JEE Chemistry Reduction Oxidation and Hydrolysis MCQs Set 01

Answer: (b) \( \ce{CH2=CH2} \) > \( \ce{RCH=CH2} \) > \( \ce{RCH=CHR} \) > \( \ce{R2C=CHR} \)

Explanation:
1. Hydrogenation needs the alkene to be adsorbed flat on the metal surface.
2. Alkyl groups on the double bond hinder this approach, so the more substituted the alkene, the slower it is hydrogenated.
3. Order: ethene > monosubstituted > disubstituted > trisubstituted.

Teacher's Note:
1. Steric hindrance, not stability, controls the rate of hydrogenation.
2. The same order is followed by the heat of hydrogenation, which is largest for the least substituted alkene.

Answer: (a) I and III

Explanation:
1. Pd/BaSO4 (poisoned palladium) is active enough only for alkynes; it stops at the alkene stage and does not touch an alkene.
2. I (cyclohexene) and III (methylenecyclopentane) are alkenes, so they are unchanged.
3. II (hex-3-yne) and IV (hex-1-yne) are alkynes and are reduced to alkenes.

Teacher's Note:
1. Count the unsaturation: C6H10 has two degrees, shared by a ring and a double bond in I and III, and by a triple bond in II and IV.
2. A poisoned catalyst is deliberately made weak so that the first reduction product can be isolated.

Answer: (d) a racemic mixture

Explanation:
1. Hydrogenation on a metal surface is a syn addition: both hydrogens arrive on the same face.
2. The product 3,4-dimethylhexane has two stereocentres. A syn addition to the isomer with the two methyl groups trans to each other gives the (R,R) and (S,S) forms, because the alkene face can be attacked from either side with equal probability.
3. The two enantiomers are produced in equal amounts, so X is a racemic mixture (the Z isomer would have given the meso compound).

Teacher's Note:
1. Compare with trans-2-butene plus a syn reagent such as OsO4, which gives a racemic diol.
2. A meso product from a symmetrical alkene needs the cis (Z) arrangement with syn addition.

Answer: (c) 1-Methylcyclohexa-1,4-diene

Explanation:
1. Birch reduction converts a benzene ring to a 1,4-cyclohexadiene (non-conjugated); the ring is not reduced to cyclohexane.
2. An electron-donating group such as methyl stays on a carbon that keeps its double bond, so the sp3 carbons are positions 3 and 6 relative to the methyl-bearing carbon.
3. This gives 1-methylcyclohexa-1,4-diene.

Teacher's Note:
1. Rule: donating groups (alkyl, alkoxy, amino) remain on sp2 carbons; withdrawing groups (NO2, CN, COOH, CHO, COR) are attached to sp3 carbons.
2. Birch reduction never gives the conjugated 1,3-diene.

Answer: (b) \( \ce{Zn-Hg/conc. HCl} \)

Explanation:
1. Zn-Hg/conc. HCl (Clemmensen reduction) converts the C=O of aldehydes and ketones into CH2.
2. Ni/H2 and LiAlH4 give alcohols (ethanol and propan-2-ol).
3. I2/NaOH gives the iodoform reaction, a degradation (acetaldehyde and acetone both give CHI3), not an alkane.

Teacher's Note:
1. Clemmensen (acid) and Wolff-Kishner (base) are the two standard C=O to CH2 reductions.
2. Choose between them by the sensitivity of the other functional groups.

Answer: (a) Aldehydes

Explanation:
1. SnCl2 and HCl reduce the nitrile to an imine hydrochloride, RCH=NH.HCl.
2. Boiling with water hydrolyses the imine to the aldehyde RCHO and ammonium chloride.

Teacher's Note:
1. Compare: LiAlH4 or H2/Ni reduce RCN to the primary amine, RCH2NH2.
2. DIBAL-H also converts RCN to RCHO.

Answer: (d) Benzyl alcohol

Explanation:
1. LiAlH4 is strong enough to reduce a carboxylic acid completely to the primary alcohol.
2. The aldehyde is an intermediate but is reduced faster than the acid, so it cannot be isolated.
3. Product: C6H5CH2OH.

Teacher's Note:
1. Only LiAlH4 (and B2H6) reduce carboxylic acids; NaBH4 does not.
2. Benzene or toluene would need the C-O bond to break, which these hydrides do not do.

Answer: (c) Meerwein-Ponndorf-Verley reduction

Explanation:
1. The ketone takes hydride from the isopropoxide through a six-membered cyclic transition state, giving the alcohol; isopropanol is oxidised to acetone.
2. This is the Meerwein-Ponndorf-Verley (MPV) reduction, the reverse of the Oppenauer oxidation.

Teacher's Note:
1. MPV is highly selective: C=C, nitro groups and esters survive.
2. Wolff-Kishner converts C=O to CH2; Stephen reduces nitriles.

Answer: (b) R-CHO

Explanation:
1. DIBAL-H adds one hydride to the C#N carbon and stops at the aluminium imine stage.
2. Hydrolysis with water converts the imine to the aldehyde RCHO.

Teacher's Note:
1. Contrast with LiAlH4, which adds two hydrides and gives RCH2NH2.
2. DIBAL-H also stops at the aldehyde stage for esters at -78 C.

Answer: (a) Ethane

Explanation:
1. Red P and HI is a strong reducing system: it cleaves C-O bonds and reduces them to C-H.
2. Each C2H5 group of the ether ends up as C2H6, giving two molecules of ethane.
3. Balance: C4H10O + 4[H] gives 2 C2H6 + H2O.

Teacher's Note:
1. Balancing atoms is a quick check: 4 carbons give two C2 units.
2. HI first cleaves the ether to ethyl iodide and ethanol, and these are reduced by more HI.

Answer: (d) R-CHO

Explanation:
1. At -78 C DIBAL-H adds a single hydride to the ester carbonyl and the tetrahedral intermediate does not collapse until the work-up.
2. On hydrolysis it releases the aldehyde RCHO and the alcohol R'OH.
3. At room temperature, or with LiAlH4, the aldehyde would be reduced further to RCH2OH.

Teacher's Note:
1. Low temperature is the key condition for stopping at the aldehyde.
2. DIBAL-H is bulky and electrophilic, so it prefers to stop at the stable aluminium hemiacetal.

Answer: (c) Alkenes and alkynes both

Explanation:
1. Baeyer's reagent hydroxylates C=C to a vicinal diol and oxidises C#C to a diketone; in both cases purple MnO4- is reduced to brown MnO2.
2. Alkanes and benzene are not attacked under these cold conditions.

Teacher's Note:
1. This is the Baeyer test for unsaturation.
2. Benzene resists the reagent because of its aromatic stability.

Answer: (b) Acetic acid (as acetate)

Explanation:
1. Alkaline KMnO4 is a strong oxidant, so the primary alcohol goes through ethanal and does not stop there.
2. Ethanal is oxidised further to acetic acid (present as the acetate in alkaline medium).

Teacher's Note:
1. Only mild reagents (PCC, PDC, Collins reagent) stop at the aldehyde stage.
2. Glyoxal and acrolein would need extra carbon-carbon changes that are not possible here.

Answer: (a) \( \ce{CH3CH2COOH + CO2 + H2O} \)

Explanation:
1. Hot KMnO4 breaks the triple bond completely.
2. The internal carbon (CH3CH2-C) becomes the carboxylic acid CH3CH2COOH.
3. The terminal carbon (#CH) becomes CO2 (and H2O).

Teacher's Note:
1. Terminal =CH2 or #CH groups always end up as CO2 with hot KMnO4.
2. Count carbons: 3 in the acid plus 1 as CO2 gives 4.

Answer: (d) 2-Methylhex-3-yne

Explanation:
1. The triple bond is broken and each carbon of it becomes COOH.
2. 2-Methylpropanoic acid is (CH3)2CH-COOH and propanoic acid is CH3CH2-COOH.
3. Joining the two COOH carbons by the triple bond gives (CH3)2CH-C#C-CH2CH3, which is 2-methylhex-3-yne (C7H12).

Teacher's Note:
1. Work backwards: replace each COOH group by a carbon and join them with a triple bond.
2. Check the formula: 3 + 2 + 2 = 7 carbons.

Answer: (c) Cold alkaline KMnO4; HCO3H followed by H3O+

Explanation:
1. Cold alkaline KMnO4 (and OsO4) add two OH groups on the same face (syn addition), giving the cis diol: R1.
2. A peroxyacid first gives the epoxide; acid-catalysed ring opening by water occurs from the back side (anti addition), giving the trans diol: R2.
3. So R1 is cold alkaline KMnO4 and R2 is HCO3H followed by H3O+.

Teacher's Note:
1. Syn: KMnO4 (cold, alkaline) and OsO4. Anti: peracid then H3O+.
2. A peracid by itself gives only the epoxide; the diol appears after the acidic ring opening.

Answer: (b) 5 and 3

Explanation:
1. Glucose, HOCH2-(CHOH)4-CHO, is cleaved at five C-C bonds: the CHO carbon and the four CHOH carbons give 5 HCOOH, the CH2OH carbon gives HCHO.
2. Fructose, HOCH2-CO-(CHOH)3-CH2OH, is also cleaved at five bonds: the three CHOH carbons give 3 HCOOH, both CH2OH carbons give 2 HCHO and the central C=O carbon gives CO2.

Teacher's Note:
1. Every CHOH or CHO carbon (with an O-bearing neighbour on both sides) ends as HCOOH; a C=O flanked by two oxygen-bearing carbons ends as CO2.
2. Terminal CH2OH always gives formaldehyde.

Answer: (a) Ketones

Explanation:
1. Hot copper catalyses dehydrogenation of the alcohol.
2. A primary alcohol gives an aldehyde, a secondary alcohol gives a ketone, and a tertiary alcohol, having no alpha H, dehydrates to an alkene.

Teacher's Note:
1. Dehydrogenation removes the alpha H and the O-H hydrogen.
2. Remember this as a method to distinguish the three classes of alcohol.

Answer: (d) \( \ce{H2/Ni} \)

Explanation:
1. Tollens', Fehling's and Schiff's reagents are oxidising (or colour-forming) tests that are given by aldehydes only.
2. Ketones are not easily oxidised, so they fail these three tests.
3. H2/Ni reduces the C=O group of both, giving ethanol and propan-2-ol.

Teacher's Note:
1. Nucleophilic addition and reduction of C=O work on both aldehydes and ketones; easy oxidation works only on aldehydes.
2. Both also give 2,4-DNP derivatives.

Answer: (c) \( \ce{HOCH2CH2CH2OH} \)

Explanation:
1. Periodic acid cleaves only C-C bonds that join two oxygen-bearing carbons (vicinal diols and alpha-hydroxy carbonyl or 1,2-dicarbonyl compounds).
2. Propane-1,3-diol has an OH group on C1 and C3, but no two adjacent carbons both carry oxygen, so it is not cleaved.
3. Dihydroxyacetone, glycerol and ethylene glycol all have adjacent oxygenated carbons.

Teacher's Note:
1. Look for adjacent O-bearing carbons: HO-C-C-OH.
2. A 1,3-diol is inert because the cyclic periodate ester cannot form easily.

Answer: (b) Benzoic acid

Explanation:
1. Acid hydrolysis converts C#N first into the amide, C6H5CONH2, and then into the acid C6H5COOH with release of NH4+.
2. The aldehyde would come from a reduction (DIBAL-H or Stephen), not from hydrolysis.

Teacher's Note:
1. Complete nitrile hydrolysis gives the carboxylic acid; stopping at the amide needs mild conditions.
2. Benzoyl chloride is made from the acid with SOCl2.

Answer: (a) Cyclohexanone and ethanol

Explanation:
1. An enol ether is protonated at the carbon of the double bond to give an oxocarbenium ion.
2. Water adds to the oxocarbenium carbon, giving a hemiketal that collapses to cyclohexanone and ethanol.

Teacher's Note:
1. Vinyl and enol ethers are hydrolysed very easily by dilute acid, unlike ordinary ethers.
2. The carbonyl compound comes from the alkene part, the alcohol from the alkyl part.

Answer: (d) Cinnamaldehyde (3-phenylprop-2-enal) and methanol

Explanation:
1. At low temperature DIBAL-H reduces the ester only as far as the aldehyde and does not attack the C=C bond.
2. The products are C6H5CH=CHCHO and CH3OH.

Teacher's Note:
1. DIBAL-H at low temperature is the standard way to go from an ester to an aldehyde.
2. Hydrogenation (H2/Pd) would instead reduce the C=C bond.

Answer: (c) No reaction occurs

Explanation:
1. NaBH4 is a mild hydride donor that reduces aldehydes, ketones and acid chlorides.
2. It does not reduce esters, acids, amides, nitriles or isolated or conjugated C=C bonds of an ester, so methyl cinnamate is unchanged.

Teacher's Note:
1. Remember the NaBH4 limit: aldehyde, ketone, acid chloride only.
2. LiAlH4 or DIBAL-H are needed for esters.

Answer: (b) Methyl butanoate

Explanation:
1. Catalytic hydrogenation reduces the C=C bond.
2. Esters are not reduced under these conditions, so the ester group remains.
3. Product: CH3CH2CH2COOCH3.

Teacher's Note:
1. Pd-C and H2 reduce alkenes, alkynes, nitro groups and C=O of aldehydes or ketones but leave esters, acids and amides alone.