GUJCET 2024 Chemistry Question Paper with Answer and Solution

(B) The given reaction is the reductive ozonolysis of $1,1'$-bicyclohexylidene.
Ozonolysis of an alkene involves the cleavage of the $C=C$ double bond to form two carbonyl compounds.
When the alkene is $1,1'$-bicyclohexylidene,the double bond connects two cyclohexane rings.
Cleavage of this double bond results in the formation of two molecules of cyclohexanone $(C_6H_{10}O)$.
Therefore,the product $X$ is two molecules of cyclohexanone.

(D) In acidic medium,$KMnO_4$ acts as an oxidising agent and gets reduced to $Mn^{2+}$. The half-reaction is: $MnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O$.
Sulphide ion $(S^{2-})$ is oxidised to elemental sulphur $(S)$: $S^{2-} \rightarrow S + 2e^-$.
To balance the electrons,we multiply the reduction half-reaction by $2$ and the oxidation half-reaction by $5$:
$2MnO_4^- + 16H^+ + 10e^- \rightarrow 2Mn^{2+} + 8H_2O$
$5S^{2-} \rightarrow 5S + 10e^-$
Adding these,the balanced equation is: $2MnO_4^- + 5S^{2-} + 16H^+ \rightarrow 2Mn^{2+} + 5S + 8H_2O$.
From the stoichiometry,$5$ moles of $S^{2-}$ react with $2$ moles of $KMnO_4$.
Therefore,$1$ mole of $S^{2-}$ reacts with $2/5$ moles of $KMnO_4$.

(A) The reaction is the electrophilic addition of $HX$ to an alkene,which follows Markovnikov's rule.
According to Markovnikov's rule,the negative part of the addendum $(X^-)$ attaches to the carbon atom of the double bond that has fewer hydrogen atoms.
The starting material is $3,3-\text{dimethylbut-1-ene}$ (or a similar branched alkene structure as shown).
Upon protonation of the double bond,a carbocation is formed. The proton $(H^+)$ adds to the terminal $CH_2$ group to form a more stable secondary or tertiary carbocation.
In this specific structure,the addition of $H^+$ to the terminal carbon creates a secondary carbocation at the $C-2$ position,which can undergo a hydride or methyl shift to form a more stable tertiary carbocation.
However,looking at the options provided,the product formed by direct Markovnikov addition to the double bond is $CH_3-C(CH_3)(X)-CH_2-CH_3$ (where $X$ is at the $C-2$ position).
Thus,the major product is $CH_3-C(CH_3)(X)-CH_2-CH_3$.

(A) The acidic strength of phenols depends on the nature of the substituent attached to the benzene ring.
Electron-withdrawing groups $(EWG)$ like $-NO_2$ increase the acidic strength by stabilizing the phenoxide ion through $-I$ and $-M$ effects.
Electron-donating groups $(EDG)$ like $-CH_3$ decrease the acidic strength by destabilizing the phenoxide ion through $+I$ and hyperconjugation effects.
Compound $(ii)$ has a $-NO_2$ group (strong $EWG$),which makes it the most acidic.
Compound $(i)$ is phenol (no substituent).
Compound $(iii)$ has a $-CH_3$ group $(EDG)$,which makes it the least acidic.
Therefore,the decreasing order of acidic strength is $(ii) > (i) > (iii)$.

(D) The reaction of an ether with $HI$ proceeds via an $S_N1$ or $S_N2$ mechanism depending on the nature of the alkyl groups attached to the oxygen atom.
In the given ether,$CH_3-CH_2-CH_2-O-C(CH_3)_2-CH_2-CH_3$,one alkyl group is a primary alkyl group ($n$-propyl) and the other is a tertiary alkyl group ($3$-methylpentan$-3-$yl group).
When an ether has a tertiary alkyl group,the reaction follows the $S_N1$ mechanism because the tertiary carbocation is highly stable.
The oxygen atom is protonated by $HI$ to form an oxonium ion,which then undergoes cleavage to form the most stable carbocation.
The tertiary carbocation,$CH_3-CH_2-C^+(CH_3)_2$,is formed,which then reacts with the iodide ion $(I^-)$ to form the tertiary alkyl iodide,$CH_3-CH_2-C(I)(CH_3)_2$.
The remaining part forms the primary alcohol,$CH_3-CH_2-CH_2-OH$.
Therefore,the major products are $CH_3-CH_2-CH_2-OH$ and $CH_3-CH_2-C(I)(CH_3)_2$.

(B) The acid-catalyzed dehydration of butan-$1$-ol $(CH_3-CH_2-CH_2-CH_2OH)$ involves the formation of a primary carbocation,which undergoes a $1,2$-hydride shift to form a more stable secondary carbocation $(CH_3-CH^+-CH_2-CH_3)$.
Elimination of a proton from this secondary carbocation yields but-$2$-ene $(CH_3-CH=CH-CH_3)$ as the major product,following Zaitsev's rule,as it is more substituted and thus more stable than but-$1$-ene $(CH_3-CH_2-CH=CH_2)$.

(A) The $pK_{a}$ value is inversely proportional to the acidity of the carboxylic acid. Stronger acids have lower $pK_{a}$ values.
Acidity is determined by the stability of the conjugate base (carboxylate ion).
Electron-withdrawing groups (EWGs) stabilize the carboxylate ion through the inductive effect ($-I$ effect),thereby increasing acidity.
Comparing the substituents:
$1$. $NO_{2}CH_{2}COOH$: The $-NO_{2}$ group is a strong electron-withdrawing group ($-I$ effect),which significantly stabilizes the carboxylate ion.
$2$. $CH_{3}COOH$: The $-CH_{3}$ group is an electron-donating group ($+I$ effect),which destabilizes the carboxylate ion.
$3$. $HCOOH$: Hydrogen has no inductive effect.
$4$. $C_{6}H_{5}COOH$: The phenyl group is electron-withdrawing by resonance but less effective than the $-NO_{2}$ group in this context.
Therefore,$NO_{2}CH_{2}COOH$ is the strongest acid among the given options and has the lowest $pK_{a}$ value.

(C) $Schiff's \ base$ is formed by the reaction of a primary amine $(R-NH_2)$ with an aldehyde or a ketone.
In the given reaction,$CH_3-CO-CH_3$ (acetone) reacts with $R$ to form a $Schiff's \ base$.
Among the given options,$CH_3-NH_2$ is a primary amine.
The reaction is: $CH_3-CO-CH_3 + CH_3-NH_2 \xrightarrow{H^{+}} CH_3-C(CH_3)=N-CH_3 + H_2O$.
Therefore,$R$ is $CH_3-NH_2$.

(D) The given reaction is the hydrogenation of an acid chloride $(RCOCl)$ to an aldehyde $(RCHO)$ using a palladium catalyst poisoned with barium sulfate $(Pd-BaSO_4)$.
This specific reaction is known as the $Rosenmund$ reduction.
Therefore,the correct option is $D$.

(C) Lactose is a disaccharide composed of two monosaccharide units.
These units are $β-D$-galactose and $β-D$-glucose.
The glycosidic linkage present in lactose is $β(1 \rightarrow 4)$ linkage.
Therefore,the correct option is $C$.

(D) The correct answer is $D$. $RNA$ contains adenine,guanine,cytosine,and uracil,not thymine. Thymine is present in $DNA$.

(D) The deficiency of Vitamin $B_{12}$ (cyanocobalamin) leads to pernicious anaemia.
In this condition,the body produces $RBC$s that are deficient in haemoglobin and are larger than normal,which impairs their ability to transport oxygen effectively.

(B) Zwitter ion is a dipolar ion that contains both a positive and a negative charge within the same molecule,resulting in a net charge of zero.
Amino acids,such as glycine $(NH_2-CH_2-COOH)$,contain both a basic amino group $(-NH_2)$ and an acidic carboxylic group $(-COOH)$.
In an aqueous solution,the proton from the $-COOH$ group is transferred to the $-NH_2$ group,forming the Zwitter ion: $^+NH_3-CH_2-COO^-$.
Therefore,the correct option is $B$.

(B) For a reaction $aA \rightarrow bB + cC$,the rate of reaction is given by:
$Rate = -\frac{1}{a} \frac{d[A]}{dt} = \frac{1}{b} \frac{d[B]}{dt} = \frac{1}{c} \frac{d[C]}{dt} = K$
Given the reaction $2A \rightarrow B + 3C$,the rate expression is:
$Rate = \frac{1}{3} \frac{d[C]}{dt} = K$
Therefore,the rate of production of $C$ is:
$\frac{d[C]}{dt} = 3 \times K$
$\frac{d[C]}{dt} = 3 \times (3.5 \times 10^{-4} \ mol \ L^{-1} \ s^{-1}) = 10.5 \times 10^{-4} \ mol \ L^{-1} \ s^{-1}$
Thus,the correct option is $B$.

(A) For a first order reaction,the integrated rate equation is $\ln[R] = -kt + \ln[R]_0$,which can be rearranged as $\ln(\frac{[R]_0}{[R]}) = kt$.
This implies that $\ln(\frac{[R]_0}{[R]})$ versus time $(t)$ is a straight line passing through the origin.
However,the graph of $\frac{[R]_0}{[R]}$ versus time $(t)$ is an exponential curve,not a straight line.
For a first order reaction,the half-life $t_{1/2} = \frac{0.693}{k}$ is independent of the initial concentration $[R]_0$.
Therefore,the graph of $t_{1/2}$ versus $[R]_0$ is a horizontal straight line parallel to the $[R]_0$ axis.
None of the provided graphs correctly represent the standard linear relationships for a first order reaction.
If we consider the plot of $\ln(\frac{[R]_0}{[R]})$ versus time,it is linear.
Given the options,if the question intended to ask for a plot that is $NOT$ linear or if there is a typo in the axis labels,the standard textbook graphs for first order are:
$1$. $\ln[R]$ vs $t$ (linear with negative slope)
$2$. $[R]$ vs $t$ (exponential decay)
$3$. $t_{1/2}$ vs $[R]_0$ (horizontal line).
Since none of the provided images match these,and assuming the question implies a specific relationship,the most common correct graph for first order kinetics is the linear plot of $\ln(\frac{[R]_0}{[R]})$ vs $t$.

(A) catalyst provides an alternative reaction pathway with lower activation energy.
It does not affect the potential energy of the reactants or products.
It does not change the equilibrium constant $(K_{eq})$ or the Gibbs free energy change $(\Delta G)$ of the reaction.
Therefore,the statement that the potential energy of reactants and products changes is incorrect.

(D) The unit of the rate constant $K$ for a reaction of order $n$ is given by the formula: $(mol \ L^{-1})^{1-n} \ S^{-1}$.
Given the unit is $mol^{-3/2} \ L^{3/2} \ S^{-1}$,we can equate the powers:
$(mol \ L^{-1})^{1-n} = mol^{-3/2} \ L^{3/2}$.
This implies $1 - n = -3/2$.
Solving for $n$: $n = 1 + 3/2 = 5/2 = 2.5$.
Therefore,the order of the reaction is $2.5$.

(D) $1$. The complex is $K[Cr(H_2O)_2(C_2O_4)_2] \cdot 3H_2O$.
$2$. Let the oxidation state of $Cr$ be $x$.
$3$. The charge on $K$ is $+1$,$H_2O$ is $0$,and $C_2O_4^{2-}$ is $-2$.
$4$. The equation is: $1 + x + 2(0) + 2(-2) = 0$.
$5$. $1 + x - 4 = 0$,which gives $x = +3$.
$6$. The coordination number is determined by the ligands attached to the central metal: $2(H_2O)$ are monodentate $(2 \times 1 = 2)$ and $2(C_2O_4^{2-})$ are bidentate $(2 \times 2 = 4)$.
$7$. Total coordination number = $2 + 4 = 6$.
$8$. Thus,the oxidation state is $+3$ and the coordination number is $6$.

(C) An optically active compound must be chiral,meaning it lacks a plane of symmetry or a center of symmetry.
In the complex $[Co(en)_3]Cl_3$,the ligand $en$ (ethylenediamine) is a bidentate ligand.
The complex $[Co(en)_3]^{3+}$ forms a tris-chelated octahedral structure that does not possess a plane of symmetry or a center of symmetry.
Therefore,it exists as a pair of non-superimposable mirror images (enantiomers),making it optically active.
The other complexes listed are either square planar or octahedral with planes of symmetry,rendering them optically inactive.

(A) In $[Ni(CO)_4]$,the oxidation state of $Ni$ is $0$. The electronic configuration of $Ni$ $(Z=28)$ is $[Ar] 3d^8 4s^2$. Since $CO$ is a strong field ligand,it causes pairing of electrons,resulting in $3d^{10} 4s^0 4p^0$. Thus,the hybridization is $sp^3$.
In $[Ni(CN)_4]^{2-}$,the oxidation state of $Ni$ is $+2$. The electronic configuration of $Ni^{2+}$ is $[Ar] 3d^8$. $CN^-$ is a strong field ligand,forcing the two unpaired electrons in the $3d$ orbitals to pair up,creating one empty $3d$ orbital. This leads to $dsp^2$ hybridization.

(B) The central metal ion is cobalt in the $+3$ oxidation state,denoted as $Co(III)$.
Ethane-$1,2$-diamine $(en)$ is a neutral bidentate ligand.
Since there are three $en$ ligands,the coordination sphere is $[Co(en)_3]^{3+}$.
Sulphate ion is $SO_4^{2-}$.
To balance the charges,we use the criss-cross method: $2$ units of $[Co(en)_3]^{3+}$ and $3$ units of $SO_4^{2-}$ are required to make the compound neutral.
Thus,the formula is $[Co(en)_3]_2(SO_4)_3$.

(A) The lanthanoid series elements typically exhibit a $+3$ oxidation state.
However,$Ce$ (Cerium,atomic number $58$) is well known to exhibit a $+4$ oxidation state because it attains a stable noble gas configuration $([Xe]4f^0)$ upon losing four electrons.
Therefore,the correct option is $A$.

(B) The spin-only magnetic moment $(\mu)$ is calculated using the formula $\mu = \sqrt{n(n+2)} \ BM$,where $n$ is the number of unpaired electrons.
$1$. For $Co^{2+}$ $(3d^7)$: $n = 3$,$\mu = \sqrt{3(3+2)} = \sqrt{15} \approx 3.87 \ BM$.
$2$. For $Mn^{2+}$ $(3d^5)$: $n = 5$,$\mu = \sqrt{5(5+2)} = \sqrt{35} \approx 5.92 \ BM$.
$3$. For $Ti^{2+}$ $(3d^2)$: $n = 2$,$\mu = \sqrt{2(2+2)} = \sqrt{8} \approx 2.83 \ BM$.
$4$. For $Fe^{2+}$ $(3d^6)$: $n = 4$,$\mu = \sqrt{4(4+2)} = \sqrt{24} \approx 4.90 \ BM$.
Comparing the values,$Mn^{2+}$ has the highest number of unpaired electrons $(n=5)$,hence it shows the highest magnetic moment.

(C) The oxidation state of chromium in $Cr_2O_3$ is $+3$.
$Cr_2O_3$ is an amphoteric oxide because it reacts with both acids and bases.
$CrO$ is basic,while $CrO_3$ is acidic in nature.

(A) In a basic medium,the reduction of $KMnO_4$ to $MnO_2$ involves the following half-reaction:
$MnO_4^- + 2H_2O + 3e^- \rightarrow MnO_2 + 4OH^-$.
From the balanced equation,$1 \ mole$ of $KMnO_4$ requires $3 \ moles$ of electrons,which is equivalent to $3 \ F$ of electricity.
Therefore,for $1.5 \ moles$ of $KMnO_4$,the electricity required is $1.5 \times 3 \ F = 4.5 \ F$.
Thus,the correct option is $A$.

(B) During the electrolysis of highly concentrated $H_2SO_4$ (approx. $50 \%$),the oxidation of $HSO_4^-$ ions occurs at the anode.
The reaction is: $2HSO_4^- \rightarrow H_2S_2O_8 + 2e^-$.
The product formed is peroxodisulphuric acid $(H_2S_2O_8)$,also known as Marshall's acid.
Therefore,the correct option is $B$.

(C) The conductivity of an electrolytic solution depends on the nature of the electrolyte,the concentration of the electrolyte,the nature of the solvent and its viscosity,and the temperature.
As the temperature increases,the kinetic energy of the ions increases and the viscosity of the solvent decreases,which leads to an increase in the conductivity of the solution.
Therefore,the statement that conductivity does not depend upon temperature is incorrect.

(D) The half-cell reaction for a hydrogen electrode is: $2H^+ (aq) + 2e^- \rightarrow H_2 (g)$.
Using the Nernst equation at $298 \ K$: $E = E^0 - \frac{0.0591}{n} \log \frac{1}{[H^+]}$.
Since $E^0 = 0 \ V$ for the standard hydrogen electrode and $n = 1$ for the reduction of $H^+$,the equation becomes: $E = 0 - 0.0591 \times pH$.
Given $pH = 10$,we have: $E = -0.0591 \times 10 = -0.591 \ V$.
Rounding to two decimal places,the potential is $-0.59 \ V$.

(A) The reactions are identified as follows:
$(i)$ $R-Cl + NaI \xrightarrow{\text{dry acetone}} R-I + NaCl$ is the Finkelstein reaction $(c)$.
$(ii)$ $CH_3-Br + AgF \xrightarrow{\Delta} CH_3-F + AgBr$ is the Swarts reaction $(a)$.
$(iii)$ $R-X + Mg$ $\xrightarrow{\text{dry ether}} R-Mg-X$ $\xrightarrow{H_2O} RH + Mg(OH)X$ involves the formation of a Grignard reagent $(d)$.
Therefore,the correct match is $i$ $\rightarrow c; ii$ $\rightarrow a; iii$ $\rightarrow d$.

(D) The reactivity of compounds in an $S_N1$ reaction depends on the stability of the carbocation intermediate formed after the departure of the leaving group $(Br^-)$.
The stability order of the carbocations is as follows:
$(I)$ $C_6H_5-CH_2^+$ (Benzyl carbocation,stabilized by one phenyl ring).
$(II)$ $(C_6H_5)_2CH^+$ (Diphenylmethyl carbocation,stabilized by two phenyl rings).
$(III)$ $C_6H_5-CH^+(CH_3)$ ($1$-Phenylethyl carbocation,stabilized by one phenyl ring and one methyl group).
$(IV)$ $(C_6H_5)_2C^+(CH_3)$ ($1$,$1$-Diphenylethyl carbocation,stabilized by two phenyl rings and one methyl group).
Comparing the stability: $(IV) > (II) > (III) > (I)$.
Since the rate of $S_N1$ reaction is directly proportional to the stability of the carbocation,the order of reactivity is $(IV) > (II) > (III) > (I)$.

(B) The van't Hoff factor $(i)$ for a strong electrolyte is equal to the total number of ions produced upon complete dissociation in an aqueous solution.
$1$. For $K_4[Fe(CN)_6]$: It dissociates as $K_4[Fe(CN)_6] \rightarrow 4K^+ + [Fe(CN)_6]^{4-}$. Total ions = $4 + 1 = 5$. Thus,$i = 5$.
$2$. For $Fe_4[Fe(CN)_6]_3$: It dissociates as $Fe_4[Fe(CN)_6]_3 \rightarrow 4Fe^{3+} + 3[Fe(CN)_6]^{4-}$. Total ions = $4 + 3 = 7$. Thus,$i = 7$.
$3$. For $[CoCl_2(en)_2]Cl$: It dissociates as $[CoCl_2(en)_2]Cl \rightarrow [CoCl_2(en)_2]^+ + Cl^-$. Total ions = $1 + 1 = 2$. Thus,$i = 2$.
Therefore,the values are $5, 7, 2$.

(A) $1$. Molar Mass Calculation:
The molecular formula of ethylene glycol is $C_2H_6O_2$.
Molar Mass $(M_2) = (2 \times 12) + (6 \times 1) + (2 \times 16) = 62 \ g \ mol^{-1}$.
$2$. Setup:
Let the mass of ethylene glycol (solute) be $w_2 \ g$.
Total mass of solution = $645 \ g$.
Mass of water (solvent),$w_1 = (645 - w_2) \ g$.
Given: $\Delta T_f = 2.25 \ K$ and $K_f = 1.86 \ K \ kg \ mol^{-1}$.
$3$. Calculation:
Using the freezing point depression formula: $\Delta T_f = \frac{K_f \times w_2 \times 1000}{M_2 \times w_1}$.
Substituting the values: $2.25 = \frac{1.86 \times w_2 \times 1000}{62 \times (645 - w_2)}$.
$2.25 = \frac{30 \times w_2}{645 - w_2}$.
$2.25(645 - w_2) = 30w_2$.
$1451.25 - 2.25w_2 = 30w_2$.
$1451.25 = 32.25w_2$.
$w_2 = \frac{1451.25}{32.25} = 45 \ g$.

(D) According to Raoult's Law,the total vapour pressure $P_{total} = P_P^0 x_P + P_Q^0 x_Q$.
Given $P_P^0 = 450 \ mm \ Hg$,$P_Q^0 = 750 \ mm \ Hg$,and $P_{total} = 600 \ mm \ Hg$.
Since $x_P + x_Q = 1$,we can write $x_Q = 1 - x_P$.
Substituting these values: $600 = 450 x_P + 750(1 - x_P)$.
$600 = 450 x_P + 750 - 750 x_P$.
$600 - 750 = -300 x_P$.
$-150 = -300 x_P$.
$x_P = 0.5$.
Therefore,$x_Q = 1 - 0.5 = 0.5$.
The mole fractions are $0.5$ and $0.5$.

(B) The molar mass of Glucose $(C_6H_{12}O_6)$ is $(6 \times 12) + (12 \times 1) + (6 \times 16) = 72 + 12 + 96 = 180 \ g/mol$.
Molality $(m)$ is defined as the number of moles of solute per kilogram of solvent.
Given $m = 0.25 \ mol/kg$.
Let the mass of the solvent (water) be $W \ kg$.
The mass of the solution is $W + \text{mass of glucose} = 2.5 \ kg$.
Mass of glucose $= \text{moles} \times \text{molar mass} = (0.25 \times W) \times 180 = 45W \ g = 0.045W \ kg$.
Substituting this into the solution mass equation: $W + 0.045W = 2.5 \ kg$.
$1.045W = 2.5 \implies W = 2.5 / 1.045 \approx 2.3923 \ kg$.
Mass of glucose $= 0.045 \times 2.3923 \ kg \approx 0.10765 \ kg = 107.65 \ g$.