Isomerism

The optical purity (OP) of an enantiomeric mixture can be calculated using the formula:
OP = ([α]_observed / [α]_pure) × 100%
Here, [α]_observed is the optical rotation measured, +12.6°, and [α]_pure is the specific rotation of the pure isomer, +30°. Substituting the given values:
OP = (12.6 / 30) × 100% = 42%
The optical purity of the mixture is 42%, which fits the expected range of 42 to 42. This confirms the accuracy of the calculation.

Dehydrohalogenation involves the elimination of HBr to form alkenes. The number of isomeric alkenes depends on the number of different -hydrogens that can be removed.
For 2-Bromopentane, two different -carbons are available, leading to the formation of multiple alkenes (e.g., pent-1-ene and pent-2-ene). Additionally, pent-2-ene can exist as cis and trans isomers, giving a total of three isomeric alkenes.
For the other options:
- 1-Bromo-2-methylbutane and 2-Bromo-3,3-dimethylpentane have only one possible elimination product.
- 2-Bromopentane gives only one product (propene).

Chirality occurs in a molecule if it has at least one chiral center, which is a carbon atom bonded to four different groups. Analyze each compound:

1. Compound 1 (Ph-CH(COOH)): The carbon attached to COOH, H, and Ph is chiral. This compound is chiral.
2. Compound 2 (A fused aromatic ring with an ester group): No carbon atom is bonded to four different groups. This compound is not chiral.
3. Compound 3 (A sugar-like compound with multiple hydroxyl groups): The molecule has several chiral centers (multiple asymmetric carbons). This compound is chiral.
4. Compound 4 (A cyclopropane derivative with two COOH groups): The cyclopropane carbon atoms are symmetrically substituted and do not have four different groups. This compound is not chiral.

Thus, the total number of chiral compounds is 2.

Step 1: Protonation of the Alkene

The reaction begins with the protonation of the double bond in 3-chloro-1-butene by HCl. Following Markovnikov’s rule, the proton adds to the carbon with more hydrogen atoms, forming a secondary carbocation:

Step 2: Carbocation Rearrangement (Hydride Shift)

The initially formed secondary carbocation undergoes a 1,2-hydride shift to form a more stable tertiary carbocation:

Step 3: Nucleophilic Attack by Chloride Ion ( )

• Attack on the secondary carbocation:
• Attack on the tertiary carbocation:
  • One meso compound
  • One pair of enantiomers
    

Total Possible Products:

• From attack on the secondary carbocation: 1 product ( )
• From attack on the tertiary carbocation: 3 products ( , including a meso compound and a pair of enantiomers)

Total number of possible isomeric products: .

Final Answer:

The total number of possible isomeric products is 4.

To determine the correct order of basicity among Piperidine, Pyridine, and Pyrrole, we need to consider the structure and electron distribution in these compounds.

1. Piperidine: It is a saturated heterocyclic amine with a nitrogen atom in a six-membered ring. The nitrogen has a lone pair of electrons that is readily available for protonation, making Piperidine a strong base.
2. Pyridine: It contains a nitrogen atom in an aromatic six-membered ring. The nitrogen in Pyridine has a lone pair of electrons that is not part of the aromatic sextet and is available for bonding. However, its availability is somewhat reduced compared to Piperidine due to resonance stabilization.
3. Pyrrole: In this compound, the nitrogen atom is part of a five-membered aromatic ring. The lone pair of electrons on the nitrogen is delocalized in the aromatic system, making it less available for protonation. This results in Pyrrole being less basic than both Piperidine and Pyridine.

Based on the availability of the lone pair of electrons for protonation, the correct order of basic strength is:

• Piperidine > Pyridine > Pyrrole

This order is due to the lone pair availability which is highest in Piperidine and lowest in Pyrrole, as explained above.

The given question involves understanding the geometrical isomers of a coordination complex.

Let's analyze the Assertion (A) and the Reason (R) provided:

1. Assertion (A): The total number of geometrical isomers shown by complex ion is three.
   • is an octahedral complex.
   • The ligand 'en' is ethylenediamine, a bidentate ligand. It can form chelates with the metal ion.
   • In an octahedral complex, can theoretically show cis and trans isomers.
   • However, for , only two geometrical isomers are possible: cis and trans.
2. Reason (R): complex ion has an octahedral geometry.
   • This is true. The coordination number is 6, which typically corresponds to an octahedral geometry.

Based on the analysis:

• The Assertion (A) is not correct as the number of geometrical isomers is overstated.
• The Reason (R) is correct as [Co(en)₂Cl₂]⁺ indeed has an octahedral geometry, but it does not justify the incorrect assertion about the number of isomers.

Therefore, the most appropriate answer is:

(A) is not correct but (R) is correct.

Formula ⇒ 2ⁿ

n → number of stereo-centers.

Here n = 5

So, Total optical isomers = 2⁵ = 32

The dipole moment is a vector quantity. In the trans isomer of 2-butene, the dipoles of substituents are equal in magnitude but opposite in direction, resulting in a net dipole moment of zero ( ). However, in the cis isomer, the dipoles do not cancel out, making the molecule polar with . Thus, the dipole moment of the cis form is greater than the trans form.

Cis isomer: , Trans isomer: .

Therefore, both the assertion and reason are true, and the reason correctly explains the assertion.

- Statement (I) is incorrect because the compounds shown are not isomeric. Isomerism refers to different compounds with the same molecular formula but different structures or functional groups.
- Statement (II) is also incorrect because and are not functional group isomers.
Functional group isomers are compounds that have the same molecular formula but differ in their functional groups.
is an amine group, while is an imine group, but they are not functional group isomers.
Therefore, both statements are false.

To determine the incorrect statements regarding geometrical isomerism, let's analyze each option:

1. Statement (A): Propene shows geometrical isomerism.
   Geometrical isomerism occurs in compounds with restricted rotation around a double bond and different groups attached to the carbons involved in the double bond. Propene has the structure CH₃-CH=CH₂, with one hydrogen and one methyl group on each of the double-bonded carbons. Since there are two identical hydrogen atoms on one of the double-bonded carbons, propene cannot show geometrical isomerism. Thus, statement (A) is incorrect.
2. Statement (B): Trans isomer has identical atoms/groups on the opposite sides of the double bond.
   This statement is correct. In trans isomers, identical groups are on opposite sides of the double bond, which differentiates them from cis isomers where identical groups are on the same side.
3. Statement (C): Cis-but-2-ene has higher dipole moment than trans-but-2-ene.
   Cis isomers generally have higher dipole moments compared to their trans counterparts due to the polarities of substituents on the same side of the double bond. Hence, statement (C) is correct for cis-but-2-ene and trans-but-2-ene.
4. Statement (D): 2-methylbut-2-ene shows two geometrical isomers.
   2-methylbut-2-ene has the structure (CH₃)₂C=CH-CH₃. As both substituents on one side of the double bond are identical (two methyl groups), there is no possibility of geometrical isomerism. Therefore, statement (D) is incorrect.
5. Statement (E): Trans-isomer has lower melting point than cis isomer.
   This statement is generally true for liquids, as trans isomers usually pack more tightly in a solid-state due to symmetry, resulting in higher melting points compared to cis isomers. However, the statement claims the opposite, so statement (E) is incorrect.

Thus, the incorrect statements are (A), (D), and (E). Therefore, the correct option is:

- Statement-I: The melting point of neopentane is indeed higher than that of n-pentane because the structure of neopentane is more compact, and hence, the intermolecular forces are stronger, leading to a higher melting point. Therefore, Statement-I is correct.

- Statement-II: Neopentane can give more than one mono-substituted product because it has multiple possible positions for substitution. Thus, Statement-II is incorrect. Hence, the correct option is (4), as Statement-I is correct, but Statement-II is incorrect.

To determine , we will analyze each complex and calculate the relevant isomer counts:

1. Analyzing :
This coordination complex is square planar ( metal center Pt(II)). Square planar complexes can exhibit geometrical isomerism.
Possible isomers:

• Cis: Adjacent NH₃ and H₂O, Br and Cl.
• Trans: Opposite NH₃ and H₂O, Br and Cl.

Thus, there are 2 geometrical isomers. Therefore, .

2. Analyzing :
This is an octahedral complex where 'ox' represents oxalate ions ( ) which are bidentate ligands. Octahedral complexes with bidentate ligands can show geometric isomerism but not optical isomerism since 'ox' is planar.
All isomers formed are optically inactive.
The complex has 2 geometrical isomers (cis and trans). Thus, .

3. Analyzing :
This complex is also octahedral. The ligands are monodentate, allowing for geometrical isomerism:

• Facial (fac): All three NO₂ groups adjacent.
• Meridional (mer): NO₂ groups are along a line.

There are 2 geometrical isomers. Thus, .

Calculating :

The computed value falls within the expected range [6,6].

Therefore, the final value of is 6.

The general combustion reaction for an alkane is:

For complete combustion, the number of moles of oxygen required is given by:

Given that 8 moles of oxygen are required for complete combustion, we set up the equation:

Multiplying both sides by 2:

Solving for :

Since the number of carbon atoms must be an integer, the alkane is approximately C₄H₁₀. Thus, the alkane has 4 carbon atoms and 10 hydrogen atoms.

Hence, the correct number of carbon and hydrogen atoms is C₄H₁₀.

The molecular formula C₅H₁₀ represents an alkane, meaning that it contains only single bonds between carbon atoms. Since Baeyer's reagent (potassium permanganate, KMnO₄) is used to test for the presence of double bonds, a compound that does not decolorize Baeyer's reagent must be an alkane, which does not have any C=C double bonds.

Now, let's consider the structural isomers of C₅H₁₀:

- The straight-chain alkane, pentane. - The branched-chain isomers: 2-methylbutane, 3-methylbutane, and isopentane.

For each isomer of C₅H₁₀, we consider the number of possible monohalo products (i.e., products obtained by replacing one hydrogen atom with a halogen atom, such as chlorine or bromine).

- For pentane, we can substitute a halogen at any of the 5 positions, giving us 5 monohalo products. - For 2-methylbutane, we can substitute a halogen at 4 different positions, giving us 4 monohalo products. - For 3-methylbutane, we can substitute a halogen at 4 different positions, giving us 4 monohalo products. - For isopentane, we can substitute a halogen at 4 different positions, giving us 4 monohalo products.

Thus, the total number of monohalo products for all isomers of C₅H₁₀ is:

Hence, the number of monohalo products is 17.