JEE Chemistry Organic Chemistry Some Basic Principles and Techniques MCQs Set 34

Answer: (a) HCl > HF > HCN > \( \ce{H2O} \) > \( \ce{NH3} \)

Explanation:
1. HCl is a strong acid (pKa about –7), HF is weak (3.2) and HCN is weaker (9.2).
2. In the last equilibrium water acts as the acid and gives OH– to NH3, so water (pKa 15.7) is a stronger acid than NH3 (pKa about 35).
3. Order: HCl > HF > HCN > H2O > NH3.

Teacher's Note:
1. Use pKa values to line up the acids.
2. Ammonia is hardly acidic at all.

Answer: (c) the anion has two equivalent resonating structures

Explanation:
1. The negative charge is shared between the two oxygen atoms through two equivalent contributing structures, H–C(=O)–O– and H–C(–O–)=O.
2. This makes both bonds identical with bond order 1.5.

Teacher's Note:
1. Equivalent resonance structures give equal bonds.
2. Hybridisation alone does not equalise a C=O and C–O bond.

Answer: (d) benzylamine

Explanation:
1. In aniline, N-methylaniline and o-toluidine the nitrogen lone pair is delocalised into the benzene ring.
2. In benzylamine a CH2 group isolates the nitrogen from the ring, so the lone pair is localised like that of an alkylamine (pKb about 4.7).

Teacher's Note:
1. A lone pair conjugated with an aryl ring is less available.
2. Benzylamine behaves like an aliphatic amine.

Answer: (c) \( \ce{HCOOH} \)

Explanation:
1. Alkyl groups donate electrons and reduce the acidity of carboxylic acids.
2. Formic acid has no alkyl group, so it is the strongest (pKa 3.75), against 4.76 for acetic acid and about 4.87 for propanoic acid.

Teacher's Note:
1. Lowest pKa means strongest acid.
2. More alkyl groups mean a weaker acid.

Answer: (c) (CH3)2CH· < (CH3)3C· < (C6H5)2CH· < (C6H5)3C·

Explanation:
1. Alkyl radicals are stabilised by hyperconjugation and the +I effect, so tertiary > secondary.
2. Aryl groups stabilise a radical by resonance, which is stronger; each phenyl ring adds delocalisation.
3. Increasing stability: isopropyl < tert-butyl < diphenylmethyl < triphenylmethyl.

Teacher's Note:
1. Resonance with phenyl rings is more effective than hyperconjugation.
2. The trityl radical is so stable that it exists in solution in equilibrium with its dimer.

Answer: (c) d < a < b < c

Explanation:
1. Isobutyric acid (d) has two methyl groups with +I effect, so it is the weakest, then acetic acid (a).
2. Methoxyacetic acid (b) has the –I effect of the alkoxy group (pKa 3.57), so it is stronger than acetic acid.
3. Trifluoroacetic acid (c) has three fluorine atoms and is the strongest (pKa 0.2).
4. Order: d < a < b < c.

Teacher's Note:
1. More –I groups near COOH mean a stronger acid.
2. CF3COOH is nearly as strong as a mineral acid.

Answer: (b) \( \ce{CCl3^-} \) > \( \ce{C6H5CH2^-} \) > \( \ce{(CH3)2CH^-} \) > \( \ce{(CH3)3C^-} \)

Explanation:
1. Three chlorine atoms stabilise CCl3– by the –I effect and by dispersal of the charge into the vacant d orbitals of chlorine.
2. The benzyl anion is delocalised into the ring.
3. Of the alkyl carbanions the secondary one is more stable than the tertiary, which is destabilised by three +I methyl groups.
4. Order: CCl3– > C6H5CH2– > (CH3)2CH– > (CH3)3C–.

Teacher's Note:
1. Electron donating groups destabilise carbanions.
2. Delocalisation and –I groups stabilise them.

Answer: (c) \( \ce{CH3CH2CH(Cl)CO2H} \)

Explanation:
1. The alpha chloro acid (2-chlorobutanoic acid, pKa 2.86) has the chlorine atom closest to COOH, so its –I effect is largest.
2. The gamma chloro acid (4-chlorobutanoic) is only slightly stronger than butanoic acid; formic acid is 3.75 and acetic acid 4.76.

Teacher's Note:
1. Distance reduces the –I effect.
2. Compare pKa values.

Answer: (a) D > C > A > B

Explanation:
1. pKa values are p-nitrophenol 7.15, m-nitrophenol 8.4, phenol 9.98 and p-cresol 10.26.
2. Order: D > C > A > B.

Teacher's Note:
1. Para nitro has –M and –I; meta nitro has only –I.
2. The methyl group lowers the acidity.

Answer: (d) III > I > II

Explanation:
1. The benzyl cation is stabilised by delocalisation of the charge over the whole benzene ring (several resonance structures).
2. The allyl cation is stabilised by two resonance structures.
3. The primary propyl cation has only hyperconjugation and the +I effect.
4. Order: III > I > II.

Teacher's Note:
1. More extensive delocalisation means more stability.
2. A primary alkyl cation is relatively unstable.

Answer: (d) x > y > z

Explanation:
1. The carboxyl group is the most acidic (pKa about 2.2).
2. The alpha ammonium group is more acidic (pKa about 9.0) than the epsilon one (about 10.5) because it is close to the electron withdrawing COOH group.
3. Order: x > y > z.

Teacher's Note:
1. –I effect from the COOH group raises the acidity of the nearer –NH3+.
2. Distance weakens the effect.

Answer: (c) sigma-p

Explanation:
1. Hyperconjugation is the delocalisation of the electrons of an alpha C–H sigma bond into an adjacent empty p orbital or pi orbital.
2. So the overlap is of the sigma bond orbital with a p (or pi) orbital.

Teacher's Note:
1. It is also called no-bond resonance.
2. pi-pi overlap is ordinary conjugation.

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

Explanation:
1. Carboxylic acids are much stronger than phenols: benzoic acid pKa 4.2 and p-toluic acid 4.4.
2. Among the phenols, p-chlorophenol (9.4) is stronger than phenol (10.0) because of the –I effect of Cl.
3. Order: III > IV > II > I.

Teacher's Note:
1. The methyl group makes p-toluic acid slightly weaker than benzoic acid.
2. Carboxylate ions are better stabilised than phenoxide ions.

Answer: (a) o-hydroxybenzoic acid

Explanation:
1. pKa values: o-hydroxybenzoic (salicylic) acid 2.98, p-hydroxybenzoic acid 4.58, p-toluic acid 4.4 and p-nitrophenol 7.15.
2. The salicylate ion is stabilised by an intramolecular hydrogen bond, so salicylic acid is the strongest.

Teacher's Note:
1. Carboxylic acids are stronger than phenols.
2. Ortho hydroxy gives extra stabilisation by hydrogen bonding.

Answer: (d) aspirin

Explanation:
1. Aspirin (acetylsalicylic acid) has a COOH group on the benzene ring.
2. Picric acid is 2,4,6-trinitrophenol (phenolic OH), barbituric acid is a cyclic imide and ascorbic acid is a lactone enediol; none has a carboxyl group.

Teacher's Note:
1. Check each structure for the C(=O)OH unit.
2. Do not confuse 'acid' in a name with a carboxyl group.

Answer: (c) p-toluidine > aniline > p-nitroaniline

Explanation:
1. The methyl group at the para position donates electrons and increases basicity (pKb 8.9).
2. Aniline has pKb 9.4.
3. The nitro group withdraws electrons by –M and –I effects and lowers basicity (pKb about 13).

Teacher's Note:
1. Electron donors raise basicity and acceptors lower it.
2. Para nitro is the strongest acceptor here.

Answer: (b) \( \ce{(CH3)2CHCOOH} \) < \( \ce{CH3CH(Br)CH2COOH} \) < \( \ce{CH3CH2CH(Br)COOH} \)

Explanation:
1. 2-Methylpropanoic acid has no electron withdrawing group and is the weakest.
2. In 3-bromobutanoic acid the bromine is at the beta position, with a moderate –I effect.
3. In 2-bromobutanoic acid the bromine is at the alpha position, giving the greatest –I effect and the strongest acid.

Teacher's Note:
1. The –I effect weakens with distance.
2. Alkyl groups reduce acidity.

Answer: (a) 4-methoxybenzoic acid < benzoic acid < 3,4-dinitrobenzoic acid

Explanation:
1. The para methoxy group donates electrons (+M), destabilising the carboxylate and weakening the acid (pKa 4.5).
2. Two nitro groups strongly withdraw electrons and stabilise the carboxylate (pKa about 2.8).
3. Increasing strength: 4-methoxy < benzoic (4.2) < 3,4-dinitro.

Teacher's Note:
1. Donors weaken benzoic acids; acceptors strengthen them.
2. Two nitro groups have a large effect.

Answer: (d) The phenoxide ion is stabilised by resonance, so phenol is acidic enough (pKa about 10) to react with NaOH, whereas hexoxide is not stabilised (pKa about 16) and hexanol does not react.

Explanation:
1. The negative charge of the phenoxide ion is delocalised into the ring.
2. Alkoxide ions have no such stabilisation and are destabilised by the +I effect of the alkyl group.
3. Hence phenol (pKa 10) reacts with NaOH but hexanol (pKa about 16) is a weaker acid than water and does not.

Teacher's Note:
1. A strong enough acid must be able to protonate OH– quantitatively.
2. Resonance stabilisation of the conjugate base increases acidity.

Answer: (c) the chlorine atom exerts a –I effect, withdrawing electrons and stabilising the chloroacetate ion

Explanation:
1. Chlorine is more electronegative than carbon and withdraws electrons through the sigma bond (–I effect).
2. This disperses the negative charge on the carboxylate ion, stabilising it, so ionisation is easier and the pKa is lower (2.86 against 4.76).

Teacher's Note:
1. A stable anion means a stronger acid.
2. Acetic acid has an electron donating methyl group instead.

Answer: (b) methylamine, being more basic than water, accepts a proton from water and gives hydroxide ions, which precipitate Fe(OH)3

Explanation:
1. \( \ce{CH3NH2 + H2O <=> CH3NH3^+ + OH^-} \) produces hydroxide ions.
2. These react with Fe3+ ions: \( \ce{Fe^3+ + 3OH^- -> Fe(OH)3 v} \) giving a reddish brown precipitate.

Teacher's Note:
1. Aliphatic amines are stronger bases than ammonia.
2. The same happens with ammonia solution.

Answer: (a) The alkyl group has a +I effect and increases the electron density on nitrogen, making the lone pair more available and stabilising the alkylammonium ion

Explanation:
1. Alkyl groups push electrons towards nitrogen (+I effect).
2. This increases the availability of the lone pair for protonation and also stabilises the positive charge of the alkylammonium ion.

Teacher's Note:
1. Think of the effect of electron donors on a lone pair.
2. Hydrogen is replaced by a donor group.