Isomerism

To determine which pair of solutions is isotonic, we need to compare their osmotic pressures. The osmotic pressure depends on the concentration of solute particles, which can be determined using the formula: where:
is the van't Hoff factor (the number of ions produced per formula unit of the solute).
is the molarity of the solution. Now, let's calculate the osmotic pressure for each solution:
1. For BaCl (0.01 M): BaCl dissociates into Ba and 2Cl , so .
Osmotic pressure = . 2. For CaCl (0.001 M):
CaCl dissociates into Ca and 2Cl , so .
Osmotic pressure = . 3. For NaCl (0.015 M):
NaCl dissociates into Na and Cl , so .
Osmotic pressure = . 4. For Al (SO ) (0.001 M):
Al (SO ) dissociates into 2Al and 3SO , so .
Osmotic pressure = . Now, comparing the osmotic pressures:
Osmotic pressure of 0.01 M BaCl = 0.03.
Osmotic pressure of 0.015 M NaCl = 0.03.
Osmotic pressures of other pairs are not equal.
Therefore, the solutions of BaCl (0.01 M) and NaCl (0.015 M) are isotonic.

The correct option is (B) : 0.01 M Bacl₂ and 0.0015 MNaCl

Functional isomers are compounds that have the same molecular formula but different functional groups.
1. Option (A): is acetic acid (carboxylic acid), and is methyl formate (ester). These are functional isomers because they have different functional groups: carboxylic acid and ester.
2. Option (B): is ethyl ether, and is methoxypropane. These compounds are not functional isomers; they have different molecular formulas and hence are not functional isomers.
3. Option (C): is ethanol (alcohol), and is dimethyl ether (ether). These are functional isomers because they have the same molecular formula but different functional groups: alcohol and ether.
4. Option (D): is nitroethane, and is glycine. These are functional isomers because they have different functional groups: nitro group and amino acid group. Thus, the correct answer is (B), as and are not functional isomers.

The correct option is (B): and

In this reaction, ethylmagnesium bromide (CH CH MgBr) reacts with methylamine (CH NH ) to form an organic compound X. The reaction proceeds as follows:

1. Nucleophilic Attack: The nucleophilic amine group (NH ) attacks the electrophilic carbon atom in the ethyl group (CH CH ) attached to the magnesium bromide, resulting in the formation of a new carbon-nitrogen bond.
2. Possibility of Isomers: The two possible isomers arise due to the two possible positions where the CH group can attach to the nitrogen atom. This results in two distinct amine compounds.

The two isomers are:
N-Ethylmethylamine: CH CH NH (ethyl group attached to nitrogen).
N-Methylmethylamine: CH NH (methyl group attached to nitrogen).

Thus, the number of possible isomers for X is 2.

To solve the problem, we need to determine the relationship between compounds A and C in the given reaction sequence starting from CH₃CHO (acetaldehyde) and leading to the final product C.

1. Understanding the First Step (CH₃CHO to A):
The starting compound is CH₃CHO (acetaldehyde), and the reagents are (i) CH₃MgBr and (ii) H₃O⁺. CH₃MgBr is a Grignard reagent, which reacts with the carbonyl group of aldehydes to form alcohols. For acetaldehyde (CH₃CHO), the carbonyl carbon is attacked by the nucleophilic CH₃⁻ from CH₃MgBr, forming a tetrahedral intermediate. After hydrolysis with H₃O⁺, this yields a secondary alcohol:
CH₃CHO + CH₃MgBr → (CH₃)₂CHOMgBr → (CH₃)₂CHOH (2-propanol, a secondary alcohol).
Thus, A is (CH₃)₂CHOH.

2. Analyzing the Second Step (A to B):
Compound A undergoes a reaction with conc. H₂SO₄ and heat (Δ). Concentrated H₂SO₄ at high temperature typically dehydrates alcohols to form alkenes via an elimination reaction (E1 or E2 mechanism). For (CH₃)₂CHOH (2-propanol), dehydration removes a water molecule, forming propene:
(CH₃)₂CHOH → CH₃CH=CH₂ + H₂O.
Thus, B is CH₃CH=CH₂ (propene).

3. Analyzing the Third Step (B to C):
Compound B reacts with (i) B₂H₆ and (ii) H₂O₂, OH⁻. This is a hydroboration-oxidation reaction, which converts alkenes to alcohols with anti-Markovnikov addition and syn orientation. For propene (CH₃CH=CH₂), the double bond is between the first and second carbons. In hydroboration, the boron from B₂H₆ adds to the less substituted carbon (the terminal carbon, CH₂), and the hydrogen adds to the more substituted carbon (CHCH₃). Subsequent oxidation with H₂O₂ and OH⁻ replaces the boron with an OH group, yielding a primary alcohol:
CH₃CH=CH₂ → CH₃CH₂CH₂OH (1-propanol, a primary alcohol).
Thus, C is CH₃CH₂CH₂OH.

4. Comparing Compounds A and C:
- A is (CH₃)₂CHOH (2-propanol), a secondary alcohol with the structure CH₃-CH(OH)-CH₃.
- C is CH₃CH₂CH₂OH (1-propanol), a primary alcohol with the structure CH₃-CH₂-CH₂-OH.
Both A and C have the same molecular formula (C₃H₈O) since they are both alcohols with three carbons. However, the position of the OH group differs: in A, the OH is on the second carbon, and in C, the OH is on the first carbon.

5. Determining the Type of Isomerism:
- Since A and C have the same molecular formula (C₃H₈O) but differ in the position of the functional group (OH), they are positional isomers.
- They are not identical (different structures).
- They are not functional isomers (both are alcohols).
- They are not optical isomers (neither has a chiral center; A has a plane of symmetry, and C is a primary alcohol with no chiral carbon).

Final Answer:
A and C are positional isomers, so the answer is (B).

The reaction involves the following steps:

CH₃CH₂OH (ethanol) reacts with CH₃MgBr (methyl magnesium bromide) to form P.

P undergoes a reaction with conc. H₂SO₄ (concentrated sulfuric acid) to form Q.

Q reacts with B₂H₆ (diborane) and H₂O₂ / OH⁻ to form R.

Analysis of compounds P and R:

Step 1: The reaction between CH₃CH₂OH and CH₃MgBr forms a Grignard reagent intermediate. This would typically lead to the formation of ethyl methyl ketone (acetone) in Step 1, which would be compound P.

Step 2: P, being an aldehyde or ketone, undergoes acidic dehydration (likely via concentrated H₂SO₄) to form an alkene. The reaction produces Q.

Step 3: The B₂H₆ and H₂O₂/OH⁻ reaction indicates a hydroboration-oxidation reaction, converting the alkene Q into R (which could be an alcohol).

Type of Isomers:

Since both P and R differ in the position of functional groups (as seen from their formation through different steps), they are position isomers.

Conclusion:

The compounds P and R are position isomers.

Thus, the correct answer is Option C: Position isomers.