Question

Cardi B prepares compound D according the scheme below. Cardi B enjoys the odor of compound D, which recalls the refreshing sTaking into account all data you have at your disposal, what statement best explains why the series of peaks around 3000 cm i

Choices: degrees of saturation: 0,1, 2, or, 3

How many degrees of unsaturation does Compound A possess? [Select ] Taking the IR data into account, can Compound A possess a

Cardi B prepares compound D according the scheme below. Cardi B enjoys the odor of compound D, which recalls the refreshing sting of ammonia or other disinfectant. SOCI 2 equiv NH LIAIH Compound A CeH1202 Compound C CgH13NO Compound D Compound B H2O Though Cardi is unwilling to provide you with the identity of D, she has provided the molecular formula and IR data of compound A and the molecula formula and H NMR spectrum of compound C IR of Compound A: L00 D- 2008 aVEMUrBER H NMR of Compound C: 1H 1
Taking into account all data you have at your disposal, what statement best explains why the series of peaks around 3000 cm in Compound A's IR spectrum 1 appear broad? C-H stretches are typically broad The molecule possesses many different C-H bonds, so together they will appear broad This series of peaks is comprised of narrow C-H stretches and a broad C O stretch This series of peaks is comprised of narrow C-H stretches and a broad O-H stretch Question 3 1 pts How many degrees of unsaturation does Compound A possess? [Select ] Taking the IR data into account, can Compound A possess a ring? [Select]
How many degrees of unsaturation does Compound A possess? [Select ] Taking the IR data into account, can Compound A possess a ring? VSelect ] Yes No, it has 0 degrees of unsaturation No, all degrees of unsaturation are accounted for by one or more C=X double bonds
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Answer #1

Question 2

This series of peaks is comprised of narrow C-H stretches and a broad O-H strech

the O-H band is broad due to hydrogen bonding interaction

Question 3

formula for degree of unsaturation is

C-H/2 +1

where C = no. of carbon

H = no. of hydrogens

6-12/2+1 = 1

degree of unsaturation = 1

therefore compound will have one double bond or ring, but according to IR, the strong band at 1700 cm-1 denote it -C=O group, as a result ring system is not available in the structure

therefore answer for final question is

No, all degree of unsaturation are accounted for by one or more C=X double bonds

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