/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 131 Methanoic acid, the first member... [FREE SOLUTION] | 91影视

91影视

Methanoic acid, the first member of carboxylic acid series, when warmed with concentrated sulphuric acid decompose in the following way and evolve carbon monoxide The driving force for this reaction lies in the fact that the \(\mathrm{HC} \equiv \mathrm{O}^{+}\) ion is very unstable acid and thus easily loses \(\mathrm{H}^{+}\). If acetic acid is replaced by triphenylacetic acid, the product formed will be (a) \(\left(\mathrm{C}_{6} \mathrm{H}_{5}\right)_{3} \mathrm{CH}+\mathrm{CO}\) (b) \(\left(\mathrm{C}_{6} \mathrm{H}_{5}\right)_{3} \mathrm{CH}+\mathrm{CO}_{2}\) (c) \(\left(\mathrm{C}_{6} \mathrm{H}_{5}\right)_{3} \mathrm{COH}+\mathrm{CO}\) (d) No reaction

Short Answer

Expert verified
The product is (b): \((C_6H_5)_3CH + CO_2\).

Step by step solution

01

Introduce the Concept of Carboxylic Acid Decomposition

Methanoic acid (HCOOH) can decompose into carbon monoxide (CO) and water (H2O) when heated with concentrated sulphuric acid. The reactivity is due to the instability of the HC鈮鈦 ion, which drives the reaction.
02

Compare Methanoic Acid with Triphenylacetic Acid

Triphenylacetic acid, \(\left(\mathrm{C}_6\mathrm{H}_5\right)_3\mathrm{C}\mathrm{COOH}\), is structurally different from methanoic acid because of the bulky phenyl groups (C6H5). Triphenylacetic acid is more stable compared to methanoic acid due to its steric hindrance, making the formation of the corresponding HC鈮鈦 ion unlikely.
03

Consider Possible Reaction Products

Since triphenylacetic acid is more stable and does not form unstable HC鈮鈦 ions readily, it will not decompose in the same way as methanoic acid. Instead, any decomposition is more likely to lead to the formation of carbon dioxide \((\mathrm{CO}_2)\) rather than carbon monoxide \((\mathrm{CO})\).
04

Match Reaction Products to Given Options

From the proposed reaction of triphenylacetic acid, the product is \((\mathrm{C}_6\mathrm{H}_5)_3\mathrm{CH} + \mathrm{CO}_2\). Therefore, the correct answer corresponds to option (b): \(\left(\mathrm{C}_6\mathrm{H}_5\right)_3 \mathrm{CH}+\mathrm{CO}_2\).

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 91影视!

Key Concepts

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

Methanoic Acid
Methanoic acid, also known as formic acid, is the simplest carboxylic acid and plays a prominent role in various chemical reactions. When discussing carboxylic acid decomposition, methanoic acid is particularly reactive due to the presence of the formyl cation, \(\mathrm{HC}\equiv \mathrm{O}^{+}\). This ion is highly unstable, making the acid prone to losing a proton.

When warmed with concentrated sulfuric acid \((\mathrm{H}_2\mathrm{SO}_4)\), methanoic acid decomposes to produce carbon monoxide \((\mathrm{CO})\) and water \((\mathrm{H}_2\mathrm{O})\). The presence of concentrated sulfuric acid acts as a dehydrating agent which facilitates this decomposition.

The chemical equation can be illustrated as follows:
  • \(\mathrm{HCOOH} \rightarrow \mathrm{CO} + \mathrm{H}_2\mathrm{O}\)
This transformation is driven by the desired stability that comes with the elimination of the unstable formyl cation.
Triphenylacetic Acid
Triphenylacetic acid, \(\left(\mathrm{C}_6\mathrm{H}_5\right)_3\mathrm{C}\mathrm{COOH}\), is another carboxylic acid but differs significantly from methanoic acid due to its bulky structure.

It contains three phenyl groups \((\mathrm{C}_6\mathrm{H}_5)\), making it considerably more stable.This stability is mainly driven by the steric hindrance caused by the large phenyl groups surrounding the carboxyl group.

Due to this structure, triphenylacetic acid cannot form the unstable formyl cation as easily as methanoic acid can.This difference in reactivity becomes key when considering its decomposition behavior. While methanoic acid readily decomposes to form carbon monoxide, triphenylacetic acid behaves differently due to its higher stability.
Chemical Stability
Chemical stability refers to the resilience of a compound to decompose or react under certain conditions. In the context of carboxylic acids, stability is significantly influenced by molecular structure and the presence of stabilizing or destabilizing groups.

Methanoic acid, being small and lacking bulky groups, is less stable and more reactive. It easily loses a proton, forming the unstable formyl cation.

Triphenylacetic acid, however, benefits from the presence of the phenyl groups that hinder internal reactions which could lead to decomposition. The steric hindrance created makes it less likely to readily lose a proton or form unstable intermediates.
  • Size and structure significantly impact stability.
  • The less hindered the acid, the more reactive it tends to be.
  • Bulky groups like phenyl rings increase stability.
Reaction Products
When considering the reaction products of carboxylic acid decomposition, it's crucial to look at the nature of the acid itself. Methanoic acid, due to its instability, primarily decomposes to form carbon monoxide \((\mathrm{CO})\).

However, triphenylacetic acid, given its structure and stability, decomposes differently.When subjected to decomposition, it tends to form carbon dioxide \((\mathrm{CO}_2)\) along with other products, as opposed to carbon monoxide.

The presence of phenyl groups ensures a different set of reaction pathways due to their stabilizing nature.
  • Methanoic Acid: Produces carbon monoxide \((\mathrm{CO})\).
  • Triphenylacetic Acid: Leads to carbon dioxide \((\mathrm{CO}_2)\) production.
Understanding these differences becomes highly relevant when predicting reaction outcomes, especially in a classroom or applied chemical analysis setting.

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

Alkyl halides and alcohols easily undergo nucleophilic substitution either through \(\mathrm{S}_{\mathrm{N}} 1\) or \(\mathrm{S}_{\mathrm{N}} 2\) mechanism. The relative case of these two processes depends upon the nature of the substrate (alkyl group as well as leaving group), nature of nucleophile and also upon the nature of solvent. \(\mathrm{S}_{\mathrm{N}} 1\) mechanism involves the formation of carbocation as intermediate while \(\mathrm{S}_{\mathrm{N}} 2\) mechanism involves the formation of a transition pentavalent state. \(\mathrm{S}_{\mathrm{N}} 1\) is the main mechanism in \(3^{\circ}\) alkyl halides and alcohols, while \(\mathrm{S}_{\mathrm{N}} 2\) mechanism is the path adopted by most of the \(1^{\circ}\) alkyl halides and \(2^{\circ}\) alkyl halides may follow \(\mathrm{S}_{\mathrm{N}} 1\) as well as \(\mathrm{S}_{\mathrm{N}} 2\). Rearrangement of alkyl groups occur when hydrogen halides react with alcohols except with most primary alcohols. The best explanation is that (a) The \(1^{\circ}\) carbocations are unstable and hence are not formed. (b) The \(1^{\circ}\) carbocations are unable to undergo rearrangement. (c) Both (a) and (b) are true (d) Both (a) and (b) are false

Methanoic acid, the first member of carboxylic acid series, when warmed with concentrated sulphuric acid decompose in the following way and evolve carbon monoxide The driving force for this reaction lies in the fact that the \(\mathrm{HC} \equiv \mathrm{O}^{+}\) ion is very unstable acid and thus easily loses \(\mathrm{H}^{+}\). What happens when acetic acid is treated with conc. \(\mathrm{H}_{2} \mathrm{SO}_{4}\) ? (a) \(\mathrm{CO}+\mathrm{H}_{2} \mathrm{O}\) (b) \(\mathrm{CH}_{4}+\mathrm{CO}_{2}\) (c) \(\mathrm{CO}+\mathrm{CH}_{4}\) (d) No reaction

Which of the following method is used for eliminating nitrogen of an amine present outside the ring? (a) Hofmann elimination (b) Cope elimination (c) Both (a) and (b) (d) Emde degradation

Alkyl halides and alcohols easily undergo nucleophilic substitution either through \(\mathrm{S}_{\mathrm{N}} 1\) or \(\mathrm{S}_{\mathrm{N}} 2\) mechanism. The relative case of these two processes depends upon the nature of the substrate (alkyl group as well as leaving group), nature of nucleophile and also upon the nature of solvent. \(\mathrm{S}_{\mathrm{N}} 1\) mechanism involves the formation of carbocation as intermediate while \(\mathrm{S}_{\mathrm{N}} 2\) mechanism involves the formation of a transition pentavalent state. \(\mathrm{S}_{\mathrm{N}} 1\) is the main mechanism in \(3^{\circ}\) alkyl halides and alcohols, while \(\mathrm{S}_{\mathrm{N}} 2\) mechanism is the path adopted by most of the \(1^{\circ}\) alkyl halides and \(2^{\circ}\) alkyl halides may follow \(\mathrm{S}_{\mathrm{N}} 1\) as well as \(\mathrm{S}_{\mathrm{N}} 2\). Neopentyl alcohol, \(\mathrm{Me}_{3} \mathrm{CCH}_{2} \mathrm{OH}\), reacts with HX according to (a) \(\mathrm{S}_{\mathrm{N}} 1\) mechanism (b) \(\mathrm{S}_{\mathrm{N}} 2\) mechanism (c) Both (a) and (b) (d) None

The decreasing order of acidic character of the compounds is \(\mathrm{CH}_{3} \mathrm{C} \equiv \mathrm{CH}, \mathrm{MeOH}, \mathrm{Me}_{2} \mathrm{CHOH}, \mathrm{Me}_{3} \mathrm{COH}, \mathrm{H}_{2} \mathrm{O}\) (a) \(\mathrm{CH}_{3} \mathrm{C} \equiv \mathrm{CH}>\mathrm{Me}_{3} \mathrm{COH}>\mathrm{Me}_{2} \mathrm{CHOH}>\mathrm{MeOH}>\mathrm{H}_{2} \mathrm{O}\) (b) \(\mathrm{MeOH}>\mathrm{Me}_{2} \mathrm{CHOH}>\mathrm{Me}_{3} \mathrm{COH}>\mathrm{H}_{2} \mathrm{O}>\mathrm{CH}_{3} \mathrm{C} \equiv \mathrm{CH}\) (c) \(\mathrm{Me}_{3} \mathrm{COH}>\mathrm{Me}_{2} \mathrm{CHOH}>\mathrm{MeOH}>\mathrm{H}_{2} \mathrm{O}>\mathrm{CH}_{3} \mathrm{C} \equiv \mathrm{CH}\) (d) \(\mathrm{MeOH}>\mathrm{H}_{2} \mathrm{O}>\mathrm{Me}_{2} \mathrm{CHOH}>\mathrm{Me}_{3} \mathrm{COH}>\mathrm{CH}_{3} \mathrm{C} \equiv \mathrm{CH}\)

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.