JEE Chemistry Organic Chemistry Some Basic Principles and Techniques MCQs Set 29

Answer: (c) displacement of sigma electrons

Explanation:
1. The inductive effect is the permanent polarisation of a sigma bond caused by an electronegative atom or group.
2. The sigma electrons are only displaced towards one atom, they are not delocalised over several atoms.

Teacher's Note:
1. Inductive effect works through sigma bonds; resonance and mesomeric effects work through pi electrons.
2. It falls off quickly and is negligible after about three carbon atoms.

Answer: (b) (CH3)3C–

Explanation:
1. +I effect increases with the number of alkyl groups on the carbon attached to the chain.
2. Order: (CH3)3C– > (CH3)2CH– > CH3CH2– > CH3–.
3. So tert-butyl has the maximum +I effect.

Teacher's Note:
1. More alkyl groups on the attached carbon push more electron density.
2. This is why tertiary carbocations are the most stable.

Answer: (d) ii > i > iv > iii

Explanation:
1. Strong –I groups: NO2 > CN.
2. Halogens are weaker: Cl is less –I than CN.
3. NH2 is the weakest of the four.
4. Order: NO2 > CN > Cl > NH2, that is ii > i > iv > iii.

Teacher's Note:
1. Remember the series NO2 > CN > COOH > F > Cl > Br > I > OH > NH2.
2. –I strength rises with electronegativity and with the positive charge or multiple bonds on the attached atom.

Answer: (b) –F > –OR > –NH2

Explanation:
1. All three groups are electron withdrawing by the –I effect.
2. The strength follows the electronegativity of the attached atom: F > O > N.
3. So –F > –OR > –NH2.

Teacher's Note:
1. Along a period, –I effect increases from left to right.
2. Compare the atoms bonded to carbon, not the whole group.

Answer: (d) All of these.

Explanation:
1. A carbocation is electron deficient, so electron donating (+I) groups reduce its positive charge and stabilise it.
2. Alkyl groups also stabilise free radicals through +I effect and hyperconjugation.
3. A carbanion carries a negative charge, so an electron withdrawing (–I) group disperses the charge and stabilises it.
4. All three statements are correct.

Teacher's Note:
1. Match the nature of the group to the nature of the intermediate: donors help cations, acceptors help anions.
2. Radicals are stabilised by donors and by delocalisation.

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

Explanation:
1. Dipole moment increases with the electron withdrawing power of the group attached to carbon.
2. –NO2 is far stronger than the halogens, and among halogens Cl > Br > I.
3. Order: nitroethane > chloroethane > bromoethane > iodoethane.

Teacher's Note:
1. Dipole moment depends on charge separation and bond length; here the electronegativity difference dominates.
2. The measured values are about 3.6 D, 2.05 D, 2.03 D and 1.91 D.

Answer: (a) identical bonding

Explanation:
1. Resonance structures must have the same positions of atoms, the same number of paired and unpaired electrons and nearly equal energy.
2. They differ in the way the pi electrons and charges are arranged, so their bonding is not identical.

Teacher's Note:
1. Only electrons are shifted between canonical forms, never atoms.
2. If the atoms move, the species are isomers, not resonance forms.

Answer: (b) \( \ce{CH2=CH-CH=O} \)

Explanation:
1. Delocalisation needs a conjugated system, that is alternate pi bond and single bond.
2. CH2=CH–CH=O (acrolein) has C=C and C=O separated by one single bond.
3. The others have an sp3 carbon breaking the conjugation or no pi bond.

Teacher's Note:
1. Look for the pattern pi bond – single bond – pi bond (or lone pair/charge).
2. An sp3 carbon interrupts conjugation.

Answer: (a) 2-butene

Explanation:
1. 1,3-Butadiene, 1,3,5-hexatriene and benzene all contain alternate single and double bonds.
2. 2-Butene has only one double bond, so no conjugated system exists.

Teacher's Note:
1. A lone double bond has nothing to conjugate with.
2. Hyperconjugation may occur in 2-butene, but that is not pi delocalisation.

Answer: (a) \( \ce{CH2=CH-N(+)(=O)O^- <-> {+}CH2-CH=N(+)(O^-)O^-} \) (the same molecule with the pi electrons shifted)

Explanation:
1. Resonance forms differ only in electron positions with identical atom positions.
2. But-2-ene and but-1-ene are position isomers.
3. Acetone and its enol are tautomers: a hydrogen atom has moved.
4. Nitromethane and methyl nitrite differ in the atom connected to carbon, so they are isomers.
5. Only the first pair differ solely in electron positions.

Teacher's Note:
1. If an atom moves, it is isomerism or tautomerism, not resonance.
2. Use curved arrows: pi electrons move towards the positive centre.

Answer: (c) resonance

Explanation:
1. The positive charge is delocalised over two carbon atoms: \( \ce{{+}CH2-CH=CH2 <-> CH2=CH-{+}CH2} \).
2. This delocalisation by resonance stabilises the cation.

Teacher's Note:
1. Allyl and benzyl cations are stabilised mainly by resonance.
2. Two equivalent contributing structures give a strong stabilisation.

Answer: (d) 7

Explanation:
1. The positive charge can sit on any one of the seven carbon atoms.
2. All seven structures are identical in energy, so seven equivalent resonance forms exist.

Teacher's Note:
1. The seven-membered ring has six pi electrons spread over seven carbons, so it is aromatic.
2. Equivalent resonance structures mean extra stability.

Answer: (d) the diradical \( \ce{{.}CH2-C#C-{.}CH2} \)

Explanation:
1. Resonance structures must have the same number of unpaired electrons.
2. Buta-1,2,3-triene has no unpaired electrons, but the diradical has two.
3. The two charge-separated forms have all electrons paired, so they are acceptable.

Teacher's Note:
1. Check paired and unpaired electron counts first.
2. A diradical is a different electronic state, not a resonance contributor.

Answer: (a) I > III > II

Explanation:
1. Structure I has no charges and complete octets, so it is the most stable.
2. In II the positive charge is on the electronegative oxygen, which is unfavourable.
3. In III the negative charge is on oxygen (electronegative), so III is more stable than II.
4. Order: I > III > II.

Teacher's Note:
1. Neutral structures with full octets are best.
2. Negative charge belongs on the more electronegative atom.

Answer: (b) II

Explanation:
1. Structure I is neutral with complete octets and is the major contributor.
2. In III every atom has an octet, with the charge separation, so it contributes moderately.
3. In II the carbon has only six electrons and carries opposite charges close together, so it is the least stable and contributes least.

Teacher's Note:
1. Incomplete octets and charge separation reduce the contribution of a structure.
2. The most stable canonical form contributes the most.

Answer: (b) y, x, y, x

Explanation:
1. (i) y has every atom with a full octet (iminium), so y is more stable.
2. (ii) Benzene is aromatic, so x is more stable than the open triene.
3. (iii) In y the cation is conjugated with the ring double bond (allylic), but in x it is not, so y is more stable.
4. (iv) The linear conjugated triene (x) is more stable than the cross-conjugated isomer (y).
5. Sequence: y, x, y, x.

Teacher's Note:
1. Look for complete octets, aromaticity and extended conjugation.
2. Cross conjugation delocalises less than linear conjugation.

Answer: (c) i > iii > iv > ii

Explanation:
1. NO2 is the strongest –M group.
2. Next is CN, then CHO, and COOH is the weakest of the four because the OH lone pair partly opposes the pull of C=O.
3. Order: NO2 > CN > CHO > COOH.

Teacher's Note:
1. Standard series: NO2 > CN > SO3H > CHO > COCH3 > COOH > COOR > CONH2.
2. –M groups have a multiple bond to an electronegative atom.

Answer: (a) i > ii > iii > iv

Explanation:
1. The negatively charged oxygen donates its lone pair most strongly.
2. Nitrogen is less electronegative than oxygen, so –NH2 donates better than –OH.
3. In –NHCOCH3 the nitrogen lone pair is shared with the carbonyl group, so it is the weakest donor.

Teacher's Note:
1. Lower electronegativity and a negative charge increase +M.
2. Acetylation of NH2 reduces its activating power.

Answer: (d) 1,4-dihydroisoquinolin-3(2H)-one

Explanation:
1. In (1) and (2) the nitrogen is bonded directly to the ring, so its lone pair is conjugated with it.
2. In (3) the C=O group is bonded directly to the ring and shows a –M effect.
3. In (4) both ring-attached atoms are CH2 groups, so neither the C=O nor the NH is conjugated with the benzene ring.

Teacher's Note:
1. The mesomeric effect needs a pi bond or lone pair directly attached to the ring.
2. A CH2 group in between breaks the conjugation.

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

Explanation:
1. NH2 is a strong +M donor, so aniline is richest.
2. CH3 donates by +I and hyperconjugation.
3. Cl withdraws by –I more than it donates by +M, so the ring is somewhat poorer.
4. NO2 withdraws strongly, so nitrobenzene is poorest.
5. Order: I > III > II > IV.

Teacher's Note:
1. Rank substituents from strong activating to strong deactivating.
2. Halogens are deactivating but still ortho and para directing.

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

Explanation:
1. Each nitro group withdraws electrons through –M and –I effects.
2. The more nitro groups, the lower the electron density.
3. Order: 0 NO2 > 1 NO2 > 2 NO2 > 3 NO2, i.e. I > II > III > IV.

Teacher's Note:
1. Electron density falls steadily as strongly withdrawing groups are added.
2. This is why picryl chloride reacts easily with nucleophiles.

Answer: (c) delocalisation of sigma electrons into an adjacent pi bond (or empty p orbital)

Explanation:
1. Hyperconjugation is the overlap of the sigma bond of C–H (alpha hydrogen) with an adjacent pi bond, empty p orbital or half-filled p orbital.
2. So sigma electrons are delocalised into the adjacent pi system.

Teacher's Note:
1. It is also called no-bond resonance or Baker–Nathan effect.
2. It needs at least one alpha C–H bond.

Answer: (b) \( \ce{C6H5-CH3} \)

Explanation:
1. Hyperconjugation needs an sp3 carbon with an alpha hydrogen next to a pi bond, empty p orbital or radical centre.
2. Toluene has CH3 attached to the ring, so it shows it.
3. Ethene has no alpha H, and in 3,3-dimethylbut-1-ene the carbon next to the double bond is quaternary with no hydrogen.
4. A carbanion has a lone pair, not an empty p orbital, so it does not show it.

Teacher's Note:
1. Count the alpha hydrogens.
2. Quaternary carbon next to the pi bond gives no hyperconjugation.

Answer: (b) propene

Explanation:
1. Number of hyperconjugation structures equals the number of alpha hydrogens.
2. Ethene has 0, propene has 3, but-1-ene has 2, and 3-methylbut-1-ene has 1.
3. Propene has the most.

Teacher's Note:
1. Alpha hydrogens are those on the carbon directly attached to the double bond.
2. Branching on that carbon reduces the count.

Answer: (d) \( \ce{(CH3)3C-CH2^{+}} \)

Explanation:
1. In neopentyl cation the positively charged carbon is attached to a quaternary carbon that has no hydrogen.
2. So there is no alpha C–H bond, and no hyperconjugation is possible.
3. The other three each have alpha hydrogens.

Teacher's Note:
1. Always inspect the carbon attached to the electron-deficient centre.
2. The methyl groups in neopentyl are beta, not alpha.

Answer: (d) hyperconjugation

Explanation:
1. The sigma C–H electrons of the methyl group overlap with the pi bond, giving the C–C bond a little double bond character.
2. This shortens the bond.

Teacher's Note:
1. Bond shortening in alkenes with methyl groups is evidence for hyperconjugation.
2. Heats of hydrogenation also show it.

Answer: (d) III > II > I

Explanation:
1. More alkyl groups on the double bond means more alpha hydrogens and more hyperconjugation.
2. 2-Butenes are more stable than 1-butene.
3. Trans is more stable than cis because of lower steric repulsion.
4. Order: trans-2-butene > cis-2-butene > 1-butene.

Teacher's Note:
1. Substituted alkenes are more stable.
2. For equal substitution, trans beats cis.

Answer: (c) I < III < II

Explanation:
1. Stability rises with the number of alpha hydrogens (hyperconjugation) and with the substitution of the double bond.
2. I has 4 alpha H on the ring, III has 4 ring plus 2 on CH2, and II has 4 ring plus 6 from two methyl groups.
3. Hence II > III > I, which is I < III < II.

Teacher's Note:
1. Tetrasubstituted alkene is more stable than trisubstituted, which is more stable than disubstituted.
2. Count alpha hydrogens to confirm.

Answer: (d) furan

Explanation:
1. Aromatic compounds are cyclic, planar, fully conjugated and have (4n + 2) pi electrons.
2. Cyclobutadiene has 4 pi electrons (antiaromatic), and cyclopentadienone has 4 (antiaromatic).
3. The cyclohexadienyl anion has an sp3 carbon, so it is not fully conjugated.
4. Furan has 6 pi electrons (two C=C plus one lone pair on O), so it is aromatic.

Teacher's Note:
1. Apply Huckel's rule, 4n + 2 pi electrons, after checking planarity and continuous conjugation.
2. Heteroatom lone pairs in a five-membered ring count in the sextet.

Answer: (a) cyclooctatetraene

Explanation:
1. Tropylium has 6 pi electrons, azulene has 10, and [18]annulene has 18 pi electrons, each satisfying 4n + 2 in a planar ring.
2. Cyclooctatetraene has 8 pi electrons and adopts a non-planar tub shape, so it is not aromatic.

Teacher's Note:
1. Both planarity and 4n + 2 are needed.
2. An 8-membered ring with 8 pi electrons avoids antiaromaticity by puckering.