11. A split peak for IR absorption due to bond stretching is observed for the ca

Question

11. A split peak for IR absorption due to bond stretching is observed for the carbonyl group in which of these compounds? 12. What can be determined from the relative abundance of the M^+. +2 peak? a) presence of a number of different elements, including S, Br, and Cl. b) presence of a number of different elements, including P, Fe, Mn. c) presence of noble gases d) none of the above 13. Predict the splitting pattern you would observe for the proton at CI of 2,3-dimethyl-2- phenylbutane. Doublet singlet quartet septet octet

Explanation / Answer

1.b

the C=O double bond absorbs infrared light at wavenumbers between approximately 16001900 cm?1. The exact location of the absorption is well understood with respect to the geometry of the molecule. This absorption is known as the "carbonyl stretch" when displayed on an infrared absorption spectrum

2. a

Important Things To Note: - The relative abundance of M+2 will tell you if a Sulfur, Chlorine, or Bromine is present. - It will only tell you if one of these is present. - Again you must make sure that your M abundance is 100% or this will not work. How Can You Use This Information: - You can use this information to determine whether there is a Sulfur, Chlorine, or Bromine. - If the M+2 relative abundance is: ~4 % then a Sulfur is present ~33% then a Chlorine is present ~100% then a Bromine is present - If the M+2 relative abundance is not near any of these numbers then there is none present. Try It On Your Own: -If you were given m/z= 102 (M, 100%), m/z=103 (6.9%), and m/z=104 (.38%) do you have a Sulfur, Chlorine, or Bromine present? (Example from pg. 108 in Lecture Supplement) Answer: There is no Sulfur, Chlorine, or Bromine present.

3.

b. singlet

Related Questions

Navigate

• Browse (All)
• Browse 1
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.