NMR 1 is what structure? A or B?NMR 2 is what strucure? A or B? ## NMR Spectroscopy Analysis**Objective:**Match each compound with its correct NMR spectrum. Integrations are indicated above each signal.### Spectral Data#### NMR Spectrum 1- **Signal Integrations:** - At approximately 7.2 ppm: (5) - At approximately 4.1 ppm: (2) - At approximately 2.4 ppm: (2) - At approximately 1.2 ppm: (3)#### NMR Spectrum 2- **Signal Integrations:** - At approximately 7.2 ppm: (5) - multiplet - At approximately 3.5 ppm: (2) - At approximately 1.3 ppm: (3)### Compounds- **Compound A:** - Structure: Phenylacetone- **Compound B:** - Structure: Phenylpropanone### Analysis- **Spectrum 1:** - The integration of 5 protons on the aromatic region (around 7.2 ppm) and distinct peaks for 2 protons and 3 protons in aliphatic regions suggest a structure that includes both an aromatic ring and aliphatic protons.- **Spectrum 2:** - Similar aromatic integration with a multiplet, and specific shifts for aliphatic protons aligning with a structure bearing phenyl and ketone functionalities typically found in phenylpropanone-like structures.### Conclusion:- **Match Spectrum 1 with Compound B.**- **Match Spectrum 2 with Compound A.**This exercise allows students to connect NMR spectral data with molecular structure, an essential skill in organic chemistry.

NMR 1 for the A compound . Since in the A compound there are two ch2 grp adjacent to the carbonyl group and thus highly deshielded , benzyllic is more deshielded and thus gives singlet near 3.5 . Whereas Ch2 adjacent to CH3 having chemical shift value 2.4 ppm gives quatret . less deshielded for CH3 proton gives triplet at 1.2 ppm . Whereas aromatic group gives multiplet in the 7 ppm region. multiplet for 5H of benzene ring at 7.2 - 7.8 with intensity ratio 2:2:1 .NMR 2 for the 2nd compound . Here only one signal for Ch2 proton gives quatret having in the deshielded region at 3.5 ppm . Whereas CH3 proton gives triplet at 1.2 ppm . And a multiplet for 5H of benzene ring at 7.2 - 7.8 with intensity ratio 2:2:1 .

Answered: Match each compound with its correct…

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Match each compound with its correct spectrum. Integrations are shown above each signal. (2) NMR 1 (3) (5) (2) 8 7 4 3 1 PPM (3) NMR 2

Organic Chemistry: A Guided Inquiry

2nd Edition

ISBN:9780618974122

Author:Andrei Straumanis

Publisher:Andrei Straumanis

ChapterL4: Proton (1h) Nmr Spectroscopy

Section: Chapter Questions

Problem 19CTQ

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Concept explainers

NMR Spectroscopy

NMR stands for Nuclear Magnetic Resonance in which the study of different molecules is done based on the interaction of radiofrequency electromagnetic radiations with the nuclei of molecules placed in a strong magnetic field. A precise study of spectra of visible light and all bands of electromagnetic radiation is called spectroscopy.

Question

NMR 1 is what structure? A or B?

NMR 2 is what strucure? A or B?

## NMR Spectroscopy Analysis

**Objective:**
Match each compound with its correct NMR spectrum. Integrations are indicated above each signal.

### Spectral Data

#### NMR Spectrum 1
- **Signal Integrations:**
- At approximately 7.2 ppm: (5)
- At approximately 4.1 ppm: (2)
- At approximately 2.4 ppm: (2)
- At approximately 1.2 ppm: (3)

#### NMR Spectrum 2
- **Signal Integrations:**
- At approximately 7.2 ppm: (5) - multiplet
- At approximately 3.5 ppm: (2)
- At approximately 1.3 ppm: (3)

### Compounds

- **Compound A:**
- Structure: Phenylacetone

- **Compound B:**
- Structure: Phenylpropanone

### Analysis

- **Spectrum 1:**
- The integration of 5 protons on the aromatic region (around 7.2 ppm) and distinct peaks for 2 protons and 3 protons in aliphatic regions suggest a structure that includes both an aromatic ring and aliphatic protons.

- **Spectrum 2:**
- Similar aromatic integration with a multiplet, and specific shifts for aliphatic protons aligning with a structure bearing phenyl and ketone functionalities typically found in phenylpropanone-like structures.

### Conclusion:
- **Match Spectrum 1 with Compound B.**
- **Match Spectrum 2 with Compound A.**

This exercise allows students to connect NMR spectral data with molecular structure, an essential skill in organic chemistry.

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