174k views
4 votes
Provide a structure for the given compound. C6H12O₂; IR: 1743 cm¯¹; ¹H NMR spectrum 2400 8 2100 17 7 Draw the compound. 1800 6 1500 J = 6.7 Hz 5 chemical shift, Hz 1200 900 2H J = 7.6 Hz 4 chemical shift, 3 ppm (8) 2H 600 2 2H 3H 300 3H J= 7.6 Hz J = 7.4 Hz 0

User Canavanin
by
7.6k points

1 Answer

4 votes

Final answer:

Exercise 4.7 involves using an online database to look up IR spectra for certain organic compounds, identifying specific absorption bands, and explaining the trend in wavelengths for C-C bonds based on vibrational frequency.

Step-by-step explanation:

Exercise 4.7 involves identifying specific IR absorption bands for various organic compounds and rationalizing the trend in wavelengths of carbon-carbon single, double, and triple bonds within the context of these compounds as 'springs'. The comparison uses IR spectra to illustrate how bond strength and mass affect the vibrational frequency of bonds.

For example, stronger bonds such as carbon-carbon triple bonds (C≡C) in 1-hexyne would have a higher frequency of vibration compared to double bonds (C=C) as seen in 4-methylcyclohexene, and even more so compared to single bonds (C-C) due to the increased bond strength. Heavier atoms bonded together would also decrease the frequency of vibration due to a higher mass.

User Sackling
by
6.8k points