JEE Chemistry Organic Chemistry Some Basic Principles and Techniques MCQs Set 58

Answer: (c) (5-ethyl-5-methyl lactone)

Explanation:
1. A ketone is more reactive than an ester towards a Grignard reagent, so the ketone carbonyl is attacked first and gives a tertiary alcohol: CH₃C(OH)(C₂H₅)CH₂CH₂COOCH₃.
2. The OH group is four atoms from the ester carbonyl, so it attacks the ester intramolecularly and expels CH₃OH, closing a five-membered lactone.
3. Product: 5-ethyl-5-methyldihydrofuran-2(3H)-one.

Teacher's Note:
1. Five- and six-membered rings form easily by intramolecular esterification of hydroxy esters.
2. Ketone is more electrophilic than ester, which gives the chemoselectivity.

Answer: (d) (i), (ii), (iii) and (iv)

Explanation:
1. CH₃MgBr gives methane with any compound that has an acidic hydrogen.
2. Terminal alkyne C-H (i), carboxylic O-H (ii), the H of HCN (iii) and the active methylene hydrogen of the 1,3-diketone (iv) are all acidic enough.
3. So all four reactions give methane.

Teacher's Note:
1. Make a list of acidic hydrogens before predicting Grignard products.
2. 1,3-Dicarbonyl compounds are acidic at the CH₂ between the carbonyl groups.

Answer: (d) PhCHO + CH₃CH₂CH(CH₃)MgBr, then H₃O⁺

Explanation:
1. The target is a secondary alcohol whose carbinol carbon carries H, a phenyl group and a sec-butyl group.
2. Benzaldehyde plus sec-butylmagnesium bromide gives exactly this. The other aldehyde routes give different carbon skeletons, and acetophenone would give a tertiary alcohol.

Teacher's Note:
1. Secondary alcohol = aldehyde + RMgX.
2. Break the C-OH carbon's bonds to find the two groups.

Answer: (a) 3 mol of (CH₃)₃COH (tert-butyl alcohol)

Explanation:
1. The first CH₃MgBr displaces one t-BuO^-, giving methyl tert-butyl carbonate-type ester CH₃COO-tBu; the second displaces the other t-BuO^- and gives acetone; the third adds to acetone.
2. Overall: 2 mol t-BuOH (leaving groups) + 1 mol (CH₃)₃COMgBr, which on hydrolysis is t-BuOH.
3. Total: 3 mol of tert-butyl alcohol.

Teacher's Note:
1. Three equivalents of RMgX convert a carbonate into R₃C-OH.
2. Here R = CH₃ so the new alcohol is again tert-butyl alcohol.

Answer: (a) benzene

Explanation:
1. The Grignard reagent takes the acidic proton of methanol: C₆H₅MgBr + CH₃OH gives C₆H₆ + CH₃OMgBr.
2. The phenyl group does not substitute the OH group.

Teacher's Note:
1. Protic compounds decompose Grignard reagents.
2. The hydrocarbon has the same skeleton as the R group.

Answer: (c) Cyclohexane

Explanation:
1. Acetone, DMSO and acetic acid all contain polar groups (C=O, S=O, O-H) and have large dipole moments.
2. Cyclohexane is a hydrocarbon with no polar bond and zero dipole moment, so it is non-polar.

Teacher's Note:
1. Polar aprotic solvents such as acetone and DMSO are polar even though they have no acidic hydrogen.
2. Hydrocarbons and benzene are the standard non-polar solvents.

Answer: (d) an optically active mixture of two compounds

Explanation:
1. Reduction of the C=O creates a new stereocentre at C-2, giving two diastereomeric tetrols: erythritol and threitol.
2. With the original centre (C-3) retained, one is meso-erythritol (inactive) and the other is optically active D-threitol.
3. So the product is a mixture of two compounds and the mixture is optically active, due to threitol.

Teacher's Note:
1. Hydride attacks both faces, giving both configurations at the new centre.
2. A meso compound is inactive, but it does not cancel the optical activity of the other component.

Answer: (c) CH₃CHO > CH₃COCH₃ > CH₃COOCH₃ > CH₃CONH₂

Explanation:
1. Aldehydes are more reactive than ketones (less +I effect, less steric hindrance).
2. In esters and amides the lone pair on O or N is donated into C=O by resonance, which reduces the electrophilicity of carbon; nitrogen is the better donor, so amides are the least reactive.

Teacher's Note:
1. Resonance donation lowers reactivity of the carbonyl carbon.
2. Order: aldehyde > ketone > ester > amide.

Answer: (b) All four: (i), (ii), (iii) and (iv)

Explanation:
1. The H-X bond weakens down the group because of poor orbital overlap, so (i) is true.
2. HF is the weakest acid, so F^- is the strongest base: (ii) is true.
3. Nucleophilicity in water rises down the group (iii), and electronegativity falls down the group (iv).
4. All four are correct.

Teacher's Note:
1. Bond strength, acid strength and basicity are all related for the halides.
2. Nucleophilicity trends depend on the solvent.

Answer: (a) tert-butoxide, ^-OC(CH₃)₃

Explanation:
1. The three bulky methyl groups around the oxygen block its approach to a carbon atom, so it cannot act as a nucleophile at carbon.
2. It can still remove a small proton, so it is a strong base.

Teacher's Note:
1. Steric hindrance lowers nucleophilicity much more than basicity.
2. Bulky bases favour elimination over substitution.

Answer: (c) NH₃

Explanation:
1. The ethyl cation is positive, AlCl₃ has an incomplete octet on Al, and SO₃ has an electron-poor sulphur.
2. NH₃ has a lone pair to donate and is a nucleophile.

Teacher's Note:
1. Electrophiles are positively charged or have vacant orbitals.
2. Neutral species with a lone pair are nucleophiles.

Answer: (b) 1-bromo-2-methylpropane, (CH₃)₂CHCH₂Br

Explanation:
1. Y is a Grignard reagent R-MgBr, and D₂O replaces MgBr by D in the carbon bearing the metal.
2. The deuterium is on the CH₂ carbon of an isobutane skeleton, so R = (CH₃)₂CHCH₂-.
3. X is therefore isobutyl bromide.

Teacher's Note:
1. R-MgBr + D₂O gives R-D, labelling the carbon that carried magnesium.
2. Match the carbon skeleton and the position of D.

Answer: (c) Quinolizidine (a fully substituted tertiary bicyclic amine with no N-H or O-H)

Explanation:
1. Acetyl chloride acylates compounds having an O-H or N-H: water gives acetic acid, benzyl alcohol gives the ester, and the secondary amine gives an N-acetyl amide.
2. A tertiary amine has no hydrogen to lose, so no stable acylation product forms (only a reversible acylammonium salt that returns to the reactants).

Teacher's Note:
1. Acylation needs a nucleophile with a replaceable hydrogen.
2. Tertiary amines only act as catalysts or bases in such reactions.

Answer: (a) H₂

Explanation:
1. Methanol, water and ammonia each have lone pairs on O or N and can donate them.
2. Molecular hydrogen has only a sigma bond pair and no lone pair, so it is not a nucleophile.

Teacher's Note:
1. Look for a lone pair or a pi bond.
2. Dihydrogen needs a catalyst before it reacts.

Answer: (c) Acetic anhydride

Explanation:
1. P₂O₅ is a strong dehydrating agent.
2. Two molecules of CH₃COOH lose one molecule of water to give (CH₃CO)₂O.

Teacher's Note:
1. Acetyl chloride needs SOCl₂, PCl₅ or PCl₃.
2. Anhydrides are made by dehydration of two acid molecules.

Answer: (c) CH₃COO^- > C₂H₅O^-

Explanation:
1. A weaker base is a better leaving group.
2. H₂O is a weaker base than OH^-, SH^- is weaker than OH^-, and I^- is weaker than Cl^-, so the first three orders are reversed.
3. The acetate ion is stabilised by resonance and is a much weaker base than ethoxide, so acetate leaves more easily.

Teacher's Note:
1. Good leaving group means a stable anion or neutral molecule.
2. Resonance delocalisation of the negative charge makes carboxylate a better leaving group than alkoxide.

Answer: (a) CH₃CH₂CONH₂

Explanation:
1. The amide ion NH₂^- is a strong nucleophile and attacks the ester carbonyl carbon.
2. The methoxide ion (a weaker base than NH₂^-) is expelled, so the ester is converted into the amide CH₃CH₂CONH₂.

Teacher's Note:
1. Substitution goes in the direction of the weaker base leaving.
2. Esters give amides with ammonia or amide ion.

Answer: (b) N-methylcyclohexanecarboxamide, C₆H₁₁-CO-NHCH₃

Explanation:
1. The amine nucleophile attacks the ester carbonyl and methanol is expelled (nucleophilic acyl substitution).
2. The acyl group keeps its carbon skeleton, so the amide carries the cyclohexyl-C=O group on N-CH₃.

Teacher's Note:
1. Esters with amines give N-substituted amides.
2. The acyl part comes from the acid, the N-part from the amine.

Answer: (a) indene

Explanation:
1. A Grignard reagent releases methane only from a sufficiently acidic C-H, N-H or O-H bond.
2. Indene has a CH₂ that is both allylic and benzylic; loss of H+ gives the aromatic indenyl anion (pKa about 20), so CH₃MgI deprotonates it.
3. Indane, cycloheptatriene (pKa about 36) and biphenyl have no C-H acidic enough.

Teacher's Note:
1. Grignard reagents are strong bases and react with acidic hydrogens to give alkanes.
2. Aromatic stabilisation of the conjugate base makes indene unusually acidic for a hydrocarbon.

Answer: (c) All of benzophenone, methyl benzoate and diethyl carbonate

Explanation:
1. Benzophenone adds PhMgBr once to give Ph₃C-OMgBr.
2. Methyl benzoate adds twice (first to give benzophenone, which then reacts) and diethyl carbonate adds three times.
3. All three finally give Ph₃COH after hydrolysis.

Teacher's Note:
1. The number of equivalents of Grignard reagent needed is 1 for a diaryl ketone, 2 for an ester and 3 for a carbonate.
2. Only formates stop at a secondary alcohol.

Answer: (c) (ii) and (iii) only

Explanation:
1. (i) is wrong: the carbanion CH₃^- is a stronger nucleophile than CH₃O^- across the period.
2. (ii) is correct: an anion is a better nucleophile than its conjugate acid. (iii) is correct: nitrogen is less electronegative than oxygen, so NH₃ donates more readily.
3. (iv) is wrong: phosphorus is larger and more polarisable, so the phosphine is the better nucleophile.

Teacher's Note:
1. Compare donor atoms: same period, same group, or charged vs neutral.
2. Larger and more polarisable donor atoms are better nucleophiles.

Answer: (b) Benzamide

Explanation:
1. Ammonia is a nucleophile and attacks the acyl carbon (nucleophilic acyl substitution).
2. Chloride leaves from the tetrahedral intermediate: PhCOCl + NH₃ gives PhCONH₂ + HCl (a second NH₃ absorbs the HCl).

Teacher's Note:
1. Acid chlorides are converted to amides by ammonia and amines.
2. The phenyl-carbonyl skeleton is retained, so acetamide is impossible.

Answer: (c) (CH₃)₂CHOH and (CH₃)₂CHOH

Explanation:
1. The formate loses its isopropoxide group and gives propan-2-ol (one product).
2. The aldehyde formed reacts with a second CH₃MgBr to give (CH₃)₂CHOH.
3. Both P and Q are propan-2-ol.

Teacher's Note:
1. A formate with excess RMgX gives R₂CHOH plus the alcohol of the ester.
2. Here R = CH₃, so both alcohols are isopropyl alcohol.

Answer: (b) sp² to sp³

Explanation:
1. The carbonyl carbon is trigonal planar (sp²).
2. After the nucleophile bonds to it the carbon has four sigma bonds and is tetrahedral (sp³).

Teacher's Note:
1. Count sigma bonds and lone pairs to find hybridisation.
2. Addition raises the number of groups on the carbon.

Answer: (d) all of these: HCN is not a strong nucleophile, it is undissociated when pure, and it cannot add to C=O on its own

Explanation:
1. Pure HCN is a weak acid and is hardly ionised, so very little CN^- is present.
2. The neutral HCN molecule is a poor nucleophile and does not add across C=O without base catalysis.
3. A trace of base makes CN^- and the reaction proceeds quickly.

Teacher's Note:
1. Add a little alkali (or use NaCN with acid) to make cyanohydrins.
2. The reaction needs the cyanide ion as the attacking nucleophile.