/*! 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.21558-14-52P Question:Answer the following qu... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Chapter 14: Q.21558-14-52P (page 564)

Question:Answer the following questions about each of the hydroxy ketones: 1-hydroxybutan-2-one (A) and 4-hydroxybutan-2-one (B).

a. What is the molecular ion in the mass spectrum?

b. What IR absorptions are present in the functional group region?

c. How many lines are observed in the 13C NMR spectrum?

d. How many signals are observed in the 1H NMR spectrum?

e. Give the splitting observed for each type of proton as well as its approximate chemical shift.

Short Answer

Expert verified

Answer

a.

b. A-1740 cm-1 for CO and 3600 cm-1 for OH

B-1730 cm-1 for CO and 3600 cm-1 for OH

c. 4 for A and B both.

d. 4 for A and B both.

e. A-(t) CH3, (q) CH2, (s) 1H, (s)CH2

B--(s) CH3, (t)CH2 , (s) 1H, (t) CH2

Step by step solution

01

 13C NMR spectrum

13C NMR spectrum gives the location of all non-equivalent carbons. The equivalent carbon has the same spectrum. The substituents affect the most in the 13C spectrum signals.

  • If the electron-withdrawing group is adjacent to the corresponding 13C,the spectra would be downfield.
  • If the electron-donating group is adjacent to the corresponding 13C, the spectra would be upfield.
02

Explanation for a and b

a. Molecular ion in the mass spectrometry is obtained by ionizing the molecule.

The molecular ions are shown as:

Molecular ion

b.

IR absorption for the ketone group in A is 1740 cm-1 , which is due to the presence of the OH group. The OH group attracts the electron towards itself and increases the stretching frequency of the ketone group.

The IR absorption for OH group in A and B is 3600cm-1 .

IR absorption for the ketone group is 1730 cm-1 due to the OH group being away from the CO. The OH group attracts the electron towards itself and increases the stretching frequency of the ketone group slightly less than A.

03

Explanation for c, d, and e

c.

The number of non-equivalent signals for both A and B are 4 – two 13C signals for twoCH2 , one CH3, and one carbon for CO.

d.

The number of non-equivalent signals for both A and B are 4 – four 1H signals for twoCH2 , one CH3, and one carbon for OH.

e.

The splitting signals for A and B are given as:

A-(t) CH3, (q) CH2, (s) 1H, (s)CH2

B--(s) CH3, (t)CH2 , (s) 1H, (t) CH2

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

Question: Using a 300 MHz NMR instrument:

a. How many Hz downfield from TMS is a signal at 2.5 ppm?

b. If a signal comes at 1200 Hz downfield from TMS, at what ppm does it occur?

c. If two signals are separated by 2 ppm, how many Hz does this correspond to?

What splitting pattern is observed for each proton in the following compounds?

a.

b.

Draw all constitutional isomers of molecular formula C3H6Cl2.

a. How many signals does each isomer exhibit in its 1HNMR spectrum?

b.How many lines does each isomer exhibit in its 13CNMR spectrum?

c. When only the number of signals in both 1Hand 13C NMR spectroscopy is considered, is it possible to distinguish all of these constitutional isomers?

Compound A exhibits two signals in its 1H NMR spectrum at 2.64 and 3.69 ppm and the ratio of the absorbing signals is 2:3. Compound B exhibits two signals in its 1H NMR spectrum at 2.09 and 4.27 ppm and the ratio of the absorbing signals is 3:2. Which compound corresponds to dimethyl succinate and which compound corresponds to ethylene diacetate?

Identify products A and B from the given 1HNMR data.

a. Treatment of CH2=CHCOCH3 with one equivalent of HCl forms compound A. A exhibits the following absorptions in its 1HNMR spectrum: 2.2 (singlet, 3 H), 3.05 (triplet, 2 H), and 3.6 (triplet, 2H) ppm. What is the structure of A?

b.Treatment of acetone [(CH3)2C=O] with dilute aqueous base forms B. Compound B exhibits four singlets in its 1HNMR spectrum at 1.3 (6 H), 2.2 (3 H), 2.5 (2 H), and 3.8 (1 H) ppm. What is the structure of B?

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.