JEE Chemistry Organic Chemistry Some Basic Principles and Techniques MCQs Set 57

Answer: (a) (CH₃)₃C-OH

Explanation:
1. The first CH₃MgBr substitutes OC₂H₅ and gives acetone, which is more reactive than the ester.
2. The second adds to acetone to give (CH₃)₃COMgBr, and hydrolysis gives tert-butyl alcohol (ethanol is the other product).

Teacher's Note:
1. Esters react twice with Grignard reagents, so the ketone cannot be isolated.
2. The alcohol has two identical R groups from the reagent.

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

Explanation:
1. Weaker bases are better leaving groups, and the weaker base is the conjugate base of the stronger acid.
2. The nitro group makes phenol more acidic, so IV is the weakest base. OCH₃ (-0.27) is a stronger donor than CH₃ (-0.17) which is stronger than H.
3. So the order of basicity is III > II > I > IV, and leaving ability is the reverse: IV > I > II > III.

Teacher's Note:
1. Leaving-group ability follows the acidity of the conjugate acid.
2. Donor substituents raise basicity; acceptor substituents lower it.

Answer: (c) R-OSO₂CH₃ > R-Cl > R-OCOCH₃ > R-OH

Explanation:
1. Good leaving groups are stable, weakly basic anions that disperse the negative charge. Sulphonates (MsO^-) are the best of these four because of strong resonance dispersal, chloride is next, carboxylate is moderate, and OH^-, OR^- and NH₂^- are poor.
2. In the first option mesylate is placed first (best in practice), then chloride, then acetate, then OH. The other options place a poor leaving group (NH₂, OCH₃) high or put the sulphonate or iodide in the wrong place.

Teacher's Note:
1. Sulphonates and triflates are better leaving groups than halides.
2. Alkoxide, hydroxide and amide are very poor leaving groups.

Answer: (d) CH₃COOH + CH₃OH (H⁺) gives CH₃COOCH₃

Explanation:
1. The aldehyde and the alkene reactions are additions, and the alcoholic KOH reaction is an elimination.
2. In esterification the OH of the acid is replaced by OCH₃: this is nucleophilic acyl substitution.

Teacher's Note:
1. Addition: two molecules become one. Elimination: a small molecule is lost to give a double bond.
2. Substitution: one group is replaced by another.

Answer: (b) CH₃MgBr (excess) + propyl cyclopentyl ketone

Explanation:
1. The target carbon carries two propyl groups, one cyclopentyl and OH. Any route must introduce exactly these groups.
2. Methylmagnesium bromide adds a methyl group, giving CH₃C(OH)(Pr)(C₅H₉), not the target.
3. The other three combinations give two propyl groups and one cyclopentyl group on the carbinol carbon.

Teacher's Note:
1. Ask which three groups end up on the carbinol carbon.
2. An ester with excess RMgX gives R₂C(OH)R'' where R'' is the acyl group.

Answer: (a) CH₃CH₂CH₂CN + CH₃MgI/ether, then H⁺, H₂O

Explanation:
1. Butanenitrile plus CH₃MgI gives an imine salt, which hydrolyses to the ketone CH₃COCH₂CH₂CH₃ (2-pentanone).
2. Butanal gives the secondary alcohol pentan-2-ol, and formaldehyde gives the primary alcohol pentan-1-ol.

Teacher's Note:
1. Nitriles give ketones with Grignard reagents.
2. The imine anion resists further addition.

Answer: (d) SH^- > OH^- > C₆H₅O^- > CH₃COO^- > H₂O

Explanation:
1. Sulphur is large and polarisable and is poorly hydrogen bonded, so SH^- is the best nucleophile in a protic solvent.
2. Among oxygen nucleophiles, nucleophilicity follows basicity: OH^- > PhO^- (charge delocalised into ring) > CH₃COO^- (charge shared by two O) > neutral H₂O.

Teacher's Note:
1. Same donor atom: more negative charge density gives better nucleophile.
2. Down the group nucleophilicity increases in protic solvents.

Answer: (b) 60

Explanation:
1. Moles of CH₄ = 1.12/22400 = 5.0 x 10^-5 mol, equal to the moles of ROH (one O-H each).
2. M = 3 x 10^-3 g / 5.0 x 10^-5 mol = 60 g/mol (propan-1-ol or propan-2-ol).

Teacher's Note:
1. Active-hydrogen (Zerewitinoff) method: one mole of CH₄ per mole of O-H.
2. Keep units consistent: mL to mol uses 22400 mL per mole.

Answer: (b) CH₃COBr

Explanation:
1. Reactivity follows the leaving-group ability: Br^- > CH₃COO^- > C₂H₅O^- > NH₂^-.
2. The acid bromide has the best leaving group and a strongly electrophilic carbonyl carbon, so it is the most reactive.

Teacher's Note:
1. Order: acid halide > anhydride > ester > amide.
2. Resonance donation from N or O reduces the carbonyl reactivity in amides and esters.

Answer: (d) HCHO

Explanation:
1. Cyanohydrin formation creates a new carbon with four different groups for CH₃CHO, PhCHO and butanone, so each gives a racemic pair of enantiomers.
2. From HCHO the product HOCH₂CN has two identical hydrogens on carbon and is achiral, so only one compound is formed.

Teacher's Note:
1. The carbonyl carbon is planar and is attacked from both faces.
2. If the product carbon is chiral, both enantiomers form in equal amounts.

Answer: (b) DMSO and acetone

Explanation:
1. Aprotic solvents have no O-H or N-H hydrogen that can hydrogen-bond.
2. DMSO and acetone qualify. Water and ethanol are protic, and methylamine has N-H bonds.

Teacher's Note:
1. Polar aprotic solvents leave anions unsolvated and favour SN2.
2. Protic solvents solvate anions by hydrogen bonding.

Answer: (c) transition state

Explanation:
1. A catalyst provides an alternative path with lower activation energy.
2. This is done by stabilising the transition state (or the intermediate leading to it) relative to the reactants.
3. The overall energy change of the reaction is not altered.

Teacher's Note:
1. Lower activation energy means a faster rate.
2. Catalysts do not change equilibrium composition.

Answer: (b) NH₃, AlCl₃, H atom (H·)

Explanation:
1. Take one example of each type: NH₃ is a nucleophile (lone pair), AlCl₃ is a Lewis acid (empty orbital) and the hydrogen atom is a radical.
2. The other groups do not contain all three types.

Teacher's Note:
1. Lewis acid = electron pair acceptor.
2. A radical has an unpaired electron.

Answer: (c) OSO₂CF₃ > OSO₂CH₃ > OCOC₂H₅ > OC₂H₅

Explanation:
1. Leaving-group ability follows the acid strength of the conjugate acid.
2. CF₃SO₃H (superacid) > CH₃SO₃H > carboxylic acid > ethanol.
3. The other sequences put the alkoxide, the worst leaving group, before stronger-acid conjugate bases.

Teacher's Note:
1. Triflate is a super leaving group.
2. Alkoxide is a very strong base and a very poor leaving group.

Answer: (c) Grignard reagent

Explanation:
1. Hexan-1-ol has an acidic O-H and reacts with RMgX to give an alkane; pentan-2-one is attacked at C=O to give a tertiary alkoxide.
2. Fehling's, Schiff's and Tollens' reagents test for aldehydes, so neither compound reacts with them.

Teacher's Note:
1. Grignard reagents react with alcohols (acid-base) and with carbonyls (addition).
2. Aldehyde tests do not respond to ketones or alcohols.

Answer: (d) an optically active ester

Explanation:
1. Fischer esterification breaks the C-O bond of the acid (O-H of the alcohol attacks the acyl carbon), so no stereocentre in either reactant is touched.
2. The product, sec-butyl 3-chlorobutanoate, is one pure stereoisomer with two stereocentres and no symmetry, so it is optically active and not meso.

Teacher's Note:
1. Check which bond breaks in the mechanism to decide stereochemistry.
2. Meso needs an internal plane of symmetry.

Answer: (a) a 50:50 mixture of D and L (racemic)

Explanation:
1. The planar C=O is attacked by CN^- equally from both faces, giving equal amounts of the two enantiomers of the cyanohydrin.
2. Hydrolysis does not touch the stereocentre, so lactic acid is obtained as the racemic (50:50) mixture.

Teacher's Note:
1. Achiral reagents on a planar achiral carbonyl give racemates.
2. Optical activity needs a chiral reagent, catalyst or enzyme.

Answer: (d) NH₃, H₂O, CN^-, I^-

Explanation:
1. Every member of the first series has a lone pair.
2. AlCl₃, BF₃ and NO₂⁺ are electron-deficient electrophiles.

Teacher's Note:
1. Lewis acids are electrophiles.
2. Anions and neutral lone-pair species are nucleophiles.

Answer: (b) CH₃COOC₂H₅

Explanation:
1. Ethyl acetate gives acetic acid and ethanol, two organic compounds.
2. Acetic acid does not hydrolyse. The amide gives acetic acid and an ammonium salt (inorganic), and the anhydride gives two molecules of the same acid.

Teacher's Note:
1. Count only organic products.
2. Symmetrical anhydrides give one acid.

Answer: (b) HCOOC₂H₅

Explanation:
1. Only formates have a hydrogen on the acyl carbon.
2. After two additions the carbon carries H, R, R and OH, which is a secondary alcohol; all other esters give tertiary alcohols.

Teacher's Note:
1. Count the groups on the carbinol carbon.
2. Formate gives R₂CHOH.

Answer: (b) II > I > III

Explanation:
1. Donor groups on the ring reduce the positive charge on the carbonyl carbon and make the cyanohydrin less favoured.
2. The NMe₂ group is a much stronger donor than OMe, so III is the lowest and I is in the middle.
3. Benzaldehyde has no donor group and has the largest K_eq.

Teacher's Note:
1. Stronger electron donation means smaller equilibrium constant for addition.
2. The +R effect of N is larger than that of O.

Answer: (d) PhMgBr + (CH₃)₃C-OH

Explanation:
1. Benzene arises when the phenyl carbanion of PhMgBr takes up a proton.
2. Only the alcohol has an acidic O-H hydrogen: PhMgBr + t-BuOH gives C₆H₆ + t-BuOMgBr.
3. Ether (which only solvates the Grignard), propane and cyclohexene have no acidic hydrogens.

Teacher's Note:
1. Grignard reagents are destroyed by water, alcohols, acids and other protic species.
2. Ethers are used as solvents because they have no acidic hydrogen.

Answer: (b) CH₃-NH₂

Explanation:
1. A nucleophile has a lone pair to donate and a complete octet.
2. Methylamine has a lone pair on nitrogen and a full octet.
3. The methyl radical has an odd electron and the carbene and nitrene have only a sextet of electrons on the reactive atom, so they are electrophilic.

Teacher's Note:
1. Species with an incomplete octet cannot act as nucleophiles.
2. Neutral amines are common nucleophiles.

Answer: (a) (i) and (iii) only

Explanation:
1. Substitution at an acyl carbon (C=O) goes through a tetrahedral intermediate: acetyl chloride with hydroxide (i) and acetic acid with PCl₅ to give acetyl chloride (iii).
2. Ethyl chloride with NaOH (ii) and isopropyl alcohol with PCl₅ (iv) take place at an sp³ carbon by SN2 or SNi.

Teacher's Note:
1. Acyl substitution = addition then elimination.
2. Alkyl substitution has no tetrahedral intermediate.

Answer: (d) (i) and (iii) only

Explanation:
1. In a polar aprotic solvent (DMSO) anions are not hydrogen bonded, so nucleophilicity follows basicity: F^- is the best.
2. In a protic solvent such as water, small anions are strongly hydrogen bonded and large polarisable ones are better nucleophiles: I^- is the best.
3. So (i) and (iii) are correct.

Teacher's Note:
1. Solvent can reverse the order of nucleophilicity.
2. Protic solvents hide small anions behind a shell of hydrogen bonds.