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Compound Z, C5H160. Deduce its structure. IR: significant peaks at 2962 and 1718 (intense) cm! 13 Mass spectrum: a few of the
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Answer #1

Firstly we can calculate the double bond equivalence, for a compound CaHbNcOd type of compound it is

ь — с b DBE - а + 1— 2 (if halogen is there count them with number of H)

So, here it is = 1.

  • At 1718 cm-1 IR peak refers to C=O stretching frequency, so, the one double bond is detected. The 2962 cm-1 is for sp3 C-H stretching, so there are only sp3 carbons in the compound and the carbonyl is ketone.
  • The overall molecular mass is 128 and the strong peak appers due to carbonyl. Now , in case of carbonyl both alpha and beta cleavage occurs. So, in the fragmentation peaks we can search for them. This will be a test for whether our predicted structure is correct or not.
  • 9 H singlet is a common phenomenon for tertiary butyl system. So, we can think of a t-butyl system there. The chemical shift value suggests it is not in the vicinity of carbonyl.
  • 3 H singlet is a common phenomenon for methyl having no neighbouring proton. Also its chemical shift value is quite high. So, we may think there is a CH3 group adjacent to carbonyl.

Up to this we got fragments,

CHз Нас. Нас- Нас

Now, we have only 2 C and 4 H left. Also, there are two 2H triplets in the 1H NMR spectrum. So, there must be a -CH2-CH2- linkage with no neighbouring H in the left or in the right. So, we can join these fragments with the linker. So, proposed molecule:

о Н2 С. Нас. CHз На Hас Hас

Now, compare mass spectrometry data and 13C NMR data:

  • Due to alpha cleavage on the right side, -CH3 group gets lost, having mass = 15. So, we'll have a peak at 128-15=113. Due to beta cleavage the left most neopentyl group gets lost, so we'll have a peak ( for -CH2-CO-CH3) at 57.
  • For 13C, the peak above 200 is due to carbonyl carbon. All other peaks are in the range of sp3 C. The alpha carbons with respect to carbonyl have a higher value near 40. Rest of the carbons come nearly at 30.
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