CUET 2026 Chemistry May 26 Shift 1 Question Paper- Download Answer Key and Solution PDF

Concept:

Mole fraction is defined as: \[ X_i = \frac{Number of moles of component i}{Total number of moles} \]

For solutions showing deviation from Raoult’s law:

Positive deviation occurs when solute-solvent interactions are weaker than solute-solute and solvent-solvent interactions.
Negative deviation occurs when solute-solvent interactions are stronger.


Methanol is a hydrogen-bonded polar compound whereas \( \mathrm{CCl_4} \) is non-polar. Mixing them weakens intermolecular attractions and hence the solution shows positive deviation from Raoult’s law.

Step 1: Calculation of mole fraction of \( \mathrm{CCl_4} \).

A 30%(w/w) solution of methanol in \( \mathrm{CCl_4} \) means:
\[ 30 g methanol + 70 g \mathrm{CCl_4} \]

Molar mass of methanol \( (\mathrm{CH_3OH}) \): \[ 12 + 4(1) + 16 = 32 g mol^{-1} \]

Number of moles of methanol: \[ n_{\mathrm{CH_3OH}} = \frac{30}{32} = 0.9375 \]

Molar mass of \( \mathrm{CCl_4} \): \[ 12 + 4(35.5) \] \[ = 12 + 142 \] \[ = 154 g mol^{-1} \]

Number of moles of \( \mathrm{CCl_4} \): \[ n_{\mathrm{CCl_4}} = \frac{70}{154} \] \[ = 0.4545 \]

Total moles: \[ 0.9375 + 0.4545 = 1.392 \]

Mole fraction of \( \mathrm{CCl_4} \): \[ X_{\mathrm{CCl_4}} = \frac{0.4545}{1.392} \] \[ = 0.326 \approx 0.33 \]

Thus, Statement I is true.

Step 2: Checking Statement II regarding deviation from Raoult’s law.

Methanol molecules are strongly hydrogen bonded in pure state.
\( \mathrm{CCl_4} \) molecules are non-polar and interact mainly through weak van der Waals forces.

When methanol and \( \mathrm{CCl_4} \) are mixed:

The strong hydrogen bonding between methanol molecules gets disrupted.
The new methanol-\( \mathrm{CCl_4} \) interactions are weaker than original methanol-methanol interactions.


Therefore: \[ A-B < A-A and B-B \]

Hence, escaping tendency of molecules increases and vapour pressure becomes greater than expected.

Therefore, the mixture shows: \[ Positive deviation from Raoult’s law \]

Thus, Statement II is also true.

Hence, both Statement I and Statement II are true.
\[ \boxed{(1) Both Statement I and Statement II are true} \] Quick Tip: For \(w/w%\) problems: \[ Mass percentage = \frac{Mass of solute}{Mass of solution} \times 100 \] Always convert masses into moles before calculating mole fraction.

Concept:

Hoffmann bromamide degradation (also called Hoffmann rearrangement) converts a primary amide into a primary amine containing one carbon less than the parent amide.

General reaction: \[ \mathrm{RCONH_2 + Br_2 + 4NaOH \rightarrow RNH_2 + Na_2CO_3 + 2NaBr + 2H_2O} \]

Important points:

Only primary amides undergo Hoffmann bromamide degradation.
Product formed is a primary amine.
One carbon atom is lost during the reaction.


Step 1: Checking Statement A.

From the balanced Hoffmann bromamide degradation reaction: \[ \mathrm{RCONH_2 + Br_2 + 4NaOH \rightarrow RNH_2 + Na_2CO_3 + 2NaBr + 2H_2O} \]

We observe:

1 mole of \( \mathrm{Br_2} \) is consumed
4 moles of NaOH are consumed


But Statement A says: \[ 2 moles of \mathrm{Br_2} \]

which is incorrect.

Hence, Statement A is false.

Step 2: Checking Statement B.

Hoffmann bromamide degradation is shown by: \[ \mathrm{RCONH_2} \]

where \(R\) can be:

Alkyl group
Aryl group


Thus, alkyl amides definitely undergo Hoffmann degradation.

Example: \[ \mathrm{CH_3CONH_2 \rightarrow CH_3NH_2} \]

Therefore, Statement B is false.

Step 3: Checking Statement C.

Primary amides on Hoffmann degradation produce primary amines.

Example: \[ \mathrm{C_6H_5CONH_2 \rightarrow C_6H_5NH_2} \]

Thus, primary amines can indeed be synthesized by this method.

Hence, Statement C is true.

Step 4: Checking Statement D.

Secondary amides do not undergo Hoffmann bromamide degradation because the reaction mechanism requires the presence of: \[ \mathrm{-CONH_2} \]

group.

Secondary amides contain: \[ \mathrm{-CONHR} \]

Therefore, secondary amides do not form secondary amines in this reaction.

Hence, Statement D is false.

Step 5: Checking Statement E.

The balanced reaction clearly shows by-products: \[ \mathrm{Na_2CO_3,\ NaBr,\ H_2O} \]

Hence, Statement E is true.

Thus, only statements C and E are true.
\[ \boxed{(3)\ C and E only} \] Quick Tip: Hoffmann bromamide degradation: \[ \mathrm{RCONH_2 \rightarrow RNH_2} \] Always remember: Only primary amides react One carbon atom is lost Product formed is always a primary amine

Concept:

Carbohydrates are polyhydroxy aldehydes or ketones or compounds that produce them upon hydrolysis.

They are classified into:

Monosaccharides
Oligosaccharides
Polysaccharides


Reducing sugars contain a free aldehydic or ketonic group capable of reducing Fehling’s solution or Tollens’ reagent.

Step 1: Checking Statement (1).

All monosaccharides possess either:

Free aldehyde group
Or ketone group capable of tautomerisation


Therefore, all monosaccharides act as reducing sugars.

Examples: \[ Glucose, fructose, galactose \]

Thus, Statement (1) is correct.

Step 2: Checking Statement (2).

Oligosaccharides may produce:

Same monosaccharide units
Different monosaccharide units


Example:

Sucrose on hydrolysis gives glucose and fructose.


Since the hydrolysis products are not always identical, the statement saying “always the same” is incorrect.

Hence, Statement (2) is incorrect.

Step 3: Checking Statement (3).

Polysaccharides contain large numbers of monosaccharide units.

Starch and cellulose are classical examples of polysaccharides.

They possess: \[ n > 10 \]

monosaccharide units and very high molecular masses.

Hence, Statement (3) is correct.

Step 4: Checking Statement (4).

Glucose exists in equilibrium between:

Open chain structure
Cyclic hemiacetal structures


This equilibrium explains phenomena like:

Mutarotation
Reducing behaviour


Therefore, Statement (4) is also correct.

Thus, the incorrect statement is:
\[ \boxed{(2)} \] Quick Tip: Remember: All monosaccharides are reducing sugars. Oligosaccharides may contain same or different monosaccharide units. Glucose shows mutarotation because cyclic and open-chain forms coexist.

Concept:

Carboxylic acids react with ammonia under heating to form amides.

Further, amides on treatment with: \[ \mathrm{Br_2/KOH} \]

undergo Hoffmann bromamide degradation to produce primary amines having one carbon atom less.

General reaction: \[ \mathrm{RCONH_2 \xrightarrow{Br_2/KOH} RNH_2} \]

Step 1: Identifying compound \(R\).

Given: \[ \mathrm{R} \]

has molecular formula: \[ \mathrm{C_6H_7N} \]

This corresponds to: \[ \mathrm{C_6H_5NH_2} \]

which is aniline.

Thus: \[ R = aniline \]

Step 2: Finding compound \(Q\).

In Hoffmann degradation: \[ \mathrm{RCONH_2 \rightarrow RNH_2} \]

If the product is aniline: \[ \mathrm{C_6H_5NH_2} \]

then the amide must be: \[ \mathrm{C_6H_5CONH_2} \]

which is benzamide.

Thus: \[ Q = benzamide \]

Step 3: Finding compound \(P\).

Benzamide is formed from benzoic acid by treatment with ammonia and heating.

Reaction: \[ \mathrm{C_6H_5COOH \xrightarrow{NH_3,\ heat} C_6H_5CONH_2} \]

Thus: \[ P = benzoic acid \]

Hence: \[ P = Benzoic acid \] \[ Q = Benzamide \] \[ R = Aniline \]

Therefore, correct option is:
\[ \boxed{(1)} \] Quick Tip: In Hoffmann bromamide degradation: \[ \mathrm{RCONH_2 \rightarrow RNH_2} \] the product amine always contains one carbon atom less than the parent amide.

Concept:

Iodoform test is given by compounds containing: \[ \mathrm{CH_3CO-} \]

group or compounds oxidisable to it.

Tollens’ reagent gives positive test mainly with aldehydes.

Acetals and ketals on hydrolysis with dilute acid regenerate carbonyl compounds.

Step 1: Analysing properties of compound \(P\).

Compound \(P\):

Gives positive iodoform test
Gives negative Tollens’ test


Therefore:

\(P\) must contain methyl ketone functionality or masked methyl ketone.
It should not contain free aldehyde group.


Step 2: Analysing compound \(Q\).

On hydrolysis with dilute acid: \[ P \rightarrow Q \]

and \(Q\):

Gives positive Tollens’ test
Gives positive iodoform test


A compound giving both tests is: \[ \mathrm{CH_3CHO} \]

that is ethanal.

Ethanal:

Gives Tollens’ test because it is an aldehyde.
Gives iodoform test because it contains:
\[ \mathrm{CH_3CHO} \]


Step 3: Identifying the structure of \(P\).

Option (4): \[ \mathrm{CH_3COC(OCH_3)_2CH_3} \]

is a ketal derivative.

On acidic hydrolysis: \[ \mathrm{CH_3COC(OCH_3)_2CH_3} \]

regenerates: \[ \mathrm{CH_3CHO} \]

which satisfies both:

Positive Tollens’ test
Positive iodoform test


Hence, option (4) is correct.
\[ \boxed{(4)} \] Quick Tip: Remember: Iodoform test detects: \[ \mathrm{CH_3CO-} \] group. Tollens’ reagent detects aldehydes. Ketals/acetals hydrolyse back to carbonyl compounds in acidic medium.

Concept:

Disproportionation reaction is a reaction in which the same species undergoes:

Oxidation
Reduction


simultaneously.

In manganate ion: \[ \mathrm{MnO_4^{2-}} \]

oxidation state of manganese is: \[ x + 4(-2) = -2 \] \[ x - 8 = -2 \] \[ x = +6 \]

Thus, manganese is in: \[ +6 \]

oxidation state.

In acidic medium, manganate ion is unstable and undergoes disproportionation.

Step 1: Understanding the oxidation and reduction processes.

During disproportionation: \[ \mathrm{Mn^{+6}} \]

undergoes:

Oxidation to \(+7\)
Reduction to \(+4\)


The corresponding compounds are:

For manganese in \(+7\) oxidation state: \[ \mathrm{MnO_4^-} \]

For manganese in \(+4\) oxidation state: \[ \mathrm{MnO_2} \]

Hence: \[ \mathrm{MnO_4^{2-} \rightarrow MnO_4^- + MnO_2} \]

Step 2: Writing the balanced disproportionation reaction.

The balanced reaction in acidic medium is: \[ 3\mathrm{MnO_4^{2-}} + 4\mathrm{H^+} \rightarrow 2\mathrm{MnO_4^-} + \mathrm{MnO_2} + 2\mathrm{H_2O} \]

This clearly shows the formation of: \[ \mathrm{MnO_4^-} \]
and \[ \mathrm{MnO_2} \]

Step 3: Checking all options carefully.

Option (1): \[ \mathrm{Mn_2O_7} \]
is manganese(VII) oxide and is not formed in this disproportionation.

Hence incorrect.

Option (2): \[ \mathrm{MnO} \]
contains manganese in \(+2\) oxidation state, not formed here.

Hence incorrect.

Option (3): \[ \mathrm{MnO_4^- \ and \ MnO_2} \]

This matches the actual disproportionation products.

Hence correct.

Option (4):
Again contains incorrect products.

Thus, the correct answer is:
\[ \boxed{(3)\ \mathrm{MnO_4^- \ and \ MnO_2}} \] Quick Tip: Remember the oxidation states of manganese: \[ \mathrm{MnO_4^-} \rightarrow +7 \] \[ \mathrm{MnO_4^{2-}} \rightarrow +6 \] \[ \mathrm{MnO_2} \rightarrow +4 \] Manganate ion (\( \mathrm{MnO_4^{2-}} \)) is unstable in acidic medium and disproportionates into permanganate and manganese dioxide.

Concept:

The magnetic behaviour of coordination compounds depends on:

Number of unpaired electrons
Strength of ligand field
Electronic configuration of metal ion


Species containing all paired electrons are:

Diamagnetic


Species containing one or more unpaired electrons are:

Paramagnetic


Step 1: Checking Statement I.

We determine the number of unpaired electrons in each complex.
\[ [\mathrm{Cu(NH_3)_4}]^{2+} \]

Copper: \[ \mathrm{Cu^{2+}} : 3d^9 \]

Electronic configuration gives: \[ 1 \]

unpaired electron.


\[ [\mathrm{Ni(NH_3)_6}]^{2+} \]

Nickel: \[ \mathrm{Ni^{2+}} : 3d^8 \]

Octahedral complex with ammonia.

Number of unpaired electrons: \[ 2 \]


\[ [\mathrm{Mn(H_2O)_6}]^{2+} \]

Manganese: \[ \mathrm{Mn^{2+}} : 3d^5 \]

Water is a weak field ligand.

Thus high-spin configuration is formed.

Hence: \[ 5 \]

unpaired electrons are present.



Therefore, among all the given complexes: \[ [\mathrm{Mn(H_2O)_6}]^{2+} \]

has maximum number of unpaired electrons.

Thus, Statement I is true.

Step 2: Checking Statement II.

First species: \[ [\mathrm{NiCl_4}]^{2-} \]

Nickel is: \[ \mathrm{Ni^{2+}} : 3d^8 \]

Chloride is weak field ligand.

Tetrahedral complex forms.

Complex is paramagnetic with: \[ 2 \]

unpaired electrons.

Thus, not diamagnetic.



Second species: \[ [\mathrm{Ni(CO)_4}] \]

Nickel oxidation state: \[ 0 \]

Electronic configuration: \[ 3d^{10} \]

All electrons are paired.

Hence diamagnetic.



Third species: \[ [\mathrm{NiO_4}]^{2-} \]

This species is diamagnetic.



Fourth species: \[ [\mathrm{O_4}]^{2-} \]

This species is also diamagnetic.

Thus, two pairs contain only diamagnetic species.

Therefore, Statement II is also true.

Hence:
\[ \boxed{(2)\ Both Statement I and Statement II are true} \] Quick Tip: Important configurations: \[ \mathrm{Mn^{2+}} = 3d^5 \] usually has maximum unpaired electrons. Also remember: \[ \mathrm{Ni(CO)_4} \] is diamagnetic because nickel attains: \[ 3d^{10} \] configuration.

Concept:

Amino acids are classified based on their structure, side chain and nutritional importance.


Essential amino acids cannot be synthesized by the human body.
Glycine is achiral because it contains two hydrogen atoms attached to the \(\alpha\)-carbon.
Histidine contains an imidazole heterocyclic ring.
Proline contains a five-membered cyclic ring.


Step 1: Check statement A

Arginine and Tryptophan are considered essential amino acids.

Hence statement A is correct.

Step 2: Check statement B

Histidine contains an imidazole ring which is heterocyclic.

Hence statement B is incorrect.

Step 3: Check statement C

Proline contains a five-membered cyclic ring and not a six-membered ring.

Hence statement C is incorrect.

Step 4: Check statement D

Glycine has structure \( \mathrm{NH_2-CH_2-COOH} \).

Since the central carbon contains two hydrogen atoms, it is achiral.

Hence statement D is correct.

Step 5: Check statement E

The side chain \( \mathrm{MeS-CH_2-CH_2-} \) belongs to Methionine and not Cysteine.

Hence statement E is incorrect.

Therefore, the correct statements are A and D only. Quick Tip: Remember: Glycine is the only achiral amino acid. Proline has a five-membered ring. Histidine contains an imidazole ring. Methionine contains sulfur in thioether form.

Concept:

Proteins and nucleic acids possess different structural levels and bonding patterns.


Enzymes lower activation energy.
Denaturation destroys higher structures of proteins.
Nucleotides are linked by phosphodiester bonds.
Quaternary structure refers to association of multiple polypeptide chains.


Step 1: Check statement A

Enzymes act as catalysts and lower the activation energy of reactions.

Hence enzyme catalysed hydrolysis has lower activation energy.

Statement A is correct.

Step 2: Check statement B

Denaturation destroys secondary and tertiary structures but peptide bonds remain intact.

Thus primary structure remains unchanged.

Statement B is correct.

Step 3: Check statement C

Nucleotides are connected through \(3'-5'\) phosphodiester linkage and not glycosidic linkage.

Hence statement C is incorrect.

Step 4: Check statement D

Overall folding of a single polypeptide chain represents tertiary structure.

Quaternary structure involves arrangement of multiple chains.

Hence statement D is incorrect.

Therefore, statements A and B only are correct. Quick Tip: Important facts: Enzymes decrease activation energy. Denaturation does not break peptide bonds. DNA nucleotides are linked by phosphodiester linkage. Tertiary structure represents overall folding of a chain.

Concept:

The reaction shown is an \(S_N2\) nucleophilic substitution reaction.

Rate depends on nucleophilicity of the attacking nucleophile.


Greater electron density implies stronger nucleophile.
Resonance decreases nucleophilicity.
Stable ions are weaker nucleophiles.


Step 1: Compare \( \mathrm{OH^-} \)

Hydroxide ion has high electron density and no resonance stabilization.

Thus it is a very strong nucleophile.

Step 2: Compare \( \mathrm{PhO^-} \)

Phenoxide ion undergoes resonance stabilization.

Hence electron density is delocalized and nucleophilicity decreases compared to \( \mathrm{OH^-} \).

Step 3: Compare \( \mathrm{CH_3COO^-} \)

Acetate ion is strongly resonance stabilized.

Thus it is weaker than phenoxide ion.

Step 4: Compare \( \mathrm{ClO_4^-} \)

Perchlorate ion is highly resonance stabilized and extremely stable.

Hence it is the weakest nucleophile.

Therefore, the correct order is
\[ \mathrm{OH^- > PhO^- > CH_3COO^- > ClO_4^-} \] Quick Tip: For oxygen nucleophiles: \[ Less resonance stabilization \Rightarrow Greater nucleophilicity \] Thus: \[ \mathrm{OH^- > PhO^- > CH_3COO^- > ClO_4^-} \]