Problem 9 - A Complete Resource Book in Chemistry for JEE Main

Question

In the reaction \(\mathrm{B}_{2} \mathrm{H}_{6}+2 \mathrm{KOH}+2 \mathrm{X} \longrightarrow 2 \mathrm{Y}+6 \mathrm{H}_{2}\) \(\mathrm{X}\) and \(\mathrm{Y}\) are respectively (a) \(\mathrm{HCl}, \mathrm{KBO}_{3}\) (b) \(\mathrm{H}_{2}, \mathrm{H}_{3} \mathrm{BO}_{3}\) (c) \(\mathrm{H}_{2} \mathrm{O}, \mathrm{KBO}_{2}\) (d) \(\mathrm{H}_{2} \mathrm{O}, \mathrm{KBO}_{3}\)

Step-by-Step Solution

Answer

Option (c): \(\mathrm{H}_{2} \mathrm{O}, \mathrm{KBO}_{2}\).

Step 1: Analyze the Reaction

Look at the given chemical equation, \( \mathrm{B}_{2} \mathrm{H}_{6} + 2 \mathrm{KOH} + 2 \mathrm{X} \longrightarrow 2 \mathrm{Y} + 6 \mathrm{H}_{2} \). We need to identify the possible compounds for \( \mathrm{X} \) and \( \mathrm{Y} \) from the given options.

Step 2: Identify the Products

The equation indicates that each molecule of \( \mathrm{B}_{2} \mathrm{H}_{6} \) reacts to produce 6 molecules of \( \mathrm{H}_{2} \), which suggests that \( \mathrm{Y} \) does not contain hydrogen. Instead, \( \mathrm{Y} \) likely contains the elements that were initially present in \( \mathrm{B}_{2} \mathrm{H}_{6} \) and\( \mathrm{KOH} \), excluding hydrogen.

Step 3: Consider Possible Compounds for Y

Given the reaction, \( \mathrm{Y} \) could be a borate because \( \mathrm{B}_{2} \mathrm{H}_{6} \) contains boron. With the presence of \( \mathrm{KOH} \), which offers potassium, \( \mathrm{Y} \) might be a potassium borate, such as \( \mathrm{KBO}_{2} \).

Step 4: Determine Suitable X

With choices narrowed for \( \mathrm{Y} \), we hypothesize that \( \mathrm{X} \) is likely \( \mathrm{H}_{2} \mathrm{O} \) because water typically participates in hydroxide reactions and there is no sign of \( \mathrm{Cl} \) or other options in the balanced equation.

Step 5: Reconciling with Balanced Equation

Reassess the balanced equation with \( \mathrm{X} = \mathrm{H}_{2} \mathrm{O} \) and \( \mathrm{Y} = \mathrm{KBO}_{2} \). Each subsequencent element count matches: Boron and Potassium from \( \mathrm{B}_{2} \mathrm{H}_{6} \) and \( \mathrm{KOH} \) end up in the \( \mathrm{Y} \), while hydrogen is released as gas.

Step 6: Final Review Against Options

Checking against available options, \( \mathrm{X} = \mathrm{H}_{2} \mathrm{O} \) and \( \mathrm{Y} = \mathrm{KBO}_{2} \) matches option (c): \( \mathrm{H}_{2} \mathrm{O}, \mathrm{KBO}_{2} \).

Key Concepts

Hydroboration

Hydroboration is a fascinating chemical reaction that involves the addition of boron and hydrogen to carbon-carbon double bonds. It's particularly useful in organic chemistry; it helps create alcohols quickly from alkenes through a straightforward, yet profound process. Hydroboration is known for being anti-Markovnikov. This means that the hydrogen from boron does not bind to the more substituted carbon atom of the double bond. Instead, it attaches to the less substituted one. This property makes hydroboration unique and valuable as it provides products that might be difficult to obtain through other reactions.
One interesting aspect of hydroboration is its preference for syn addition. Here, both the boron and hydrogen add on the same side of the molecule, maintaining the geometric structure. This trait is used in designing complex molecules with specific shapes and properties. Through an understanding of hydroboration, students can better grasp how reactions can be harnessed to build larger and more advanced organic structures.

Reaction Analysis

Analyzing chemical reactions involves understanding the transformation of reactants into products. Its goal is to comprehend why and how different substances mix and change. In the exercise, we analyzed the reaction of diborane (\( \mathrm{B}_{2} \mathrm{H}_{6} \)) with potassium hydroxide and another reactant, leading to the formation of specific products.

To conduct a thorough reaction analysis:

• First, identify possible reactants and reaction conditions. This helps predict what interactions might occur among the substances involved.
• Next, determine any intermediate steps that lead to the final products. This step helps illustrate how atoms rearrange during the reaction.
• Finally, hypothesize the structure of any resultant compounds and the energy changes during the reaction. Such predictions assist in understanding the stability and feasibility of the product formation.

Analyzing a chemical reaction is like solving a puzzle where the atoms must fit perfectly with each other following nature’s rules. Through reaction analysis, one can not only identify products but also gain insights into the reaction mechanism.

Balancing Equations

Balancing equations is a critical skill in chemistry that ensures that the laws of conservation of mass and matter are respected. A balanced chemical equation has equal numbers of each type of atom on both sides of the equation, reflecting that matter is neither created nor destroyed.

To balance chemical equations, one should follow these helpful tips:

• Start by writing down the number of atoms of each element on both sides of the equation.
• Identify any imbalances, and adjust the coefficients (the numbers in front of the chemical formulas) to equalize the quantities of each element on both sides.
• Re-evaluate the whole equation after each adjustment to ensure that adding or changing coefficients did not inadvertently unbalance another part of the equation.
• Repeat these steps until the representation rules of molecule numbers and species are satisfied on each side of the equation.

This practice of checking and adjusting coefficients helps solidify the understanding of chemical reactions. Additionally, balanced equations are crucial for calculations involving reactants and products in stoichiometry. Balancing ensures that any predictions or calculations about the amount of material involved are accurate. It's not just an academic exercise but a practical tool necessary for laboratory and industrial chemistry as well.