JEE Chemistry Reduction Oxidation and Hydrolysis MCQs Set 02

Answer: (a) 3

Explanation:
1. Pd-BaSO4 gives cis-but-2-ene by syn addition. Syn addition of D2 across the cis alkene gives a single meso compound, CH3CHD-CHDCH3 (x = 1).
2. Na/NH3 gives trans-but-2-ene. Syn addition of D2 to the trans alkene gives the (R,R) and (S,S) pair, which are two stereoisomers (y = 2).
3. x + y = 1 + 2 = 3.

Teacher's Note:
1. syn + cis gives meso; syn + trans gives a racemic pair, for symmetrically substituted alkenes.
2. Count stereoisomers, not just compounds, when the question asks for the number of products.

Answer: (d) 3

Explanation:
1. Ketohexose HOCH2-CO-(CHOH)3-CH2OH: both terminal CH2OH carbons give HCHO, so 2 mol.
2. Aldohexose OHC-(CHOH)4-CH2OH: only the terminal CH2OH carbon gives HCHO (the CHO carbon becomes HCOOH), so 1 mol.
3. Total = 2 + 1 = 3.

Teacher's Note:
1. Only a CH2OH next to a CHOH or C=O carbon is cleaved off as formaldehyde.
2. The CHO carbon of an aldose ends up as formic acid.

Answer: (c) 7

Explanation:
1. Reaction (1) is esterification (a condensation that forms water), not hydrolysis.
2. Reactions (2) to (8) each split a molecule with water: ester, amide, nitrile, ketal, imine, enol ether and acetal.

Teacher's Note:
1. Hydrolysis breaks a bond using H2O; esterification is its reverse.
2. Check each substrate for a C-O, C-N or C=N bond that water can cleave in acid.

Answer: (b) Cinnamonitrile or methyl cinnamate

Explanation:
1. DIBAL-H reduces a nitrile to the imine, which hydrolyses to the aldehyde.
2. At low temperature it also reduces an ester only to the aldehyde stage.
3. The C=C bond is untouched, so both compounds give cinnamaldehyde.

Teacher's Note:
1. DIBAL-H is the reagent to stop at the aldehyde from both esters and nitriles.
2. LiAlH4 would go on to the alcohol or amine.

Answer: (a) II and III

Explanation:
1. Enol ethers hydrolyse to a carbonyl compound and an alcohol.
2. I gives propanal (no CH3CO group, iodoform negative) and ethanol: not both positive.
3. II gives ethanal (CH3CHO) and propan-2-ol (CH3CH(OH)-): both positive.
4. III gives acetone (CH3CO-) and ethanol (CH3CH(OH)-): both positive.
5. So II and III both qualify.

Teacher's Note:
1. Iodoform test: CH3CO- or CH3CH(OH)- groups.
2. Draw the carbonyl compound from the vinyl part and the alcohol from the alkyl part.

Answer: (d) RCOONa and glycerol

Explanation:
1. Alkali hydrolyses (saponifies) each of the three ester links.
2. Each acid part leaves as the sodium carboxylate (soap), RCOONa.
3. The alcohol part is propane-1,2,3-triol (glycerol).

Teacher's Note:
1. Saponification gives soap plus glycerol.
2. Aldehydes cannot form because no reduction is involved.

Answer: (c) cis-Hex-3-ene

Explanation:
1. Pd-BaSO4 is a poisoned catalyst: hydrogenation stops at the alkene stage.
2. Syn addition of the two hydrogens puts the ethyl groups on the same side: cis-hex-3-ene.

Teacher's Note:
1. Poisoned Pd (Lindlar, Pd-BaSO4) and Ni2B give cis alkenes; Na/NH3 gives trans.
2. A full-strength catalyst would give hexane.

Answer: (b) A racemic mixture of (3R,4R)- and (3S,4S)-hexane-3,4-diol

Explanation:
1. Na/NH3 reduces the alkyne by anti addition, so F is trans-hex-3-ene.
2. Cold alkaline KMnO4 adds two OH groups on the same face (syn addition).
3. Syn hydroxylation of a trans alkene gives the (R,R) and (S,S) diols, formed equally: a racemic mixture.

Teacher's Note:
1. syn + cis gives meso; syn + trans gives racemic.
2. The meso form would have come from the cis alkene.

Answer: (a) meso-Hexane-3,4-diol

Explanation:
1. Pd-BaSO4 gives cis-hex-3-ene (C).
2. Syn hydroxylation of a cis alkene gives the meso diol (3R,4S), which has an internal mirror plane.
3. The racemic pair would arise from the trans alkene.

Teacher's Note:
1. Meso diols have a mirror plane and are optically inactive.
2. syn + cis gives meso.

Answer: (d) Two moles of propanal

Explanation:
1. D is hexane-3,4-diol, a vicinal diol.
2. HIO4 cleaves the C3-C4 bond, each carbon becoming a carbonyl carbon.
3. Each half is CH3CH2-CHO, so two moles of propanal are formed.

Teacher's Note:
1. HIO4 cuts a 1,2-diol into two carbonyl compounds; secondary carbinol carbons become aldehydes only when they carry H.
2. No acid forms because periodic acid does not oxidise the aldehyde further.

Answer: (c) Oxidation by heating with copper followed by reaction with Fehling's solution

Explanation:
1. Heating with copper dehydrogenates 1-propanol to propanal (aldehyde) and 2-propanol to acetone (ketone).
2. Propanal reduces Fehling's solution (red Cu2O); acetone does not.
3. Strong oxidants (KMnO4, dichromate) turn propanol into propanoic acid, which gives a negative Fehling test, so those methods fail; concentrated H2SO4 only dehydrates.

Teacher's Note:
1. To distinguish classes of alcohol, stop at the carbonyl stage with Cu/heat or PCC.
2. Strong oxidants oxidise the aldehyde further to an acid.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. Dimethyl sulphide (Me2S) removes an oxygen atom from the ozonide, giving the aldehydes or ketones and Me2SO (DMSO).
2. DMSO is water soluble and so is easily separated, and any excess Me2S (boiling point about 38 C) evaporates, leaving a clean product. So the Reason explains the Assertion.

Teacher's Note:
1. Reductive work-up (Zn/H2O or Me2S) prevents over-oxidation of aldehydes to acids.
2. Oxidative work-up uses H2O2.

Answer: (a) 4-Hydroxyphenylacetaldehyde

Explanation:
1. A positive Tollens' test needs an aldehyde group; a positive FeCl3 test needs a phenolic OH.
2. The para-substituted C8H8O2 compound with both is HO-C6H4-CH2-CHO (4-hydroxyphenylacetaldehyde).
3. 4-Methoxybenzaldehyde has no phenol group; 4-hydroxyacetophenone has no aldehyde group; 4-methylbenzoic acid has neither.

Teacher's Note:
1. Match each observation with one functional group before assigning the structure.
2. Check the formula: C6H4 + CH2CHO + OH gives C8H8O2.

Answer: (d) C = 4-hydroxyacetophenone, D = 4-methylbenzoic acid, E = 4-acetoxyphenol (4-hydroxyphenyl acetate)

Explanation:
1. Iodoform test needs CH3CO-: 4-HOC6H4COCH3.
2. Extraction into NaHCO3 shows a carboxylic acid: 4-CH3C6H4COOH.
3. Hydrolysis to hydroquinone and acetic acid requires an acetate ester of hydroquinone: 4-HOC6H4OCOCH3.
4. All three have the formula C8H8O2.

Teacher's Note:
1. Only carboxylic acids (not phenols) dissolve in aqueous NaHCO3.
2. Esters hydrolyse to give both an acid and an alcohol or phenol.

Answer: (c) Fehling's solution

Explanation:
1. P gives acetone and acetic acid; Q gives propanal and acetic acid.
2. Propanal is an aldehyde and reduces Fehling's solution; acetone does not.
3. 2,4-DNP and NaHSO3 react with both aldehydes and ketones; Lucas reagent is for alcohols.

Teacher's Note:
1. To distinguish aldehydes from ketones use Tollens', Fehling's, Benedict's or Schiff's.
2. Enol esters hydrolyse to the carbonyl compound (the tautomer of the enol) plus the acid.

Answer: (b) \( \ce{Li/NH3/C2H5OH} \)

Explanation:
1. Alkali metals in liquid ammonia reduce alkynes to trans alkenes via anti addition of hydrogen.
2. The other options give cis alkene (poisoned Pd), the alkane, or no reaction (NaBH4).

Teacher's Note:
1. Li or Na in NH3 gives trans; Lindlar gives cis.
2. NaBH4 does not reduce C#C.

Answer: (a) 0

Explanation:
1. Cleaving each of the four C=C bonds gives two molecules of ethanal and two molecules of 2-methylpropanedial, OHC-CH(CH3)-CHO.
2. Each carbon that was a stereocentre is flanked by two identical CHO groups in the product, so it is no longer a stereocentre.
3. All products are achiral: the number of optically active products is 0.

Teacher's Note:
1. Check each product for four different groups on any carbon.
2. A carbon with two identical -CHO substituents cannot be chiral.

Answer: (d) \( \ce{NaBH4 in C2H5OH} \)

Explanation:
1. NaBH4 reduces aldehydes (and ketones) only; esters, acids and epoxides are untouched.
2. LiAlH4 reduces the ester, the acid and also opens the epoxide.
3. BH3 reduces the carboxylic acid selectively (and the aldehyde); Raney Ni/H2 can open the epoxide and reduce several groups.

Teacher's Note:
1. Choose the mildest reagent that reduces the target group.
2. BH3 reduces acids faster than most other groups, so it would modify the COOH.

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

Explanation:
1. Catalytic hydrogenation over Pd adds H2 across the C=C bond.
2. Zn/HCl and Sn/HCl generate nascent hydrogen, but this does not reduce an isolated alkene; Zn-Hg alone is not a reagent for this.

Teacher's Note:
1. Alkenes need catalytic hydrogenation (Pd, Pt, Ni) for reduction.
2. Metal-acid reductions work on nitro groups, carbonyls and halides.

Answer: (b) Prop-2-en-1-ol (allyl alcohol)

Explanation:
1. LiAlH4 reduces the carboxylic acid group to the primary alcohol.
2. The isolated or conjugated C=C bond of the acid is not attacked under the usual conditions, so the double bond remains.
3. Product: CH2=CH-CH2OH.

Teacher's Note:
1. LiAlH4 normally leaves C=C alone (it delivers hydride to polar C=O groups).
2. The aldehyde is not isolable because it is reduced faster than the acid.

Answer: (a) Butan-2-one

Explanation:
1. Clemmensen reduction (Zn-Hg, HCl) reduces the C=O group of aldehydes and ketones to CH2.
2. Butan-2-one is a ketone and gives butane.
3. Esters, amides and acids are not reduced under Clemmensen conditions.

Teacher's Note:
1. Clemmensen works for aldehydes and ketones only.
2. Acid-sensitive groups (alcohols, alkenes, ethers) are not compatible.

Answer: (d) Ethanal (CH3CHO)

Explanation:
1. Ozonolysis of but-2-ene gives an ozonide that splits at the double bond.
2. Reductive work-up (Zn/H2O) gives two molecules of ethanal.
3. B is CH3CHO.

Teacher's Note:
1. But-2-ene is symmetrical, so both fragments are the same.
2. Zn/H2O stops at the aldehyde (no peroxide to oxidise it).

Answer: (c) But-2-ene

Explanation:
1. An aldehyde of molar mass 44 is ethanal, CH3CHO (C2H4O = 44).
2. A symmetrical alkene that gives two ethanal molecules is CH3CH=CHCH3, but-2-ene.

Teacher's Note:
1. Mass 44: CH3CHO (propanal would be 58, formaldehyde 30).
2. Ethene would give formaldehyde.

Answer: (b) A vinyl group

Explanation:
1. A terminal =CH2 carbon is converted to HCHO on ozonolysis.
2. A vinyl group (-CH=CH2) therefore indicates formaldehyde; isopropyl groups and internal bonds do not give HCHO.

Teacher's Note:
1. Formaldehyde needs a terminal =CH2.
2. Two double bonds would not necessarily give HCHO.