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

A compound of molecular weight 180 is acetylated to give a compound of molecular weight \(390 .\) The number of amino groups in the compound is (a) 6 (b) 5 (c) 4 (d) 3

Short Answer

Expert verified
The compound has 5 amino groups.

Step by step solution

01

Determine the Weight Difference

First, calculate the molecular weight increase upon acetylation. Subtract the original molecular weight from the acetylated molecular weight: \( 390 - 180 = 210 \). This is the increase in weight due to acetylation.
02

Calculate the Weight Per Amino Group

An acetyl group has the chemical formula \( CH_3CO \) and a molecular weight of 43. Therefore, each amino group being acetylated adds 43 units to the molecular weight.
03

Compute the Number of Amino Groups

Divide the total increase in molecular weight by the weight added per amino group: \( \frac{210}{43} \approx 4.88 \). Since the number of amino groups must be an integer, round this to the closest whole number, which is 5.

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.

Molecular Weight Calculation
Understanding how to calculate molecular weight is key to many areas of chemistry. The molecular weight is essentially the sum of the atomic masses of all atoms in a molecule. For example, if you have a molecule composed of two hydrogen atoms and one oxygen atom, you sum their atomic masses to get the molecular weight.
To calculate this, you would look at the periodic table for the atomic masses:
  • Hydrogen (H) has an atomic mass of about 1.01 u (atomic mass units).
  • Oxygen (O) has an atomic mass of about 16.00 u.
Adding these together for water (H2O) gives: \[ (2 \times 1.01) + 16.00 = 18.02 \text{ u} \] This process applies to any molecule, regardless of complexity, enabling us to understand and predict its behavior in chemical reactions.
Acetylation Process
The acetylation process is a chemical reaction that introduces an acetyl group into a molecule. An acetyl group is characterized by the formula \(CH_3CO\). This process is especially relevant in biochemistry and organic chemistry.
Acetylation can modify the properties of molecules significantly. When applied to amino groups in a compound, it can change solubility, stability, and reactivity.
During acetylation, every amino group that is modified increases the molecular weight by 43 due to the addition of the acetyl group. This is crucial when trying to determine the number of amino groups, as seen in the exercise.
Chemical Formula Analysis
Chemical formula analysis involves interpreting the symbols in a chemical formula to understand the composition of the compound. It tells us which elements are present and in what proportions.
For instance, in the formula \(CH_3CO\), we observe:
  • One carbon atom (C)
  • Three hydrogen atoms (H) combined into a methyl group (CH3)
  • Another carbon atom (C) linked to an oxygen atom (O) representing the carbonyl group (C=O)
Understanding chemical formulas allows you to do more than just calculate molecular weight; it helps you predict reactions and interactions with other compounds. These reactions and interactions are vital for designing experiments and interpreting results accurately.

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

In the reaction \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{C} \equiv \mathrm{CH}$$\stackrel{\mathrm{H}_{2} \mathrm{SO}_{4} \mathrm{HgSO}_{4}}{\longrightarrow}[\mathrm{A}]\) The compound [A] is (a) \(\mathrm{CH}_{3}-\mathrm{CH}_{2}-\mathrm{CH}_{2}-\mathrm{COOH}\) (b) \(\mathrm{CH}_{3}-\mathrm{CH}_{2}-\mathrm{CH}_{2}-\mathrm{CHO}\) (c) \(\mathrm{CH}_{3}-\mathrm{CH}_{2}-\mathrm{CO}-\mathrm{CH}_{3}\) (d) none of these

Treatment of propionaldehyde with dil. \(\mathrm{NaOH}\) solution gives (a) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COCH}_{2} \mathrm{CH}_{2} \mathrm{CHO}\) (b) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CHOHCH}_{2} \mathrm{CH}_{2} \mathrm{CHO}\) (c) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CHOHCH}\left(\mathrm{CH}_{3}\right) \mathrm{CHO}\) (d) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COOCH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{3}\)

The work of concentration. \(\mathrm{H}_{2} \mathrm{SO}_{4}\) in esterification process is as (a) dehydrating agent and catalyst (b) dehydrating agent (c) hydrolyzing agent (d) catalyst

Consider the following substances: 1\. HCHO 2\. \(\mathrm{CH}_{3} \mathrm{CHO}\) 3\. \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COCH}_{3}\) 4\. \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COCH}_{2} \mathrm{CH}_{3}\) Correct order of reactivity towards nucleophillic addition reaction is: (a) \(1>4>2>3\) (b) \(1>2>4>3\) (c) \(1>3>2>4\) (d) \(1>2>3>4\)

On mixing ethyl acetate with aqueous sodium chloride, the composition of the resultant solution is (a) \(\mathrm{CH}_{3} \mathrm{COOC}_{2} \mathrm{H}_{5}+\mathrm{NaCl}\) (b) \(\mathrm{CH}_{3} \mathrm{COONa}+\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\) (c) \(\mathrm{CH}_{3} \mathrm{COCl}+\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}+\mathrm{NaOH}\) (d) \(\mathrm{CH}_{3} \mathrm{Cl}+\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{COONa}\)

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