JEE Chemistry Organic Chemistry Some Basic Principles and Techniques MCQs Set 64

Answer: (c) 3-hydroxy-4-methylpentanoic acid

Explanation:
1. The principal chain contains the COOH group and has 5 carbons: pentanoic acid.
2. Numbering from COOH gives OH at C3 and CH₃ at C4; substituents are named alphabetically: 3-hydroxy-4-methylpentanoic acid.

Teacher's Note:
1. Number so that the principal functional group gets the lowest locant, then list prefixes alphabetically.

Answer: (b) CH₃COCH₂COCH₃

Explanation:
1. Acetylacetone is a 1,3-diketone: its enol form is stabilised by intramolecular hydrogen bonding and conjugation.
2. In the ester and amide the carbonyl next to the CH₂ is already in resonance with OEt or NH₂, so they enolise less.

Teacher's Note:
1. 1,3-Diketones have a higher enol content than β-keto esters or amides.

Answer: (c) 2-chloro-1-methyl-4-nitrobenzene

Explanation:
1. Chloro, methyl and nitro are all prefixes, so numbering gives the lowest set of locants.
2. The set (1, 2, 4) is lowest, and the names are listed alphabetically: chloro, methyl, nitro.

Teacher's Note:
1. Benzene with only prefix groups: lowest locant set first, alphabetical order second.

Answer: (b) differential adsorption of the substances on the solid phase

Explanation:
1. The mixture is placed on a column of solid adsorbent such as silica gel or alumina.
2. Components are adsorbed to different extents and so move at different rates down the column.

Teacher's Note:
1. Adsorption (surface) is the principle, not absorption.

Answer: (b) Higher Rf value means higher adsorption

Explanation:
1. Rf = distance moved by the compound/distance moved by the solvent, which is always less than 1.
2. A more strongly adsorbed compound travels less, so it has a lower Rf; the statement that higher Rf means higher adsorption is incorrect.

Teacher's Note:
1. Higher Rf means weaker adsorption on the stationary phase.

Answer: (b) (2) < (3) < (1) < (4)

Explanation:
1. For similar attacking atoms, nucleophilicity follows basicity of the conjugate base.
2. Acid strength: CH₃SO₃H > CH₃CO₂H > H₂O, so basicity is H₂O < CH₃SO₃⁻ < CH₃CO₂⁻ < OH⁻.

Teacher's Note:
1. A weaker acid has a stronger conjugate base and so a better nucleophile.

Answer: (b) phosphorus

Explanation:
1. Phosphorus is oxidised by Na₂O₂ to sodium phosphate, which gives phosphoric acid on boiling with HNO₃.
2. Ammonium molybdate gives the yellow precipitate (NH₄)₃PO₄·12MoO₃.

Teacher's Note:
1. The yellow ammonium phosphomolybdate precipitate is the test for phosphorus.

Answer: (d) 3,5-dimethyl-4-propylhept-1-en-6-yne

Explanation:
1. The longest chain containing both the double and triple bonds has 7 carbons.
2. Number so that the multiple bonds as a set get the lowest locants, 1-ene and 6-yne; the substituents are methyl at 3 and 5 and propyl at 4: 3,5-dimethyl-4-propylhept-1-en-6-yne.

Teacher's Note:
1. With both a double and a triple bond, number the chain to give the lowest locants to the pair; ene gets the lower number only on a tie.

Answer: (d) CH(CN)₃

Explanation:
1. A stronger acid gives a more stable conjugate base.
2. The carbanion of CH(CN)₃ is stabilised by the strong −I and −R effects of three CN groups.

Teacher's Note:
1. Strongly electron-withdrawing groups that delocalise the negative charge increase acidity.

Answer: (a) DCM and water will stay as lower and upper layers respectively in the separating funnel

Explanation:
1. DCM (density 1.33 g cm⁻³) is heavier than water (1 g cm⁻³) and is immiscible with it.
2. So DCM forms the lower layer and water the upper layer.

Teacher's Note:
1. The denser immiscible solvent forms the lower layer.

Answer: (c) γ-hydrogen

Explanation:
1. Light promotes free radical substitution.
2. The γ-hydrogens are allylic, and the allyl radical formed is resonance stabilised.

Teacher's Note:
1. Allylic and benzylic hydrogens are replaced most easily in radical bromination.

Answer: (d) 4-bromo-3-methylpent-2-ene

Explanation:
1. The longest chain with the double bond has 5 carbons; the double bond gets the lowest locant, so it starts at the CH₃–CH= end: pent-2-ene.
2. Methyl is at C3 and bromine at C4.

Teacher's Note:
1. Number from the end that gives the double bond the lowest number; the double bond has priority over a halogen.

Answer: (d) C₆H₈N₂

Explanation:
1. Carbon: 6 mol CO₂ gives 6 C.
2. Hydrogen: 4 mol H₂O gives 8 H. Nitrogen: 1 mol N₂ gives 2 N.
3. So the formula is C₆H₈N₂.

Teacher's Note:
1. Count atoms: each mole of H₂O has 2 H and each mole of N₂ has 2 N.

Answer: (a) A-Q, B-R, C-S

Explanation:
1. Sugar from water is obtained by crystallisation.
2. Aniline is steam volatile and insoluble in water, so steam distillation separates it.
3. Toluene is not steam volatile in this sense and is insoluble in water, so differential extraction is used.

Teacher's Note:
1. Choose the method from volatility and solubility of the components.

Answer: (c) HC≡CH > CH₃–C≡CH > CH₂=CH₂

Explanation:
1. sp carbon has the highest s-character and electronegativity, so alkynes are more acidic than the sp² alkene.
2. The electron-donating CH₃ group decreases the acidity of propyne relative to ethyne.

Teacher's Note:
1. Acidity falls with decreasing s-character and with electron-donating groups.

Answer: (b) aniline

Explanation:
1. Kjeldahl's method converts nitrogen to ammonium sulphate.
2. It does not work for nitro, azo or diazo compounds or for nitrogen in a ring, so only aniline is suitable.

Teacher's Note:
1. Kjeldahl's method fails for N in a ring and for nitro, azo and diazo groups.

Answer: (d) 4,4-dimethylcyclohexa-2,5-dienone

Explanation:
1. Benzene, furan (4n + 2 = 6 π electrons) and pyridine are aromatic.
2. The dienone is not aromatic because the sp³ C(CH₃)₂ carbon breaks the cyclic conjugation.

Teacher's Note:
1. Aromatic compounds follow Huckel's rule and have extra resonance stabilisation.

Answer: (c) distillation under reduced pressure

Explanation:
1. Glycerol has a high boiling point (about 290°C) and may decompose on heating.
2. Reducing the pressure lowers the boiling point so it can be distilled safely.

Teacher's Note:
1. Use reduced pressure for liquids that decompose near their boiling points.

Answer: (a) 24

Explanation:
1. Mass of Br = (80/188) × 141 mg = 60 mg.
2. Percentage = 60/250 × 100 = 24.

Teacher's Note:
1. Percent of Br = (80/188) × (mass of AgBr/mass of compound) × 100.