JEE Chemistry Organic Chemistry Some Basic Principles And Techniques MCQs Set 53

Answer: (c) I > II > III

Explanation:
1. In (I) both C-Cl bonds are on one carbon, 120 degrees apart, so they add to 2cos30 = 1.73 times one C-Cl moment, and the two methyl groups on the other carbon push in the same direction.
2. In (II) the C-Cl bonds are on adjacent carbons on the same side, so the resultant is 2cos60 = 1.0 times one C-Cl moment.
3. In (III) the trans arrangement makes the bond moments cancel, so the dipole moment is about zero.
4. Order: I > II > III.

Teacher's Note:
1. Add bond moments as vectors, not as plain numbers.
2. A symmetrical trans alkene with identical substituents has zero dipole moment.

Answer: (a) Trans-1-chloroprop-1-ene

Explanation:
1. 1,4-Dichlorobenzene is symmetrical, so the two C-Cl moments cancel.
2. E-1,2-Dibromoethene has a centre of symmetry, so its moments cancel.
3. Propadiene is non-polar because its two CH2 ends lie in perpendicular planes and cancel.
4. In trans-1-chloroprop-1-ene the substituents (Cl and CH3) are different, so a net moment remains.

Teacher's Note:
1. Look for a centre or plane of symmetry that cancels the vectors.
2. Different groups on the two alkene carbons usually leave a net moment.

Answer: (d) \( \ce{CH3Cl} \)

Explanation:
1. Dipole moment = charge x bond length.
2. The C-F bond is the most polar but also the shortest, while the C-I bond is long but weakly polar.
3. Observed values: CH3Cl 1.87 D, CH3F 1.85 D, CH3Br 1.81 D, CH3I 1.62 D.
4. So CH3Cl has the highest value.

Teacher's Note:
1. The dipole moments of methyl halides do not follow electronegativity alone.
2. Remember the order CH3Cl > CH3F > CH3Br > CH3I.

Answer: (b) Geometrical isomers always have the same dipole moment.

Explanation:
1. Dipole moment is charge multiplied by distance and has direction, so it is a vector.
2. Enantiomers are identical in all scalar physical properties, including dipole moment.
3. Cis and trans isomers generally differ (for example cis-1,2-dichloroethene is polar, the trans isomer has zero dipole moment).
4. Hence the false statement is that geometrical isomers have the same dipole moment.

Teacher's Note:
1. Enantiomers differ only in optical rotation.
2. Diastereomers (including cis-trans isomers) differ in physical properties.

Answer: (b) toluene < phenol < nitrobenzene < 4-nitrotoluene

Explanation:
1. Toluene has a small moment (about 0.4 D) because CH3 is only weakly electron releasing.
2. Phenol is about 1.5 D.
3. Nitrobenzene is about 4.2 D because of the strong -M and -I effects of NO2.
4. In 4-nitrotoluene the electron-releasing CH3 acts in the same direction as the NO2 pull, so the moment (about 4.4 D) is higher than that of nitrobenzene.

Teacher's Note:
1. In para-disubstituted benzenes the moments add when one group donates and the other withdraws.
2. They cancel only when both groups act in opposite directions.

Answer: (d) Azulene

Explanation:
1. Naphthalene, phenanthrene and anthracene are non-polar hydrocarbons with zero or negligible moment.
2. Azulene (a fused five- and seven-membered ring) has a dipolar structure with a tropylium-like seven ring and a cyclopentadienyl-like five ring, so charge separation gives it a dipole moment of about 0.8 D.

Teacher's Note:
1. Both rings of azulene attain a 6 pi-electron aromatic sextet in the dipolar form.
2. That charge separation is the source of its dipole moment.

Answer: (a) Sodium 4-nitrophenoxide

Explanation:
1. The negative charge on oxygen in the phenoxide is delocalised into the ring and onto the para NO2 group, giving a large separation of charge along the molecular axis.
2. In the benzoate the charge sits on the COO group and is not conjugated into the ring, so the effect is smaller.
3. The phenoxide with the para nitro group therefore has the largest dipole moment.

Teacher's Note:
1. A strong donor and strong acceptor para to each other maximise the moment.
2. The carboxylate group is a poor pi donor to the ring.

Answer: (c) vacuum distillation

Explanation:
1. Glycerol decomposes at its normal boiling point (290 degrees C).
2. Reducing the pressure lowers the boiling point so that it distils without decomposition.
3. Hence vacuum (reduced pressure) distillation is used.

Teacher's Note:
1. Use reduced pressure for liquids that decompose below their boiling point.
2. Steam distillation suits liquids immiscible with water.

Answer: (d) a separating funnel

Explanation:
1. Immiscible liquids form two layers because of their different densities.
2. The denser layer is run off through the stopcock of a separating funnel.

Teacher's Note:
1. Miscible liquids with close boiling points need fractional distillation.
2. Sublimation is for solids that vaporise directly.

Answer: (b) changes directly into the vapour form on heating

Explanation:
1. In sublimation the solid passes directly into vapour without melting, and the vapour condenses back to a solid on cooling.
2. Camphor, naphthalene and ammonium chloride are purified this way.

Teacher's Note:
1. Sublimation is a physical change; no new substance forms.
2. Used to separate a sublimable solid from a non-volatile impurity.

Answer: (a) phenol < 1,2- < 1,3- < 1,4-dihydroxybenzene

Explanation:
1. Phenol (182 degrees C) has only one OH for intermolecular hydrogen bonding.
2. Catechol (245 degrees C) has intramolecular hydrogen bonding that lowers the intermolecular association.
3. Resorcinol (277 degrees C) and hydroquinone (287 degrees C) are strongly associated by intermolecular hydrogen bonding, the para isomer most of all.
4. Order: phenol < 1,2 < 1,3 < 1,4.

Teacher's Note:
1. More intermolecular hydrogen bonding means a higher boiling point.
2. Intramolecular hydrogen bonding lowers it.

Answer: (c) w > z > x > y

Explanation:
1. Cyclopentanol (about 140 degrees C) has strong O-H hydrogen bonding.
2. Piperidine (about 106 degrees C) has weaker N-H hydrogen bonding.
3. Tetrahydropyran (about 88 degrees C) is an ether with no hydrogen bonding.
4. N-Methylpyrrolidine (about 80 degrees C) is a tertiary amine with no N-H, hence the lowest.
5. Order: w > z > x > y.

Teacher's Note:
1. Hydrogen bonding: O-H > N-H > none.
2. Among non-hydrogen-bonded compounds, compare dipole moment and molecular mass.

Answer: (c) iodobenzene > bromobenzene > chlorobenzene > fluorobenzene

Explanation:
1. For halobenzenes, boiling point rises with molecular mass and polarisability (van der Waals forces).
2. Observed: C6H5I 188, C6H5Br 156, C6H5Cl 132, C6H5F 85 degrees C.

Teacher's Note:
1. Larger halogen, larger molecule, stronger London forces.
2. The same trend holds for alkyl halides.

Answer: (a) \( \ce{CH3I > CH3Br > CH3Cl > CH3F} \)

Explanation:
1. The boiling points are CH3I 42.4, CH3Br 3.6, CH3Cl -24 and CH3F -78 degrees C.
2. With increasing size of the halogen, the polarisability and van der Waals forces increase, so the boiling point rises.

Teacher's Note:
1. Do not confuse the boiling point trend (follows size) with the dipole moment trend.
2. Compare molecular mass first for similar compounds.

Answer: (d) Increase in branching of the carbon chain

Explanation:
1. A longer chain, stronger hydrogen bonding and larger molecular mass all increase intermolecular attraction.
2. Branching makes the molecule more spherical, reducing the surface area of contact and so the van der Waals forces; the boiling point falls.

Teacher's Note:
1. Straight chain isomers boil higher than branched ones.
2. n-Pentane 36 degrees C, neopentane 9.5 degrees C.

Answer: (b) n-propyl > isopropyl > ethyl > methyl formate

Explanation:
1. These are isomer-like esters with increasing carbon number: methyl 32, ethyl 54, n-propyl 81 degrees C.
2. Isopropyl formate (68 degrees C) is branched, so it boils lower than the straight-chain n-propyl formate but above ethyl formate.

Teacher's Note:
1. Boiling point rises with molecular mass in a homologous series.
2. Branching lowers it relative to the straight chain isomer.

Answer: (b) p-Dibromobenzene

Explanation:
1. Para isomers are symmetrical and pack best in the crystal lattice.
2. Melting points: p-dibromobenzene 87, o- 7, m- -7, bromobenzene -31 degrees C.

Teacher's Note:
1. Melting point depends on symmetry of packing.
2. Boiling point depends mainly on polarity and size.

Answer: (d) 1,4-Dihydroxybenzene

Explanation:
1. Hydroquinone (1,4-) is symmetrical and has extensive intermolecular hydrogen bonding, melting at about 172 degrees C.
2. Catechol melts at about 105, resorcinol at about 110 and p-cresol at about 35 degrees C.

Teacher's Note:
1. Symmetry plus hydrogen bonding gives the highest melting point.
2. Compare para, meta and ortho isomers.

Answer: (a) 3 > 1 > 2

Explanation:
1. 2-Methylpropan-2-ol is highly symmetrical and nearly spherical, so it packs well and melts at about 25 degrees C.
2. 2-Methylpropan-1-ol melts at about -108 and butan-2-ol at about -115 degrees C.
3. Order: 3 > 1 > 2.

Teacher's Note:
1. Branched, spherical molecules have higher melting points but lower boiling points.
2. Compare symmetry for isomers.

Answer: (c) Tetracene (linear four-ring fused hydrocarbon)

Explanation:
1. For fused aromatic hydrocarbons, melting point rises with molecular mass and with planar, symmetrical packing.
2. Naphthalene 80, phenanthrene 100, anthracene 216 and tetracene about 357 degrees C (with decomposition).

Teacher's Note:
1. Linear fusion packs better than angular fusion.
2. More rings increase the van der Waals forces.

Answer: (d) \( \ce{C6H5CH2CH2OH} \)

Explanation:
1. Solubility in water depends on hydrogen bonding with water against the size of the hydrophobic part.
2. Ethylene glycol and propanol are miscible; 2-methylpropan-1-ol is fairly soluble.
3. The phenyl group in 2-phenylethanol is a large hydrophobic part, so it is least soluble.

Teacher's Note:
1. Large non-polar parts reduce solubility.
2. More OH groups increase solubility.

Answer: (b) Alcohol

Explanation:
1. Alcohols form strong hydrogen bonds with water.
2. Ethers can only accept hydrogen bonds, and alkanes and alkenes cannot hydrogen bond at all.

Teacher's Note:
1. Hydrogen bond donor plus acceptor gives best solubility.
2. Alcohol > ether > alkene > alkane.

Answer: (a) Malonic acid

Explanation:
1. Malonic acid is a small dicarboxylic acid with two COOH groups close together, highly soluble (about 760 g per litre).
2. Succinic acid is much less soluble (about 83 g per litre) because of better crystal packing.
3. Salicylic acid and phthalic acid carry a large benzene ring and are sparingly soluble.

Teacher's Note:
1. Small polar molecules dissolve well.
2. Aromatic ring lowers solubility.