Structural Representations Of Organic Compounds

We need to know that the "right hand" and "left hand" classifications can be utilized to name the enantiomers of a chiral compound. In that event, the bolt is focused in a counterclockwise way, and the design of the stereo-center is visible as S.

We also need to know that, at the alphabet E, the level of strokes are largely on a similar side; in the E isomer, the higher need more bunches which are present on the inverse sides. In the Z shape, the even strokes are on opposite sides; in the Z isomer, the gatherings are on similar sides.

The less needed molecule is in every case away from the watcher. The need is seen based on nuclear numbers furthermore if nuclear numbers are the same based on nuclear mass. Find in the request for higher to bring down need.

The stereochemistry structure of the compound is given,

On the off chance, that will be clockwise, this is R.

On the off chance, that will be in anticlockwise, this is S.

So, The complete name of that particle is (R) 1 - bromo-1 - chloroethane.

Hence, the correct option is (A).

Step 1: Analyze the Geometry of Each Compound

• A. (Ammonia):
  • Has a trigonal pyramidal structure.
  • Nitrogen is the central atom with three bonded hydrogen atoms and one lone pair.
• B. (Bromine Pentafluoride):
  • Has a square pyramidal geometry.
  • Bromine is the central atom with five fluorine atoms and one lone pair.
• C. (Xenon Tetrafluoride):
  • Has a square planar geometry.
  • Xenon is the central atom with four fluorine atoms and two lone pairs.
• D. (Sulfur Hexafluoride):
  • Has an octahedral geometry.
  • Six fluorine atoms surround the central sulfur atom.

Step 2: Match the Compounds with the Correct Geometry

• A-I: has trigonal pyramidal geometry.
• B-IV: has square pyramidal geometry.
• C-II: has square planar geometry.
• D-III: has octahedral geometry.

Step 3: Conclusion

The correct match is:

A-I, B-IV, C-II, D-III

Aniline, a primary aromatic amine, indeed does not undergo the Friedel-Crafts alkylation reaction. The reason for this lies in the neutralization of the Lewis acid catalyst (such as AlCl₃) used in the Friedel-Crafts reaction by the lone pair of electrons on the nitrogen atom of aniline, forming a complex that deactivates the catalyst. Hence, Statement I is true.

Regarding Statement II, Gabriel synthesis is a method used to prepare primary amines from alkyl halides. In the Gabriel synthesis, phthalimide is first alkylated, and then hydrolyzed to produce primary amines. However, this method is not suitable for preparing aryl amines like aniline, because aryl halides (such as bromobenzene) do not readily undergo nucleophilic substitution reactions that are fundamental to this synthesis. Therefore, Statement II is also true.

Given these explanations, both statements are correct: Both Statement I and Statement II are true.

Step 1: Write the Reaction

The neutralization reaction between NaOH and HCl is:

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Step 6: Conclusion

The mass of unreacted NaOH is 250 mg.

To comprehend the given statements, let's first define some essential terms in coordination chemistry:

• Homoleptic Complex: A complex compound where the central metal ion is coordinated with only one type of ligand.
• Heteroleptic Complex: A complex compound where the central metal ion is coordinated with more than one type of ligand.

Now evaluate the given cobalt complexes:

• Statement I Analysis:
  The complex [Co(NH₃)₆]³⁺ has six ammine (NH₃) ligands, all of which are the same type, making it a homoleptic complex.
  Conversely, [Co(NH₃)₄Cl₂]⁺ contains both ammine (NH₃) and chloride (Cl^-) ligands, making it heteroleptic. Thus, Statement I is true.
• Statement II Analysis:
  Complex [Co(NH₃)₆]³⁺ indeed contains only one kind of ligand (NH₃), affirming the statement.
  Meanwhile, the complex [Co(NH₃)₄Cl₂]⁺ includes two types of ligands: NH₃ and Cl^-. Therefore, Statement II is also true.

Hence, the correct choice is: Both Statement I and Statement II are true.