Structure Of Amines

Step 1: Concept
Aniline's group has a +R effect.
Step 2: Analysis
Lone pair electrons delocalize over the benzene ring.
Step 3: Conclusion
Electrons are less available for protonation.
Final Answer: (A)

Step 1: Formula
Lateral shift . Given and .
Step 2: Expansion
.
Step 3: Solve
, so .
Final Answer: (b)

Step 1: Understand the reaction.
When sodium nitrite ( ) reacts with amines and urea in the presence of hydrochloric acid, nitrogen gas ( ) is liberated. This is a typical reaction known as the diazotization reaction, which produces nitrogen gas as a byproduct.

Step 2: Analyze the options.
- A. CH CH NH : Ethylamine reacts with sodium nitrite to liberate gas.
- B. Urea: Urea reacts with sodium nitrite to produce gas.
- C. CH CONH : Acetamide reacts with sodium nitrite, also liberating gas.
- D. C H NH : Aniline does not react in this manner to produce gas under typical conditions.

Step 3: Conclusion.
The correct answer is (1), as ethylamine, urea, and acetamide all liberate nitrogen gas when reacted with sodium nitrite and hydrochloric acid.

Step 1: Understanding the reactivity of primary amines.
Primary amines undergo a unique reaction with chloroform and alcoholic KOH, known as the Hofmann degradation reaction, which results in the formation of a primary amine.

Step 2: Explanation of the options.
- (1) Reaction with HNO2: Both primary and secondary amines can react with nitrous acid (HNO2), producing diazonium salts.
- (2) Reaction with chloroform and alcoholic KOH: This reaction is specific to primary amines, where they react to form an isocyanate, followed by further degradation.
- (3) Reaction with acetyl chloride: Acetyl chloride reacts with both primary and secondary amines, forming amides.
- (4) Reaction with Grignard reagent: This reaction generally involves alkyl or aryl Grignard reagents and reacts with carbonyl compounds, but is not specific to primary amines.

Step 3: Conclusion.
The correct answer is (2) Reaction with chloroform and alcoholic KOH.

Step 1: Analyze the structure. The compound is: - Nitrogen has two substituents: ethyl and methyl - Functional group is amide (methanamide base)

Step 2: Correct IUPAC name. The correct name should be:

Step 3: Match with options. - (1) Incorrect - (2) Correct name but mismatch in given options context - (3) Incorrect - (4) Correct as per given choices Conclusion: Since the correct IUPAC name is not properly matching the listed options, the answer is (4) None of the above.

Step 1: Understand the structure of primary amines.
Primary amines are compounds where the nitrogen atom is attached to one alkyl group and two hydrogen atoms. For , we need to determine the different possible arrangements of the alkyl groups to form isomers.

Step 2: Draw the possible isomers.
For , we can have:
1. Butylamine ( ),
2. Isobutylamine ( ),
3. Sec-butylamine ( ),
4. Tert-butylamine ( ).

Step 3: Conclusion.
The number of isomeric primary amines is 4, which is option (2).

Step 1: Understanding the Concept:
Gadolinium (Z=64) is a lanthanide. It has exceptional configuration due to half-filled f-subshell stability.
Step 2: Detailed Explanation:
Expected: [Xe] . Actual: [Xe] .
Half-filled (7 electrons, all unpaired) is extra stable. 5d electron is promoted from 4f.
Step 3: Final Answer:
Thus, Gd = [Xe] .

Step 1: Understanding the Concept:
PBrCl has phosphorus as the central atom. Structure depends on arrangement.
Step 2: Detailed Explanation:
PBrCl can have different isomers. If all Cl atoms are in equatorial positions and Br in axial (or vice versa), the molecule may have zero dipole if symmetric.
From the given options in image, the isomer with symmetric distribution has no dipole.
Step 3: Final Answer:}
Thus, the isomer with trigonal bipyramidal geometry having all similar atoms in one plane has no dipole moment.

Step 1: Understanding the Concept:
Potassium superoxide (KO ) reacts with CO and moisture to produce O .
Step 2: Detailed Explanation:
4KO + 2CO 2K CO + 3O
4KO + 4CO + 2H O 4KHCO + 3O
Thus, it absorbs CO and releases O .
Step 3: Final Answer:
Thus, KO absorbs CO and releases O .

Step 1: Understanding the Concept:
Ozonolysis of alkenes gives carbonyl compounds. For dienes, each double bond is cleaved separately.
Step 2: Detailed Explanation:
Products: 2HCHO (formaldehyde) and CHO-CHO (glyoxal).
CH =CH-CH=CH (1,3-butadiene) on ozonolysis:
Terminal CH = gives HCHO; internal CH=CH gives CHO-CHO.
Thus, A = 1,3-butadiene.
Step 3: Final Answer:
Thus, A = CH =CH-CH=CH .

Concept: Primary amines are formed by reduction of nitriles, oximes, or by reductive amination. Some reactions form derivatives instead.

Step 1: Analyze option (A): Reductive amination gives primary amine.

Step 2: Option (B): Nitrile reduction gives primary amine (R–CH NH ).

Step 3: Option (C): Oxime reduction also gives primary amine.

Step 4: Option (D): Hydrazine forms hydrazone, not a primary amine.
Conclusion:
Thus, option (D) does not yield a primary amine.