/*! 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 43 Compare the rate of reduction wi... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 4: Problem 43

Compare the rate of reduction with \(\mathrm{NaBH}_{4}\). CC(=O)c1ccccc1 O=Cc1ccc([N+](=O)[O-])cc1 O=Cc1ccccc1 (I) (II) (III) (A) I > II > III (B) II > III > I (C) II > I > III (D) \(\mathrm{III}>\mathrm{II}>\mathrm{I}\)

Short Answer

Expert verified
The rate of reduction with NaBH鈧� can be compared by examining the electron-donating or withdrawing groups' effect on the carbonyl group reactivity. Acetophenone (I) has an electron-donating methyl group (-CH3) making it more susceptible to reduction; 4-Nitrobenzaldehyde (II) has an electron-withdrawing nitro group (-NO鈧�), making it less susceptible; Benzaldehyde (III) has no additional groups, lying between the other two. Thus, the correct order of rate of reduction is I > III > II or (A) I > II > III.

Step by step solution

01

Convert SMILES notations to chemical structures

Use an appropriate tool or software to draw the structures of the compounds from the given SMILES notations: (I) CC(=O)c1ccccc1 - Acetophenone (II) O=Cc1ccc([N+](=O)[O-])cc1 - 4-Nitrobenzaldehyde (III) O=Cc1ccccc1 - Benzaldehyde
02

Examine the functional groups

Take note of the functional groups present in each compound, which are affected by reduction using NaBH鈧�. The primary functional group that undergoes reduction in all three structures is the carbonyl group, C=O.
03

Analyze the effect of different groups on the rate of reduction

The electron-donating or withdrawing groups on the aromatic ring have a significant influence on the rate of reduction. Electron-donating groups activate the carbonyl group and make it more susceptible to reduction. Conversely, electron-withdrawing groups deactivate the carbonyl group and make it less susceptible to reduction. In the given compounds: - Acetophenone (I) has an electron-donating methyl group (-CH3) at the alpha position. - 4-Nitrobenzaldehyde (II) has an electron-withdrawing nitro group (-NO鈧�) at the para position. - Benzaldehyde (III) has no additional electron-donating or withdrawing groups.
04

Determine the relative rates of reduction

Compare the structures considering the electron-donating or withdrawing groups present: - Acetophenone (I) has an electron-donating group, making it easier to reduce. - 4-Nitrobenzaldehyde (II) has an electron-withdrawing group, making it harder to reduce. - Benzaldehyde (III) lies somewhere in between as it has neither an electron-donating nor an electron-withdrawing group. Thus, the correct order for the rate of reduction is Acetophenone > Benzaldehyde > 4-Nitrobenzaldehyde, or I > III > II.
05

Choose the correct option

Based on our analysis, the correct order of the rate of reduction is I > III > II. Therefore, the correct option is (A)I > II > III.

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!

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

How many moles of \(\mathrm{CH}_{3} \mathrm{MgCl}\) is required to completely react with 1 mole of 3,3 -dimethyl penta-1, 4 -dyne?

The most appropriate reagent for the following conversion is CCOC1C=CC=CC1O Oc1ccccc1 (A) \(\mathrm{Cu} / 300^{\circ} \mathrm{C}\) (B) conc. \(\mathrm{H}_{2} \mathrm{SO}_{4}\) (C) Red P/HI (D) NaI in Acetone

Compound which loses its optical activity upon standing an enantiomerically pure form of it in a solution of \(\mathrm{NaOEt} ?\) (A) CC1=CC(=O)C(C)(C)CC1 (B) CC(=O)C1CC=C(C)CC1 (C) CC(C)C1CCC=CC1=O (D) Both B and C

$$ \mathrm{CH}_{3}-\mathrm{CH}=\mathrm{CH}-\mathrm{CH}_{3} \longrightarrow \mathrm{CH}_{3}-\mathrm{CHO} $$ ' \(\mathrm{X}\) ' can be : (A) \(\mathrm{O}_{3}, \mathrm{Zn}\) (B) \(\mathrm{KMnO}_{4} / \mathrm{OH}, \Delta\) (C) (i) \(\mathrm{HCO}_{3} \mathrm{H}\), (ii) \(\mathrm{HIO}_{4}\) (D) Both (A) and (C)

The incorrect option regarding the following reaction sequence is (A) Reaction is reversible and catalysed by acid (B) \(\mathrm{Z}\) can be alkyl, aryl, \(\mathrm{OH}, \mathrm{NHCONH}_{2}\) (C) Product is favoured because of rapid dehydration of (P) (D) If \(\mathrm{NHZ}_{2}\) is used in place of \(\mathrm{H}_{2} \mathrm{~N}-\mathrm{Z}\), then imine will be obtained as (Q)
See all solutions

Recommended explanations on Chemistry Textbooks

Nuclear Chemistry

Read Explanation

Organic Chemistry

Read Explanation

Chemistry Branches

Read Explanation

Chemical Analysis

Read Explanation

Physical Chemistry

Read Explanation

Making Measurements

Read Explanation
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.