
JEE Main 2024 Jan 29 Shift 2 Chemistry Question Paper with Solution pdf is available for download here. Students found easy and hard. carried the highest weightage and overall difficulty level was moderate.
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The ascending acidity order of given H atoms is:
The relative acidities follow from the stability of their conjugate bases in the order C < D < B < A.
Match List I (Bio Polymer) with List II (Monomer):
A. Starch | I. Nucleotide
B. Cellulose | II. α-glucose
C. Nucleic acid | III. β-glucose
D. Protein | IV. α-amino acid
Starch is α-glucose based, Cellulose is β-glucose, Nucleic acids are nucleotides, and Proteins are α-amino acids.
Match List I (Compound) with List II (pKa value):
A. Ethanol | II. 15.9
B. Phenol | I. 10.0
C. m-Nitrophenol | IV. 8.3
D. p-Nitrophenol | III. 7.1
Ethanol=15.9, Phenol=10.0, m-Nitrophenol=8.3, p-Nitrophenol=7.1.
Which reaction is correct?
HI adds Markovnikov-style, attaching I to less hydrogenated carbon.
IUPAC name of the compound?
The double bond and OH are numbered for lowest locants => 2-en-1-ol.
Correct IUPAC name of K₂MnO₄ is?
Mn is in +6 oxidation => “tetraoxidomanganate(VI).”
Reagent giving red ppt with Ni²⁺ in basic medium?
DMG + Ni²⁺ => red ppt of Ni(dmg)₂ in basic medium.
Phenol + CHCl₃ + NaOH => acid hydrolysis => product= ? => 2-hydroxybenzaldehyde
Reimer-Tiemann reaction forms salicylaldehyde (2-hydroxybenzaldehyde).
Match H spectral series vs region: Lyman/Balmer/Paschen/Pfund => UV/Visible/IR/IR
Lyman=UV, Balmer=visible, Paschen & Pfund=IR.
A brown ppt with Nessler’s reagent => gas=? => NH₃
Nessler’s reagent + ammonia => brown precipitate (Millon's base).
The product A in a diazotization + Sandmeyer reaction => chlorobenzene
Diazonium salt + Cu₂Cl₂ => chlorobenzene.
Identify reagents for a certain conversion => DIBAL-H, NaOH(alc), Zn/HCl => correct set?
DIBAL-H => partial ester reduction, NaOH(alc)=> aldol steps, Zn/HCl=> Clemmensen.
Strong reducing agent among Ce=58, Eu=63, Gd=64, Lu=71 => ? => Eu²⁺
Eu²⁺ easily oxidizes to Eu³⁺ => strong reducing agent.
Which chromatography is based on differential adsorption? => Column & TLC => (A,B only)
Column & Thin Layer rely on adsorption, Paper uses partition.
Statements re: Zn, Cd, Hg => correct ones? => B & D only
Zn & Cd no variable O.S., Hg does +1,+2; all are soft metals.
Highest first I.E. among Si, Al, N, C => ? => N
N’s half-filled 2p³ => highest I.E.
Alkyl halide => alkyl isocyanide => reagent=? => AgCN
AgCN + R–X => R–NC due to covalent nature.
Which has geometrical isomerism? => CH₃CH=CHBr
Different substituents => restricted rotation => cis/trans forms.
I: F has most negative EGE in group => false. II: O has least negative EGE in group => true => so (I false, II true)
Cl has more negative EGE than F; O is smallest negative in its group.
Oxygen’s anomaly => small size + high electronegativity
O's unique properties arise from small radius & strong EN.
Total antibonding MOs from 2s,2p in a diatomic => 4
1 antibonding from 2s*, plus 3 from 2p* => total=4.
Brown ring test complex => Fe in +1 state
[Fe(H₂O)₅(NO)]²⁺ => Fe=+1 once charges are tallied.
NH₃ formation => [N₂]=2×10⁻², [H₂]=3×10⁻², [NH₃]=1.5×10⁻² => Kc=? =>417
Kc= [NH₃]² / ([N₂][H₂]³)= (1.5×10⁻²)² /((2×10⁻²)(3×10⁻²)³)=417.
0.8 M H₂SO₄, density=1.06 => molality=? => 815×10⁻³ m
Mass(1 L)=1060 g, moles acid=0.8 => ~78.4 g => solute in 981.6 g solvent => molality=0.08 mol/kg => 0.815 => 815×10⁻³.
50 mL NaOH neutralizes 50 mL 0.5 M oxalic => mass(NaOH)=4 g
Eq’s => n(oxalic)= 0.5×0.05=0.025 => n(NaOH)= 2×0.025=0.05 => mass=2g/0.05? => 4g total.
2-formylhex-4-enoic acid => total σ+π bonds=22
Summing 16 sigma, 6 pi => 22 total.
Bromine-82 half-life=36 h => fraction left after 24 h=? =>0.63
Using N/N₀= (1/2)^(24/36)= (1/2)^(2/3)=~0.63.
ΔH(vap)(CCl₄)=30.5 kJ/mol => 284 g => heat=? =>56 kJ
Moles=284/154=1.84 => total=1.84×30.5=56 kJ.
AuCl₄⁻ electrolyzed => cathode mass gained 1.314 g => total charge=2 F
Q= n(e⁻)×F => from mass & eq.wt => 2 F total.
Zero dipole among CH₄,BF₃,H₂O,HF,NH₃,CO₂,SO₂ => total=? =>3
CH₄,BF₃,CO₂ have zero net dipole => total 3.
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