UP Board Class 12 Chemistry Question Paper 2024 (Code 347 GD) Available- Download Here with Solution PDF

Step 1: Osmotic pressure is directly proportional to the number of dissolved particles in a solution.

Electrolytes dissociate into multiple ions, increasing the number of solute particles in solution.

Gold solution is a colloidal solution and has the lowest osmotic pressure as it contains fewer dissolved ions compared to ionic compounds like potassium chloride or magnesium chloride. Quick Tip: Osmotic pressure follows the relation \( \pi = iMRT \), where \( i \) is the van't Hoff factor, representing the number of particles a solute dissociates into.

Step 1: Acidity is determined by the stability of the conjugate base formed after losing a proton.

Phenol (\( C_6H_5OH \)) is more acidic than water and ethanol due to resonance stabilization of its conjugate base (\( C_6H_5O^- \)).

Ether (\( CH_3OCH_3 \)) does not release protons easily and is least acidic. Quick Tip: The acidity of organic compounds can be compared by evaluating the stability of their conjugate bases.

Step 1: Lucas reagent is a solution of concentrated hydrochloric acid and anhydrous zinc chloride (\( ZnCl_2 \)).

It is used to distinguish primary, secondary, and tertiary alcohols based on their reactivity in the Lucas test.

Tertiary alcohols react immediately, secondary alcohols show turbidity after some time, and primary alcohols do not react under normal conditions. Quick Tip: Lucas reagent is used for classifying alcohols based on their reactivity with \( ZnCl_2/HCl \), following the SN1 mechanism.

Step 1: Schiff’s reagent is used to distinguish between aldehydes and ketones.

Aldehydes react with Schiff’s reagent to give a pink color, whereas ketones do not react.

Since acetaldehyde is an aldehyde and acetone is a ketone, Schiff’s reagent can differentiate between them. Quick Tip: Schiff’s reagent test is a qualitative test to detect aldehydes, giving a pink/magenta color due to the formation of a complex.

Step 1: The carbylamine test is given by primary amines (both aliphatic and aromatic), but not by secondary or tertiary amines.

Step 2: \( CH_3NHCH_3 \) is a secondary amine, which does not give the carbylamine reaction.

All other options contain primary amines and will give a positive test. Quick Tip: The carbylamine reaction is used to detect primary amines, where they react with chloroform and alcoholic KOH to form isocyanides with a foul odor.

Step 1: Vitamin A is essential for vision, immune function, and cell growth.

Step 2: Its deficiency primarily leads to night blindness (nyctalopia), which affects the ability to see in low light conditions.

Other symptoms may include dry skin, weakened immunity, and delayed growth. Quick Tip: Vitamin A plays a crucial role in vision by forming rhodopsin, a pigment needed for low-light vision.

Step 1: The osmotic pressure (\( \pi \)) is related to molarity (\( C \)) using the formula:
\[ \pi = CRT \]

where \( R = 0.0821 \) L atm mol\(^{-1}\)K\(^{-1}\) and \( T = 293 \) K.

Step 2: Molarity \( C \) is given by:
\[ C = \frac{mass of solute}{molar mass \times volume (L)} \]

Substituting values,
\[ C = \frac{45}{342 \times 1} = 0.1316 mol/L \]

Step 3: Now, solving for \( C \),
\[ 3.2 = (0.0821 \times 293) C \]
\[ C = \frac{3.2}{0.0821 \times 293} = 0.1332 mol/L \] Quick Tip: Osmotic pressure calculations use the formula \( \pi = CRT \). Ensure correct unit conversions for temperature and gas constant.

Step 1: The formula for molar conductivity (\( \Lambda_m \)) is:
\[ \Lambda_m = \frac{\kappa \times 1000}{C} \]

where \( \kappa = 0.0248 \) S cm\(^{-1}\) and \( C = 0.20 \) M.

Step 2: Substituting values:
\[ \Lambda_m = \frac{0.0248 \times 1000}{0.20} = 124 S cm^2 mol^{-1} \] Quick Tip: Molar conductivity is calculated using \( \Lambda_m = \frac{\kappa \times 1000}{C} \), where \( \kappa \) is conductivity and \( C \) is concentration.

Step 1: For a zero-order reaction, the rate equation is:
\[ Rate = k [A]^0 \]

Step 2: The unit of velocity (rate of reaction) is:
\[ mol L^{-1} s^{-1} \]

Step 3: The unit of rate constant for a zero-order reaction is:
\[ mol L^{-1} s^{-1} \] Quick Tip: The unit of a rate constant depends on the reaction order: for a zero-order reaction, it is \( mol L^{-1} s^{-1} \).

Step 1: Identify the longest carbon chain containing the aldehyde group.

The parent chain contains 6 carbons, making it a hexanal.

Step 2: Number the chain starting from the aldehyde functional group.

A methyl group is attached to carbon 3.

Step 3: The IUPAC name is:
\[ 3-methylhexanal \]



(B) \( CH_3COCH_2COCH_3 \)



Step 1: Identify the longest chain containing the ketone groups.

Step 2: The parent name is pentane with two ketone groups at positions 2 and 4.

Step 3: The IUPAC name is:
\[ 2,4-pentanedione \] Quick Tip: In aldehydes, numbering starts from the carbonyl carbon, and substituents are named accordingly.
For diketones, use the suffix ‘dione’ and specify ketone positions in the longest chain.

Step 1: The rate of reaction is given by:
\[ Rate \propto [C]^n \]

where \( n \) is the order of the reaction.

Step 2: Since the volume is doubled, the concentration is halved:
\[ [C]_{new} = \frac{[C]}{2} \]

Step 3: The rate is also halved:
\[ \frac{Rate_{new}}{Rate} = \left(\frac{1}{2}\right)^n = \frac{1}{2} \]

Step 4: Comparing,
\[ \left(\frac{1}{2}\right)^n = \frac{1}{2} \]
\[ n = 1 \]

Thus, the order of reaction is **1**. Quick Tip: The order of reaction can be determined using the rate equation: \( R \propto [C]^n \). Compare rate changes with concentration changes.

1. Variable oxidation states: Transition metals exhibit multiple oxidation states due to the involvement of d-electrons in bonding.

2. Formation of colored compounds: Transition metal ions form colored solutions due to d-d electronic transitions. Quick Tip: Transition metals exhibit unique properties such as variable oxidation states, catalytic behavior, and paramagnetism due to unpaired d-electrons.

Step 1: The given reaction sequence suggests dehydration followed by reduction.

- Dehydration of acetamide (\( CH_3CONH_2 \)) using \( P_2O_5 \) forms methyl isocyanide (\( CH_3NC \)).

\[ CH_3CONH_2 \xrightarrow{P_2O_5, \Delta} CH_3NC \]

So, (A) = Methyl isocyanide (\( CH_3NC \)).

Step 2: The second step is catalytic reduction using Sn/HCl.

- Reduction of methyl isocyanide leads to the formation of **methyl amine** (\( CH_3CH_2NH_2 \)).

\[ CH_3NC \xrightarrow{4[H], Sn/HCl} CH_3CH_2NH_2 \]

So, (B) = Methyl amine (\( CH_3CH_2NH_2 \)). Quick Tip: Dehydration of amides with \( P_2O_5 \) forms isocyanides, which can be reduced to primary amines using catalytic hydrogenation.

Step 1: The integrated rate law for a first-order reaction is:
\[ t = \frac{2.303}{k} \log \frac{[A]_0}{[A]} \]

Step 2: For half-life, at \( t = t_{1/2} \), \( [A] = \frac{[A]_0}{2} \):
\[ t_{1/2} = \frac{0.693}{k} \]

Step 3: When 3/4 of the reaction is complete, \( [A] = \frac{[A]_0}{4} \):
\[ t = \frac{2.303}{k} \log \frac{[A]_0}{[A]_0/4} \]
\[ t = \frac{2.303}{k} \log 4 \]
\[ t = \frac{2.303}{k} \times 0.602 \]
\[ t = 2 t_{1/2} \]

Thus, the time taken for 3/4 completion is twice the half-life. Quick Tip: For a first-order reaction, the fraction of reactant remaining can be analyzed using the integrated rate equation \( t = \frac{2.303}{k} \log \frac{[A]_0}{[A]} \).

Step 1: Complex compounds are used in qualitative analysis due to their selective reactivity with metal ions.

Step 2: Example - Ammonia Complexation with Copper:
\[ Cu^{2+} + 4NH_3 \rightarrow [Cu(NH_3)_4]^{2+} \]

This forms a deep blue solution, aiding in the detection of \( Cu^{2+} \) ions. Quick Tip: Complex formation in qualitative analysis enhances selective detection of metal ions through distinct color changes.

\[ CH_3CH_2CH_2CH_2OH + HI \rightarrow CH_3CH_2CH_2CH_2I + H_2O \]

(B) 1-chlorobutane
\[ CH_3CH_2CH_2CH_2Cl + NaI \xrightarrow{acetone} CH_3CH_2CH_2CH_2I + NaCl \]

(C) But-1-ene
\[ CH_3CH_2CH=CH_2 + HI \rightarrow CH_3CH_2CH_2CH_2I \] Quick Tip: 1-Iodobutane can be synthesized via nucleophilic substitution or addition reactions involving iodide sources.

Step 1: The given reaction suggests the Hoffmann bromamide degradation.

Step 2: The probable reactants and products are:

- Compound (A): Benzamide (\( C_6H_5CONH_2 \))
- Compound (B): Aniline (\( C_6H_5NH_2 \))
- Compound (C): Phenyl isocyanide (\( C_6H_5NC \))

Reactions:
\[ C_6H_5CONH_2 + NH_3 \rightarrow C_6H_5NH_2 + H_2O \]
\[ C_6H_5NH_2 + Br_2 + KOH \rightarrow C_6H_5NC + KBr + H_2O \] Quick Tip: The Hoffmann bromamide reaction converts amides to amines, and further treatment with \( Br_2/KOH \) can yield isocyanides.

Step 1: The acidified \( KMnO_4 \) acts as a strong oxidizing agent and reacts with various reducing agents.

Step 2: Reactions:

(A) With ferrous sulphate:
\[ 2 KMnO_4 + 10 FeSO_4 + 8 H_2SO_4 \rightarrow 5 Fe_2(SO_4)_3 + 2 MnSO_4 + K_2SO_4 + 8 H_2O \]

(B) With \( SO_2 \):
\[ 2 KMnO_4 + 5 SO_2 + 2 H_2O \rightarrow 2 MnSO_4 + K_2SO_4 + 2 H_2SO_4 \]

(C) With oxalic acid:
\[ 2 KMnO_4 + 5 (COOH)_2 + 3 H_2SO_4 \rightarrow 2 MnSO_4 + 10 CO_2 + 8 H_2O + K_2SO_4 \]

(D) With sodium thiosulphate:
\[ 2 KMnO_4 + 5 Na_2S_2O_3 + 6 H_2SO_4 \rightarrow 2 MnSO_4 + 5 Na_2SO_4 + 3 S + K_2SO_4 + 3 H_2O \] Quick Tip: Acidified \( KMnO_4 \) acts as an oxidizing agent, converting reducing agents into their oxidized forms while itself being reduced to \( Mn^{2+} \).

Step 1: Identify the ligands, oxidation state, and coordination number.

Step 2: The correct IUPAC names are:

(A) Hexaamminecobalt(C) chloride

(B) Hexaaquatitanium(C) ion

(C) Potassium tetracyanonickelate(B)

(D) Diammine dichloroplatinum(B) Quick Tip: In coordination compounds, ligands are named first in alphabetical order, followed by the central metal with oxidation state in Roman numerals.

(A) 1-bromopropane into 2-bromopropane
\[ CH_3CH_2CH_2Br \xrightarrow{alc. KOH} CH_3CH=CH_2 \xrightarrow{HBr, Peroxide} CH_3CHBrCH_3 \]

(B) 2-bromopropane into 1-bromopropane
\[ CH_3CHBrCH_3 \xrightarrow{alc. KOH} CH_3CH=CH_2 \xrightarrow{HBr} CH_3CH_2CH_2Br \]

(C) Chlorobenzene into Diphenyl
\[ C_6H_5Cl + 2Na + C_6H_5Cl \xrightarrow{Dry Ether} C_6H_5-C_6H_5 + 2NaCl \]

(D) Chlorobenzene into Phenol
\[ C_6H_5Cl + NaOH \xrightarrow{573K, 300 atm} C_6H_5OH + NaCl \] Quick Tip: Wurtz reaction is used for diphenyl preparation, while nucleophilic substitution helps convert chlorobenzene to phenol.

(A) Glucose into Glucose Phenylhydrazone
\[ C_6H_{12}O_6 + C_6H_5NHNH_2 \rightarrow C_6H_{11}O_5CH=NNHC_6H_5 + H_2O \]

(B) Glucose into n-Hexane
\[ C_6H_{12}O_6 \xrightarrow{HI} C_6H_{14} \] Quick Tip: Glucose forms phenylhydrazone derivatives with phenylhydrazine, while reduction with HI converts it to n-hexane.

Step 1: The elevation in boiling point is given by:
\[ \Delta T_b = K_b \times m \times i \]

where \( m \) is the molality and \( i \) is the van't Hoff factor.

Step 2: Calculate molality:
\[ m = \frac{Mass of solute (g)}{Molar mass (g/mol) \times Mass of solvent (kg)} \]
\[ m = \frac{12.48}{208.34 \times 1} = 0.0599 mol/kg \]

Step 3: Calculate van't Hoff factor \( i \):
\[ \Delta T_b = 373.0832 - 373 = 0.0832 \]
\[ i = \frac{\Delta T_b}{K_b \times m} = \frac{0.0832}{0.52 \times 0.0599} = 2.67 \]

Step 4: Degree of dissociation \( \alpha \) is calculated as:
\[ i = 1 + \alpha(n - 1) \]

For BaCl\(_2\), \( n = 3 \):
\[ 2.67 = 1 + \alpha(3 - 1) \]
\[ \alpha = \frac{2.67 - 1}{2} = 0.835 \]

Thus, the degree of dissociation is **0.835**. Quick Tip: The van't Hoff factor (\( i \)) accounts for ionization in colligative properties and is used to determine dissociation in solutions.

Step 1: The van't Hoff factor (\( i \)) is determined using:
\[ \pi = i M R T \]

Since the solutions are isotonic, their osmotic pressures are equal:
\[ i \times M_{NaCl} = M_{glucose} \]

Step 2: Given:
\[ M_{NaCl} = \frac{1.2}{58.5} = 0.0205 mol/L \]
\[ M_{glucose} = \frac{7.2}{180} = 0.04 mol/L \]

Step 3: Solve for \( i \):
\[ i = \frac{M_{glucose}}{M_{NaCl}} = \frac{0.04}{0.0205} = 1.95 \]

Thus, the van't Hoff factor for NaCl is 1.95. Quick Tip: For isotonic solutions, osmotic pressure equations can be equated to determine the van't Hoff factor.

(A) Bromination of phenol into 2,4,6-tribromophenol

Reagent: \( Br_2 \) (Bromine Water)

(B) Conversion of benzyl alcohol into benzoic acid

Reagent: \( K_2Cr_2O_7 \) / \( H_2SO_4 \) (Acidic Potassium Dichromate)

(C) Dehydration of propan-2-ol into propene

Reagent: \( H_2SO_4 \) at \( 443K \) (Concentrated Sulfuric Acid)

(D) Conversion of butan-2-one into butan-2-ol

Reagent: \( NaBH_4 \) (Sodium Borohydride) or \( LiAlH_4 \) (Lithium Aluminium Hydride)

(v) Oxidation of primary alcohol into carboxylic acid

Reagent: \( KMnO_4 \) / \( H_2SO_4 \) (Acidified Potassium Permanganate) Quick Tip: Reagents used in organic transformations are crucial for oxidation, reduction, and substitution reactions.

Step 1: Reimer-Tiemann reaction is used to introduce a formyl (-CHO) group into phenol.

Step 2: Reaction:
\[ C_6H_5OH + CHCl_3 + NaOH \rightarrow o-hydroxybenzaldehyde + p-hydroxybenzaldehyde \]

Step 3: The reaction mechanism involves the formation of a dichlorocarbene intermediate, which undergoes electrophilic substitution at the ortho- and para-positions of phenol.


Reimer-Tiemann reaction is an electrophilic substitution reaction introducing an aldehyde (-CHO) group into phenol.

(B) Williamson’s synthesis


Step 1: Williamson’s synthesis is used for the preparation of ethers.

Step 2: Reaction:
\[ R-O^- + R'-X \rightarrow R-O-R' + X^- \]

where \( R'X \) is an alkyl halide and \( R-O^- \) is an alkoxide ion.

Step 3: Example:
\[ C_2H_5ONa + CH_3I \rightarrow C_2H_5OCH_3 + NaI \] Quick Tip: Williamson’s synthesis involves an SN2 reaction mechanism and is widely used for asymmetric ether synthesis.

Step 1: The cell reaction is:
\[ Cu^{2+} + 2e^- \rightarrow Cu \quad (E^0 = +0.34V) \]
\[ Ag^+ + e^- \rightarrow Ag \quad (E^0 = +0.80V) \]

Step 2: The overall cell reaction:
\[ Cu + 2Ag^+ \rightarrow Cu^{2+} + 2Ag \]

Step 3: The e.m.f. of the cell is calculated as:
\[ E_{cell} = E^0_{cathode} - E^0_{anode} \]
\[ E_{cell} = 0.80 - 0.34 = 0.46V \]

Thus, the e.m.f. of the cell is **0.46V**. Quick Tip: The standard e.m.f. of a galvanic cell is calculated using \( E_{cell} = E^0_{cathode} - E^0_{anode} \).

Step 1: Gibbs free energy is related to e.m.f. using:
\[ \Delta G^0 = -n F E^0_{cell} \]

where \( n = 2 \) (electrons transferred), \( F = 96500 \) C/mol, and \( E^0_{cell} = 1.1V \).

Step 2: Calculate:
\[ \Delta G^0 = - (2 \times 96500 \times 1.1) \]
\[ \Delta G^0 = -212300 J/mol = -212.3 kJ/mol \]

Thus, the Gibbs free energy is **-212.3 kJ/mol**. Quick Tip: The Gibbs free energy of a cell reaction is calculated using \( \Delta G^0 = -nFE^0_{cell} \).

(A) Mixture of calcium acetate and calcium formate is dry distilled?
\[ (COO)_2Ca + HCOO_2Ca \xrightarrow{\Delta} CH_3CHO + CaCO_3 \]

(B) Acetone is distilled with conc. \( H_2SO_4 \)?
\[ 2 CH_3COCH_3 \xrightarrow{H_2SO_4} CH_3CH=CH_2 + CH_3COOH \]

(C) Acetone is heated with chloroform in presence of alkali?
\[ CH_3COCH_3 + CHCl_3 + NaOH \rightarrow CH_3CCl_3OH + NaCl \]

(D) Formic acid is heated with Fehling’s solution?
\[ HCOOH + 2Cu^{2+} + 5OH^- \rightarrow CO_2 + Cu_2O + 3H_2O \]

(v) Formaldehyde is heated with conc. NaOH solution?
\[ HCHO + NaOH \rightarrow HCOONa + H_2 \] Quick Tip: Dry distillation of salts, aldol condensation, and haloform reactions follow unique mechanistic pathways.

Step 1: In benzoin condensation, two aldehyde molecules condense in presence of cyanide ion.

Step 2: Reaction:
\[ 2 C_6H_5CHO + CN^- \rightarrow C_6H_5CH(OH)C(=O)C_6H_5 \]

Benzoin is formed.


Benzoin condensation is a nucleophilic addition reaction involving aldehydes.

(B) Perkin’s reaction


Step 1: Perkin’s reaction involves condensation of aromatic aldehydes with acid anhydrides.

Step 2: Reaction:
\[ C_6H_5CHO + (CH_3CO)_2O \xrightarrow{NaOAc} C_6H_5CH=CHCOOH \]

It forms cinnamic acid.


Perkin’s reaction is used for the synthesis of α,β-unsaturated carboxylic acids.


(C) Cross Cannizzaro reaction
Solution:

Step 1: Cross Cannizzaro reaction occurs between two different aldehydes, one undergoing oxidation and the other reduction.

Step 2: Example:
\[ HCHO + C_6H_5CHO + NaOH \rightarrow HCOONa + C_6H_5CH_2OH \] Quick Tip: Cross Cannizzaro reaction involves disproportionation of non-enolizable aldehydes in strong base.