JEE Chemistry Reduction Oxidation and Hydrolysis MCQs Set 03

Answer: (d) Ammoniacal silver nitrate (AgNO3 in ammonia)

Explanation:
1. Propyne is a terminal alkyne and has an acidic sp C-H, so it forms a white precipitate of silver propynide with ammoniacal AgNO3.
2. Propene has no such hydrogen and gives no precipitate.
3. Br2/CCl4, dilute KMnO4 and concentrated H2SO4 react with both unsaturated compounds.

Teacher's Note:
1. Terminal alkynes (R-C#C-H) give metal acetylides with Ag+ or Cu+.
2. Bromine water and Baeyer's reagent only show unsaturation.

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

Explanation:
1. An alkyne is adsorbed on the catalyst much more strongly than any alkene, so it is hydrogenated fastest.
2. Among the alkenes, the more substituted one is hydrogenated more slowly: the tetrasubstituted alkene IV is slowest.
3. cis-But-2-ene is more strained and less stable than the trans isomer, so it is hydrogenated a little faster.

Teacher's Note:
1. Rate order: alkyne > unhindered alkene > hindered alkene.
2. cis alkenes react faster than trans alkenes.

Answer: (b) The compound with the CN group on an sp3 carbon of a 1,4-dihydro ring (A)

Explanation:
1. Birch reduction of naphthalene reduces one ring at positions 1 and 4, giving a 1,4-dihydro product with a non-conjugated C=C.
2. An electron-withdrawing group (CN) sits at the sp3 ipso position; the reduced ring is the one carrying CN.
3. This gives 1,4-dihydronaphthalene-1-carbonitrile, option A.

Teacher's Note:
1. Withdrawing groups end on the sp3 carbon in Birch reduction.
2. Only one ring is reduced, and the product is a 1,4-diene, never a fully saturated ring.

Answer: (a) \( \ce{C6H5CH=CHCH2OH} \)

Explanation:
1. LiAlH4 delivers hydride to the carbonyl carbon (1,2-addition), converting CHO to CH2OH.
2. Under normal conditions the conjugated C=C bond is left intact, so cinnamyl alcohol is formed.

Teacher's Note:
1. Hydrogenation, not LiAlH4, would reduce the C=C.
2. The product must keep the same carbon skeleton; options with an extra CH2 are wrong.

Answer: (d) \( \ce{Ph2CD(OH)} \)

Explanation:
1. The deuteride (D-) from LiAlD4 is added to the carbonyl carbon.
2. Water (ordinary H3O+) protonates the oxygen, giving O-H.
3. Product: Ph2CD-OH.

Teacher's Note:
1. C gets its isotope from the metal hydride; O gets it from the work-up.
2. Use D3O+ or D2O in the work-up to get O-D.

Answer: (c) \( \ce{H2/Pd-BaSO4} \)

Explanation:
1. The conversion stops at the aldehyde stage: this is the Rosenmund reduction, H2 over Pd-BaSO4 (poisoned).
2. NaBH4, LiAlH4 and Na/EtOH reduce the acid chloride all the way to the alcohol.

Teacher's Note:
1. Rosenmund reduction needs a poisoned catalyst to prevent over-reduction.
2. Strong hydrides over-reduce because the aldehyde is more reactive than the acid chloride.

Answer: (b) Benzaldehyde

Explanation:
1. This is the Rosenmund reduction: the Pd catalyst is poisoned with BaSO4 so that reduction stops at the aldehyde.
2. The C-Cl bond is replaced by C-H: C6H5COCl becomes C6H5CHO.

Teacher's Note:
1. Benzoic acid would come from hydrolysis, not reduction.
2. Phenol is unrelated (the ring is untouched).

Answer: (a) Any one of NH2NH2/KOH, Zn-Hg/HCl or red P + HI

Explanation:
1. Wolff-Kishner (hydrazine, KOH), Clemmensen (Zn-Hg, HCl) and red P/HI each convert C=O to CH2.
2. Nothing else in the molecule needs to be spared, so all three can be used.

Teacher's Note:
1. The three C=O to CH2 reductions are interchangeable if other groups are unaffected.
2. Choose by the acid/base sensitivity of the rest of the molecule.

Answer: (d) Ethyl 4-hydroxypentanoate; pentane-1,4-diol

Explanation:
1. NaBH4 reduces only the ketone: a = CH3CH(OH)CH2CH2COOEt.
2. LiAlH4 reduces both the ketone and the ester: b = CH3CH(OH)CH2CH2CH2OH (with ethanol released).

Teacher's Note:
1. Mild versus strong hydride: NaBH4 leaves esters alone.
2. The ester gives two alcohols: the new primary alcohol and EtOH.

Answer: (c) Propanone and methanal

Explanation:
1. The alkene that gives only ethanal is but-2-ene, CH3CH=CHCH3 (C4H8).
2. An isomer with the same formula is 2-methylpropene, (CH3)2C=CH2.
3. Its ozonolysis gives propanone from the (CH3)2C= end and methanal from the =CH2 end.

Teacher's Note:
1. Isomers keep the molecular formula C4H8.
2. =CH2 becomes HCHO, =C(CH3)2 becomes acetone.

Answer: (b) 2,2-Dimethyloxirane (isobutene oxide)

Explanation:
1. Alcoholic KOH removes HCl (E2 elimination) to give 2-methylpropene, X.
2. A peroxyacid converts the C=C into an epoxide, giving 2,2-dimethyloxirane (Y).

Teacher's Note:
1. alc. KOH means elimination; aqueous KOH means substitution.
2. Peracids epoxidise alkenes.

Answer: (a) \( \ce{C2H4} \)

Explanation:
1. Alkaline KMnO4 oxidises the C=C bond of ethene, so the purple colour disappears and brown MnO2 forms.
2. Propane, methane and carbon tetrachloride are saturated and are not attacked.

Teacher's Note:
1. Decolourisation of cold KMnO4 is a test for unsaturation.
2. CCl4 has no C-H or C=C to oxidise.

Answer: (d) A cold, dilute alkaline potassium permanganate solution

Explanation:
1. Baeyer's reagent is cold, dilute, alkaline (or neutral) KMnO4.
2. It is used for the test of unsaturation and hydroxylates C=C to a diol.
3. Acidified permanganate and bromine water are different reagents.

Teacher's Note:
1. Cold and dilute gives diols; hot and concentrated cleaves the C=C.
2. Bromine water also tests for unsaturation but is not Baeyer's reagent.

Answer: (c) A and C

Explanation:
1. Hydrolysis of benzamide gives benzoic acid (A) and ammonia (E, as ammonium).
2. LiAlH4 reduces benzoic acid to benzyl alcohol (B).
3. Acidified KMnO4 oxidises benzyl alcohol back to benzoic acid (C), so A and C are identical.

Teacher's Note:
1. Track each compound by functional group: amide, acid, alcohol, acid.
2. P is the amide, which is different from A.

Answer: (b) The epoxide of the side-chain C=C, with the benzene ring unchanged

Explanation:
1. Peroxyacids epoxidise isolated and conjugated C=C bonds.
2. The benzene ring is aromatic and is not epoxidised.
3. So only the C=C between the phenyl and cyclohexyl groups is epoxidised.

Teacher's Note:
1. Aromatic rings do not react with peracids under ordinary conditions.
2. Epoxidation is a syn addition that retains the alkene geometry.

Answer: (a) \( \ce{FeSO4 + H2O2} \)

Explanation:
1. Fenton's reagent is a mixture of ferrous sulphate and hydrogen peroxide.
2. Fe2+ catalyses the decomposition of H2O2 to give hydroxyl radicals, which hydroxylate aromatic rings.

Teacher's Note:
1. Fe(II) plus H2O2 is the hydroxyl radical source.
2. Remember: Fenton = Fe2+ + H2O2.

Answer: (d) Grignard reagent

Explanation:
1. Grignard reagents add to the carbonyl carbon of any aldehyde or ketone (to form secondary or tertiary alcohols).
2. Tollens', Schiff's and Fehling's reagents are given only by aldehydes.

Teacher's Note:
1. Nucleophilic addition works for both; easy oxidation works only for aldehydes.
2. Acetone is a methyl ketone, so it does not reduce Tollens' reagent.

Answer: (c) Dimethyl malonate, \( \ce{CH2(COOCH3)2} \)

Explanation:
1. Hydrolysis of a diester gives the diacid Q and methanol; the formula C5H8O4 with CH3O- groups fits CH2(COOCH3)2.
2. Q is malonic acid, which loses CO2 on strong heating to give acetic acid R.
3. Red P/HI reduces acetic acid to ethane.

Teacher's Note:
1. Count atoms: CH2 + 2 COOCH3 = C5H8O4.
2. Diethyl malonate would release ethanol and has the formula C7H12O4.

Answer: (b) \( \ce{Cu2O} \)

Explanation:
1. The aldehyde is oxidised to the carboxylate while Cu2+ (in the tartrate complex) is reduced to Cu+.
2. Cu+ precipitates as red copper(I) oxide, Cu2O.

Teacher's Note:
1. Red precipitate: Cu2O; black: CuO.
2. Benzaldehyde does not give this test.

Answer: (a) C6H5CH(OH)CH3 and CH3CH=CHCH2OH are both oxidised

Explanation:
1. Activated MnO2 oxidises selectively allylic and benzylic alcohols.
2. C6H5CH(OH)CH3 is benzylic and CH3CH=CHCH2OH is allylic, so both are oxidised.
3. C6H5CH2CH2OH and CH2=CHCH2CH2OH have the OH group away from the ring or double bond and are not oxidised.

Teacher's Note:
1. The activating feature is the adjacent pi system.
2. Saturated alcohols are not oxidised by MnO2.

Answer: (d) \( \ce{CH3C#CCH3 ->[Wilkinson's catalyst] CH3CH=CHCH3} \)

Explanation:
1. In homogeneous reduction the catalyst is soluble in the reaction medium.
2. Wilkinson's catalyst, RhCl(PPh3)3, dissolves in the solvent.
3. Ni, Pd/CaCO3 and Pd/BaSO4 are solid, insoluble catalysts (heterogeneous).

Teacher's Note:
1. Wilkinson's catalyst is the standard homogeneous hydrogenation catalyst.
2. A solid metal catalyst always indicates heterogeneous catalysis.

Answer: (c) Any of NaBH4/EtOH, LiAlH4/THF or Al(OiPr)3/iPrOH

Explanation:
1. NaBH4 and LiAlH4 reduce ketones to secondary alcohols.
2. Al(OiPr)3 in isopropyl alcohol (MPV reduction) does the same.
3. No other functional group is present, so all three work.

Teacher's Note:
1. Three standard ketone-to-alcohol reductions: NaBH4, LiAlH4, MPV.
2. They differ only in selectivity.

Answer: (b) \( \ce{NaBH4} \)

Explanation:
1. Only the aldehyde has to be reduced while the ester is retained.
2. NaBH4 reduces aldehydes but not esters.
3. LiAlH4 and DIBAL-H also attack the ester; Na/NH3 is a dissolving-metal reduction of aromatic rings and alkynes.

Teacher's Note:
1. Selective reduction of CHO in the presence of an ester: NaBH4.
2. LiAlH4 is non-selective.

Answer: (a) \( \ce{LiAlH4} \)

Explanation:
1. LiAlH4 reduces carboxylic acids directly to primary alcohols.
2. NaBH4 is too weak to reduce a carboxylic acid; Na/alcohol is for esters; Zn-Hg/HCl is for ketones and aldehydes.

Teacher's Note:
1. Strong hydride for acids: LiAlH4 (or B2H6).
2. Remember the strength order LiAlH4 > DIBAL-H > NaBH4.

Answer: (d) \( \ce{CH3CH2CH(OD)CH2CH2CH3} \)

Explanation:
1. Hydride (H-) from NaBH4 adds to the carbonyl carbon, so the carbon keeps ordinary H.
2. The oxygen anion is protonated (deuterated) by D2O, giving O-D.

Teacher's Note:
1. Isotope on carbon comes from the reagent; isotope on oxygen from the solvent.