JEE Main 2019 Chemistry Question Paper with Answer and Solution

(A) Glucose and Galactose are epimers at the $C-4$ position.
In $D$-glucose,the hydroxyl group $(-OH)$ at $C-4$ is in the equatorial position (in the Haworth projection,it is below the plane).
In $D$-galactose,the hydroxyl group $(-OH)$ at $C-4$ is in the axial position (in the Haworth projection,it is above the plane).
Therefore,they differ in configuration only at the $C-4$ carbon atom.

(A) $RNA$ is typically single-stranded and does not form a double-stranded $\alpha - \text{helix}$ structure like $DNA$. Therefore,the statement that it always has a double-stranded $\alpha - \text{helix}$ structure is incorrect.

(A) The $pK_b$ value is inversely proportional to the basic strength of the compound.
Basic strength depends on the availability of the lone pair on the nitrogen atom.
Electron-donating groups $(EDG)$ increase basicity (decrease $pK_b$),while electron-withdrawing groups $(EWG)$ decrease basicity (increase $pK_b$).
The substituents on the phenyl ring are: $(A) -F$ ($EWG$ by inductive effect),$(B) -OCH_3$ ($EDG$ by resonance),$(C) -NO_2$ (strong $EWG$),$(D) -CH_3$ ($EDG$ by hyperconjugation).
Basicity order: $(B) > (D) > (A) > (C)$.
Since $pK_b$ is inversely proportional to basicity,the increasing order of $pK_b$ is: $(C) < (A) < (D) < (B)$.

(C) The oxidizing power of a species is directly proportional to its standard reduction potential $(E^o)$.
Comparing the given values:
$Bi^{3+} (E^o = +0.20 \ V) < Ce^{4+} (E^o = +1.61 \ V) < Pb^{4+} (E^o = +1.67 \ V) < Co^{3+} (E^o = +1.81 \ V)$.
Therefore,the order of increasing oxidizing power is $Bi^{3+} < Ce^{4+} < Pb^{4+} < Co^{3+}$.

(C) Step $1$: Oxidation of the primary alcohol group with $CrO_3$ yields a carboxylic acid group. The phenolic $-OH$ group remains unaffected under these conditions.
Step $2$: Treatment with $SOCl_2/\Delta$ converts the carboxylic acid into an acid chloride $(-COCl)$.
Step $3$: Heating $(\Delta)$ induces a Friedel-Crafts acylation reaction. The acid chloride group undergoes intramolecular cyclization with the ortho-position of the phenol ring to form a cyclic ketone (indanone derivative).
The final product is a $5$-hydroxy-indanone derivative.

(B) In an octahedral complex,the ligands are present along the $x, y,$ and $z$ axes.
When the two ligands along the $z-$ axis are removed,the complex becomes square planar.
The ligands now exert repulsion only along the $x$ and $y$ axes.
Consequently,the orbitals with components along the $x$ and $y$ axes ($d_{x^2-y^2}$ and $d_{xy}$) will experience higher repulsion and thus have higher energy.
The $d_{z^2}$ orbital,having no ligand interaction along the $z-$ axis,will decrease significantly in energy.
The $d_{xz}$ and $d_{yz}$ orbitals will also decrease in energy but remain higher than $d_{z^2}$ due to their interaction with the $x$ and $y$ ligands.
The correct splitting pattern for a square planar complex is $d_{x^2-y^2} > d_{xy} > d_{z^2} > d_{xz}, d_{yz}$,which corresponds to the pattern shown in image $821-b1769$.

(A) $m$-Cresol reacts with $K_2CO_3$ to form the phenoxide ion ($m$-methylphenoxide).
This phenoxide ion acts as a nucleophile in an $S_N2$ reaction with propargyl bromide $(HC \equiv C-CH_2Br)$.
The oxygen atom of the phenoxide ion attacks the electrophilic carbon of the propargyl bromide,displacing the bromide ion.
This results in the formation of an ether linkage,yielding $3$-methylphenyl prop-$2$-ynyl ether as the major product.

(C) $(CH_3)_3CCl$ undergoes ionization to form a stable $tert$-butyl carbocation $(CH_3)_3C^+$ and a chloride ion $Cl^-$.
The $Cl^-$ ion reacts with $Ag^+$ ions from $AgNO_3$ to form a white precipitate of $AgCl$.
Other compounds like $CH_2=CHCl$ (vinyl chloride) have partial double bond character due to resonance,and $CHCl_3$ and $CCl_4$ are covalent and do not easily ionize to give $Cl^-$ ions under these conditions.

(B) When benzene diazonium chloride reacts with aniline in the presence of dilute $HCl$ at a low temperature $(273-278 \ K)$,an electrophilic substitution reaction occurs at the para-position of the aniline ring. This process is known as coupling reaction,which yields $p$-aminoazobenzene,a yellow-colored dye.

(D) The coordination number is defined as the number of ligand donor atoms to which the metal is directly bonded.
$en$ (ethane$-1,2-$diamine) is a bidentate ligand,meaning it occupies $2$ coordination sites.
In $[Co(Cl)(en)_2]Cl$,the coordination number of $Co$ is $1(Cl) + 2(2 \times en) = 5$.
$C_2O_4^{2-}$ (oxalate) is also a bidentate ligand.
In $K_3[Al(C_2O_4)_3]$,the coordination number of $Al$ is $3(C_2O_4) \times 2 = 6$.
Therefore,the coordination numbers are $5$ and $6$.

(D) Glycogen is a highly branched polysaccharide,not a straight chain polymer. It is often referred to as animal starch. It consists of $\alpha -D-$glucose units linked by $\alpha -1,4-$glycosidic bonds in the linear chains and $\alpha -1,6-$glycosidic bonds at the branch points. Therefore,the statement that it is a straight chain polymer similar to amylose is $INCORRECT$.

(A) In a simple cubic unit cell,the number of atoms is $1$.
In a body-centered cubic $(BCC)$ unit cell,the number of atoms is $2$.
In a face-centered cubic $(FCC)$ unit cell,the number of atoms is $4$.
Therefore,the ratio is $1 : 2 : 4$.

(D) The thermal decomposition of potassium permanganate $(KMnO_4)$ at $513 \ K$ is given by:
$2KMnO_4 (X) \xrightarrow{513 \ K} K_2MnO_4 (Y) + MnO_2 + O_2 (g)$
Thus,$X = KMnO_4$ and $Y = K_2MnO_4$.
$MnO_2$ reacts with $NaCl$ and concentrated $H_2SO_4$ to produce chlorine gas,which is a pungent gas:
$MnO_2 + 4NaCl + 4H_2SO_4 \to MnCl_2 + 4NaHSO_4 + 2H_2O + Cl_2 (Z)$
Thus,$Z = Cl_2$.
Therefore,the correct sequence is $X = KMnO_4, Y = K_2MnO_4, Z = Cl_2$.

(D) Vinyl halides $(CH_2=CH-X)$ do not undergo nucleophilic substitution reactions easily because of the partial double bond character between the carbon and the halogen atom due to resonance.
The reason provided is incorrect because the intermediate carbocation formed from vinyl halides is highly unstable,not stabilized,and the $C-X$ bond is strong due to $sp^2$ hybridization and resonance,making cleavage difficult.
Therefore,$(A)$ is a correct statement,but $(R)$ is a wrong statement.

(A) . $EDTA$ (ethylene diamine tetraacetate) is used for the treatment of lead poisoning.
$B$. $Cis-platin$ is used as an anti-cancer drug.
$C$. $D-penicillamine$ is used for copper poisoning.
$D$. $Desferrioxime \ B$ is used for iron poisoning.

(D) Due to the lanthanoid contraction, the elements of the $5d$ series have atomic radii very similar to those of the corresponding elements of the $4d$ series. $Mo$ ($4d$ series) and $W$ ($5d$ series) exhibit this phenomenon, resulting in similar atomic radii.

(A) In the leaching of bauxite $(Al_2O_3 \cdot 2H_2O)$,the ore contains impurities like $SiO_2$,$Fe_2O_3$,and $TiO_2$.
When treated with concentrated $NaOH$ solution,$Al_2O_3$ dissolves to form sodium aluminate $(Na[Al(OH)_4])$ and $SiO_2$ dissolves to form sodium silicate $(Na_2SiO_3)$.
Thus,option $A$ is correct.
Option $B$ is incorrect because the Hall-Heroult process is used only for aluminium.
Option $C$ is incorrect because the blistered appearance of copper is due to the evolution of $SO_2$ gas,not $CO_2$.
Option $D$ is incorrect because cast iron is obtained from pig iron.

(A) The given reaction is $2N_2O_{5(g)} \to 4NO_{2(g)} + O_{2(g)}$.
Rate of reaction is given by: $\text{Rate} = -\frac{1}{2} \frac{\Delta [N_2O_5]}{\Delta t} = \frac{1}{4} \frac{\Delta [NO_2]}{\Delta t}$.
Change in concentration of $N_2O_5$ is $\Delta [N_2O_5] = 2.75 - 3.00 = -0.25 \, mol \, L^{-1}$.
Time interval $\Delta t = 30 \, min$.
Rate of disappearance of $N_2O_5 = -\frac{\Delta [N_2O_5]}{\Delta t} = -\frac{-0.25}{30} = 8.333 \times 10^{-3} \, mol \, L^{-1} \, min^{-1}$.
From the stoichiometry,$\frac{1}{4} \frac{\Delta [NO_2]}{\Delta t} = -\frac{1}{2} \frac{\Delta [N_2O_5]}{\Delta t}$.
Therefore,$\frac{\Delta [NO_2]}{\Delta t} = 2 \times (-\frac{\Delta [N_2O_5]}{\Delta t}) = 2 \times 8.333 \times 10^{-3} = 1.667 \times 10^{-2} \, mol \, L^{-1} \, min^{-1}$.

(D) The given structure is the repeating unit of a polymer formed from the monomer isobutylene $(CH_2=C(CH_3)_2)$.
Upon polymerization,the double bond breaks to form a chain,resulting in the polymer polyisobutylene.

(A) Step $1$: Dehydrohalogenation of $2-$chloro$-1-$phenylbutane with $EtOK/EtOH$ (a strong base) proceeds via the $E2$ mechanism to form the more stable conjugated alkene,$1-$phenylbut$-1-$ene $(X)$,as the major product.
Step $2$: The reaction of $1-$phenylbut$-1-$ene $(X)$ with $Hg(OAc)_2/H_2O$ followed by $NaBH_4$ is an Oxymercuration-Demercuration $(OMDM)$ reaction.
Step $3$: $OMDM$ follows the Markovnikov addition rule,where the $-OH$ group attaches to the more substituted carbon atom of the double bond.
Step $4$: In $1-$phenylbut$-1-$ene $(Ph-CH=CH-CH_2-CH_3)$,the carbon at position $1$ is attached to a phenyl group,making it more substituted than the carbon at position $2$. Therefore,the $-OH$ group adds to the $C1$ position,yielding $1-$phenylbutan$-1-$ol as the major product $Y$.

(B) Colloidal particles are aggregates of molecules or ions with diameters ranging from $1 \ nm$ to $1000 \ nm$.
Because colloidal particles are much larger than the solute particles in a true solution,the number of particles in a given mass of colloid is much smaller than in a true solution.
Since osmotic pressure is a colligative property and depends on the number of particles,the osmotic pressure of a colloidal solution is of a lower order than that of a true solution at the same concentration.
Therefore,the statement in option $B$ is $INCORRECT$.

(C) The osmotic pressure $\pi$ is given by the formula $\pi = CRT$,where $C$ is the total molarity of the solution.
First,calculate the moles of urea: $n_{\text{urea}} = \frac{0.6 \ g}{60 \ g \ mol^{-1}} = 0.01 \ mol$.
Next,calculate the moles of glucose: $n_{\text{glucose}} = \frac{1.8 \ g}{180 \ g \ mol^{-1}} = 0.01 \ mol$.
Total moles of solute $= 0.01 + 0.01 = 0.02 \ mol$.
The volume of the solution is $100 \ mL = 0.1 \ L$.
Total molarity $C = \frac{0.02 \ mol}{0.1 \ L} = 0.2 \ M$.
Temperature $T = 27 \ ^oC = 27 + 273 = 300 \ K$.
Now,calculate osmotic pressure: $\pi = 0.2 \ mol \ L^{-1} \times 0.08206 \ L \ atm \ K^{-1} \ mol^{-1} \times 300 \ K$.
$\pi = 0.2 \times 0.08206 \times 300 = 4.9236 \ atm$.

(B) The starting material is $3$-benzoylpropanenitrile. It contains both a ketone and a nitrile functional group.
When treated with excess $CH_3MgBr$,the Grignard reagent attacks both the carbonyl carbon of the ketone and the carbon of the nitrile group.
$1$. The ketone group reacts with one equivalent of $CH_3MgBr$ to form a tertiary alcohol after workup.
$2$. The nitrile group reacts with another equivalent of $CH_3MgBr$ to form an imine intermediate,which upon hydrolysis $(H_3O^+)$ yields a ketone.
Therefore,the final product is $5$-hydroxy-$5$-phenylhexan-$2$-one.