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

HCHO reacts with \(\mathrm{CH}_{3} \mathrm{MgI}\) to give (a) \(\mathrm{CH}_{3} \mathrm{CH}_{2}-\mathrm{OH}\) (b) \(\mathrm{CH}_{3} \mathrm{OH}\) (c) \(\mathrm{CH}_{3} \mathrm{CH}_{2}-\mathrm{I}\) (d) \(\mathrm{CH}_{3} \mathrm{I}\)

Short Answer

Expert verified
The correct answer is (a) \(\mathrm{CH}_{3} \mathrm{CH}_{2}-\mathrm{OH}\).

Step by step solution

01

Identify the Reactants

The exercise states that formaldehyde (HCHO) is reacting with methylmagnesium iodide \(\text{(CH}_{3} \text{MgI)}\). These are the two reactants involved in the Grignard reaction.
02

Understand Grignard Reaction Basics

The Grignard reagent \(\text{CH}_3\text{MgI}\) will add to the carbonyl carbon of formaldehyde (HCHO) by donating its \(\text{CH}_3^-\) group, forming a new carbon-carbon bond. This forms an alkoxide intermediate.
03

Form the Alkoxide Intermediate

When \(\text{CH}_{3}\text{MgI}\) reacts with HCHO, the \(\text{CH}_3^-\) group attacks the carbonyl carbon of formaldehyde, creating \(\text{CH}_{3}\text{CH}_{2}\text{O}^-\), which is an alkoxide ion.
04

Protonation in the Acidification Step

The alkoxide \(\text{CH}_{3}\text{CH}_{2}\text{O}^-\) is then protonated, typically with water or acid, to produce the alcohol \(\text{CH}_{3}\text{CH}_{2}\text{OH}\) (ethanol) as the final product.
05

Select the Correct Option

From the products \((a) \text{CH}_{3}\text{CH}_{2}-\text{OH}\), \((b) \text{CH}_{3}\text{OH}\), \((c) \text{CH}_{3}\text{CH}_{2}-\text{I}\), \((d) \text{CH}_{3}\text{I}\), the formed product is \((a) \text{CH}_{3}\text{CH}_{2}-\text{OH}\).

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.

Formaldehyde Introduction
Formaldehyde, with the chemical formula HCHO, is a simple aldehyde. It contains a carbonyl group (C=O) attached to a hydrogen atom. This structure makes it highly reactive, especially in nucleophilic addition reactions. In Grignard reactions, formaldehyde's carbonyl carbon is the electrophilic center. This means it attracts nucleophiles, like the \( \text{CH}_3^- \) ion from methylmagnesium iodide. This reactivity paves the way for the formation of new compounds, such as alcohols in synthesis reactions.
Alkoxide Formation
Alkoxide ions are intermediates formed during Grignard reactions. When formaldehyde reacts with methylmagnesium iodide, the \(\text{CH}_3^-\) ion attacks the carbonyl carbon of the formaldehyde. This creates a carbon-carbon bond and disrupts the double-bond between carbon and oxygen. The resulting species is an alkoxide ion, specifically \(\text{CH}_3\text{CH}_2\text{O}^-\). This ion is stabilized by the presence of magnesium species in the mixture. Alkoxide formation is a crucial step before the eventual conversion to alcohol during the reaction.
Methylmagnesium Iodide and Its Role
Methylmagnesium iodide (\(\mathrm{CH}_3\mathrm{MgI}\)) is a Grignard reagent. It's a powerful nucleophile and acts as a source of nucleophilic \(\text{CH}_3^-\) ions. In the context of the reaction with formaldehyde, the key role of \(\mathrm{CH}_3\mathrm{MgI}\) is to provide a methyl group that readily attacks electrophilic centers. This addition results in the initial transformation of formaldehyde into the alkoxide ion intermediate. Without Grignard reagents like methylmagnesium iodide, the formation of carbon-carbon bonds in this specific manner would be much more challenging.
Protonation of the Alkoxide
Protonation is the step where the alkoxide ion is converted into an alcohol. This happens after the alkoxide ion \(\text{CH}_3\text{CH}_2\text{O}^-\) is formed. By adding water or acid, the alkoxide ion receives a proton (\(\text{H}^+\)) to form ethanol (\(\text{CH}_3\text{CH}_2\text{OH}\)). This step is necessary to neutralize the negative charge of the alkoxide ion, stabilizing the molecule into a neutral alcohol. Protonation is a key step that completes the synthesis of alcohol in Grignard reactions.
Alcohol Synthesis Conclusion
Alcohol synthesis through Grignard reactions is an essential process in organic chemistry. In this specific exercise involving formaldehyde and methylmagnesium iodide, the production of ethanol demonstrates the transformation of simple reactants into more complex structures. By following the reaction steps - formation of alkoxide ion via nucleophilic addition, followed by subsequent protonation - a straightforward and predictable synthesis pathway is highlighted. This mechanism showcases how Grignard reactions play a significant role in building alcohols, which are pivotal in various industrial and laboratory applications.

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

To prepare 3 -ethylpentan-3-ol, the reagents needed are (a) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{MgBr}+\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COCH}_{2} \mathrm{CH}_{3}\) (b) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{MgBr}+\mathrm{CH}_{3} \mathrm{COCH}_{2} \mathrm{CH}_{3}\) (c) \(\mathrm{CH}_{3} \mathrm{MgBr}+\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{COCH}_{2} \mathrm{CH}_{3}\) (d) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{MgBr}+\mathrm{CH}_{3} \mathrm{COCH}_{2} \mathrm{CH}_{3}\)

Which of the following ethers cannot be synthesized by directly williamson's ether synthesis?

n-propyl alcohol and isopropyl alcohol can be chemically distinguished by (a) reduction (b) \(\mathrm{PCl}_{5}\) (c) ozonolysis (d) oxidation with potassium dichromate

Which of the following reactions will yield 2-propanol? (1) \(\mathrm{CH}_{2}=\mathrm{CH}-\mathrm{CH}_{3}+\mathrm{H}_{2} \mathrm{O} \stackrel{\mathrm{H}^{\prime}}{\longrightarrow}\) (2) \(\mathrm{CH}_{3} \mathrm{CHO} \stackrel{\text { (i) } \mathrm{CH}_{3} \mathrm{MgI}, \text { (ii) } \mathrm{H}_{2} \mathrm{O}}{\longrightarrow}\) (3) \(\mathrm{CH}_{2} \mathrm{O} \quad\) (i) \(\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{Mgl}\), (ii) \(\mathrm{H}_{2} \mathrm{O}\) (4) \(\mathrm{CH}_{2}=\mathrm{CH}-\mathrm{CH}_{3} \stackrel{\text { Neutral } \mathrm{KMnO}_{4}}{\longrightarrow}\) (a) 1 and 3 (b) 2 and 4 (c) 1 and 2 (d) 3 and 4

The reaction of elemental sulphur with Grignard reagent followed by acidification leads to the formation of (a) sulphoxide (b) mercaptan (c) sulphonic acid (d) thioester

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