyperconjugation interaction and stabilisation of the cation.
2. I is 3°, II is 2°, III is 1° and IV is the methyl cation.
3. Thus, the stability order is I > II > III > IV, i.e. IV < III < II < I.

Teacher's Note:
1. More α-H atoms → more hyperconjugation structures → more stable carbocation.
2. CH₃⁺ has no α-H at all and is the least stable.

Answer: (b) I > II > III

Explanation:
1. I is the most stable: it has more covalent bonds, each carbon and oxygen atom has a complete octet and there is no separation of opposite charges.
2. II has charge separation but the negative charge is on the more electronegative atom (O) and the positive charge on the more electropositive atom (C).
3. III hardly contributes since oxygen (more electronegative) has the positive charge and carbon has the negative charge; hence it is least stable.
4. Order: I > II > III.

Teacher's Note:
1. Rules: more bonds and complete octets > charge separation; negative charge on the more electronegative atom is preferred.
2. A structure with + on O and − on C is a very minor contributor.

Answer: (a) I shows +R-effect, whereas II shows −R-effect

Explanation:
1. Aniline (I) shows +R-effect, whereas nitrobenzene (II) shows −R-effect.
2. In +R-effect, the transfer of electrons is away from the atom or substituent group (lone pair of —NH₂) attached to the conjugated system; this increases electron density at the ortho and para positions. This effect in aniline is shown as:
3. −R-effect is observed when the transfer of electrons is towards the atom or substituent group attached to the conjugated system, e.g. in nitrobenzene this electron displacement can be depicted as:

Teacher's Note:
1. Groups with a lone pair on the atom attached to the ring (—NH₂, —OH, —OR, halogens) show +R.
2. Groups with a multiple bond to an electronegative atom (—NO₂, —CHO, —COOH, —CN) show −R.

Answer: (b)

Explanation:
1. (a), (c) and (d) are resonance structures of the benzyl radical, obtained by moving single electrons (half arrows) with the odd electron at the benzylic, ortho or para carbon.
2. In (b), there are two odd electrons (on the CH₂ carbon and on a ring carbon) and the bonds do not correspond to the other structures; the number of unpaired electrons is changed.
3. Resonance structures must have the same number of paired and unpaired electrons, so (b) is incorrect.

Teacher's Note:
1. In a radical, the odd electron can appear only at the benzylic, ortho and para positions.
2. Resonance structures differ only in the position of electrons, never in the number of unpaired electrons.

Answer: (c) hyperconjugative effect of the Me group

Explanation:
1. Me₂C=CH₂ has two CH₃ groups attached to the double bond, i.e. 6 α-H atoms, so 6 hyperconjugative structures.
2. MeCH₂CH=CH₂ has only one CH₂ group attached to the double bond, i.e. 2 α-H atoms, so 2 hyperconjugative structures.
3. Therefore, Me₂C=CH₂ is more stable than MeCH₂CH=CH₂ due to the hyperconjugative effect.

Teacher's Note:
1. More α-H atoms next to a C=C bond means more hyperconjugation and a more stable alkene.
2. This is why more substituted alkenes are generally more stable.

Answer: (b) σ-p

Explanation:
1. Hyperconjugation involves overlapping of the σ-p orbitals. It is a general stabilising interaction.
2. It involves delocalisation of σ-electrons of the C—H bond of an alkyl group directly attached to an atom of an unsaturated system or to an atom with an unshared p-orbital.
3. The σ-electrons of the C—H bond enter into partial conjugation with the attached unsaturated system or with the unshared p-orbital. It is a permanent effect.

Teacher's Note:
1. Hyperconjugation is also called ‘no bond resonance’.
2. It needs C—H bonds on the carbon next to the C⁺, radical centre or double bond.

Answer: (c) +I effect

Explanation:
1. Due to +I effect, electron density increases, which results in a decrease in acidic character.
2. Acetic acid, CH₃—COOH, has a —CH₃ group that shows +I effect, whereas no such group is present in formic acid (HCOOH).
3. The +I effect of CH₃ destabilises the acetate ion, so formic acid is a stronger acid than acetic acid.

Teacher's Note:
1. Electron-releasing (+I) groups decrease acid strength; electron-withdrawing (−I) groups increase it.
2. pK_a: HCOOH ≈ 3.75, CH₃COOH ≈ 4.76.

Answer: (a) (CH₃)₃C— < (CH₃)₂CH— < CH₃CH₂—

Explanation:
1. The inductive effect of alkyl groups on a saturated carbon chain follows the order (CH₃)₃C— > (CH₃)₂CH— > CH₃CH₂—.
2. However, when an alkyl group is attached to an unsaturated system such as a double bond or a benzene ring, the order of electron release is reversed.
3. This is due to hyperconjugation (Baker-Nathan effect): CH₃CH₂— has 2 α-H, (CH₃)₂CH— has 1 α-H and (CH₃)₃C— has none.
4. So the order is (CH₃)₃C— < (CH₃)₂CH— < CH₃CH₂—.

Teacher's Note:
1. Baker-Nathan order of electron release follows the number of α-H atoms: CH₃— > CH₃CH₂— > (CH₃)₂CH— > (CH₃)₃C—.
2. On a saturated chain, the +I order is the reverse.

Answer: (b) Ether

Explanation:
1. Electromeric effect implies complete transfer of π-electrons in the presence of a reagent.
2. Simple ethers do not contain any multiple bond, hence they do not show electromeric effect.
3. Alkenes (C=C), aldehydes and ketones (C=O) have π bonds and show electromeric effect.

Teacher's Note:
1. Electromeric effect needs a multiple (π) bond and an attacking reagent.
2. It is a temporary effect that vanishes when the reagent is removed.

Answer: (a) A-1, B-2, C-3

Explanation:
1. A. The carbanion ⁻CH₂—NO₂ is stabilised by resonance: the negative charge is delocalised onto the oxygen atoms of the nitro group → 1.
2. B. In F₃C⁺, the −I effect of the three F atoms creates further electron deficiency at the carbon carrying the positive charge, so it is destabilised → 2.
3. C. In CH₃—C⁺H—CH₃, C⁺ is attached to two carbons and is stabilised by hyperconjugation → 3.

Teacher's Note:
1. Electron-withdrawing groups stabilise carbanions but destabilise carbocations.
2. Isopropyl cation has 6 α-H atoms available for hyperconjugation.

Answer: (d)

Explanation:
1. In all the given carboxylate ions, the negative charge is dispersed by resonance (+R-effect / conjugation of the carboxylate group), which is common to all four.
2. The overall dispersal of negative charge therefore depends on the number of halogen atoms and their electronegativity (−I-effect).
3. In (a), the methyl group (+I) increases the density on C-atom and makes the ion least stable.
4. F has the highest electronegativity and two F atoms are present in option (d); thus, dispersal of negative charge is maximum and F₂CH—COO⁻ is the most stable.

Teacher's Note:
1. More and stronger −I groups near the COO⁻ group spread the charge and stabilise the anion.
2. This is why difluoroacetic acid is a much stronger acid than acetic acid.

Answer: (a) *CH₃—CH₂—Cl

Explanation:
1. Electronegativity of Cl, Br, C and Mg follows the order Cl > Br > C > Mg.
2. *CH₃ → CH₂ → Cl (−I effect of Cl); *CH₃ ← CH₂ ← Mg⁺Cl⁻ (electron releasing Mg); *CH₃ → CH₂ → Br (−I effect); *CH₃—CH₂—CH₃ (+I effect).
3. −I effect of Cl > Br.
4. Hence, in CH₃—CH₂—Cl the marked carbon has the greatest positive charge.

Teacher's Note:
1. The inductive effect weakens with distance, but the more electronegative the atom, the more positive charge is relayed.
2. In a Grignard reagent, carbon bonded to Mg carries partial negative charge.

Answer: (a) α-elimination

Explanation:
1. CHCl₃ + K(t-butoxide) → :CCl₂ (dichlorocarbene).
2. Here the dominating groups, i.e. hydrogen and chlorine, are both lost from the same carbon atom.
3. So, it is an example of α-elimination.

Teacher's Note:
1. α-elimination: both groups leave from the same carbon (gives carbenes); β-elimination: from adjacent carbons (gives alkenes).
2. Dichlorocarbene is a neutral electrophile with only six valence electrons on carbon.

Answer: (d)

Explanation:
1. Reactivity of carbonyl compounds towards nucleophilic addition reactions depends on the presence of substituent groups.
2. Electron withdrawing (−I, −M) groups increase the reactivity, as they make the carbonyl carbon more positive.
3. Order: p-nitrobenzaldehyde (−I, −M) > benzaldehyde > p-tolualdehyde (+I) > acetophenone (+R of CH₃ and ketone).
4. So, p-nitrobenzaldehyde is the most reactive.

Teacher's Note:
1. Aldehydes are more reactive than ketones towards nucleophiles (less steric hindrance, fewer electron-releasing groups).
2. An NO₂ group para to CHO pulls electrons from the carbonyl carbon and increases its electrophilic character.

Answer: (c) 2° carbocation

Explanation:
1. When the electrophile (H⁺) attacks CH₃—CH=CH₂, delocalisation of electrons can take place in two possible ways.
2. CH₃—CH=CH₂ + H⁺ → CH₃—C^⊕H—CH₃ (2° carbocation) or CH₃—CH₂—C^⊕H₂ (1° carbocation).
3. As the 2° carbocation is more stable than the 1° carbocation, the first addition is more feasible.
4. Note: stability of carbocations is the basis of Markovnikov’s rule.

Teacher's Note:
1. The electrophile adds to the carbon that already has more H atoms, giving the more stable carbocation.
2. That is why HBr adds to propene to give 2-bromopropane.

Answer: (d) CH₃—CH₂—CH(CH₃)—CH₂Br + NH₃ → CH₃—CH₂—CH(CH₃)—CH₂NH₂

Explanation:
1. In (d), the nucleophile NH₃ replaces the Br⁻ ion (leaving group) from the alkyl halide.
2. CH₃CH₂CH(CH₃)CH₂Br + NH₃ → CH₃CH₂CH(CH₃)CH₂NH₂.
3. It is an example of nucleophilic substitution reaction.
4. (a) and (c) are nucleophilic additions and (b) is an electrophilic addition.

Teacher's Note:
1. Substitution means one group replaces another; addition means atoms add across a multiple bond.
2. Alkyl halides typically undergo nucleophilic substitution.

Answer: (b) Elimination

Explanation:
1. Dehydration of alcohol involves the loss of two atoms or groups (H and OH) from the adjacent carbon atoms.
2. Hence, it is an example of β-elimination reaction, which forms an alkene.

Teacher's Note:
1. Elimination reactions convert a saturated compound into an unsaturated one.
2. Example: (CH₃)₃COH → (CH₃)₂C=CH₂ + H₂O on heating with acid.