What is the structure of the unknown? Determine it by analyzing the spectroscopic data given in IR, MS, 1H NMR, 13C NMR (Broadband decoupled with carbon types) and fully substantiate your answer with key supporting data from each spectrum. IR Spectrum(liquid film)40003000200016001200800v (cm1)V (ст100Mass SpectrumUV Spectrum105802 max = (log10ɛz 2.5)(1o910260nmƐz2.5)6040M*'= 134204080120160200240280m/e% of base peak 13C NMR Spectrum(100.0 MHz, CDCI, solution)DEPT CH CH, сн!solventproton decoupled20016012080408 (ppm)1H NMR Spectrum(400 MHz, CDCI, solution)expansionsexpansionexpansion7.57.3 ppm2.82.7 ppm1.81.7 ppm1.451.35 ppm1.05 0.95TMS109.874318 (ppm)5

The structure of the unknown compound is shown below according to the given spectroscopic data - IR,…

Answered: What is the structure of the unknown?…

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

See similar textbooks

Related questions

Question

What is the structure of the unknown? Determine it by analyzing the spectroscopic data given in IR, MS, 1H NMR, 13C NMR (Broadband decoupled with carbon types) and fully substantiate your answer with key supporting data from each spectrum.

IR Spectrum
(liquid film)
4000
3000
2000
1600
1200
800
v (cm1)
V (ст
100
Mass Spectrum
UV Spectrum
105
80
2 max = (log10ɛz 2.5)
(1o910
260nm
Ɛz2.5)
60
40
M*'= 134
20
40
80
120
160
200
240
280
m/e
% of base peak

13C NMR Spectrum
(100.0 MHz, CDCI, solution)
DEPT CH CH, сн!
solvent
proton decoupled
200
160
120
80
40
8 (ppm)
1H NMR Spectrum
(400 MHz, CDCI, solution)
expansions
expansion
expansion
7.5
7.3 ppm
2.8
2.7 ppm
1.8
1.7 ppm
1.45
1.35 ppm
1.05 0.95
TMS
10
9.
8
7
4
3
1
8 (ppm)
5

Expert Solution

Trending now

This is a popular solution!

Step by step

Solved in 2 steps with 1 images

SEE SOLUTION Check out a sample Q&A here

Knowledge Booster

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, chemistry and related others by exploring similar questions and additional content below.

Similar questions