**Infrared Spectroscopy: Distinguishing Molecules**### Which choice best explains why the molecules below can be distinguished by IR?![Molecules](molecules.jpg)- **Molecule A:** Contains a benzene ring with an aldehyde functional group.- **Molecule B:** Contains a benzene ring with an acetyl group (a ketone substituted with a methyl group).### Multiple Choice Explanation:1. **Molecule A has a peak just above 3000 cm⁻¹ and Molecule B does not.**2. **Molecule A has a peak for an aldehyde around 2200 cm⁻¹ and Molecule B does not.**3. **Molecule B has a carbonyl peak around 1750 cm⁻¹ and Molecule A does not.**4. **Molecule A and Molecule B can't be easily distinguished by IR.**### Correct Answer: **Molecule B has a carbonyl peak around 1750 cm⁻¹ and Molecule A does not.**#### Explanation:Infrared (IR) spectroscopy is a powerful technique used to identify certain functional groups based on their characteristic absorption bands. Molecule A (which contains an aldehyde) and Molecule B (which contains a ketone) can be distinguished based on the specific absorption frequencies characteristic to these groups.- **Aldehyde (Molecule A)**: Typically shows a distinctive absorption around 1725 cm⁻¹ corresponding to the C=O stretch of the aldehyde.- **Ketone (Molecule B)**: Typically shows a strong absorption around 1705-1750 cm⁻¹ for the carbonyl (C=O) stretch.Thus, **option 3** correctly identifies that Molecule B has a carbonyl peak around 1750 cm⁻¹, which Molecule A does not have, allowing easy distinction using IR spectroscopy.### Note:IR spectroscopy peaks are critical for identifying different functional groups in chemical molecules, making it an essential tool in molecular analysis in organic chemistry.

Answered: Image /qna-images/answer/ddc37c93-49ed-436a-94e5-1d195ac907aa.jpg

Which choice best explains why the molecules below can be distinguished by IR. H Molecule A O Molecule A has a peak just above 3000 cm-1 and Molecule B does not. Molecule A has a peak for an aldehyde around 2200 cm and Molecule B does not. -1 Molecule B has a carbonyl peak around 1750 cm1 and Molecule A does not. Molecule A and Molecule B can't be easily distinguished by IR. OOOO .CH3 Molecule B

Which choice best explains why the molecules below can be distinguished by IR. H Molecule A O Molecule A has a peak just above 3000 cm-1 and Molecule B does not. Molecule A has a peak for an aldehyde around 2200 cm and Molecule B does not. -1 Molecule B has a carbonyl peak around 1750 cm1 and Molecule A does not. Molecule A and Molecule B can't be easily distinguished by IR. OOOO .CH3 Molecule B

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

Similar questions

Choose the best explanation for why -OH and -NH peaks are often broad singlets in a ¹H NMR spectrum. A) The greater electronegativity difference in the N-H or O-H bond compared to a C-H bond makes the peak broader. B) The greater mass involved in the N-H or O-H bond compared to a C-H bond makes the peak broader. C) The hydrogen nuclei on the carbon next to the -OH or -NH are too far away to impact the multiplicity. D) Protons in OH or NH groups are acidic enough to rapidly exchange between different molecules, so the coupling to vicinal hydrogens is not clearly observed.
Please don't provide handwriting solution
help me please. thanks
Need help thx
answer the following questions about the 1-Butanol-,3-methyl-acetate H NMR 1- identify the number of the chemical shift of peaks, signals, and integration of the NMR 2-Justify the identity of the compound using the NMR (how does this NMR represent 1-Butanol-,3-methyl-acetate?)
Based on this portion of an IR spectra is there a ketone that is present?
None
2. Which C4H,Br isomers give rise to the two spectra shown below? b. 8 C. 8 7 7 6 5 5 34 4 8 (ppm) 33 32 PPM 4 8 (ppm) 2H t I 2H d 3 ند 3 2H 2H quint sext NH 2 1H m 2 6H d I C 3H 1.0 PPM 0 0
A compound and its ¹H-NMR spectrum are shown below. Assign peak 1 and peak 2 in the spectrum to the hydrogen(s) labeled in red. i. a H3C- 10 C4H8O₂ Compound C b c -O—CH,CH3 2006 Brooks/Cole Thomson Submit Answer C 9 ↑ Which labeled hydrogen(s) correspond to each peak? peak 1: peak 2: 8 7 6 2H 5 Chemical Shift (8) Peak 3 Retry Entire Group 6 more group attempts remaining 3. 3H Peak 2 3H Peak 1 0 ppm
Determine splitting for all CHx groups using n+1 rule then match to appropriate spectrum drawing CH or OH group above each of the peaks.
Which of these choices best describes the interpretation of the following peak that may be recorded in a 'H NMR spectrum? 2.5 8 (2H, d). The underlined hydrogen atom is intended to be the one producing the peak that we are interpreting. O=C-CH,CO2H O=C-CH2CHX2 O=C-CH2CH2X O=C-CH2CH3 None of these interpretations describes this peak.