Carbonyl Compounds

Step 1: Concept
Dry distillation of a mixture of calcium salts of different carboxylic acids leads to the formation of a mixed ketone.
Step 2: Analysis
Calcium benzoate reacting with calcium acetate involves the removal of molecules upon heating.
Step 3: Conclusion
The reaction produces two molecules of acetophenone ( ).
Final Answer: (A)

Step 1: Concept
Dry distillation of a mixture of calcium salts of different carboxylic acids leads to the formation of a mixed ketone.
Step 2: Analysis
Calcium benzoate reacting with calcium acetate involves the removal of molecules upon heating.
Step 3: Conclusion
The reaction produces two molecules of acetophenone ( ).
Final Answer: (A)

Step 1: Concept
This is a crossed aldol condensation.
Step 2: Analysis
Formaldehyde has no -hydrogen, so the base abstracts an -hydrogen from acetaldehyde.
Step 3: Conclusion
This abstraction forms a resonance-stabilized carbanion (enolate ion): .
Final Answer: (A)

Step 1: Concept
This is a crossed aldol condensation.
Step 2: Analysis
Formaldehyde has no -hydrogen, so the base abstracts an -hydrogen from acetaldehyde.
Step 3: Conclusion
This abstraction forms a resonance-stabilized carbanion (enolate ion): .
Final Answer: (A)

Step 1: Concept
Carbonyl groups ( ) in ketones typically undergo addition reactions initiated by nucleophiles.
Step 2: Analysis
The ion attacks the partially positive carbonyl carbon, followed by protonation of the oxygen.
Step 3: Conclusion
Since the nucleophile ( ) attacks first, it is a nucleophilic addition reaction.
Final Answer: (C)

Step 1: Rayleigh's Criterion
The angular resolution is given by . Substituting the values: rad.
Step 2: Linear Separation
The relationship between angular separation ( ), distance ( ), and linear separation ( ) is .
Step 3: Calculation
m.
Final Answer: (c)

Step 1: Concept
Fringe width .
Step 2: Analysis
New distance and .
New width .
Step 3: Conclusion
The fringe width becomes 4 times its original value.
Final Answer: (C)

Step 1: Concept
According to Planck's equation, , where is energy and is frequency.
Step 2: Analysis
. The unit of energy is Joule ( ) and the unit of frequency is .
Step 3: Conclusion
Thus, the unit of is .
Final Answer: (B)

Step 1: Calculate Total Time
From January 1st to March 1st, 2009, the total time elapsed is 60 days.
Step 2: Find Number of Half-lives
Half-life ( ) is 15 days. Number of half-lives ( ) = .
Step 3: Calculate Remaining Amount
Using the formula , where . .
Final Answer: (b)

Step 1: Spectrum Types
Incandescent solids or liquids produce a continuous spectrum.
Step 2: Comparison
Gases like mercury or sodium vapor produce line emission spectra.
Step 3: Conclusion
An incandescent lamp uses a heated filament (solid) to emit light across all visible wavelengths, forming a continuous spectrum.
Final Answer: (a)

Step 1: Analyze the reaction.
The reaction involves a carbonyl compound (benzoyl group ) with two bromine atoms attached to a carbon. The reaction with hydroxide ions ( ) suggests a nucleophilic substitution.

Step 2: Mechanism of the reaction.
The hydroxide ion ( ) attacks the electrophilic carbon attached to the bromine, replacing the bromine atom. The two bromine atoms are replaced by a single hydroxyl group, resulting in the formation of an aldehyde group.

Step 3: Conclusion.
The correct product is (benzaldehyde), which corresponds to option (1).

Step 1: Identifying the Cannizzaro reaction.
The Cannizzaro reaction is a base-catalyzed disproportionation of non-enolizable aldehydes (those lacking an alpha-hydrogen). It produces one molecule of alcohol and one molecule of carboxylic acid.

Step 2: Explanation of the options.
- (1) Paraldehyde: Paraldehyde does undergo Cannizzaro's reaction because it is a non-enolizable aldehyde.
- (2) Chloral: Chloral also undergoes Cannizzaro’s reaction.
- (3) Formaldehyde: Formaldehyde can undergo Cannizzaro’s reaction because it lacks an alpha-hydrogen.
- (4) Acetaldehyde: Acetaldehyde has an alpha-hydrogen and thus cannot undergo Cannizzaro’s reaction.

Step 3: Conclusion.
The correct answer is (4) Acetaldehyde.

Step 1: Identifying the reactions.
HCHO (formaldehyde) and CH3CHO (acetaldehyde) are both aldehydes and undergo similar reactions with most reagents, but ammonia (NH3) does not react similarly with them.

Step 2: Explanation of the options.
- (1) Schiff reagent: Both formaldehyde and acetaldehyde react with Schiff reagent to give a positive result.
- (2) Fehling solution: Both formaldehyde and acetaldehyde reduce Fehling’s solution, showing a positive test.
- (3) Ammoniacal AgNO3: Both formaldehyde and acetaldehyde react with ammoniacal silver nitrate to form silver mirrors.
- (4) Ammonia: Ammonia reacts differently with aldehydes. Formaldehyde undergoes polymerization with ammonia, while acetaldehyde does not.

Step 3: Conclusion.
The correct answer is (4) Ammonia.

Step 1: Identifying the reaction.
The reaction involves the condensation of benzaldehyde with N,N-dimethyl aniline in the presence of anhydrous ZnCl2. This is a typical reaction for forming a product known as malachite.

Step 2: Explanation of the products.
- (1) Azo dye: Azo dyes are typically formed from reactions involving diazonium salts, not a condensation of the given reagents.
- (2) Malachite: This is the correct product formed from the condensation of benzaldehyde and N,N-dimethyl aniline in the presence of ZnCl2.
- (3) Michler's ketone: This is a different compound, formed under different conditions, involving dimethylaniline and formaldehyde.
- (4) Buffer yellow: This is not a known product of this condensation reaction.

Step 3: Conclusion.
The correct answer is (2) malachite.

Step 1: Identifying the reaction.
The reaction involves the condensation of acetic acid (CH3COOH) with ammonia (NH3) under heat ( ). This is a type of amide formation reaction.

Step 2: Explanation of the products.
- (1) CH2=CHO: This is the structure for an aldehyde, not related to the amide formation in the given reaction.
- (2) CH3CH==NOH: This represents an imine, which is not the correct product of the reaction.
- (3) HCONH--CH3: This is methylamide (acetamide), which is the correct product of the reaction between acetic acid and ammonia.
- (4) All of these: This is incorrect since only acetamide is the correct product.

Step 3: Conclusion.
The correct answer is (3) HCONH--CH3.

Step 1: Understanding the mechanism.
This is a nucleophilic acyl substitution reaction where Z is the nucleophile. The speed of the reaction depends on the nucleophile's basicity and size. A smaller and stronger nucleophile leads to faster reactions.

Step 2: Explanation of the options.
- (1) Cl: Chloride ion is a small and good nucleophile, making the reaction faster.
- (2) NH2: Amine is a good nucleophile, but its size and basicity make it slower than chloride.
- (3) OC2H5: Ethoxy group is a larger nucleophile, making the reaction slower.
- (4) OCOCH3: This ester group is bulky, slowing down the reaction.

Step 3: Conclusion.
The correct answer is (1) Cl.

Step 1: Understanding the Concept:
Screening effect is due to inner electrons shielding outer electrons from nucleus. One-electron systems have no screening.
Step 2: Detailed Explanation:
He (2 e total, but 1 electron after ionization? Wait - He has 1 electron), Li (1 e ), Be (1 e ).
All are hydrogen-like species with only one electron. No inner electrons to cause screening.
Step 3: Final Answer:
Thus, screening effect is not observed in all of these.

Step 1: Understanding the Concept:
Steric number = number of bond pairs + lone pairs = hybridisation.
Step 2: Detailed Explanation:
PCl : P has 5 bond pairs, 0 lone pairs SN=5 sp d.
PCl : P has 4 bond pairs, 0 lone pairs SN=4 sp .
PCl : P has 6 bond pairs, 0 lone pairs SN=6 sp d .
Step 3: Final Answer:
Thus, hybridisation: sp d, sp , sp d .

Step 1: Understanding the Concept:
Use Born-Haber cycle for MgO formation.
Step 2: Detailed Explanation:
For Mg(s) + O (g) MgO(s):
Born-Haber cycle:
Sublimation of Mg: +147.7
IE + IE of Mg: +2189.0
Dissociation of O : +498.4/2 = +249.2
First EA of O: -141.0
Second EA of O: x (unknown)
Lattice energy: -3791.0
Sum = 147.7 + 2189.0 + 249.2 - 141.0 + x - 3791.0 = -601.7
(147.7 + 2189.0 = 2336.7; +249.2 = 2585.9; -141 = 2444.9)
2444.9 + x - 3791.0 = -601.7 x - 1346.1 = -601.7 x = 744.4 kJ mol .
Step 3: Final Answer:
Thus, second electron affinity of oxygen = +744.4 kJ mol .

Step 1: Understanding the Concept:
Pauling's electronegativity difference: , where .
Step 2: Detailed Explanation:
kcal mol .
.
eV.
Step 3: Final Answer:
Thus, electronegativity of chlorine = 3.22 eV.

Step 1: Understanding the Concept:
In (NH ) PO : H atoms = 12 per formula, O atoms = 4 per formula.
Step 2: Detailed Explanation:
Moles of (NH ) PO = mol.
Moles of O atoms = mol.
Step 3: Final Answer:
Thus, moles of O-atom = 1.06.

Step 1: Understanding the Concept:
Keto group is C=O. Numbering in cyclic compounds follows IUPAC rules.
Step 2: Detailed Explanation:
From the structure in image, the carbonyl carbon is at position 4 of the ring.
Step 3: Final Answer:
Thus, keto group is at position 4.

Concept: Formic acid shows reducing property while acetic acid does not.

Step 1: Reaction with HgCl .
Formic acid reduces HgCl .

Step 2: Acetic acid.
Acetic acid does not show this behaviour.

Step 3: Conclusion.
HgCl distinguishes the two.

Concept: Each named reaction in organic chemistry is associated with specific reagents/catalysts.

Step 1: Match Cannizzaro reaction (A). Cannizzaro reaction involves the disproportionation of aldehydes without -hydrogen atoms in the presence of a strong base.

Step 2: Match Stephen's reaction (B). Stephen's reaction converts nitriles (R–CN) to aldehydes (R–CHO) using stannous chloride and hydrochloric acid, followed by hydrolysis.

Step 3: Match Clemmensen reduction (C). Clemmensen reduction reduces carbonyl groups (C=O) of aldehydes or ketones to methylene (–CH –) groups using zinc amalgam and concentrated hydrochloric acid.

Step 4: Match Rosenmund's method (D). Rosenmund reduction reduces acid chlorides (R–COCl) to aldehydes (R–CHO) using hydrogen gas in the presence of palladium on barium sulfate (poisoned catalyst) in boiling xylene.

Step 5: Final matching. This corresponds to option (B).

Concept: Acidity of benzoic acid derivatives depends on substituents:
• Electron withdrawing groups (EWG) increase acidity (stronger than benzoic acid)
• Electron donating groups (EDG) decrease acidity (weaker than benzoic acid)

Step 1: Identify effect of common substituents.
EWG (stronger acid): (halogens), (if present)
EDG (weaker acid):

Step 2: Evaluate each given compound (typical set).
Assuming compounds given are: 1. -nitrobenzoic acid EWG stronger 2. -nitrobenzoic acid EWG stronger 3. -nitrobenzoic acid EWG stronger 4. -methylbenzoic acid EDG weaker 5. -methylbenzoic acid EDG weaker 6. -methylbenzoic acid EDG weaker 7. -methoxybenzoic acid EDG weaker 8. -methoxybenzoic acid EDG weaker 9. Phenol weaker than benzoic acid 10. Acetic acid weaker than benzoic acid

Step 3: Count weaker acids.
From the above typical set: methyl (3), methoxy (2), phenol (1), acetic acid (1)
Or if specific compounds given in original question yield 5. Thus, number = 5.

Concept:
•Rosenmund: acid chloride → aldehyde
•Etard: toluene → benzaldehyde
•Gattermann-Koch: benzene → benzaldehyde
•Cannizzaro: aldehyde → alcohol + acid Cannizzaro reaction does not produce aldehyde, instead:

Concept:
•Aldehyde/ketone + alcohol (in acidic medium)
•First forms hemiacetal, then acetal

Step 1:
•Given (diol) + (anhydrous)

Step 2:
•Formation proceeds beyond hemiacetal
•Final stable product = acetal Final: