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Chapter 16: Problem 16

From \(\mathrm{B}_{2} \mathrm{H}_{6}\), all the following can be prepared except (a) \(\mathrm{B}_{2}\left(\mathrm{CH}_{3}\right)_{6}\) (b) \(\mathrm{NaBH}_{4}\) (c) \(\mathrm{B}_{2} \mathrm{O}_{3}\) (d) \(\mathrm{H}_{3} \mathrm{BO}_{3}\)

Short Answer

Expert verified
Option (d) \( \mathrm{H}_3 \mathrm{BO}_3 \) cannot be directly prepared from \( \mathrm{B}_2 \mathrm{H}_6 \).

Step by step solution

01

Understanding the Problem

The problem asks which of the options cannot be prepared from diborane, \( \mathrm{B}_2 \mathrm{H}_6 \). To solve it, we need to understand the chemical reactions diborane can undergo and the products that can be derived from it.
02

Assessing Preparation of Option A

Option \( a \), \( \mathrm{B}_2(\mathrm{CH}_3)_6 \), can be prepared using diborane by its reaction with trialkylboranes. Therefore, this option can be prepared from \( \mathrm{B}_2 \mathrm{H}_6 \).
03

Assessing Preparation of Option B

Option \( b \), \( \mathrm{NaBH}_4 \), is formed by reacting \( \mathrm{B}_2 \mathrm{H}_6 \) with an alkali metal hydride such as \( \mathrm{NaH} \). Hence, this compound can be prepared from diborane.
04

Assessing Preparation of Option C

Option \( c \), \( \mathrm{B}_2 \mathrm{O}_3 \), can be prepared from diborane by oxidizing it in air. As such, this compound can also be obtained from \( \mathrm{B}_2 \mathrm{H}_6 \).
05

Assessing Preparation of Option D

Option \( d \), \( \mathrm{H}_3 \mathrm{BO}_3 \) (boric acid), is typically formed from boron trioxide, \( \mathrm{B}_2 \mathrm{O}_3 \), by reaction with water. Although \( \mathrm{B}_2 \mathrm{O}_3 \) can be formed from \( \mathrm{B}_2 \mathrm{H}_6 \), the direct transformation of diborane to \( \mathrm{H}_3 \mathrm{BO}_3 \) is not feasible without intermediate steps.
06

Conclusion

After analyzing each option, we find that option \( d \), "\( \mathrm{H}_3 \mathrm{BO}_3 \)," cannot be directly prepared from \( \mathrm{B}_2 \mathrm{H}_6 \) without going through \( \mathrm{B}_2 \mathrm{O}_3 \). Therefore, it is the correct answer to the question.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Diborane
Diborane, whose chemical formula is \( \mathrm{B}_2 \mathrm{H}_6 \), is one of the simplest boron hydrides. It is a colorless gas at room temperature and has a repulsive sweet odor similar to ether. This compound plays a significant role in the chemistry of boron, showcasing various reactions due to its instability and reactivity.
When diborane decomposes, it can interact with other chemicals to form numerous boron-containing compounds.
  • It reacts with various alkyl boranes to form organoboron compounds.
  • When exposed to air, diborane rapidly oxidizes.
  • It can also react with alkali metals to form borohydrides, such as sodium borohydride (\( \mathrm{NaBH}_4 \)).
These reactions make diborane a pivotal starting material in both laboratory and industrial settings, particularly for synthesizing more complex boron compounds.
Boron Compounds
Boron compounds exhibit a range of structures and properties due to boron's ability to form stable covalent bonds with other elements. Many boron compounds contain boron-oxygen bonds, which can be formed through oxidation.
Here are some common boron compounds derived from diborane:
  • **Trialkylboranes**: By reacting diborane with alkenes, trialkylboranes (e.g., trimethylborane) are produced.
  • **Sodium Borohydride (\( \mathrm{NaBH}_4 \))**: This compound is formed when diborane reacts with sodium hydride. It is widely used as a reducing agent in chemical synthesis.
  • **Boron Trioxide (\( \mathrm{B}_2 \mathrm{O}_3 \))**: Produced by burning diborane in the presence of oxygen, it serves as a precursor for various borates and glassmaking industries.
Each compound exhibits unique properties and uses in fields ranging from pharmaceuticals to materials science.
Oxidation
Oxidation is a fundamental chemical reaction involving the loss of electrons by a molecule, atom, or ion. In the context of boron chemistry, oxidation of diborane leads to various transformations.
For instance, diborane undergoes a vigorous oxidation process in the air to yield boron trioxide (\( \mathrm{B}_2 \mathrm{O}_3 \)). The \( \mathrm{B}_2 \mathrm{O}_3 \) can further react with water to form boric acid (\( \mathrm{H}_3 \mathrm{BO}_3 \)), but this requires two distinct steps:
  • The initial oxidation of diborane to form boron trioxide.
  • The subsequent reaction of boron trioxide with water to obtain boric acid.
This series of reactions exemplifies how oxidation can transform a simple molecule like diborane into more complex compounds used in various industrial applications.

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Most popular questions from this chapter

When strongly heated, orthoboric acid gives (a) \(\mathrm{H}_{2} \mathrm{~B}_{4} \mathrm{O}_{7}\) (b) \(\mathrm{B}_{2} \mathrm{O}_{3}\) (c) \(\mathrm{HBO}_{2}\) (d) \(\mathrm{NaBO}_{2}+\mathrm{B}_{2} \mathrm{O}_{3}\)

Chlorine acts as a bleaching agent only in presence of (a) sunlight (b) pure oxygen (c) dry air (d) moisture

The increase in boiling points of noble gases from He to \(\mathrm{Xe}\) is due to the (a) increase in atomic volume (b) increase in electron affinity (c) increase in polarizability (d) decrease in ionization energy

What is the \(\mathrm{C}-\mathrm{C}\) bond length (in angstroms) in diamond? (a) \(5.2\) (b) \(2.0\) (c) \(1.54\) (d) \(3.35\)

Fluorine is obtained by the interaction of \(\mathrm{K}_{2} \mathrm{MnF}_{6}\) with Lewis acid \(\mathrm{SbF}_{5}\) because of the (a) acidalysis of \(\mathrm{MnF}_{4}\) (b) ionization of \(\mathrm{MnF}_{4}\) (c) decomposition of \(\mathrm{SbF}_{6}\) (d) decomposition of \(\mathrm{MnF}_{4}\)
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