JEE Chemistry Organic Chemistry Some Basic Principles and Techniques MCQs Set 44

Answer: (d) It is the polarisation of \(\sigma\) bond electrons.

Explanation:
1. The I effect is only a partial shift of \(\sigma\) electrons, so no full charge is created and no electron leaves its orbital.
2. Its magnitude decreases rapidly with distance.
3. So only the statement about polarisation of \(\sigma\) bonds is correct.

Teacher's Note:
1. The electrons never leave their original atomic orbital in the inductive effect.
2. Partial charges appear, not full charges.

Answer: (d) -COO^-

Explanation:
1. A group that pushes electron density into the chain shows +I effect.
2. The negatively charged carboxylate group is electron releasing.
3. Br, COOH and OCH\(_3\) all withdraw electrons by -I effect.

Teacher's Note:
1. Groups bearing a negative charge, and alkyl groups, are +I groups.
2. Hydrogen is taken as the reference with zero inductive effect.

Answer: (b) (CH\(_3\))\(_3\)C-

Explanation:
1. The +I effect increases with the number of alkyl branches on the carbon attached to the chain.
2. The tert-butyl group has three methyl groups, so its +I effect is the largest.

Teacher's Note:
1. Order of +I: tertiary > secondary > primary > methyl.
2. More branching means more electron release.

Answer: (b) -NO\(_2\) > -CN > -F > -NH\(_2\)

Explanation:
1. The standard order of -I strength is NO\(_2\) > CN > F > NH\(_2\).
2. Nitro has a formal positive charge on nitrogen, which makes it the strongest.
3. Among the others, fluorine is more electronegative than nitrogen of NH\(_2\).

Teacher's Note:
1. Learn the -I series: -NR\(_3\)^+ > -NO\(_2\) > -CN > -COOH > -F > -Cl > -Br > -I > -OH > -OR.
2. Charged and multiply bonded groups are strong -I groups.

Answer: (c) -N^+(CH\(_3\))\(_3\)

Explanation:
1. A group carrying a full positive charge withdraws electrons most strongly.
2. The trimethylammonium group is positively charged on nitrogen, so it has the strongest -I effect.

Teacher's Note:
1. Positively charged groups are the strongest -I groups.
2. Fluorine is the strongest among the neutral atoms listed.

Answer: (b) \(\pi\) electrons (and lone pairs) through a conjugated system

Explanation:
1. In resonance the \(\pi\) electrons and lone pairs spread over the conjugated atoms, while the positions of nuclei stay fixed.
2. Delocalisation of \(\sigma\) electrons towards a p orbital is hyperconjugation, not resonance.

Teacher's Note:
1. Resonating structures differ only in the arrangement of electrons.
2. Resonance needs p orbitals that overlap side by side.

Answer: (c) Their energies are always equal.

Explanation:
1. Resonating structures differ in energy; the more stable ones contribute more to the hybrid.
2. Net charge and the numbers of paired and unpaired electrons must remain the same.

Teacher's Note:
1. The resonance hybrid is more stable than any single contributing structure.
2. Contributing structures are imaginary, only the hybrid is real.

Answer: (a) an equal contribution to the resonance hybrid

Explanation:
1. Contributing structures differ in stability, so their contributions to the hybrid are usually unequal.
2. Atom positions, paired electron count and net charge cannot change between contributing structures.

Teacher's Note:
1. Equal contribution happens only when the structures are identical in energy, as in benzene.
2. Stable structures contribute more.

Answer: (c) CH\(_3\)-CH=CH\(_2\)

Explanation:
1. Propene has one double bond with only sp\(^3\) carbon attached, so there is nothing to conjugate with (only hyperconjugation occurs).
2. Acrolein and butadiene have alternate double and single bonds, and vinyl chloride has a lone pair adjacent to the double bond.

Teacher's Note:
1. A conjugated system needs a \(\pi\) bond, lone pair, charge or odd electron attached to a \(\pi\) bond through a single bond.
2. Lone pair next to a double bond counts as conjugation.

Answer: (d) -NO\(_2\)

Explanation:
1. The nitro group withdraws electrons through both \(\sigma\) bond and \(\pi\) system, so it shows -I and -M.
2. OH, NH\(_2\) and F donate by +M effect.

Teacher's Note:
1. -M groups usually contain a multiple bond to an electronegative atom.
2. Nitro is the classic deactivating group.

Answer: (c) -O^-

Explanation:
1. The oxide ion has a lone pair to donate (+M) and a negative charge that pushes electrons through \(\sigma\) bonds (+I).
2. OH and NHCH\(_3\) show +M but -I; tert-butyl shows only +I.

Teacher's Note:
1. Negatively charged atoms show +I as well.
2. Neutral O and N donors show -I with +M.

Answer: (b) -OH

Explanation:
1. A hydroxyl group has lone pairs on oxygen to donate (+M), while oxygen withdraws through the \(\sigma\) bond (-I).
2. The other three contain a carbonyl or cyano group and show -M.

Teacher's Note:
1. +M groups have a lone pair on the atom attached to the ring or chain.
2. -M groups have a \(\pi\) bond to an electronegative atom.

Answer: (b) a \(\sigma\) orbital and a p (or pi) orbital

Explanation:
1. In hyperconjugation the C-H \(\sigma\) bond electrons of an alpha carbon overlap with an adjacent empty p orbital, a pi* orbital or a half-filled p orbital.
2. Hence it is called \(\sigma\)-p (no-bond) resonance.

Teacher's Note:
1. Hyperconjugation is also called no bond resonance.
2. It needs at least one alpha hydrogen.

Answer: (b) (CH\(_3\))\(_3\)C-CH\(_2\)^+ (neopentyl cation)

Explanation:
1. The positive carbon of the neopentyl cation is attached to a carbon that bears three methyl groups and no hydrogen, so there is no alpha C-H bond.
2. The other species each have alpha hydrogens next to a radical, cation centre or double bond.

Teacher's Note:
1. Count the hydrogens on carbons directly attached to the unsaturated or electron-deficient carbon.
2. A quaternary alpha carbon gives no hyperconjugation.

Answer: (b) CH\(_3\)-CH=CH\(_2\)

Explanation:
1. Number of structures equals the number of alpha hydrogens: ethene 0, propene 3, but-1-ene 2, 3-methylbut-1-ene 1.
2. Propene has the maximum.

Teacher's Note:
1. Only hydrogens on the carbon attached to the double bond count.
2. Alpha hydrogens are the C-H bonds next to the \(\pi\) bond.

Answer: (b) CH\(_3\)-CH=CH-CO-CH\(_3\)

Explanation:
1. Pent-3-en-2-one has a C=O group (-I and -M), a C=C in conjugation with it, and alpha hydrogens on the methyl groups.
2. CH\(_3\)Cl lacks a \(\pi\) system, propene lacks a conjugating group, butadiene lacks alpha hydrogens.

Teacher's Note:
1. Check each effect separately for each molecule.
2. Hyperconjugation needs alpha hydrogens, mesomeric effect needs conjugation.

Answer: (a) Naphthalene

Explanation:
1. Naphthalene has 10 \(\pi\) electrons in two fused planar rings and contains only carbon and hydrogen.
2. Cyclobutadiene is antiaromatic, and furan and pyridine are heterocycles, not hydrocarbons.

Teacher's Note:
1. Hydrocarbon means only carbon and hydrogen.
2. Check Huckel's rule 4n+2 for aromaticity.

Answer: (c) a relatively high percentage of carbon

Explanation:
1. Aromatic compounds are carbon rich.
2. The air supply is not enough for complete combustion of so much carbon, so unburnt carbon particles produce a sooty flame.

Teacher's Note:
1. The carbon to hydrogen ratio of benzene is 1:1.
2. Saturated alkanes burn with a clean flame.

Answer: (b) (CH\(_3\))\(_3\)C^-

Explanation:
1. Alkyl groups release electrons and intensify the negative charge, so the tertiary carbanion is the least stable.
2. The ethynyl carbanion has 50% s character, and the trityl anion is stabilised by resonance with three rings.
3. Methyl anion has no electron-releasing group.

Teacher's Note:
1. Carbanion stability: s character and electron withdrawing groups help.
2. Alkyl groups destabilise carbanions.

Answer: (b) HC\(\equiv\)C^- and H\(_2\)C=CH^-

Explanation:
1. Stability of a carbanion increases with s character: sp > sp\(^2\) > sp\(^3\).
2. So ethynyl anion is more stable than vinyl anion.
3. In the other pairs the first anion is the less stable (more alkyl substitution, or sp\(^3\) against sp\(^2\)).

Teacher's Note:
1. Alkyl substitution destabilises a carbanion.
2. Higher s character holds the lone pair closer to the nucleus.

Answer: (d) benzyl > allyl > 2\(^\circ\) > 1\(^\circ\)

Explanation:
1. The benzyl radical is delocalised over the ring (more resonating structures) and is more stable than allyl.
2. Both resonance stabilised radicals are more stable than simple alkyl radicals, and 2\(^\circ\) is more stable than 1\(^\circ\) because of hyperconjugation.

Teacher's Note:
1. Resonance stabilisation outweighs hyperconjugation.
2. More resonating structures mean more stability.

Answer: (b) (CH\(_3\))\(_3\)C\(\bullet\) > (CH\(_3\))\(_2\)CH\(\bullet\) > CH\(_3\)CH\(_2\)\(\bullet\) > CH\(_3\)\(\bullet\)

Explanation:
1. Stability follows the number of alpha hydrogens available for hyperconjugation and the +I effect of alkyl groups.
2. Tertiary has 9, secondary 6, primary 3 and methyl none, so the order is tertiary > secondary > primary > methyl.

Teacher's Note:
1. Alkyl radicals are electron deficient, alkyl groups stabilise them.
2. Same order as for carbocations.

Answer: (d) ^+CH\(_2\)-CHO

Explanation:
1. The carbonyl group next to the positive carbon withdraws electrons by -I and cannot donate to the cation, so the charge is strongly destabilised.
2. Ethyl cation is stabilised by hyperconjugation, the methoxymethyl cation by +M of oxygen, and in the chloroethyl cation Cl is one carbon further away.

Teacher's Note:
1. -I and -M groups attached directly destabilise a carbocation.
2. +M groups stabilise it.

Answer: (b) p\(\pi\)-d\(\pi\) overlap

Explanation:
1. The positive carbon overlaps with the adjacent \(\pi\) bond (p\(\pi\)-p\(\pi\) resonance) and with the C-H bonds of the methyl group (hyperconjugation).
2. No d orbitals are involved because there are only C and H atoms.

Teacher's Note:
1. d orbitals matter only for elements like S, P or Si.
2. Allylic cations have resonance as well as hyperconjugation.

Answer: (b) ^-CH\(_2\)NO\(_2\) > ^-CH\(_2\)CN > ^-CH\(_2\)Cl > ^-CH\(_2\)CH\(_3\)

Explanation:
1. Nitro stabilises the carbanion by both -I and strong -M; cyano by -I and -M (weaker); chlorine only by -I.
2. The ethyl group destabilises through +I.
3. So NO\(_2\) > CN > Cl > CH\(_3\).

Teacher's Note:
1. -M groups delocalise the negative charge, so they are the best stabilisers.
2. The -M strength is NO\(_2\) > CN.

Answer: (c) Nitrobenzene

Explanation:
1. The nitro group has strong -I and -M effects, which gives a large dipole moment of about 4.2 D.
2. Chlorobenzene is about 1.7 D, aniline about 1.5 D and toluene about 0.4 D.

Teacher's Note:
1. Dipole moment depends on the magnitude and direction of electron displacement.
2. -I and -M act in the same direction in the nitro group.

Answer: (c) p-nitrobenzyl anion

Explanation:
1. The nitro group at the para position withdraws electrons by -I and -M and delocalises the negative charge onto its oxygen atoms.
2. Methoxy is a donor, which destabilises the carbanion.

Teacher's Note:
1. -M groups at ortho and para positions stabilise benzylic carbanions.
2. Donor groups stabilise carbocations, not carbanions.