Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 13: Problem 24

Calculate the index of hydrogen deficiency of these compounds. (a) Aspirin, \(\mathrm{C}_{9} \mathrm{H}_{8} \mathrm{O}_{4}\) (b) Ascorbic acid (vitamin \(\mathrm{C}\) ), \(\mathrm{C}_{6} \mathrm{H}_{g} \mathrm{O}_{6}\) (c) Pyridine, \(\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{~N}\) (d) Urea, \(\mathrm{CH}_{4} \mathrm{~N}_{2} \mathrm{O}\) (e) Cholesterol, \(\mathrm{C}_{2} \mathrm{H}_{45} \mathrm{O}\) (f) Dopamine, \(\mathrm{C}_{8} \mathrm{H}_{11} \mathrm{NO}_{2}\)

Short Answer

Expert verified
a) Aspirin b) Ascorbic acid c) Pyridine d) Urea e) Cholesterol f) Dopamine Answer: a) Aspirin: IHD = 10 b) Ascorbic acid: IHD = 6 c) Pyridine: IHD = 6 d) Urea: IHD = 2 e) Cholesterol: IHD = 28 f) Dopamine: IHD = 8.5

Step by step solution

01

Find the number of carbon, hydrogen, and other atoms.

In aspirin, there are 9 carbons (n=9), 8 hydrogens (H=8), and 4 oxygen atoms. No other atoms are present.
02

Use the IHD formula.

Using the formula, we get: \(IHD = \frac{2(9) + 2 - 8 + 0 - 0}{2} = \frac{20}{2} = 10\). #a) Aspirin: IHD = 10# #b) Ascorbic acid: C6H9O6#
03

Find the number of atoms.

In ascorbic acid, there are 6 carbons (n=6), 8 hydrogens (H=8), and 6 oxygen atoms. No other atoms are present.
04

Use the IHD formula.

Using the formula, we get: \(IHD = \frac{2(6) + 2 - 8 + 0 - 0}{2} = \frac{12}{2} = 6\). #b) Ascorbic acid: IHD = 6# #c) Pyridine: C5H5N#
05

Find the number of atoms.

In pyridine, there are 5 carbons (n=5), 5 hydrogens (H=5), and 1 nitrogen atom(N=1).
06

Use the IHD formula.

Using the formula, we get: \(IHD = \frac{2(5) + 2 - 5 + 0 + 1}{2} = \frac{12}{2} = 6\). #c) Pyridine: IHD = 6# #d) Urea: CH4N2O#
07

Find the number of atoms.

In urea, there is 1 carbon (n=1), 4 hydrogens (H=4), 2 nitrogen atoms (N=2), and 1 oxygen atom.
08

Use the IHD formula.

Using the formula, we get: \(IHD = \frac{2(1) + 2 - 4 + 0 + 2}{2} = \frac{4}{2} = 2\). #d) Urea: IHD = 2# #e) Cholesterol: C27H45O#
09

Find the number of atoms.

In cholesterol, there are 27 carbons (n=27), 45 hydrogens (H=45), and 1 oxygen atom. No other atoms are present.
10

Use the IHD formula.

Using the formula, we get: \(IHD = \frac{2(27) + 2 - 45 + 0 - 0}{2} = \frac{56}{2} = 28\). #e) Cholesterol: IHD = 28# #f) Dopamine: C8H11NO2#
11

Find the number of atoms.

In dopamine, there are 8 carbons (n=8), 11 hydrogens (H=11), 1 nitrogen atom (N=1), and 2 oxygen atoms.
12

Use the IHD formula.

Using the formula, we get: \(IHD = \frac{2(8) + 2 - 11 + 0 + 1}{2} = \frac{17}{2} = 8.5\). #f) Dopamine: IHD = 8.5# #a) Aspirin: IHD = 10 #b) Ascorbic acid: IHD = 6 #c) Pyridine: IHD = 6 #d) Urea: IHD = 2 #e) Cholesterol: IHD = 28 #f) Dopamine: IHD = 8.5

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.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

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

When placed in a powerful magnetic field, there is a small population bias for the \({ }^{1} \mathrm{H}\) and \({ }^{13} \mathrm{C}\) nuclei to be aligned with the magnetic field, and they precess.

According to the \((n+1)\) rule, if a hydrogen has \(n\) hydrogens nonequivalent to it but equivalent among themselves on the same or adjacent atom(s), its \({ }^{1} \mathrm{H}-\mathrm{NMR}\) signal will be split into \((n+1)\) peaks. \- Splitting patterns are commonly referred to as singlets (s), doublets (d), triplets \((t)\), quartets \((q)\), quintets, and multiplets ( \(m\) ). \- The relative intensities of peaks in a multiplet can be predicted from an analysis of spin combinations for adjacent hydrogens or from the mnemonic device called Pascal's triangle. \- A coupling constant \((J)\) is the distance between adjacent peaks in a multiplet and is reported in hertz \((\mathrm{Hz})\). The value of \(J\) depends only on internal fields within a molecule and is independent of the spectrometer field.

When a hydrogen nucleus is coupled to more than one set of adjacent hydrogen nuclei, the couplings combine. \- In the general case, if a hydrogen nucleus is coupled to a set of \(n\) hydrogen nuclei on one side and a set of \(m\) hydrogen nuclei on the other, the signal will be split into a maximum of \((n+1)(m+1)\) peaks. \- In molecules that are rigid, for example alkenes or cyclic molecules, all the \((n+1)(m+1)\) peaks can often be seen. \- However, because coupling constants can be similar, especially in flexible molecules, this splitting can simplify to a number of observed peaks that is equal to the number of adjacent H atoms \(A 1\), regardless of patterns of equivalence.

Compound M, molecular formula \(\mathrm{C}_{5} \mathrm{H}_{10} \mathrm{O}\), readily decolorizes \(\mathrm{Br}_{2}\) in \(\mathrm{CCl}_{4}\) and is converted by \(\mathrm{H}_{2} / \mathrm{Ni}\) into compound \(\mathrm{N}\), molecular formula \(\mathrm{C}_{3} \mathrm{H}_{12} \mathrm{O}\). Following is the \({ }^{1} \mathrm{H}-\mathrm{NMR}\) spectrum of compound \(\mathrm{M}\). The \({ }^{19} \mathrm{C}-\mathrm{NMR}\) spectrum of compound \(\mathrm{M}\) shows signals at \(\delta 146.12,110.75,71.05\), and \(29.88\). Deduce the structural formulas of compounds \(M\) and N.

The experimental conditions required to cause nuclei to resonate are affected by the local chemical and magnetic environments. \- Electrons around a hydrogen or carbon create local magnetic fields that shield the nuclei of these atoms from the applied field. \- Any factor that increases the exposure of nuclei to an applied field is said to deshield them and shifts their signal downfield to a larger \(\delta\) value. \- Conversely, any factor that decreases the exposure of nuclei to an applied field is said to shield them and shifts their signal upfield to a smaller \(\delta\) value.
See all solutions

Recommended explanations on Chemistry Textbooks

Ionic and Molecular Compounds

Read Explanation

The Earths Atmosphere

Read Explanation

Physical Chemistry

Read Explanation

Chemical Analysis

Read Explanation

Making Measurements

Read Explanation

Inorganic Chemistry

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.

Sign-up for free