GUJCET 2021 Chemistry Question Paper with Answer and Solution

(D) The reaction is the electrophilic addition of $HCl$ to allylbenzene $(C_6H_5-CH_2-CH=CH_2)$.
According to Markovnikov's rule,the electrophile $(H^+)$ adds to the carbon atom of the double bond that has more hydrogen atoms.
In allylbenzene,the double bond is between $C_2$ and $C_3$ (where $C_1$ is attached to the phenyl ring: $C_6H_5-CH_2-CH=CH_2$).
$H^+$ adds to the terminal $CH_2$ group,forming a stable benzylic carbocation intermediate $(C_6H_5-CH_2-CH^+-CH_3)$.
Then,the nucleophile $(Cl^-)$ attacks this carbocation to form the final product: $C_6H_5-CH_2-CHCl-CH_3$ ($1$-phenyl$-2-$chloropropane).
Therefore,the correct option is $D$.

(D) The $pK_a$ value is inversely proportional to the acidity of the compound. Higher acidity corresponds to a lower $pK_a$ value.
Nitro groups $(-NO_2)$ are strong electron-withdrawing groups $(EWG)$ that increase the acidity of phenol by stabilizing the phenoxide ion through $-I$ and $-M$ effects.
In $p-O_2N-C_6H_4-OH$,$m-O_2N-C_6H_4-OH$,and $o-O_2N-C_6H_4-OH$,the $-NO_2$ group increases acidity compared to phenol $(C_6H_5OH)$.
Since $C_6H_5OH$ lacks any electron-withdrawing substituent,it is the least acidic among the given compounds.
Therefore,$C_6H_5OH$ has the maximum $pK_a$ value.

(A) The reaction of a Grignard reagent $(R-MgX)$ with methanal $(HCHO)$ followed by hydrolysis yields a primary alcohol with one more carbon atom than the Grignard reagent.
To obtain $2-$methylpropan$-1-$ol $(CH_3-CH(CH_3)-CH_2-OH)$,the Grignard reagent must be isobutylmagnesium halide $(CH_3-CH(CH_3)-CH_2-MgX)$.
The reaction is:
$CH_3-CH(CH_3)-CH_2-MgX + HCHO \rightarrow CH_3-CH(CH_3)-CH_2-CH_2-OMgX$
$CH_3-CH(CH_3)-CH_2-CH_2-OMgX + H_2O \rightarrow CH_3-CH(CH_3)-CH_2-CH_2-OH + Mg(OH)X$
Wait,the target product is $2-$methylpropan$-1-$ol $(C_4H_{10}O)$. The reaction of $R-MgX + HCHO$ adds one carbon. Thus,$R$ must be an isopropyl group $(CH_3-CH(CH_3)-)$.
$CH_3-CH(CH_3)-MgX + HCHO$ $\rightarrow CH_3-CH(CH_3)-CH_2-OMgX$ $\xrightarrow{H_2O} CH_3-CH(CH_3)-CH_2-OH$.
Therefore,the correct Grignard reagent is isopropylmagnesium halide,which is option $A$.

(D) is the correct answer.
Acidic strength of benzoic acid derivatives is increased by electron-withdrawing groups $(EWG)$ and decreased by electron-donating groups $(EDG)$.
$4-$Nitrobenzoic acid contains a $-NO_2$ group,which is a strong electron-withdrawing group ($-I$ and $-M$ effect),thereby stabilizing the carboxylate anion and increasing acidity.
$2-$Methoxybenzoic acid and $4-$Methoxybenzoic acid contain a $-OCH_3$ group,which acts as an electron-donating group ($+M$ effect),decreasing the acidic strength compared to benzoic acid.
Therefore,$4-$Nitrobenzoic acid has the maximum acidic strength among the given options.

(C) When glucose is heated with prolonged $HI$,it forms $n$-hexane $(CH_3-(CH_2)_4-CH_3)$.
This reaction confirms that all six carbon atoms in glucose are linked in a straight chain.

(C) For a first order reaction,the rate equation is given by $k = \frac{2.303}{t} \log \frac{[R]_0}{[R]}$.
Given that the concentration reduces to $1/16^{th}$ of its initial value,$[R] = \frac{[R]_0}{16}$,so $\frac{[R]_0}{[R]} = 16$.
Substituting the values: $t = \frac{2.303}{k} \log(16)$.
$t = \frac{2.303}{60} \log(2^4) = \frac{2.303 \times 4 \times 0.3010}{60}$.
$t = \frac{2.303 \times 1.204}{60} \approx 0.0462 \text{ s}$.
Therefore,$t = 4.6 \times 10^{-2} \text{ s}$.

(B) The unit of the rate constant $K$ for a reaction of order $n$ is given by $(mol \ L^{-1})^{1-n} \ s^{-1}$.
For a second-order reaction $(n = 2)$,the unit is $(mol \ L^{-1})^{1-2} \ s^{-1} = (mol \ L^{-1})^{-1} \ s^{-1} = L \ mol^{-1} \ s^{-1}$.
Since the given unit is $L \ mol^{-1} \ s^{-1}$,the reaction is of the second order.

(B) The given isomers are $[Co(NH_3)_5(SO_4)]Br$ and $[Co(NH_3)_5Br]SO_4$.
These compounds produce different ions in an aqueous solution.
$[Co(NH_3)_5(SO_4)]Br$ gives $Br^-$ ions,while $[Co(NH_3)_5Br]SO_4$ gives $SO_4^{2-}$ ions.
This type of isomerism,where the counter ion in the coordination sphere is exchanged with a ligand,is known as $Ionisation$ isomerism.

(A) $1$. The central metal ion is Cobalt in the $+3$ oxidation state,denoted as $Co(III)$.
$2$. The ligand 'pentaammine' corresponds to $5$ ammonia molecules,$(NH_3)_5$.
$3$. The ligand 'carbonato' corresponds to the carbonate ion,$CO_3^{2-}$.
$4$. The coordination sphere is $[Co(NH_3)_5(CO_3)]$.
$5$. To balance the charge: $Co$ is $+3$,$CO_3$ is $-2$,and $NH_3$ is $0$. The total charge on the coordination sphere is $(+3) + (-2) = +1$.
$6$. To neutralize the $+1$ charge of the complex,one chloride ion $(Cl^-)$ is required outside the coordination sphere.
$7$. Therefore,the correct formula is $[Co(NH_3)_5(CO_3)]Cl$.

(D) The element with atomic number $26$ is Iron $(Fe)$.
Its electronic configuration is $[Ar] 3d^6 4s^2$.
The divalent ion $Fe^{2+}$ has the configuration $[Ar] 3d^6$.
In the $3d$ subshell, there are $4$ unpaired electrons.
The magnetic moment $(\mu)$ is calculated using the formula $\mu = \sqrt{n(n+2)} \ BM$, where $n$ is the number of unpaired electrons.
Substituting $n = 4$, we get $\mu = \sqrt{4(4+2)} = \sqrt{24} \approx 4.90 \ BM$.

(A) In a cell representation, the anode (oxidation) is written on the left and the cathode (reduction) on the right.
$1$. Oxidation half-reaction: $Mg_{(s)} \rightarrow Mg^{2+}_{(aq)} + 2e^-$. This occurs at the anode: $Mg_{(s)} \mid Mg^{2+}_{(aq)}$.
$2$. Reduction half-reaction: $Cl_{2(g)} + 2e^- \rightarrow 2Cl^-_{(aq)}$. This occurs at the cathode. Since $Cl_2$ is a gas, an inert electrode like $Pt$ is used: $Cl_{2(g)} \mid Cl^-_{(aq)} \mid Pt_{(s)}$.
$3$. Combining these with the salt bridge $(\parallel)$, the representation is $Mg_{(s)} \mid Mg^{2+}_{(aq)} \parallel Cl_{2(g)} \mid Cl^-_{(aq)} \mid Pt_{(s)}$.

(C) The strength of a reducing agent is inversely proportional to its standard reduction potential $(E^0_{red})$. $A$ lower (more negative) $E^0_{red}$ value indicates a stronger reducing agent.
Comparing the given values:
$E^0_{MnO_4^-/Mn^{2+}} = 1.51 \text{ V}$
$E^0_{Cl_2/Cl^{-}} = 1.36 \text{ V}$
$E^0_{Cr_2O_7^{2-}/Cr^{3+}} = 1.33 \text{ V}$
$E^0_{Cr^{3+}/Cr} = -0.74 \text{ V}$
Since $E^0_{Cr^{3+}/Cr}$ has the lowest (most negative) value of $-0.74 \text{ V}$,the species $Cr$ acts as the strongest reducing agent.

(C) During the electrolysis of an aqueous solution of sodium chloride ($NaCl$ (aq)),the following reactions occur:
At the cathode: $2H_2O(l) + 2e^- \rightarrow H_2(g) + 2OH^-(aq)$.
At the anode: $2Cl^-(aq) \rightarrow Cl_2(g) + 2e^-$.
The remaining ions in the solution are $Na^+$ and $OH^-$,which combine to form $NaOH(aq)$.
Therefore,the products obtained are $NaOH, Cl_2$ and $H_2$.

(D) The rate of $S_N 1$ reaction depends on the stability of the carbocation intermediate formed during the rate-determining step.
$S_N 1$ reactivity order is: $3^{\circ} > 2^{\circ} > 1^{\circ} > CH_3-X$.
$A$: $2-$bromo-$3-$methylbutane is a $2^{\circ}$ halide.
$B$: $2-$chloro-$3-$methylbutane is a $2^{\circ}$ halide.
$C$: Chloromethane is a $1^{\circ}$ halide.
$D$: $2-$bromo-$2-$methylpropane is a $3^{\circ}$ halide.
Since $3^{\circ}$ carbocations are the most stable,$2-$bromo-$2-$methylpropane will undergo $S_N 1$ reaction the fastest.

(C) The given compound is $CH_3-CH=CH-C(Cl)(CH_3)_2$.
In this structure,the $-Cl$ atom is attached to a carbon atom that is directly bonded to a carbon-carbon double bond $(C=C)$.
Specifically,the carbon atom bearing the $-Cl$ is adjacent to the $sp^2$ hybridized carbon of the double bond.
Such a halide,where the halogen is attached to an $sp^3$ hybridized carbon atom next to a carbon-carbon double bond,is classified as an allylic halide.

(C) The process of removing salt from sea water to make it potable is known as desalination.
Reverse osmosis is the most common method used for this purpose.
In this process,a semi-permeable membrane is used to separate salt from water by applying pressure greater than the osmotic pressure.
Therefore,the correct option is $C$.

(B) The molar mass of ethanoic acid $(CH_3COOH)$ is $(2 \times 12) + (4 \times 1) + (2 \times 16) = 24 + 4 + 32 = 60 \ g/mol$.
Number of moles of ethanoic acid = $\frac{\text{mass}}{\text{molar mass}} = \frac{3.0 \ g}{60 \ g/mol} = 0.05 \ mol$.
Mass of solvent (benzene) = $50 \ g = 0.05 \ kg$.
Molality $(m)$ = $\frac{\text{moles of solute}}{\text{mass of solvent in kg}} = \frac{0.05 \ mol}{0.05 \ kg} = 1.0 \ mol/kg$.
Therefore,the correct option is $B$.

(A) The elevation in boiling point is a colligative property,which depends on the van't Hoff factor $(i)$ and the concentration of the solute particles. The formula is $\Delta T_b = i \times K_b \times m$.
Since the molality $(m)$ is the same for all solutions $(0.1 \ M)$,the solution with the minimum van't Hoff factor $(i)$ will have the minimum boiling point elevation,and thus the minimum boiling point.
For Urea (non-electrolyte),$i = 1$.
For $K_2SO_4$ $(2K^+ + SO_4^{2-})$,$i = 3$.
For $NaCl$ $(Na^+ + Cl^-)$,$i = 2$.
For $FeCl_3$ $(Fe^{3+} + 3Cl^-)$,$i = 4$.
Comparing the values,Urea has the lowest $i$ value $(1)$,therefore $0.1 \ M$ Urea has the minimum boiling point.