0.56 g 1,4‐dimethoxybenzene, 0.9 mL 2‐chloro‐2‐methylpropane, 5 mL nitrobenzene and 0.52 g graphite were combined, and the mixture was refluxed for 1.5 hours. It was then cooled to room temperature and filtered. The filtered graphite was washed with 15 mL hexane. The filtrate was distilled under vacuum to remove any remaining 1,4‐dimethoxybenzene, 2‐chloro‐2‐ methylpropane and nitrobenzene (6.3 g), yielding 0.44 g 1,4‐di‐tert‐butyl‐2,5‐dimethoxybenzene (43.4% yield). What is the Atom economy, E factor, and effective mass yield? **Reaction Scheme:****Reactants:**- A compound with the molecular formula C6H4(OMe)2, where "OMe" represents methoxy groups attached to a benzene ring at the 1,4-positions.- A chloro compound, (CH3)3CCl, where a tertiary carbon is bonded to a chlorine atom.**Reaction Conditions:**- The presence of graphite and nitrobenzene is indicated as part of the reaction conditions. The circle with an "X" (⊗) usually denotes a specific experimental condition or special setup requiring protective measures.**Products:**- A compound with the molecular formula C6H3(OMe)2C(CH3)3, where the benzene ring now has three substituents: - Two methoxy groups in a 1,4-relationship. - A tert-butyl group (C(CH3)3) introduced to the benzene ring. This chemical reaction depicts the synthesis of a tert-butyl derivative of the dimethoxybenzene compound through the action of graphite and a chloro precursor in nitrobenzene solvent.

The given problem is related to green chemistry. Given: Reactants- Mass of 1,4‐dimethoxybenzene-…

Answered: 0.56 g 1,4‐dimethoxybenzene, 0.9 mL…

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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0.56 g 1,4‐dimethoxybenzene, 0.9 mL 2‐chloro‐2‐methylpropane, 5 mL nitrobenzene and 0.52 g graphite were combined, and the mixture was refluxed for 1.5 hours. It was then cooled to room temperature and filtered. The filtered graphite was washed with 15 mL hexane. The filtrate was distilled under vacuum to remove any remaining 1,4‐dimethoxybenzene, 2‐chloro‐2‐ methylpropane and nitrobenzene (6.3 g), yielding 0.44 g 1,4‐di‐tert‐butyl‐2,5‐dimethoxybenzene (43.4% yield). What is the Atom economy, E factor, and effective mass yield?

**Reaction Scheme:**

**Reactants:**
- A compound with the molecular formula C6H4(OMe)2, where "OMe" represents methoxy groups attached to a benzene ring at the 1,4-positions.
- A chloro compound, (CH3)3CCl, where a tertiary carbon is bonded to a chlorine atom.

**Reaction Conditions:**
- The presence of graphite and nitrobenzene is indicated as part of the reaction conditions. The circle with an "X" (⊗) usually denotes a specific experimental condition or special setup requiring protective measures.

**Products:**
- A compound with the molecular formula C6H3(OMe)2C(CH3)3, where the benzene ring now has three substituents:
- Two methoxy groups in a 1,4-relationship.
- A tert-butyl group (C(CH3)3) introduced to the benzene ring.

This chemical reaction depicts the synthesis of a tert-butyl derivative of the dimethoxybenzene compound through the action of graphite and a chloro precursor in nitrobenzene solvent.

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