GUJCET 2022 Chemistry Question Paper with Answer and Solution

(C) $1$. Identify the principal functional group: The ether group $(-OC_2H_5)$ is present.
$2$. Number the cyclohexane ring: To give the substituents the lowest possible locants,we start numbering at the carbon attached to the ethoxy group as $1$. Then,we proceed towards the dimethyl-substituted carbon to give it the lowest possible number,which is $1$. Thus,the carbon with two methyl groups is at position $1$.
$3$. The $IUPAC$ name is $2-$ethoxy$-1, 1-$dimethylcyclohexane. Therefore,the correct option is $C$.

(B) The conversion of phenol to salicylic acid is achieved using the $Kolbe$ reaction.
In this reaction,phenol is treated with sodium hydroxide $(NaOH)$ to form sodium phenoxide,which then reacts with carbon dioxide $(CO_2)$ under pressure followed by acidification to yield $2$-hydroxybenzoic acid,commonly known as salicylic acid.

(D) . $C_6H_5COONa + NaOH/CaO \xrightarrow{\Delta} C_6H_6 + Na_2CO_3$ (Benzene is formed).
$B$. $C_6H_5N_2^{+}Cl^{-} + H_3PO_2 + H_2O \longrightarrow C_6H_6 + H_3PO_3 + HCl + N_2$ (Benzene is formed).
$C$. $C_6H_5OH + Zn \xrightarrow{\Delta} C_6H_6 + ZnO$ (Benzene is formed).
$D$. $C_6H_5OH + H_2CrO_4 \xrightarrow{[O]} \text{Benzoquinone}$ (Benzene is $NOT$ formed).
Therefore,the correct option is $D$.

(B) The given reaction is a $Cross-Aldol$ condensation reaction between $benzaldehyde$ $(C_6H_5CHO)$ and $acetaldehyde$ $(CH_3CHO)$ in the presence of a dilute base $(OH^-)$ and heat $(\Delta)$.
$1.$ $Acetaldehyde$ contains $\alpha$-hydrogens,so it forms an enolate ion in the presence of a base.
$2.$ This enolate ion attacks the carbonyl carbon of $benzaldehyde$ (which has no $\alpha$-hydrogens) to form a $\beta$-hydroxy aldehyde intermediate $(C_6H_5CH(OH)CH_2CHO)$.
$3.$ Upon heating $(\Delta)$,this intermediate undergoes dehydration (loss of $H_2O$) to form the stable $\alpha,\beta$-unsaturated aldehyde,which is $cinnamaldehyde$ $(C_6H_5CH=CHCHO)$.
Therefore,the main product is $C_6H_5CH=CHCHO$.

(D) The "Iodoform" test is given by compounds containing the $CH_3CO-$ group or the $CH_3CH(OH)-$ group.
$A$. $CH_3CHO$ (Acetaldehyde) contains the $CH_3CO-$ group,so it gives a positive "Iodoform" test.
$B$. $CH_3CH(OH)CH_2CH_3$ (Butan-$2$-ol) contains the $CH_3CH(OH)-$ group,which is oxidized to a $CH_3CO-$ group by "Sodium hypoiodite",hence it gives a positive "Iodoform" test.
$C$. $CH_3CH_2COCH_3$ (Butan-$2$-one) contains the $CH_3CO-$ group,so it gives a positive "Iodoform" test.
$D$. $CH_3CH_2COCH_2CH_3$ (Pentan-$3$-one) does not contain the $CH_3CO-$ group or the $CH_3CH(OH)-$ group. Therefore,it will not give the "Iodoform" test.
Thus,the correct option is $D$.

(D) The acidic strength of carboxylic acids depends on the electron-withdrawing effect of substituents. Electron-withdrawing groups $(EWG)$ increase acidity by stabilizing the carboxylate anion,while electron-donating groups decrease it.
$A$. $CH_2FCH_2CH_2COOH < CH_3CHFCH_2COOH$: Correct. The $F$ atom is closer to the $COOH$ group in the second molecule,exerting a stronger $-I$ effect.
$B$. $CH_2ClCOOH < CH_2FCOOH$: Correct. $F$ is more electronegative than $Cl$,so $CH_2FCOOH$ is more acidic.
$C$. $CH_3COOH < CH_2ClCOOH$: Correct. $Cl$ is an $EWG$,making the chloroacetic acid more acidic than acetic acid.
$D$. $HCOOH < C_6H_5COOH$: Incorrect. Formic acid $(HCOOH)$ is more acidic than benzoic acid $(C_6H_5COOH)$ because the phenyl group in benzoic acid is electron-donating by resonance and has a weaker $-I$ effect compared to the hydrogen atom in formic acid. Thus,the correct order is $C_6H_5COOH < HCOOH$.

(B) Amylopectin is a branched-chain polymer of $\alpha$-$D$-glucose units.
It consists of a linear chain formed by $C_1-C_4$ glycosidic linkages.
Additionally,it contains branching points formed by $C_1-C_6$ glycosidic linkages.
Therefore,the correct answer is $B$.

(C) Statement $(i)$ is False because $\text{Pentose sugar} + \text{base} \rightarrow \text{Nucleoside}$.
Statement $(ii)$ is False because $\text{Nucleoside} + \text{phosphate} \rightarrow \text{Nucleotide}$.
Statement $(iii)$ is True as $DNA$ contains Adenine $(A)$,Guanine $(G)$,Cytosine $(C)$,and Thymine $(T)$.
Statement $(iv)$ is True as $RNA$ contains Adenine $(A)$,Guanine $(G)$,Cytosine $(C)$,and Uracil $(U)$.
Thus,the sequence is $FFTT$.

(A) The rate law for the reaction is given by $r_1 = k[A]^1[B]^2$.
When the concentrations of both $A$ and $B$ are doubled,the new concentrations become $[A]' = 2[A]$ and $[B]' = 2[B]$.
The new rate $r_2$ is calculated as:
$r_2 = k[2A]^1[2B]^2$
$r_2 = k \times 2[A] \times 4[B]^2$
$r_2 = 8 \times k[A][B]^2$
$r_2 = 8r_1$.
Therefore,the rate increases by $8$ times.

(B) For a first-order reaction,the integrated rate equation is given by:
$K = \frac{2.303}{t} \log \frac{[R]_0}{[R]}$
Rearranging this equation to the form $y = mx + c$:
$\log \frac{[R]_0}{[R]} = \frac{K}{2.303} \times t$
Comparing this with the equation of a straight line $y = mx$,where $y = \log \frac{[R]_0}{[R]}$,$x = t$,and the slope $m = \frac{K}{2.303}$.
Therefore,the slope of the graph is $\frac{K}{2.303}$.

(B) An ambidentate ligand is a ligand that can coordinate to a central metal atom through two different donor atoms.
$(Q) = NO_2^-$ can coordinate through $N$ (nitro) or $O$ (nitrito).
$(R) = CN^-$ can coordinate through $C$ (cyano) or $N$ (isocyano).
$(S) = SCN^-$ can coordinate through $S$ (thiocyanato) or $N$ (isothiocyanato).
$(P) = NO_3^-$ is a monodentate ligand that coordinates only through the oxygen atom and does not exhibit ambidentate behavior.
Therefore,$(P)$ is not an ambidentate ligand.

(D) An ambidentate ligand is a ligand that can coordinate to a central metal atom through two different donor atoms.
In the given list:
$P: NO_3^-$ is not ambidentate.
$Q: NO_2^-$ can coordinate through $N$ (nitro) or $O$ (nitrito),so it is ambidentate.
$R: CN^-$ can coordinate through $C$ (cyano) or $N$ (isocyano),so it is ambidentate.
$S: SCN^-$ can coordinate through $S$ (thiocyanato) or $N$ (isothiocyanato),so it is ambidentate.
Among the given options,$Q$,$R$,and $S$ are ambidentate. The combination $Q$ and $R$ is listed in option $D$.

(A) The complex $[Pt(NH_3)(Br)(Cl)(Py)]$ is of the type $[Mabcd]$,where $M = Pt$,$a = NH_3$,$b = Br$,$c = Cl$,and $d = Py$.
For a square planar complex of the type $[Mabcd]$,the number of geometrical isomers is $3$.
These isomers are formed by fixing one ligand (e.g.,$NH_3$) and arranging the other three ligands ($Br$,$Cl$,$Py$) in the $trans$ positions relative to the fixed ligand.

(A) Transition elements are defined as elements that have incompletely filled $d$-orbitals in their ground state or in any of their oxidation states.
$Zinc$ ($Zn$,atomic number $30$) has the electronic configuration $[Ar] 3d^{10} 4s^2$.
In its common oxidation state of $+2$,it forms $Zn^{2+}$ with the configuration $[Ar] 3d^{10}$.
Since the $d$-orbital is completely filled in both the ground state and the $+2$ oxidation state,$Zinc$ is not considered a transition element and does not exhibit variable oxidation states.

(D) The correct answer is $D$.
Due to the lanthanoid contraction,the atomic radii of elements in the $5d$ series are nearly the same as those of the corresponding elements in the $4d$ series.
Therefore,the statement that the atomic sizes of the $5d$ series are greater than the $4d$ series is incorrect.

(A) In a nickel-cadmium cell,the chemical reactions are as follows:
At the anode: $Cd(s) + 2OH^-(aq) \rightarrow Cd(OH)_2(s) + 2e^-$
At the cathode: $NiO_2(s) + 2H_2O(l) + 2e^- \rightarrow Ni(OH)_2(s) + 2OH^-(aq)$
In the cathode reaction,$NiO_2$ (often represented as $Ni(OH)_3$ in some contexts or hydrated nickel oxide) undergoes reduction,meaning it acts as the oxidising agent.
Therefore,the correct option is $A$.

(B) The cell reaction is: $Zn_{(s)} + Cu_{(0.02 \ M)}^{2+} \rightarrow Zn_{(x \ M)}^{2+} + Cu_{(s)}$.
According to the Nernst equation: $E_{cell} = E_{cell}^0 - \frac{0.059}{2} \log \frac{[Zn^{2+}]}{[Cu^{2+}]}$.
Substituting the values: $E_{cell} = E_{cell}^0 - 0.0295 \log \frac{x}{0.02}$.
To increase $E_{cell}$,the term $\log \frac{x}{0.02}$ must be negative.
This occurs when $\frac{x}{0.02} < 1$,which implies $x < 0.02 \ M$.

(A) The reduction half-reaction for $Cr_2O_7^{2-}$ in an acidic medium is given by:
$Cr_2O_7^{2-} + 14H^{+} + 6e^{-} \rightarrow 2Cr^{3+} + 7H_2O$
From the balanced equation,$1 \ mol$ of $Cr_2O_7^{2-}$ requires $6 \ mol$ of electrons,which is equivalent to $6 \ F$ of electricity.
Therefore,for the reduction of $2 \ mol$ of $Cr_2O_7^{2-}$,the total electricity required is $2 \times 6 \ F = 12 \ F$.

(D) The rate of an $S_N2$ reaction depends primarily on the steric hindrance around the electrophilic carbon atom.
Less sterically hindered substrates react faster in $S_N2$ reactions.
The order of reactivity for $S_N2$ is: $\text{primary} > \text{secondary} > \text{tertiary}$.
In the given options:
$A$ is a secondary alkyl halide.
$B$ is a secondary alkyl halide (with two bulky phenyl groups).
$C$ is a tertiary alkyl halide.
$D$ $(C_6H_5CH_2Br)$ is a primary benzylic halide.
Since $D$ is a primary halide,it experiences the least steric hindrance,making it the most reactive towards $S_N2$ substitution.
Therefore,the correct option is $D$.

(A) The dissociation of $CaCl_2$ is represented as: $CaCl_2 \rightarrow Ca^{2+} + 2Cl^-$.
Initially,let the number of moles be $1$.
After $80\%$ dissociation,the moles are:
$CaCl_2 = 1 - 0.8 = 0.2$
$Ca^{2+} = 0.8$
$Cl^- = 2 \times 0.8 = 1.6$
Total moles after dissociation = $0.2 + 0.8 + 1.6 = 2.6$.
The Van't Hoff factor $(i)$ is defined as the ratio of total moles after dissociation to the initial moles.
$i = \frac{2.6}{1} = 2.6$.
Therefore,the correct option is $A$.

(C) Henry's law states that the solubility of a gas in a liquid is proportional to the partial pressure of the gas above the liquid,given by the equation $P = K_H \cdot x$,where $K_H$ is Henry's constant.
As the temperature increases,the kinetic energy of the gas molecules increases,making it easier for them to escape from the solution.
Consequently,the solubility of the gas decreases.
Since $K_H = P / x$,for a constant partial pressure $P$,an increase in temperature leads to a decrease in the mole fraction $x$ of the dissolved gas.
Therefore,the value of $K_H$ increases with an increase in temperature.

(B) $1 \ m$ solution means $1 \ \text{mole}$ of solute (urea) is dissolved in $1 \ \text{kg}$ $(1000 \ \text{g})$ of solvent (water).
Number of moles of urea $(n_{\text{urea}})$ = $1 \ \text{mol}$.
Number of moles of water $(n_{\text{water}})$ = $\frac{1000 \ \text{g}}{18 \ \text{g/mol}} \approx 55.56 \ \text{mol}$.
Mole fraction of urea $(x_{\text{urea}})$ = $\frac{n_{\text{urea}}}{n_{\text{urea}} + n_{\text{water}}}$.
$x_{\text{urea}} = \frac{1}{1 + 55.56} = \frac{1}{56.56} \approx 0.01768$.