Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 13: Problem 35

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.

Short Answer

Expert verified
In summary, compound M has the structural formula \(\mathrm{CH}_{3}\mathrm{OCH}_{2}\mathrm{CH}_{2}\mathrm{CH}=\mathrm{CH}_{2}\), which contains an alkene group and is consistent with the given \({ }^{19} \mathrm{C}-\mathrm{NMR}\) data. Compound N, formed by hydrogenation of M, has the structural formula \(\mathrm{CH}_{3}\mathrm{OH}\) (methanol). These structures account for the given molecular formulas, reaction data, and the observed \({ }^{19} \mathrm{C}-\mathrm{NMR}\) data for compound M.

Step by step solution

01

Analyze the reactions of compound M

Compound M readily decolorizes \(\mathrm{Br}_{2}\) in \(\mathrm{CCl}_{4}\). This indicates that compound M has an alkene group (C=C) in its structure, since alkenes react with bromine to form vicinal dibromides, removing the color of the bromine solution. The reaction of compound M with \(\mathrm{H}_{2} / \mathrm{Ni}\) converts it into compound N with the molecular formula \(\mathrm{C}_{3} \mathrm{H}_{12} \mathrm{O}\). This reaction suggests that compound M undergoes hydrogenation, where the double bond (alkene group) is reduced to a single bond (alkane group). The change in molecular formula also implies that after hydrogenation, two of the carbons from M are no longer a part of the N molecule.
02

Interpret the \({ }^{1} \mathrm{H}-\mathrm{NMR}\) data for compound M

The information provided does not give enough detail about the peaks found in the \({ }^{1} \mathrm{H}-\mathrm{NMR}\) data for compound M. However, this information will not be critical to deducing the structure of the molecule.
03

Interpret the \({ }^{19} \mathrm{C}-\mathrm{NMR}\) data for compound M

The \({ }^{19} \mathrm{C}-\mathrm{NMR}\) data for compound M shows signals at \(\delta 146.12,110.75,71.05\), and \(29.88\). These chemical shift values suggest the presence of the following carbon environments in the molecule: - \(\delta 146.12\): a carbon within the alkene group (C=C) - \(\delta 110.75\): a carbon adjacent to the alkene group (C-C=C) - \(\delta 71.05\): a carbon with a heteroatom (hydroxyl, ether, or ester) - \(\delta 29.88\): a saturated carbon (C-C)
04

Deduce the structure of compound M

Based on the above analysis, we can now determine the structure of compound M. The presence of an alkene group and four distinct carbon environments in the \({ }^{19} \mathrm{C}-\mathrm{NMR}\) data lead us to the following possible structure for M: C(CH2)=CHCH2OCH3, as simplified in: \(\mathrm{CH}_{3}\mathrm{OCH}_{2}\mathrm{CH}_{2}\mathrm{CH}=\mathrm{CH}_{2}\) This structure explains each of the carbon environments observed in the \({ }^{19} \mathrm{C}-\mathrm{NMR}\) data, as well as the molecular formula \(\mathrm{C}_{5} \mathrm{H}_{10} \mathrm{O}\). It also confirms that compound M contains an alkene group.
05

Deduce the structure of compound N

From the given reaction data, we know that compound M undergoes hydrogenation to form compound N with the molecular formula \(\mathrm{C}_{3} \mathrm{H}_{12} \mathrm{O}\). The hydrogenation reaction removes the double bond found in M's alkene group, and eliminates two carbons from the molecule. Based on this information, the structure of compound N is derived as follows: - Hydrogenation: \(\mathrm{CH}_{3}\mathrm{OCH}_{2}\mathrm{CH}_{2}\mathrm{CH}_{3}\) - Elimination of two carbons (breaking at the ether linkage): \(\mathrm{CH}_{3}\mathrm{OH}\) and \(\mathrm{CH}_{3}\mathrm{CH}_{2}\mathrm{CH}_{3}\) Compound N has the structure: \(\mathrm{CH}_{3}\mathrm{OH}\) (methanol)

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

\({ }^{13} \mathbf{C}-\mathrm{NMR}\) is like \({ }^{1} \mathrm{H}-\mathrm{NMR}\), except the nuclear spins of \({ }^{13} \mathrm{C}\) nuclei are being analyzed. \- \({ }^{13}\) C-NMR spectra are commonly recorded in a hydrogen-decoupled instrumental mode. In this mode, all \({ }^{13} \mathrm{C}\) signals appear as singlets. \- The number of different signals in a \({ }^{13} \mathrm{C}-\mathrm{NMR}\) spectrum tell you how many nonequivalent carbon atoms are in a molecule. \- \({ }^{13}\) CNMR chemical shifts tell you what kind of carbon atoms are present.

Compound \(\mathrm{K}\), molecular formula \(\mathrm{C}_{8} \mathrm{H}_{14} \mathrm{O}\), readily undergoes acid-catalyzed dehydration when warmed with phosphoric acid to give compound L, molecular formula \(\mathrm{C}_{6} \mathrm{H}_{12}\), as the major organic product. The \({ }^{1} \mathrm{H}\)-NMR spectrum of compound \(\mathrm{K}\) shows signals at \(\delta 0.90(\mathrm{t}, 6 \mathrm{H}), 1.12(\mathrm{~s}, 3 \mathrm{H}), 1.38(\mathrm{~s}, 1 \mathrm{H})\), and \(1.48(\mathrm{q}, 4 \mathrm{H})\). The \({ }^{19} \mathrm{C}-\mathrm{NMR}\) spectrum of compound \(\mathrm{K}\) shows signals at \(\delta 72.98,33.72,25.85\), and 8.16. Deduce the structural formulas of compounds \(\mathrm{K}\) and \(\mathrm{L}\).

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.

Groups of atoms in which substitution of one atom by an isotope creates an achiral molecule are called homotopic. \- Homotopic groups always have identical chemical shifts. Those in which such substitution produces a chiral molecule are enantiotopic. Enantiotopic groups have identical chemical shifts, except in a chiral environment.

Following is the \({ }^{1} \mathrm{H}\)-NMR spectrum of compound \(\mathrm{O}\), molecular formula \(\mathrm{C}_{7} \mathrm{H}_{12}\) Compound \(\mathrm{O}\) reacts with bromine in carbon tetrachloride to give a compound with the molecular formula \(\mathrm{C}_{7} \mathrm{H}_{12} \mathrm{Br}_{2}\). The \({ }^{13} \mathrm{C}-\mathrm{NMR}\) spectrum of compound \(\mathrm{O}\) shows signals at \(\delta 150.12,106.43,35.44,28.36\), and \(26.36\). Deduce the structural formula of compound \(\mathrm{O}\).
See all solutions

Recommended explanations on Chemistry Textbooks

Physical Chemistry

Read Explanation

Kinetics

Read Explanation

Organic Chemistry

Read Explanation

Nuclear Chemistry

Read Explanation

Making Measurements

Read Explanation

Chemical Analysis

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