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

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

Short Answer

Expert verified

Answer

2,6-Dimethylheptan-4-ol has molecular weight 144. Highest mass peak at 126 is not the molecular ion peak. It is due to the loss of water molecule from molecular ion.

The peak at m/z = 111 is loss of methyl fragment from fragment of m/z 126. This is known as allylic cleavage which generates secondary allylic carbocation.

Peak at m/z= 87 results from the fragmentation on one side of alcohol.

Step by step solution

01

The peak at m/z 126

2,6-Dimethylheptan-4-ol has molecular weight 144. Highest mass peak at 126 is not the molecular ion peak but it is due to the loss of water molecules from the molecular ion.

Peak at m/z 126

02

The peak at m/z 111

The peak at m/z 111 is formed due to the following fragmentation:

03

The peak at m/z 87

The peak at m/z 87 is formed due to the following fragmentation:

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!

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

Predict the masses and the structures of the most abundant fragments observed in the mass spectra of the following compounds

a. 2-methylpentane

b. 3-methylhex-2-ene

c. 4-methylpentane-2-ol

d. 2-methyl-1-phenylpropane

e. Cyclohexylisopropy; ether [cyclohexyl--O-CH(CH3)2 ]

f. CH3CH2CH2NHC(CH3)2

g.

h.3-bromo-2-methylpentane


Question: Four infrared spectra are shown, corresponding to four of the following compounds. For each spectrum, determine the structure and explain how the peaks in the spectrum correspond to the structure you have chosen

Question: A C-D (carbonโ€“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?

These five structures all have distinguishing absorption in the IR. Match each structure with its characteristic absorption.

Structure 1

Structure 2

Structure 3

Structure 4

Structure 5

(a)sharp, 2254 cm-1 (b) very broad, centered about3330cm-1 (c) strong, slightly broadened, 1645cm-1 (d) broad with spikes at 3367cm-1and 3392cm-1(e) strong, sharp 1717cm-1

Question: (A true story.) While organizing the undergraduate stockroom, a new chemistry professor found a half-gallon jug containing a cloudy liquid (bp 100โ€“105 ยฐC), marked only โ€œSTUDENT PREP.โ€ She ran a quick mass spectrum, which is printed below. As soon as she saw the spectrum (without even checking the actual mass numbers), she said, โ€œI know what it is.โ€

(a) What compound is the โ€œstudent prepโ€? Any uncertainty in the structure?

(b) Suggest structures for the fragments at 136, 107, and 93. Why is the base peak (at m/z 57) so strong?

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