Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

What will happen if \(\mathrm{LiAlH}_{4}\) is added to an ester? (a) one unit of alcohol and one unit of acid is obtained (b) two units of alcohol are obtained (c) two units of acids are obtained (d) none of these

Short Answer

Expert verified
(b) two units of alcohol are obtained

Step by step solution

01

Understanding the Reaction with LiAlH4

Lithium aluminium hydride (\(\mathrm{LiAlH}_4\)) is a strong reducing agent used to reduce esters. When \(\mathrm{LiAlH}_4\) reacts with an ester, it reduces the ester to form primary alcohols.
02

Identifying the Products of Ester Reduction

When \(\mathrm{LiAlH}_4\) is added to an ester, the ester is reduced in two stages. The first stage reduces one part of the ester to an aldehyde, and in the second stage, the aldehyde is further reduced to a primary alcohol. Meanwhile, the other ester moiety is transformed into an alcohol as well.
03

Resulting Products

As a result of the reduction process, for each ester molecule, two alcohol molecules are produced. Therefore, two units of alcohol are obtained.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

LiAlH4
Lithium aluminium hydride, commonly abbreviated as \( \text{LiAlH}_4 \), is a powerful chemical compound mainly used in organic chemistry. It serves as a reducing agent and is known for its effectiveness in transforming a variety of organic compounds into simpler, often less oxidized forms.
This compound consists of lithium, aluminum, and hydride ions. The high reactivity of \( \text{LiAlH}_4 \) stems from its ability to donate hydride ions \((\text{H}^-)\), which play a crucial role in the reduction reactions.
In the context of reducing esters, \( \text{LiAlH}_4 \) can convert esters into alcohols through a two-stage reaction mechanism. Initially, an ester is converted into an aldehyde; subsequently, the aldehyde is further reduced to form a primary alcohol. This exceptional capability of \( \text{LiAlH}_4 \) makes it indispensable in synthetic organic chemistry, especially when aiming to reduce complex molecules efficiently.
Primary Alcohols
Primary alcohols are organic compounds characterized by the presence of a hydroxyl group \((\text{-OH})\) attached to a primary carbon atom, meaning a carbon atom bonded to only one other carbon. This structural feature gives primary alcohols distinct chemical properties.
One notable aspect of primary alcohols is their ability to undergo a wide range of chemical transformations. These compounds serve as a starting point for synthesizing various chemicals, including esters and ethers.
In the reaction with esters and \( \text{LiAlH}_4 \), primary alcohols emerge as the final product. As \( \text{LiAlH}_4 \) reduces an ester, the resulting aldehyde then quickly transforms into a primary alcohol. In the case of ester reduction, two such alcohol molecules are produced for each ester molecule, highlighting the efficiency of this reaction in generating primary alcohols.
Reducing Agents
Reducing agents, also known as reductants, are substances essential in chemical reactions that involve the gain of electrons, commonly referred to as reduction. Their primary function is to donate electrons to another substance, consequently reducing it. This process often results in the reductant being oxidized.
In organic synthesis, reducing agents like \( \text{LiAlH}_4 \) are particularly valuable. They facilitate the transformation of more oxidized functional groups, such as esters, into less oxidized forms like alcohols.
Reducing agents play a central role in various industrial and research applications, enabling the production of simpler and more versatile chemical structures. For esters, \( \text{LiAlH}_4 \) excises carbonyl groups, effectively "breaking down" the complex structure to yield cleaner alcohols. Through this key function, reducing agents become pivotal in both laboratory settings and large-scale chemical processes.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

In the following sequence of reactions: Tolouene \(\stackrel{\mathrm{KMnO}_{4}}{\longrightarrow} \mathrm{A} \stackrel{\mathrm{SOCl}_{2}}{\longrightarrow} \mathrm{B} \frac{\mathrm{H}_{2} / \mathrm{Pd}}{\mathrm{BaSO}_{4}} \mathrm{C}\), the product \(\mathrm{C}\) is (a) \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COOH}\) (b) \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}_{3}\) (c) \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}_{2} \mathrm{OH}\) (d) \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CHO}\)

The compound which reacts with hydroxyl amine but does not react with Tollen's reagent is (a) \(\mathrm{CH}_{3} \mathrm{CHO}\) (b) HCHO (c) \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\) (d) \(\mathrm{CH}_{3} \mathrm{COCH}_{3}\)

Compound 'A' (molecular formula \(\mathrm{C}_{3} \mathrm{H}_{8} \mathrm{O}\) ) is treated with acidified potassium dichromate to form a product 'B' (molecular formula \(\mathrm{C}_{3} \mathrm{H}_{6} \mathrm{O}\) ). 'B' forms a shining silver mirror on warming with ammonical silver nitrate. 'B' when treated with an aqueous solution of \(\mathrm{H}_{2} \mathrm{NCONHNH}_{2}\) HCl and sodium acetate gives a product 'C'. Identify the structure of ' \(\mathrm{C}^{\prime}\). (a) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}=\mathrm{NNHCONH}_{2}\) (b) \(\mathrm{CH}_{3}-\mathrm{C}=\mathrm{NNHCONH}_{2}\)

Place the following acid chlorides in the decreasing order of reactivity to hydrolysis. (1) \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COCl}\) (2) \(\mathrm{p}-\mathrm{O}_{2} \mathrm{NC}_{6} \mathrm{H}_{4} \mathrm{COCl}\) (3) \(\mathrm{p}-\mathrm{CH}_{3} \mathrm{OC}_{6} \mathrm{H}_{4} \mathrm{COCl}\) (a) \(2>1>3\) (b) \(3>1>2\) (c) \(1>2>3\) (d) \(1>3>2\)

The cyanohydrin of a compound \(\mathrm{X}\) on hydrolysis gives lactic acid; the \(\mathrm{X}\) is (a) \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CO}\) (b) \(\mathrm{CH}_{3} \mathrm{CHO}\) (c) \(\mathrm{HCHO}\) (d) \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}_{2} \mathrm{CHO}\)

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free