JEE Chemistry Organic Chemistry Some Basic Principles and Techniques MCQs Set 59

Answer: (b) HOCH₂CH₂C(OH)(CH₃)CH₂COOH (3,5-dihydroxy-3-methylpentanoic acid)

Explanation:
1. Breaking the ester C(=O)-O bond gives COOH at one end and CH₂OH (from the ring oxygen) at the other end.
2. The tertiary OH and methyl remain on C-3: HOCH₂CH₂-C(OH)(CH₃)-CH₂-COOH.

Teacher's Note:
1. Hydrolysis of a lactone is the reverse of intramolecular esterification.
2. The other substituents are not changed.

Answer: (a) 2

Explanation:
1. Acyl substitution works when the incoming nucleophile is the stronger base (or similar in basicity) compared with the leaving group.
2. (a) HO^- displaces Cl^- and (d) NH₂^- displaces C₂H₅O^-: both proceed.
3. (b) I^- is a much weaker base than NH₂^- and cannot displace it; (c) is simply an acid-base reaction (the acid is deprotonated).
4. So two reactions.

Teacher's Note:
1. Compare the basicity of Nu and L.
2. An acid with an alkoxide just undergoes proton transfer.

Answer: (a) All four

Explanation:
1. NaH is a strong base (H^-) and removes any acidic proton, giving H₂.
2. The carboxylic acid, the amide N-H (pKa about 17), the terminal alkyne (pKa about 25) and ethanol (pKa about 16) all have such a proton.
3. So all four liberate hydrogen.

Teacher's Note:
1. H^- + H-Z gives H₂ + Z^- for any acidic H-Z.
2. Amides are weak acids but NaH is strong enough.

Answer: (b) 1-phenylpropan-1-ol, CH₃CH₂CH(OH)Ph

Explanation:
1. The phenyl group of PhMgBr adds to the carbonyl carbon of propanal and gives CH₃CH₂CH(OMgBr)Ph.
2. Hydrolysis gives the secondary alcohol 1-phenylpropan-1-ol.

Teacher's Note:
1. Aldehyde plus RMgX gives a secondary alcohol (primary only for HCHO).
2. The new group attaches to the former carbonyl carbon.

Answer: (c) 6

Explanation:
1. Electrophiles are electron-deficient: H⁺, Br⁺, AlCl₃ and BH₃ (empty orbital), the acylium ion CH₃CO⁺ and NO₂⁺.
2. CN^- and NH₃ have lone pairs and are nucleophiles.
3. So six are electrophiles.

Teacher's Note:
1. Cations and species with an incomplete octet are electrophilic.
2. Anions and lone-pair molecules are nucleophilic.

Answer: (c) C₆H₅-CH(OD)-CH₃

Explanation:
1. LiAlH₄ delivers H^- to the carbonyl carbon, so the C-H comes from the reagent.
2. The oxygen is protonated (here deuterated) by the work-up reagent D₂O, so the O-H becomes O-D.

Teacher's Note:
1. The hydride gives the C-H and the work-up solvent gives the O-H.
2. Deuterium on carbon needs LiAlD₄ or NaBD₄.

Answer: (a) 4

Explanation:
1. Active-hydrogen groups give CH₄: both phenolic OH groups, the COOH and the SO₃H group.
2. The methyl group and the ring C-H bonds are not acidic enough, so the total is 4.

Teacher's Note:
1. Phenols, carboxylic acids and sulphonic acids all react with Grignard reagents as acids.
2. Count groups, not hydrogens, if the question asks for groups.

Answer: (c) (b) and (c) only

Explanation:
1. In (b) Cl is replaced by NH₂ and in (c) I is replaced by OH, so both are substitutions.
2. (a) is addition of H₂ to a C=C bond, and (d) is addition of HCN to C=O.

Teacher's Note:
1. A substitution replaces one atom or group by another with no change in unsaturation.
2. Additions change a pi bond to two sigma bonds.

Answer: (b) C₆H₅-CD(OH)-CH₃

Explanation:
1. D^- from LiAlD₄ attacks the carbonyl carbon, giving C-D.
2. Protonation of the alkoxide by H₂O gives O-H.

Teacher's Note:
1. Hydride source decides the C-H or C-D.
2. Work-up reagent decides the O-H or O-D.

Answer: (c) (CH₃CH₂)₂C(OH)CH₃, from CH₃COOEt and C₂H₅MgBr

Explanation:
1. With excess RMgX an ester R''COOEt gives R''C(OH)R₂, a tertiary alcohol whose carbinol carbon carries two identical groups R from the reagent.
2. 3-Methylpentan-3-ol has two identical ethyl groups and one methyl group (from the acetate), so it fits.
3. Butan-2-ol is secondary and butan-1-ol is primary, which an ordinary ester cannot give; the phenyl alcohol has three different groups.

Teacher's Note:
1. Look for two identical alkyl groups on the carbinol carbon.
2. Only formate esters give secondary alcohols.

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

Explanation:
1. Electron-withdrawing groups (-I and -M) increase the positive charge on the carbonyl carbon and speed up addition; electron-donating groups slow it.
2. NO₂ is the strongest acceptor, Cl is weakly withdrawing, H is the reference, and CH₃ donates electrons.
3. Order: III > II > I > IV.

Teacher's Note:
1. Rate of nucleophilic addition rises with electrophilicity of C=O.
2. The equilibrium constant follows the same order.

Answer: (d) I^- > Br^- > Cl^- > H₂O

Explanation:
1. Down a group, the donor atom becomes larger and more polarisable, so nucleophilicity increases in a protic solvent: I^- > Br^- > Cl^-.
2. The neutral water molecule is a weaker nucleophile than any of the anions.

Teacher's Note:
1. Nucleophilicity increases with size of the donor atom in a group.
2. Anions beat neutral conjugate acids.

Answer: (b) (A) and (B) only

Explanation:
1. Leaving-group ability rises with stability of the anion.
2. (A) is true: ArSO₃H is a strong acid, RCOOH a weak acid. (B) is true: fluorine is more electron-withdrawing than chlorine, so the triflate is more stable.
3. (C) is false: HI is a much stronger acid than HCN, so I^- is the better leaving group. (D) is false: NH₂^- is a stronger base than OH^-.

Teacher's Note:
1. Stronger conjugate acid means better leaving group.
2. Amide ion and hydroxide are poor leaving groups.

Answer: (a) Acetic acid, formamide and ammonia only

Explanation:
1. A protic solvent has a hydrogen bonded to O or N that it can donate (O-H or N-H).
2. Acetic acid (O-H), formamide (N-H) and ammonia (N-H) qualify.
3. THF, acetonitrile, acetone, DMSO, DMF (no N-H) and cyclohexane have no such hydrogen and are aprotic.

Teacher's Note:
1. DMF has N-CH₃ groups only, so it is aprotic.
2. Formamide, HCONH₂, has N-H bonds and is protic.

Answer: (b) 1-hydroxycyclopentane-1-carbonitrile (the cyanohydrin)

Explanation:
1. The cyanide ion adds to the carbonyl carbon and the alkoxide formed is protonated by the acid.
2. The product is 1-hydroxycyclopentanecarbonitrile, C₅H₈(OH)CN; no hydrolysis occurs under these mild conditions.

Teacher's Note:
1. Cyanohydrins contain C(OH)(CN).
2. Acid hydrolysis of the CN group needs heating with strong acid.

Answer: (a) HS^-, C₂H₅OH and (CH₃)₃N

Explanation:
1. HS^-, the oxygen of the alcohol and the nitrogen of the tertiary amine each have a lone pair to donate.
2. BF₃ has an empty p orbital on boron and the carbene :CH₂ has a sextet of electrons on carbon, so both act as electrophiles.

Teacher's Note:
1. A carbene has only six valence electrons around carbon.
2. Lone pair plus complete octet means nucleophile.

Answer: (a) Formaldehyde, HCHO

Explanation:
1. C₂H₅MgCl adds to HCHO to give C₂H₅CH₂OMgCl, and hydrolysis gives CH₃CH₂CH₂OH.
2. Acetaldehyde would give butan-2-ol, acetone gives 2-methylbutan-2-ol, and CO₂ gives propanoic acid.

Teacher's Note:
1. Formaldehyde with RMgX gives a primary alcohol with one more carbon.
2. Any other aldehyde gives a secondary alcohol.

Answer: (c) CH₃D and Mg(OD)I

Explanation:
1. The carbanion-like CH₃ group takes a deuteron from D₂O to form CH₃D.
2. The remaining OD^- combines with MgI⁺ as Mg(OD)I.

Teacher's Note:
1. Grignard reagent plus D₂O is a way of introducing one deuterium atom.
2. Only one D ends up in the hydrocarbon.

Answer: (d) It exists as two diastereomers (cis and trans), each optically inactive

Explanation:
1. Both C=O groups add HCN to give C(OH)(CN) at C-1 and C-4.
2. The ring has a plane of symmetry in both the cis and the trans arrangement of the two OH groups, so each is achiral (optically inactive).
3. The cis and trans forms are diastereomers (not mirror images), so the product is a mixture of two diastereomers.

Teacher's Note:
1. 1,4-Disubstituted cyclohexanes show cis-trans isomerism.
2. A plane of symmetry means no optical activity.

Answer: (d) cyclopentyl-CH(OH)-CN

Explanation:
1. CN^- adds to the aldehyde carbon; protonation of the oxygen gives the cyanohydrin C₅H₉-CH(OH)-CN.

Teacher's Note:
1. Aldehyde cyanohydrins are secondary alcohols bearing CN.
2. The new C-C bond is to carbon of CN.

Answer: (d) phenol, C₆H₅¹⁸OH

Explanation:
1. Acid hydrolysis of an ester follows the addition-elimination path with cleavage of the acyl-oxygen bond (C(=O)-O).
2. The ¹⁸O atom stays attached to the phenyl ring and the new C-OH oxygen in the acid comes from water.
3. So C₆H₅¹⁸OH is formed along with unlabelled benzoic acid.

Teacher's Note:
1. Ester hydrolysis breaks C-OR', not O-R'.
2. The alkoxy oxygen ends up in the alcohol.

Answer: (b) CH₃NC and CH₃CN

Explanation:
1. The cyanide ion is an ambident nucleophile that can attack through carbon or through nitrogen.
2. KCN is ionic, so the free CN^- attacks through the more nucleophilic carbon and gives the nitrile CH₃CN.
3. AgCN is largely covalent, the carbon being bound to silver, so the nitrogen lone pair attacks and the isocyanide CH₃NC is formed.

Teacher's Note:
1. Ionic cyanides give nitriles; covalent cyanides (AgCN, CuCN) give isocyanides.
2. Other ambident ions: NO₂^- and enolates.

Answer: (b) OCH₃ > CH₃ > Cl > NO₂

Explanation:
1. Groups that push electrons (+I, +M) raise the electron density on oxygen and increase nucleophilicity; electron-withdrawing groups lower it.
2. OCH₃ (+M) and CH₃ (+I, hyperconjugation) are donors; Cl is withdrawing (-I); NO₂ is strongly withdrawing (-I, -M).
3. The order is OCH₃ > CH₃ > Cl > NO₂.

Teacher's Note:
1. Nucleophilicity of the same donor atom follows its basicity here.
2. Nitro gives the most stable (least basic) phenoxide.

Answer: (a)

Explanation:
1. The alcohol oxygen attacks the acyl carbon and chloride is expelled: PhCOCl + C₆H₁₁OH gives PhCOOC₆H₁₁ + HCl.
2. The product is cyclohexyl benzoate.

Teacher's Note:
1. Alcohol plus acid chloride gives an ester.
2. Esterification keeps the alcohol carbon skeleton unchanged.

Answer: (b) (b) and (c) only

Explanation:
1. In (b) two bromine atoms are removed and a C=C bond is formed; in (c) HCl is lost and propene forms. Both are eliminations.
2. (a) and (d) replace OH by Cl and OH by OCH₃: substitutions.

Teacher's Note:
1. Eliminations create a double bond by losing two atoms or groups from adjacent carbons.
2. Zinc dehalogenation of a vicinal dibromide is an elimination.

Answer: (d) C₆H₅-CD(OD)-CH₃

Explanation:
1. BD₄^- transfers D^- to carbon, giving C-D.
2. The alkoxide is then protonated by the solvent EtOD, giving O-D.

Teacher's Note:
1. Track the two hydrogens: one from the hydride source, one from the solvent.
2. NaBD₄ in H₂O would give C-D and O-H.