CBSE Sample Papers for Class 12 Chemistry Set 10 with Solutions

Students must start practicing the questions from CBSE Sample Papers for Class 12 Chemistry with Solutions Set 10 are designed as per the revised syllabus.

CBSE Sample Papers for Class 12 Chemistry Set 10 with Solutions

Time Allowed: 3 Hours
Maximum Marks: 70

General Instructions:

• There are 35 questions in this question paper with internal choices.
• Section A consists of 18 multiple-choice questions carrying 1 mark each.
• Section B consists of 7 very short answer questions carrying 2 marks each.
• Section C consists of 5 short answer questions carrying 3 marks each.
• Section D consists of 2 case-based questions carrying 4 marks each.
• Section E consists of 3 long answer questions carrying 5 marks each.
• All questions are compulsory.
• Use of log tables and calculators are not allowed.

SECTION – A (18 Marks)
(The following questions are multiple-choice questions with one correct answer.
Each question carries 1 mart There is no internal choice in this section.)

Question 1.
Name the elements of lanthanoids series which is well known for its + 4 oxidation state. [1]
(a) Ce
(b) Pr
(c) Nd
(d) Pu
Answer:
(a) Ce

Explanation:
The electronic configuration of Ce is [Xe]4f¹5d¹6s². Lanthanides are very less electronegative. Their electronegative value is nearly equal to s-block elements. So they can lose electrons to form a cation. When Cerium loses 4 e-, and it will acquire fully field electronic configuration of Xenon (2, 8, 18, 18, 8). As fully filled or half filled electronic configuration possesses extra stability.

Question 2.
Standard electrode potentials of three metals X. Y and Z are shown in graph. [1]

The reducing power of these metals will be:
(a) Y > Z > X
(b) Y > X > Z
(c) Z > X > Y
(d) X > Y > Z
Answer:
(c) Z > X > Y

Explanation:
The reducing agent is stronger when it has a more negative reduction potential and weaker when it has a more positive reduction potential.

Question 3.
Compound A is reacted with alcoholic KOH to give propene and propene reacts with reagent B to give 1-bromopropane. Identify A and B. [1]
(a) 2-bromopropane, benzoyl peroxide
(b) 3-bromopropane, water
(c) 2-bromopropane, peroxide
(d) Bromopropane, alkylhalide
Answer:
(a) 2-bromopropane, benzoyl peroxide

Explanation:
Conversion of 2-bromopropane to 1-bromopropane occurs in two steps. First conversion of 2-bromopropane to propene is done with the help of alc KOH and then it converts to 1-bromoropane with the reagent of benzoyl peroxide.

Question 4.
Which of the following shall form an octahedral complex? [1]
(a) (Low spin) d⁴
(b) (High spin) d⁸
(c) (Low spin) d⁶
(d) (Low spin) d⁹
Answer:
(c) (Low spin) d⁶

Explanation:
An octahedral complex will be formed by d5 low spin. Out of 5d orbitals, 3 will contain electrons and 2 will be empty. These 2 empty d-orbitals can undergo sp³d² hybridisation (or d²sp³ hybridisation) to form octahedral complex.

Question 5.
CH₃CHO and C₆H₅CH₂CHO can be distinguished chemically by: [1]
(a) Fehling’s test
(b) Tollen’s test
(c) Iodoform test
(d) Ferric chloride test
Answer:
(c) Iodoform test

Explanation:
CH₃CHO₃ contains CH₃C = O group hence gives positive Iodoform test.

Question 6.
Identify the correct IUPAC name: [1]
(a) (CH₃CH₂)₂NCH₃ : N-ethyl-N-methylethanamine
(b) (CH₂)₂CNH₂ : 2-methylpropan-2-amine
(c) CH₃NHCH (CH₃)₂ : N-methylpropan-2-amine
(d) (CH₃)₂CHNH₂ : 2, 2-dimethyl-N-propanamine
Answer:
(a) (CH₃CH₂)₂NCH₃ : N-ethyl-N-methylethanamine

Explanation:
The IUPAC naming rules are:
(1) To determine the longest carbon chain.
(2) The carbon nearest to the substituent gets the number 1.
(3) One methyl and one ethyl group are attached with N, so their position is written as ‘N-‘ itself.
So, the IUPAC name of (CH₃CH₂)₂—N —CH₃ is, N-ethyl-N-methylethanamine.

Related Theory:
Structure of N-ethyl-N-methylethanamine.

This is an example of tertiary amine as it Contains three R groups.

Question 7.
Which acid is the strongest? [1]
(a) Cl₂CH.COOH
(b) ClCH₂COOH
(c) CH₃COOH
(d) Cl₃C.COOH
Answer:
(d) Cl₃C.COOH

Explanation:
Due to the presence of -I effect chlorine atom increases the acidic nature by withdrawing electrons. The correct order is:
Cl₃CCOOH > Cl₂CHCOOH > Cl-CH₂-COOH > CH₃COOH

Related Theory:
Carboxylic acids contain carboxylic group which has two equivalent resonating structures.

Electron withdrawing substituents like -Cl, -F, -NO₂ etc., attached to the carboxyl group increases the acidity of the carboxylic acids whereas electron donating substituents —CH₃—OCH₃ decreases the acidity of carboxylic acids.

Question 8.
Which of the following reagent would not be a good choice for reducing an aryl nitro compound to an amine? [1]
(a) H₂(excess)/Pt
(b) LiAlH₄ in ether
(c) Fe and HCl
(d) Sn and HCl
Answer:
(b) LiAlH₄ in ether

Explanation:
All other reagents react with nitro benzene to give aniline.

Question 9.
Reduction of CH₃CH₂NC with hydrogen in presence of Ni or Pt as catalyst gives: [1]
(a) CH₃CH₂NH₂
(b) CH₃CH₂NHCH₃
(c) CH₃CH₂NHCH₂CH₃
(d) (CH₃)₃N
Answer:
(b) CH₃CH₂NHCH₃

Explanation:
Reduction occurs in the following manner.
Cyanide or nitrile gets reduced to primary amines as shown in the following manner: ethane nitrile gets reduced to ethylamine:
CH₃CN + 4[H] → CH₃CH₂NH₂

Question 10.
The reaction of the formation of cyanohydrins from a ketone is a type of: [1]
(a) Nucleophilic addition reaction
(b) Nucleophilic substitution reaction
(c) Electrophilic addition reaction
(d) Electrophilic substitution reaction
Answer:
(a) Nucleophilic addition reaction.

Explanation:
The conversion of the ketones to cyanohydrins can be shown by the mechanism given below:

Question 11.
Which of the following is not affected by temperature? [1]
(a) Molarity
(b) Molality
(c) Formality
(d) Mole fraction
Answer:
(b) Molality

Explanation:
Molality can be calculated from the formula given below:
Molality = m = \(\frac{\text { moles of solute }}{\text { weight of solvent }(\mathrm{kg})}\)
The molality is number of moles of solute divided by the mass of the solvent in kg, here the mass of the solvent is independent of the temperature since mass does not depends on the temperature.

Related Theory:
Molarity of a given solution is defined as the total number of moles of solute per litre of solution.
M = n/V
Here, M is the molarity of the solution that is to be calculated, n is the number of moles of the solute and V is the volume of solution given in terms of litres. Molarity is the temperature dependent parameter since the volume of a solution increases with an increase in temperature due to increase in intermolecular spaces and vice versa.

Question 12.
Which of the following is not a colligative property? [1]
(a) Osmotic pressure
(b) Boiling point
(c) Vapour pressure
(d) Electrical conductivity
Answer:
(d) Electrical conductivity

Explanation:
The osmotic pressure, boiling point and vapour pressure all depends on the number of moles and mass of the substance, compound taken and the properties which depends on the number of moles are known as colligative properties whereas in case of electrical conductivity which is the measure of the reciprocal of resistivity is not dependent on the number of moles and is not regarded as the colligative property.

Question 13.
What reactant and reagent would be utilised to produce 2, 4-dimenthyl pentan-3-ol? [1]
(a) Propanol and propyl magnesium bromide
(b) 3-methylbutanal and 2-methyl magnesium iodide
(c) 2-dimethylpropanone and methyl magnesium iodide
(d) 2-methylpropanal and isopropyl magnesium iodide
Answer:
(d) 2-methylpropanal and isopropyl magnesium iodide

Explanation:

Question 14.
Which of the following alcohols reacts with Lucas reagent the easiest?
(a) CH₃CH₂CH₂OH
(b)

(c)

(d)

Answer:
(c)

Explanation:
Lucas reagent is a mixture of conc. HCl and anhydrous zinc chloride. The reactivity of alcohols towards Lucas reagent follows the order: tertiary > secondary > primary. When a tertiary alcohol, reacts with Lucas reagent, an oily layer (turbidity) is formed at once in cold.

In the following questions, a statement of assertion (A) is followed by a statement of reason (R). Mark the correct choice as:
(a) Both (A) and (R) are true and (R) is the correct explanation of (A).
(b) Both (A) and (R) are true but (R) is not the correct explanation of (A).
(c) (A) is true but (R) is false
(d) (A) is false but (R) is true

Question 15.
Assertion: The boiling point of alkyl halide decreases in the order Rl > RBr > RCl > RF
Reason: The boiling point of the alkyl halides is higher than that of hydrocarbons of comparable molecular mass [1]
Answer:
(b) Both (A) and (R) are true but (R) is not the correct explanation of (A).

Explanation:
The boiling points of alkyl halides decrease in the order RI>RBr>RCl>RF. This is because as the size of halogen increases, the magnitude of vander waals forces increases, and hence the boiling point increases. Thus, assertion and reason both are correct statements, but reason is not the correct explanation of the assertion.

Question 16.
Assertion: Glycerol is high viscous in nature.
Reason: Glycerol is an monosaccharides. [1]
Answer:
(c) (A) is true but (R) is false

Explanation:
Glycerol is high viscous in nature as glycerol is an alcohol.

Question 17.
Assertion: [Cu(en)₂]²⁺ is more stable than [CU(NH₃)₂]²⁺.
Reason: Ethylene diamine is a n-bonded complex. [1]
Answer:
(d) (A) is false but (R) is true

Explanation:
[Cu(en)₂]²⁺ is achelate while [Cu (NH₃)₂]²⁺ is not as amine is a unidentate ligand. Thus assertion is wrong statement, but reason is correct statement.

Question 18.
Assertion: The chemical reaction of the cell is reversed and the current flows in opposite direction the opposing emf is slightly greater than that of the cell.
Reason: The opposing emf is slightly smaller than that of the cell. [1]
Answer:
(c) (A) is true but (R) is false

Explanation:
The chemical reaction of the cell is reversed and the current flows in opposite direction the opposing emf is slightly greater than that of the cell when the opposing emf is slightly greater than that of the cell.

Section – B (14 Marks)
(The following questions are very short answer type with internal choice in two questions and carry 2 marks each.)

Question 19.
Fora chemical reaction 2A + B → 2C + 3D, the rate of disappearance of A is 0.10 mol L^-1s^-1. Calculate the rate of reaction and the rate of appearance of D. [2]
Answer:
2A + B → 2C + 3D

Question 20.
Draw the structure of following amino acids: [2]
(A) L-alanine
Answer:

(B) D-glycine
Answer:

Question 21.
What do you understand by chelate effect? Give the IUPAC name of the complex [Ni(NH₃)₆]Cl₂.
OR
Describe the shape and magnetic behaviour of the complex [CO(NH₃)₆]³⁺. [2]
Answer:
When a bidentate or a polydentate ligand contains donor atoms positioned in such a way that when they coordinate with the central metal ion, a five or a six membered ring is formed. This effect is called chelate effect. As a result, the stability of the complex increases.

Example:
the complex of Ni²⁺ with ‘+en’ is more stable than NH₃.
IUPAC name: Hexaamminenickel (II) chloride.

OR

Hybridization: d²sp³
Shape: Octahedral
Magnetic behaviour: Diamagnetic (absence of unpaired electrons)

Question 22.
Explain the following using suitable chemical equations:
(A) Allyl chloride can be distinguished from vinyl chloride by NaOH and silver nitrate test. Comment.
(B) Grignard reagent should be prepared under anhydrous conditions.
OR
Distinguish between the following pair of organic compounds with the help of suitable chemical reactions:
(A) Chlorobenzene and Benzyl chloride.
(B) Bromoethane and Ethyl alcohol. [2]
Answer:
(A) Allyl chloride reacts with alcoholic silver nitrate to give allyl nitrate and white precipitate of AgCl is formed.

But vinyl chloride is stabilized by resonance due to the presence of partial double bond (CH₂ = CH—Cl), it will therefore not react with AgNO₃
Therefore, vinyl chloride cannot be distinguished with alc. silver nitrate.

Related Theory:
Resonance in vinyl chloride -is due to Cl which haslone pair so it donates to next carbon atom.

(B) Grignard reagents are very reactive. In the presence of moisture, they react to give al¬kanes.
RMgX + H₂O → RH + Mg(OH)X

OR

(A) Benzyl chloride (Ph—CH₂ Cl) on reaction with AgNO₃ form AgBr

Chlorobenzene on reaction with AgNO₃→ no reaction

(B) Ethyl alcohol on iodoform test gives yellow precipitate.
C₂H₅OH + NaOl → CHI₃ (yellow ppt) + HCOONa
Bromoethane gives no reaction.

Question 23.
Define the following with the help of suitable example. [2]
(A) Peptide linkage.
Answer:
Peptide linkage: A peptide linkage is an amide linkage formed between —COOH group of one a-amino acid and NH₂ group of the other a-amino acid by loss of a molecule of water.

(B) Glycosidic linkage.
Answer:
Glycosidic linkage: The two monosaccharide units are joined together through an etheral or oxide Linkage formed by Loss of a molecule of water. Such a linkage between two monosaccharide units through oxygen atom is called glycosidic linkage.

Question 24.
18 g glucose, C₆H₁₂O₆ (Molar mass – 180 g mol^-1) is dissolved in 1 kg of water in a sauce pan. At what temperature will this solution boil? (K_b for water = 0.52 K kg mol^-1, boiling point of pure water = 373.15 K).
Answer:
Elevation of boiling point ∆T_b
Given: W_B = 18 g
M_B = Formula of glucose is
C₆H₁₂O₆
= 6 × 12 + 12 + 6 × 16
= 180
Wt. of solvent = 1 kg or 1000 g,
k_b = 0.52 K kg mol^-1
Hence, ∆T_b = k_b × m = 0.52 K
∴ B.P of the solution
= 373.15 + 0.052
= 373.202 K

Question 25.
Which would undergo S_N² reaction faster in the following pair? Give reason to support your answer.

Answer:
C₆H₅CH₂CH₂Br will undergo S_N² reaction faster.
As this is a primary alkyl halide and reactivity in S_N² is 1⁰ > 2⁰ > 3⁰.

Section – C (15 Marks)
(The following questions are short answer type with internal choice in two questions and carry 3 marks each.)

Question 26.
Write down any three significances of the salt bridge. [3]
Answer:
Significance of the salt bridge are as follows:
1. It completes the cell circuit by joining the solutions of two half-cells.

2. It stops the solutions from diffusing or transferring from one half-cell to the other.

3. It maintains electrical neutrality for the solutions in two half-cells. Positive ions enter the solution in an anodic half-cell, accumulating additional positive charge in the solution near the anode to block the passage of electrons away from the anode. Because salt bridge provides negative ions, this does not occur. Similar to this, in a cathodic half-cell, positive ions will de¬posit via reduction, causing negative ions to build up near the cathode. The salt bridge supplies enough positive ions to sufficiently neutralise these negative ions. SaItbridge preserves electrical neutrality as a result.

Question 27.
(A) Express the relation between mole fraction and molality.
(B) Determine how much cane sugar and water are needed to make 250g of a 25% cane sugar solution. [3]
Answer:
(A) \(\frac{X_B \times 1000}{\left(1-X_B\right) m}\) = m

(B) (B) Mass percentage of cane sugar = 25
We know that,
Mass percentage
= \(\frac{\text { Mass of solute }}{\text { Mass of solution }}\) × 100
So, 25 = \(\frac{\text { Mass of cane sugar }}{250}\)
Mass of cane sugar
= \(\frac{25 \times 250}{100}\) = 62.5 g
Mass of water = (250 – 62.5) = 187.5 g

Question 28.
(A) Identify A and B in each of the following processes:

(B) Why is an alkyl amine more basic than ammonia?
OR
Arrange the following in increasing order of property specified:
(A) Aniline, ethanamine, 2-ethylethanamine (solubility in water)
(B) Ethanoic acid, ethanamine, ethanol (boiling point).
(C) Methanamine, N, N dimethyl methanamine and N- methyl methanamine [3]
(basic strength in aqueous phase)
Answer:
(A)

(B) Due to electron releasing inductive effect (+I) of alkyl group, the electron density on the nitrogen atom increases and thus, it can donate the lone pair of electrons more easily than ammonia.

OR

(A) Aniline, N-ethylethanamine, Etanamine The given compounds can be arranged in their increasing order of solubility in water as:
Aniline < N-ethylethanamine < Ethanamine.
The above trend is because the Primary amines are less soluble than tertiary amines as primary amines can form hydrogen bonds with water but tertiary amines cannot.

(B) Ethanamine, ethanol, ethanoic acid
The given compounds can be arranged in their increasing order of boiling point as: Ethanamine < ethanol < ethanoic acid.

(C) N, N dimethylmethanamine, methanamine, N-methylmethanamine
The given compounds can be arranged in their increasing order of basic strength in aqueous phase) as: N, N dimethylmethanamine, methanamine, N-methylmethanamine.

Question 29.
(A) Write one difference between α-helix and β-pleated sheet structure of proteins.
(B) Write the reaction of glucose with HI. What does it suggest about the structure of glucose? [3]
Answer:

Alpha Helix	Beta pleated
Alpha helix is a polypeptide chain that is rod-shaped and coiled in a spiral/spring like structure.	Beta pleated sheets are made up of beta stands connected laterally.
Held by intramolecular hydrogen bonds	Held by intermolecular hydrogen bonds forming a backbone.
Eg: hemoglobin	Eg: silk

(B) The result of reaction between glucose and HI is n-Hexane. This suggests that glucose has 6 membered chain structure.

Question 30.
Complete the following reactions: [3]
(A)

Answer:

(B)

Answer:

(C)

Answer:

Section – D [8 Marks]
(The following questions are case-based questions. Each question has an internal choice and carries 4 (1+1+2) marks each. Read the passage carefully and answer the questions that follow.)

Question 31.
Strong and weak electrolytes behave differently in solution because of the difference in their degree of ionization. The conductance behaviour and effect of concentration on conductance is different for both strong and weak electrolytes. Molar conductivity is the product of specific conductance of the electrolyte and the volume of the solution containing one gram mole of the electrolyte. The variation of molar conductivity with concentration for strong and weak electrolytes is shown in the graph given below:

(A) Name a weak electrolyte? [1]
(B) Who gave the relation between molar conductance at any concentration and molar conductance at infinite dilution. [1]
(C) Conductivity of 2.5 × 10^-4 M methanoic acid is 5.25 × 10^-5 S cm^-1. Calculate its molar conductivity and degree of dissociation.
Given:
λ°_(H⁺) = 349.5 Scm² mol^-1
and λ°(HCOO^–) = 50.5S cm²mol^-1.
OR
A voltaic cell is set up at 25°C with the following half cells:
Al | AL³⁺ (0.001M) and Ni | Ni²⁺ (0.50M)
There action occurs when the cell generate san electric current. Determine the cell potential.
E°_Ni²⁺/Ni = – 0.25V, E°_Al³⁺/Al = – 1.66V) [2]
Answer:
(A) NH₄OH

Explanation:
NH₄OH is a weak electrolyte with dissociation constant equal to 1.76 × 10^-5. It dissociates partially into ions in an aqueous solution.

Related Theory
A strong electrolyte is a solution/solute that completely, ionizes or dissociates in a solution. These ions are good conductors of electric current in the solution.

(B) Debye Huckel

Explanation:
Debye-Huckel Onsager equation gives a relation between molar conductivity, ∧_m at a particular concentration and molar conductivity ∧_m at infinite dilution.

Related Theory:
The concentrated solutions of strong electrolytes have significant interionic attractions which reduce the speed of ions and lower the value of conductivity.

The increase of temperature decreases inter-ionic attractions and increases kinetic energy of ions and their speed thus increasing conductivity.

(C) ∧_m = \(\frac{1000 \times K}{M}\) S cm² mol^-1
∧_m = \(\frac{1000 \times 5.25 \times 10^{-5}}{2.5 \times 10^{-4}}\) S cm² mol^-1
= 210 S cm² mol^-1
∧⁰_m(HCOOH)
= λ(⁰HCOO^–) + λ(⁰H⁺)
= (50.5 + 349.5) S cm² mol^-1
= 400 S cm² mol^-1
α = \(\frac{\Lambda_m}{\Lambda_m^{\circ}}\)
α = \(\frac{210}{400}\) = 0.525 or 52.5%

Applying Nernst equation to the above cell reaction.

Question 32.
The existence of coordination compounds with the same formula but different arrangements of the Ligands was crucial in the development of coordination chemistry. Two or more compounds with the same formula but different arrangements of the atoms are called isomers. Because isomers usually have different physical and chemical properties, it is important to know which isomer we are dealing with if more than one isomer is possible. As we will see, coordination compounds exhibit the same types of isomers as organic compounds, as well as several kinds of isomers that are unique. Isomers are compounds with the same molecular formula but different structural formulas and do not necessarily share similar properties. There are many different classes of isomers, like stereoisomers, enantiomers, and geometrical isomers.
(A) How many different geometric isomers of the tetrahedral complex [Ni(CO)₄] are conceivable? [1]
(B) Write I.U.P.A.C. name of Tollens’ reagent. [1]
(C) The following compounds reflect what kind of structural isomerism?
(i) [Mn(CO)₅(SCN)] and [Mn(CO)₅ (NCS)]
(ii) [Co(NH₃)₅(NO₃)]SO₄
OR
Draw the isomers of:
(i) [CoCl₂(en)₂]⁺
(ii) [Co(en)₃]³⁺ [2]
Answer:
(A) No isomer because the unidentate ligands linked to the core metal atom are in the same relative positions to one another.

(B) Ammoniacal silver nitrate.

(C) (i) Linkage isomerism
(ii) Ionization isomerism

OR

Section – E (15 Marks)
(The following questions are long answer type and carry 5 marks each. Two questions have an internal choice.)

Question 33.
(A) A first order reaction takes 40mins for 30% decomposition. Calculate its half¬life period.
(B) For the decomposition of azoisoprene to hexane and nitrogen at 543K, the following data is obtained:

Calculate the rate constant of A into product which has value of k as 4.5 × 10³ s^-1 at 10°C and energy of activation is 60kJ mol^-1. At what temperature would k be 1.5 × 10⁴ s^-1 ?
OR
(A) Define order of the reaction.
For a reaction R → P, half-life (t_½) is observed to be independent of the initial concentration of reactants. What is the order of reaction?
(B) A first order gas phase reaction: A₂B₂(g) → 2A(g) + 2B(g) at the temperature 400°C has the rate constant
k = 2.0 × 10^-4 sec^-1. What percentage of A₂B₂ is decomposed on heating for 900 seconds? (Antilog 0.0781 = 1.197)
(C) Explain the nature of the reaction from the graph given below :

Answer:
(A) For the first order reaction,

(B) The decomposition reaction is of gaseous nature and expression of the rate equation for the reaction is:

OR

(A) Order of the reaction: The sum of powers to which the concentration term is raised in the rate law equation.
For a first order reaction t_½ = 0.693/k
Since the t_½ of a first order reaction is independent of initial concentration of reactants.

(B) Since the reaction is of the first order:

(C) The given graph is of the zero order where the rate of the reaction is independent of the initial concentration of the reactants involved in the reaction.

Question 34.
Account for the following:
(A) Zr and Hf have almost similar radii.
(B) There is an increase in density from titanium to copper.
(C) Mn shows the highest oxidation state of +7 with oxygen but with fluorine it shows the highest oxidation state of +4. [5]
Answer:
(A) In case of post lanthanide elements like Hf 4fsubshell is filled and it is not very effective at shielding the outer shell electrons. There occurs lanthanide contraction. Therefore, Zr and Hf have almost similar atomic radii.

(B) The increase in density from Ti to Zn is because the shielding effect of d-electrons is less, therefore there is a gradual increase in attraction from nucleus and decrease in metallic radii. This decrease in metallic radii causes decrease in size of an atom from Ti to Cu with increase in atomic radii.
Since, density = \(\frac{\text { mass }}{\text { volums }}\)
hence with increase in mass and decrease in volume, density increases.

(C) This is because of ability of oxygen atom to form multiple bonding (pπ-dπ) with the element. However, the highest oxidation state exhibited with fluorine is +4 (e.g., MnF₄) since no multiple bonding is possible with the element.

Question 35.
(A) Give suitable chemical equation to explain the following name reactions:
(i) HVZ reaction
(ii) Rosemund Reduction
(B) Write the product in the following reactions:
(i) CH₃ — CH = CH — CN + DIBAl— H + H₃O →
(ii) C₆H₅COONa + NaOH + CaO →
(iii) CH₃COCH₃ + HCN →
OR
(A) Give the reagents used in the following reactions:

(B) An unsaturated organic compound X with molecular formula, C₈H₈O which does not decolourise bromine water, it forms an orange-red precipitate with 2, 4-DNP reagent and gives yellow precipitate on heating with iodine in the presence of sodium hydroxide. It neither reduces Tollen’s or Fehling’s reagent, nor does it decolourise bromine water or Baeyer’s reagent. On drastic oxidation with chromic acid, it gives a carboxylic acid Y having molecular formula, C₇H₆O₂. Identify the compounds X and Y and explain the reactions involved. [5]
Answer:
(A) (i) HYZ reaction or Hell-Volhard-Zelin- sky reaction: In this reaction aliphatic carboxylic acids containing a hydrogen react with chlorine or bromine in the presence of a small amount of red phosphorous to give a haloacids.

(ii) Rosenmund reduction: Is a hydrogenation process in which an acyl chloride is selectively reduced to an aldehyde.

(B)
(i)

(ii) C₆H₅COONa + NaOH + CaO → C₆H₆ + CaO₂ + NaO₂

(iii) CH₃COCH₃ + HCN → CH₃(OH)(CN)CH₃

OR

(A) (i) Anhy AlCl₃ is the reagent used and this is Friedal Craft Acylation.
(ii) KMnO₄ is the reagent in this oxidation reaction.

(B) (1) ‘A’ forms an orange colour precipitate with 2, 4-dintrophenylhydrazine. It means that A is an aldehyde or ketone.
(2) Given ‘A’ does not reduce tollen or fehling’s reagent.
A must be a ketone.
(3) It is given that A gives iodoform test, so it must have CH₃—CO group. A is unsaturated, but also given A does not decolorise bromine water hence, having benzene ring.
(4) ‘A’ forms an orange colour precipitate with 2, 4-dintrophenylhydrazine. It means that A is an aldehyde or Ketone. It is given that ‘A’ on drastic oxidation changes to carboxylic acid ‘B’ As A is a ketone B will have one less carbon than A.
Hence, A is Acetophenone
[C₆H₅—CO—CH₃]
B is Benzoic acid [C₇H₆O₂].