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

When methyl benzoate is nitrated with \(\mathrm{HNO}_{3} / \mathrm{H}_{2} \mathrm{SO}_{4^{\prime}}\) the meta product is the major product. This is because (a) The \(-\mathrm{CO}_{2} \mathrm{CH}_{3}\) group is a meta director. (b) The \(-\mathrm{OCH}_{3}\) is a meta director. (c) The \(-\mathrm{NO}_{2}\) group is deactivating and a meta director. (d) Nitration usually occurs at the meta position.

Short Answer

Expert verified
(a) The \(-\mathrm{CO}_{2}\mathrm{CH}_{3}\) group is a meta director.

Step by step solution

01

Understanding the Groups Involved

Methyl benzoate has the ester group \(-\mathrm{CO}_{2}\mathrm{CH}_{3}\). During nitration, this group's properties affect the reaction. The ester group is an electron-withdrawing group, making it a deactivating group that directs incoming electrophiles to the meta position during electrophilic aromatic substitution reactions.
02

Review of Electrophilic Aromatic Substitution

In electrophilic aromatic substitution, an electrophile attacks an aromatic ring. The substituents already on the ring influence where the new substituent will attach either by donating or withdrawing electrons. Electron-withdrawing groups, like \(-\mathrm{CO}_{2}\mathrm{CH}_{3}\), tend to direct electrophiles to the meta position on the ring.
03

Evaluate Options Based on Electron Direction

Since methyl benzoate's ester group \(-\mathrm{CO}_{2}\mathrm{CH}_{3}\) is deactivating and directs new groups to the meta position, option (a) is supported. Option (b) refers to a methoxy group \-\mathrm{OCH}_{3}\, which is not present here. Option (c) refers to the behavior of the nitro group after nitration, but does not address why the major product is meta initially. Option (d) suggests a general statement not specific or accurate to all nitration conditions.

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.

Methyl Benzoate
Methyl benzoate is an aromatic compound characterized by a benzene ring bonded to an ester group, specifically \(-\mathrm{CO}_{2}\mathrm{CH}_{3}\) group. This structural arrangement significantly influences its behavior in chemical reactions. In aromatic substitution reactions, specific substituents can change how the reaction proceeds. Here, the ester group plays a crucial role.

When mixed with strong nitrating agents, the benzene ring in methyl benzoate undergoes a reaction called nitration. However, the positioning of the substituents on the benzene ring is not random. This ester group is what determines the position and type of reaction taking place. Understanding methyl benzoate's structural properties helps us predict the outcomes of its chemical interactions.
Meta Director
In chemical terms, a meta director refers to a substituent on an aromatic ring that directs incoming electrophiles to attach to the meta position. For methyl benzoate, the ester group \(-\mathrm{CO}_{2}\mathrm{CH}_{3}\) acts as a meta director. This means it makes the meta position more attractive for the nitro group during a nitration process.

Meta directors are typically electron-withdrawing groups. These groups pull electron density away from the aromatic ring through resonance or inductive effects, reducing the reactivity of the ortho and para positions. By decreasing the electron density at these sites, they render them less reactive and promote the meta location as a preferable site for substitution.
Electron-Withdrawing Group
An electron-withdrawing group (EWG) is an essential concept in aromatic chemistry. These groups, such as the ester group in methyl benzoate, pull electrons toward themselves, decreasing the electron density within the aromatic ring.

The impact of an electron-withdrawing group is crucial, as it deactivates the aromatic ring and predisposes it to favor substitution at the meta position. Deactivating nature means that such rings are less reactive toward electrophiles, guiding them specifically to the meta position.
  • Deactivates the aromatic ring
  • Directs new groups to attach at the meta position
  • Makes the molecule less reactive overall
Understanding the role of EWGs is vital for predicting the outcomes of reactions such as nitration in aromatic systems like methyl benzoate.
Nitration Reaction
Nitration is an electrophilic aromatic substitution reaction where a nitro group \(-\mathrm{NO}_{2}\) is introduced into an aromatic ring. For methyl benzoate, this involves reacting with a mixture of nitric acid (\(\mathrm{HNO}_{3}\)) and sulfuric acid (\(\mathrm{H}_{2}\mathrm{SO}_{4}\)). The acids work together to generate the nitronium ion (\(\mathrm{NO}_{2}^{+}\)), a powerful electrophile that attacks the benzene ring.

Given methyl benzoate's structure and the presence of the electron-withdrawing ester group, the nitration predominantly occurs at the meta position. This reaction highlights how the activating and directing effects of substituents govern the orientation of electrophilic attacks on aromatic compounds. It's a perfect example of how seemingly small changes in structure can significantly affect chemical reactivity and product formation.
  • Involves strong acids to generate the nitronium ion
  • Meta substitution is preferred due to electron-withdrawing ester group
  • Key for synthesizing nitro-aromatic compounds

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

\(\frac{\text { (i) } \mathrm{BH}_{3}}{\text { (ii) } \mathrm{NaOH}, \mathrm{H}_{2} \mathrm{O}_{2}}\) Identify reactant

Which of the following gives positive Fehling's solution test (a) Ph-CHO (b) \(\mathrm{CH}_{3} \mathrm{CHO}\) (c) \(\mathrm{CCl}_{3} \mathrm{CHO}\) (d) All of them

Match Column I with Column II. Column I (a) Aldol condensation (b) Cannizzaro reaction (c) Reformatsky reaction (d) Benzoin condensation Column II (p) Hydride ion transfer (q) Zinc enolate (r) \(\overline{\mathrm{C}} \mathrm{N}\) as catalyst (s) Enolate ion

Identify correct method of preparation of acetaldehyde from reaction of cyanide (a) \(\mathrm{Me}-\mathrm{C} \equiv \mathrm{N} \frac{\text { (i) DIBAL }}{\text { (ii) } \mathrm{H}_{3} \mathrm{O}^{+}}\) (b) \(\mathrm{Me}-\mathrm{C} \equiv \mathrm{N} \frac{\text { (i) } \mathrm{SnCl}_{2}+\mathrm{HCl}}{\text { (ii) } \mathrm{H}_{3} \mathrm{O}^{+}}\) (c) \(\mathrm{Me}-\mathrm{C} \equiv \mathrm{N} \frac{\text { (i) Conc. } \mathrm{H}_{2} \mathrm{SO}_{4}}{\text { (ii) dill. } \mathrm{NaOH}}\) (d) \(\mathrm{Me}-\mathrm{C} \equiv \mathrm{N} \frac{\text { (i) } \mathrm{Pd} / \mathrm{BaSo}_{4} / \mathrm{H}_{2}}{\text { (ii) } \mathrm{H}_{3} \mathrm{O}^{\oplus}}\)

In general, ketones are more reactive towards nucleophiles than esters because (a) The \(\alpha\) -protons of a ketone are more acidic than those of an ester. (b) The alkyl group in a ketone is an electron donating group due to hyperconjugation. (c) Alkoxy (RO-) groups are sterically larger than the related alkyl group. (d) Alkoxy (RO-) groups are stronger electron donating than alkyl groups via resonance.

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