JEE Chemistry Organic Chemistry Some Basic Principles and Techniques MCQs Set 17

Answer: (b) aqueous sodium bicarbonate

Explanation:
1. A carboxylic acid is stronger than carbonic acid and liberates \(\ce{CO2}\) with \(\ce{NaHCO3}\) (brisk effervescence), but phenol is weaker than carbonic acid and gives no effervescence.
2. Sodium metal and sodium hydroxide react with both, and ammoniacal silver nitrate reacts with neither.

Teacher's Note:
1. Sodium bicarbonate test separates carboxylic acids from phenols.
2. Both react with Na and with NaOH.

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

Explanation:
1. The better the leaving group, the weaker the base it is, so the stronger its conjugate acid.
2. Acidity order: HBr > acetic acid > phenol > ethanol.
3. So the leaving group ability is Br- > acetate > phenoxide > ethoxide.

Teacher's Note:
1. Good leaving groups are the conjugate bases of strong acids.
2. Weaker base means better leaving group.

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

Explanation:
1. The acidic strength of a cation is opposite to the basic strength of its amine.
2. p-Nitroaniline is the weakest base, so its cation is the strongest acid. Aniline is weaker than methylamine, and guanidine is a very strong base (the cation is resonance stabilised), so the guanidinium ion is the weakest acid.
3. Order: IV > II > I > III.

Teacher's Note:
1. A weaker base has a stronger conjugate acid.
2. Guanidine is a very strong organic base.

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

Explanation:
1. Reactivity depends on ring electron density.
2. In 3-methoxyphenol both OH and OMe donate electrons into the same positions, so it is more reactive than phenol.
3. Chlorine withdraws electrons by -I and nitro by -M and -I, so these phenols are less reactive, nitro the least. Order: I > IV > II > III.

Teacher's Note:
1. Two donor groups reinforce each other when they are meta to each other.
2. Nitro is the strongest deactivator.

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

Explanation:
1. Cyclohexylamine is aliphatic and is the strongest base.
2. Aniline has its lone pair delocalised in the ring and is much weaker.
3. Benzamide has the lone pair delocalised onto the carbonyl and is weaker still, and in phthalimide the lone pair is shared with two carbonyl groups, so it is the weakest. Order: IV > II > I > III.

Teacher's Note:
1. Delocalisation of the lone pair lowers basicity.
2. Imides are weaker than amides.

Answer: (a) II > I > III

Explanation:
1. Bond length depends on bond order. Each C - C bond in benzene is a resonance hybrid of single and double bonds (140 pm), the longest of the three.
2. In cyclohexa-1,3-diene the double bonds are conjugated, which lengthens them slightly (about 135 pm).
3. In cyclohexene the double bond is isolated (about 134 pm). Order: II > I > III.

Teacher's Note:
1. Conjugation lengthens a double bond.
2. Benzene has bond length between single and double.

Answer: (b) 4-nitrobenzoic acid

Explanation:
1. The nitro group withdraws electrons by both -M and -I and stabilises the benzoate ion most.
2. Chlorine withdraws only by -I (partly cancelled by +M), while the methyl and hydroxyl groups donate electrons and weaken the acid.
3. So 4-nitrobenzoic acid is strongest.

Teacher's Note:
1. The nitro group is the strongest common electron-withdrawing group on a ring.
2. Para donors weaken benzoic acid.

Answer: (c) a structure with a nitrogen bearing two N = O double bonds (\(\ce{O=N+=O}\)) and an sp3 ring carbon carrying \(\ce{O-}\)

Explanation:
1. Nitrogen has only four orbitals in its valence shell, so it cannot form five bonds.
2. The \(\ce{O=N+=O}\) form in (c) would give nitrogen two double bonds plus a bond to carbon (five bonds), which is impossible.
3. The others are valid structures obtained by moving pi electrons from the phenoxide oxygen into the ring and nitro group.

Teacher's Note:
1. Second-period atoms obey the octet rule.
2. Check the valence of N in resonance structures.

Answer: (c) \(\ce{CH3-CO-CH2^-}\)

Explanation:
1. The enolate (c) places the negative charge on oxygen, the most electronegative atom, in a resonance structure.
2. The imine analogue (d) places it on nitrogen, which is less electronegative.
3. The allyl carbanion is delocalised only over carbon atoms, and the ethyl carbanion is not delocalised.

Teacher's Note:
1. Charge on a more electronegative atom is more stable.
2. Enolates are far more stable than allyl carbanions.

Answer: (c) inductive effect

Explanation:
1. The inductive effect is the permanent displacement of sigma electrons along a chain of atoms because of a difference in electronegativity.
2. The mesomeric effect involves pi electrons or lone pairs, the electromeric effect is temporary and involves pi bonds, and hyperconjugation involves sigma C - H electrons delocalised into a pi system.

Teacher's Note:
1. Inductive effect is the sigma bond effect.
2. It decreases with distance.

Answer: (c) \(\ce{SO3}\), \(\ce{H2O}\), \(\ce{CH3-}\)

Explanation:
1. A nucleophile has a lone pair or is an anion that can donate an electron pair.
2. \(\ce{SO3}\) is electron deficient at sulphur and is an electrophile, so the set in (c) is not correct.
3. All the species in the other sets have a lone pair to donate.

Teacher's Note:
1. Neutral molecules with lone pairs (water, amines) are nucleophiles.
2. Electron-deficient species such as SO3 are electrophiles.

Answer: (a) \(\ce{CH3-}\) and \(\ce{CH3+}\)

Explanation:
1. In heterolysis the bonding pair goes to the more electronegative atom.
2. Carbon is more electronegative than copper, so \(\ce{CH3-Cu}\) gives a carbanion \(\ce{CH3-}\) and \(\ce{Cu+}\).
3. Chlorine is more electronegative than carbon, so \(\ce{CH3-Cl}\) gives \(\ce{CH3+}\) and \(\ce{Cl-}\).

Teacher's Note:
1. The more electronegative atom takes the electron pair.
2. Organometallic compounds behave as carbanion sources.

Answer: (a) a = b

Explanation:
1. The nitro group is a resonance hybrid of two equivalent structures, \(\ce{-N+(=O)O-}\) and \(\ce{-N+(O-)=O}\).
2. Each N - O bond therefore has the same bond order of 1.5 and the same length.

Teacher's Note:
1. Equivalent resonance structures give equal bond lengths.
2. The same is true for the carboxylate ion.

Answer: (d) \(\ce{CH3-Cl}\)

Explanation:
1. The electromeric effect is the temporary shift of pi electrons to one atom when an attacking reagent approaches.
2. It needs a multiple bond.
3. Acetonitrile, ethanal and ethene all have pi bonds, but chloromethane has only sigma bonds.

Teacher's Note:
1. No pi bond means no electromeric effect.
2. It is temporary and appears only in the presence of a reagent.

Answer: (c) formation of acetaldehyde cyanohydrin

Explanation:
1. In the +E effect the pi electrons move toward the atom to which the attacking electrophile bonds.
2. Hydrohalogenation, hydroboration and hydration of alkenes and alkynes start with attack of an electrophile (H+ or BH3) on the pi bond.
3. In cyanohydrin formation a nucleophile attacks the carbonyl carbon and the pi electrons move to oxygen, which is the -E effect.

Teacher's Note:
1. Electrophile attack gives +E; nucleophile attack gives -E.
2. Cyanide adds to the carbonyl carbon.

Answer: (c) 1,2,3,4-tetrahydroisoquinoline

Explanation:
1. In tetrahydroisoquinoline the nitrogen is a secondary aliphatic amine, and its lone pair is not part of the benzene ring, so it is available for protonation.
2. Pyridine has an sp2 nitrogen (less basic), and in tetrahydroquinoline and aniline the nitrogen is attached directly to the benzene ring, so its lone pair is delocalised.

Teacher's Note:
1. Aliphatic amines are more basic than aromatic amines or pyridine.
2. Check whether the N lone pair is conjugated.

Answer: (b) 3-methylbutan-2-one

Explanation:
1. Keto-enol tautomerism needs an alpha hydrogen on a carbon next to the carbonyl group.
2. 3-Methylbutan-2-one has alpha hydrogens on both sides.
3. Di-tert-butyl ketone and benzophenone have no alpha hydrogen, and cyclopentadienone has only sp2 carbons next to C = O.

Teacher's Note:
1. No alpha hydrogen means no keto-enol tautomerism.
2. Tautomers are interconverted by H shifts.

Answer: (a) benzyl > allyl > 2 degree > 1 degree

Explanation:
1. Radical stability increases with delocalisation of the unpaired electron.
2. The benzyl radical is delocalised over the ring (several resonance structures), the allyl radical over three carbons only.
3. Alkyl radicals gain only hyperconjugation, so 2 degree is more stable than 1 degree.

Teacher's Note:
1. More resonance structures means a more stable radical.
2. Benzyl is more stable than allyl.

Answer: (a) the cyclopentadiene-type ring with a saturated \(\ce{NH2+}\) (no lone pair on N)

Explanation:
1. An antiaromatic ring is cyclic, planar, fully conjugated and has \(4n\) pi electrons.
2. Cyclopentadienone (4 pi electrons), 1,4-dihydropyrazine (8 pi electrons) and oxirene (4 pi electrons) all qualify.
3. In (a) the nitrogen is sp3 with four bonds (\(\ce{NH2+}\)), so the ring is not fully conjugated and is non-aromatic.

Teacher's Note:
1. Antiaromatic needs a continuous conjugated ring of 4n pi electrons.
2. An sp3 atom breaks the conjugation.

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

Explanation:
1. Piperidine is a saturated amine whose lone pair is in an sp3 orbital and is fully available.
2. The lone pair of pyridine is in an sp2 orbital (more s-character, less basic). Aniline has its lone pair delocalised into the ring, and phenol (oxygen) is the weakest base.
3. Order: I > II > III > IV.

Teacher's Note:
1. sp3 amine > sp2 nitrogen > aromatic amine.
2. Oxygen is less basic than nitrogen.

Answer: (b) I, II and III

Explanation:
1. Groups that withdraw electrons by -I or -M deactivate the ring and direct incoming groups to the meta position.
2. The \(\ce{-CCl3}\), \(\ce{-NO2}\) and \(\ce{-NH3+}\) groups are all electron withdrawing, so the nitronium ion goes meta in I, II and III.
3. The phenoxide ion is a strong donor and directs ortho and para.

Teacher's Note:
1. Electron-withdrawing groups are meta directing (halogens excepted).
2. Donor groups direct ortho and para.

Answer: (a) \(\ce{CH3COCH2COOC2H5}\) and \(\ce{CH3C(OH)=CHCOOC2H5}\)

Explanation:
1. Resonance structures differ only in the position of electrons, not atoms.
2. The keto and enol forms in (a) differ in the position of a hydrogen atom, so they are tautomers and not resonance forms.
3. The other pairs have the same atoms in the same positions.

Teacher's Note:
1. Tautomers differ in the position of an atom (H).
2. Resonance forms differ only in electron positions.

Answer: (c) \(\ce{CH3CH=CH-CO-CH3}\)

Explanation:
1. Hyperconjugation needs an alpha C - H bond next to a pi system, which is present in the methyl groups of (b) and (c).
2. The mesomeric effect needs a conjugated pi system, as in the C = C - C = O unit of (c).
3. The inductive effect is present because of the polar C = O group.
4. Only (c) has all three effects.

Teacher's Note:
1. Look for methyl next to a pi bond (hyperconjugation) and a conjugated unit (mesomeric).
2. Inductive effect is always present when atoms differ in electronegativity.