/*! 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} Q.3P For each hydrocarbon spectrum, d... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

For each hydrocarbon spectrum, determine whether the compound is an alkane, an alkene, an alkyne, or an aromatic hydrocarbon, and assign the major peaks above (to the left of) 1600 cm-1 . More than one unsaturated group may be present.

Short Answer

Expert verified

Answer

The first spectrum represents alkenes.

The second spectrum represents alkanes.

The third spectrum represents aromatic hydrocarbon

Step by step solution

01

IR spectrum

The functional group present in a molecule can be determined by using its IR spectrum.

For the compounds with bond, the characteristic peak is observed at 1700cm-1 .

For alkanes, the characteristic stretch is observed at 3000cm-1 .

For alkenes the characteristic stretching peak is observed at 1500cm-1 .

02

First spectrum:Alkenes

A thin absorption at about 1700 cm-1is observed pointing downward. It is considered as carbonyl group (C=O). Alkenes (C=C) stretch is observed around 1500 cm-1. To be sure about alkene, there is unsaturated hydrocarbon at 3000 cm-1(C=C-H).

03

Second Spectrum:Alkanes

Alkanes have stretches at about 3000 cm-1 (C-H), C-H scissoring at about 1470 cm-1, methyl rock at about and long chain methyl rock at about.

04

Third Spectrum: Aromatic Hydrocarbon

Aromatic ring absorption spectra consist of series of bumps (1650 cm-1to 2000 cm-1). Number of bumps depends upon substitution of benzene ring.

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 91Ó°ÊÓ!

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

The mass spectrum of n-octane shows a prominent molecular ion peak (m/z 114). There is also a large peak at m/z 57, but it is not the base peak. The mass spectrum of 3,4-dimethylhexane shows a smaller molecular ion, and the peak at mass 57 is the base peak. Explain these trends in abundance of the molecular ions and the ions at mass 57 and predict the intensities of the peaks at masses 57 and 114 in the spectrum of 2,2,3,3-tetramethylbutane.


Question:Spectra are given for three compounds. Each compound has one or more of the following functional groups: alcohol, amine, ketone, aldehyde, and carboxylic acid. Determine the functional group(s) in each compound, and assign the major peaks above 1600cm-1.

Account for the peaks at m/z 87, 111, and 126 in the mass spectrum of 2,6-dimethylheptan-4-ol.

Question: A C-D (carbon–deuterium) bond is electronically much like a C-H bond, and it has a similar stiffness, measured by the spring constant, k. The deuterium atom has twice the mass (m) of a hydrogen atom, however.

(a) The infrared absorption frequency is approximately proportional tokm , when one of the bonded atoms is much heavier than the other, and m is the lighter of the two atoms (H or D in this case). Use this relationship to calculate the IR absorption frequency of a typical C-D bond. Use as a typical C-H absorption frequency.

(b) A chemist dissolves a sample in deuterochloroform (CDCl3) and then decides to take the IR spectrum and simply evaporates most of theCDCl3 . What functional group will appear to be present in this IR spectrum as a result of theCDCl3 impurity?

Show the fragmentations that give rise to the peaks at m/z 43, 57, and 85 in the mass spectrum of 2,4-dimethylpentane (Figure 12-17).

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.