JEE Chemistry Organic Chemistry Some Basic Principles and Techniques MCQs Set 33

Answer: (c) –NO2, –CN, –SO3H

Explanation:
1. –NO2, –CN and –SO3H withdraw electrons by resonance and have no lone pair to donate through the pi system.
2. Groups such as –OR, –Cl, –NH2, –OH, –NR2 and –SR have lone pairs and show +M.

Teacher's Note:
1. –M groups have a multiple bond or positive centre next to the ring.
2. Groups with lone pairs on the attached atom show +M.

Answer: (c) –Cl

Explanation:
1. Chlorine is electronegative and withdraws electrons through the sigma bond (–I effect).
2. It also has lone pairs that can be donated into a pi system (+M effect).
3. –CHO and –NO2 show –M and –I, and –CH3 shows +I and hyperconjugation.

Teacher's Note:
1. Halogens are the classic +M, –I groups.
2. The –I effect dominates, which is why halobenzenes are deactivated but ortho-para directing.

Answer: (d) (c), (d) and (e)

Explanation:
1. The inductive effect is a permanent polarisation of sigma bonds, so (c) is true and (a) is false.
2. The resonance (mesomeric) effect is the delocalisation of pi electrons or lone pairs, so (d) is true and (b) is false.
3. Hyperconjugation is the delocalisation of the sigma electrons of an alpha C–H bond into an adjacent pi system or empty p orbital, so (e) is true.

Teacher's Note:
1. Inductive: sigma. Resonance: pi and lone pairs. Hyperconjugation: sigma C–H.
2. Only (c), (d) and (e) are true.

Answer: (c) 2,3-dimethylbut-2-ene

Explanation:
1. Hyperconjugation transfers sigma electron density from alpha C–H bonds into the C=C pi system, which gives the double bond some single bond character and lengthens it.
2. The more alpha hydrogens, the greater the effect: 2,3-dimethylbut-2-ene has 12 alpha hydrogens, but-2-ene 6, propene 3 and ethene none.

Teacher's Note:
1. Count the alpha C–H bonds on the carbons next to the double bond.
2. More hyperconjugation means a longer double bond.

Answer: (b) The C1–C2 bond is shorter than the C2–C3 bond.

Explanation:
1. In naphthalene the positions 1, 4, 5 and 8 are the alpha positions and 2, 3, 6, 7 the beta positions.
2. The resonance structures put a double bond between C1 and C2 more often (in two of the three structures) than between C2 and C3, so the C1–C2 bond has more double bond character.
3. The measured lengths are 136 pm for C1–C2 and 142 pm for C2–C3, so the C1–C2 bond is the shorter.

Teacher's Note:
1. Bond order is the average over the contributing structures.
2. Naphthalene is aromatic but its bonds are not all equal, unlike benzene.

Answer: (a) they contain a relatively high percentage of carbon

Explanation:
1. Aromatic hydrocarbons have a high carbon to hydrogen ratio, for example benzene is C6H6.
2. The available oxygen does not burn all this carbon completely, and the unburnt carbon appears as soot.

Teacher's Note:
1. The same happens with other unsaturated compounds such as ethyne.
2. A sooty flame is often used as a simple test for aromatic character.

Answer: (d) \( \ce{CH3^-} \) and \( \ce{CH3CH2^+} \) ions

Explanation:
1. Heterolysis of a C–C bond gives one cation and one anion (not radicals, and not two cations).
2. The more stable carbocation and the less destabilised carbanion are formed: the ethyl cation is more stable than the methyl cation, and the methyl carbanion is more stable than the ethyl carbanion (the +I effect of CH3 destabilises it).
3. So the products are CH3– and CH3CH2+.

Teacher's Note:
1. Homolysis gives radicals; heterolysis gives ions of opposite charge.
2. Choose the pair with the more stable ions.

Answer: (c) free radical

Explanation:
1. A free radical has an unpaired electron and so is paramagnetic.
2. A carbocation (six electrons, all paired) and a carbanion (a lone pair) have no unpaired electrons and are diamagnetic.

Teacher's Note:
1. Paramagnetism needs unpaired electrons.
2. Carbenes in the triplet state are also paramagnetic.

Answer: (c) \( \ce{C6H5-CH=CH-C^+(CH3)2} \)

Explanation:
1. The cation in (c) is tertiary, and it is also conjugated with both the C=C bond and the phenyl ring, so the positive charge is delocalised over a large system.
2. (a) is a primary allylic cation with one extra alkyl group, and (b) is the simple allyl cation.
3. So (c) is the most stable.

Teacher's Note:
1. Both substitution and extended conjugation increase carbocation stability.
2. The tertiary allylic cation with a phenyl group beats simpler cations.

Answer: (b) \( \ce{HC#C^-} \) and \( \ce{H2C=CH^-} \)

Explanation:
1. (a) CCl3– is more stable than CF3– because of the polarisable chlorine (and d orbital effects); the first is not more stable.
2. (b) The sp carbanion has 50% s-character and is more stable than the sp2 vinyl anion: the first is more stable.
3. (c) The phenyl anion has its lone pair in an sp2 orbital, so it is more stable than the sp3 cyclohexyl anion; the first is not.
4. (d) The cyclopentadienyl anion is aromatic (six pi electrons) and much more stable than the cycloheptatrienyl anion (eight pi electrons, antiaromatic-like).

Teacher's Note:
1. The more s-character the carbanion orbital has, the more stable it is.
2. Aromatic anions are specially stable.

Answer: (c) acetamidine

Explanation:
1. On protonation acetamidine gives an amidinium ion in which the positive charge is delocalised equally over both nitrogen atoms.
2. Acetamide and thioacetamide are delocalised amides with very weak basicity, and the enamine is a weak base because its lone pair conjugates with the double bond.

Teacher's Note:
1. Resonance stabilisation of the conjugate acid makes amidines strong bases.
2. Amides are essentially neutral.

Answer: (b) hexane-2,4-dione

Explanation:
1. A 1,3-dicarbonyl compound has a very acidic CH2 group between two C=O groups.
2. Its enol is stabilised by conjugation with the other C=O and by a strong intramolecular hydrogen bond.
3. Hexane-2,4-dione is the 1,3-diketone, so its enol content is much higher than that of the other compounds.

Teacher's Note:
1. Acetylacetone type diketones are largely enolised.
2. 1,4- and 1,2-diketones do not gain this stabilisation.

Answer: (a) –CH3

Explanation:
1. Electron withdrawing groups (–NO2, –CN, –CHO) stabilise the phenoxide ion and raise acidity.
2. The methyl group is electron donating (+I and hyperconjugation), which destabilises the phenoxide ion and lowers the acidity.

Teacher's Note:
1. Donating groups reduce the acidity of phenols and acids.
2. Withdrawing groups increase it.

Answer: (d) ethanol

Explanation:
1. pKa values: formic acid 3.75, cyanoacetic acid 2.5, phenol 10, ethanol 16.
2. The maximum pKa corresponds to the weakest acid, which is ethanol.

Teacher's Note:
1. Larger pKa means a weaker acid.
2. Alcohols are weaker acids than phenols.

Answer: (c) o-cresol

Explanation:
1. pKa values: p-nitrophenol 7.15, p-chlorophenol 9.4, phenol 9.98, o-cresol 10.3.
2. The methyl group of o-cresol donates electrons, so it is the weakest acid.

Teacher's Note:
1. Donating groups weaken phenols.
2. –Cl increases acidity slightly through its –I effect.

Answer: (b) \( \ce{^+H3N-(CH2)2-COOH} \)

Explanation:
1. The –NH3+ group is a very strong –I group and withdraws electrons from the carboxylate even through two carbon atoms.
2. 3-Fluoropropanoic acid has pKa about 4.0, succinic acid pKa1 about 4.2, and butanoic acid 4.8, while the ammonium acid has pKa about 3.6.
3. So the ammonium substituted acid is the strongest.

Teacher's Note:
1. A charged –I group has a greater effect than F.
2. Alkyl groups reduce the acidity.

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

Explanation:
1. The first dissociation constants (pKa1) are oxalic 1.25, malonic 2.85, succinic 4.2 and adipic 4.4.
2. The nearer the second –COOH group (with its –I effect), the stronger the acid.
3. So Ka increases: adipic < succinic < malonic < oxalic.

Teacher's Note:
1. As the CH2 chain lengthens, the second COOH influence fades.
2. Larger Ka means smaller pKa.

Answer: (c) III < II < I

Explanation:
1. The para isomer is the most acidic because the –CHO group withdraws electrons by –M and –I effects (pKa 7.6).
2. In the ortho isomer the same effects operate but intramolecular hydrogen bonding between OH and CHO reduces the acidity (pKa 8.4).
3. The meta isomer has only the –I effect (pKa 9.0).
4. Increasing pKa: III < II < I.

Teacher's Note:
1. –M acts only from ortho and para positions.
2. Hydrogen bonding in the ortho isomer weakens the acid.

Answer: (d) III > I > II

Explanation:
1. Basicity is highest in propylamine and falls as the carbon next to the CH2NH2 group changes from sp3 to sp2 to sp, because of the increased –I effect of more s-rich carbons.
2. Basicity: II > I > III. The pKb value is highest for the weakest base, so the order of decreasing pKb is III > I > II.

Teacher's Note:
1. Larger pKb means a weaker base.
2. More s-character on the neighbouring carbon gives a stronger –I effect.

Answer: (c) aniline > m-nitroaniline > p-nitroaniline > o-nitroaniline

Explanation:
1. The conjugate acid pKa values are aniline 4.6, m-nitroaniline 2.5, p-nitroaniline 1.0 and o-nitroaniline –0.3.
2. The meta isomer shows only –I; the para isomer shows –I and –M; the ortho isomer shows –I and –M at the shortest distance and also has intramolecular hydrogen bonding that locks the lone pair.

Teacher's Note:
1. The ortho isomer is the weakest base of all nitroanilines.
2. –M works only from ortho and para positions.

Answer: (c) p < m < o

Explanation:
1. Conjugate acid pKa values: p-toluidine 5.1, m-toluidine 4.7, o-toluidine 4.4, so the basicity is p > m > o.
2. The ortho isomer is weakest because of steric hindrance to protonation and solvation (ortho effect).
3. The pKb is smallest for the strongest base, so pKb increases in the order p < m < o.

Teacher's Note:
1. pKb falls as the base gets stronger.
2. A methyl group at the para position gives the largest +I and hyperconjugation effect.

Answer: (b) pKa = 3.0

Explanation:
1. Convert to pKb using pKa + pKb = 14 for a conjugate pair.
2. Options: 9.40, 2.88, 11.0 and 3.32.
3. The weakest base has the highest pKb, 11.0, that is the one with pKa = 3.0.

Teacher's Note:
1. The pKa given here is that of the conjugate acid.
2. A high pKb means a weak base.

Answer: (c) phenol + NaHCO3 giving sodium phenoxide + H2O + CO2

Explanation:
1. Bicarbonate reacts only with acids stronger than carbonic acid (pKa 6.4).
2. Benzoic acid and benzenesulfonic acid are stronger, so (a) and (b) occur.
3. Phenol (pKa 10) is weaker than carbonic acid, so (c) does not occur.
4. p-Nitrobenzoic acid is stronger than benzoic acid, so it displaces it from its salt: (d) occurs.

Teacher's Note:
1. A stronger acid displaces a weaker acid from its salt.
2. NaHCO3 test: carboxylic acids give CO2, phenols do not.

Answer: (b) p-methylaniline > p-chloroaniline > p-aminoacetophenone

Explanation:
1. (a) Sodium hydroxide is a strong base and ionises completely, so methylamine is weaker.
2. (b) –CH3 raises basicity, –Cl lowers it and the –M group of acetyl lowers it more: true.
3. (c) Pyridine (sp2 N) is a much weaker base than dimethylamine.
4. (d) The nitro group (–M, –I) lowers basicity more than the –CHO group, so p-nitroaniline is weaker than p-aminobenzaldehyde; the order is wrong.

Teacher's Note:
1. Compare substituent effects on the para position.
2. Strong bases such as NaOH are fully dissociated.