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please find the integration and tell what each peak represent

H3C CH3 CH3 camphor

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Answer #1

In camphor there are total 16 proton. At 0 ppm ( that is for reference TMS) integration is 1.0 TMS molecule contain 12 proton , so we have to multiply 12 with each integration value given . Around 1 ppm integration is .67, that signal is for 0.67*12= 8 hydrogen , between 2.4-1.6 ppm integration is 0.21, means there are 0.21*12= around 3 hydrogen and greater than 7 ppm integration is .04 means there are .04*12=0.48 hydrogen . Proton which are shieled gives lower ppm value and deshieled proton gives higher ppm value. Carbonyl oxygen has much electron density around it ,so protons far mostfrom the oxygen are most deshieled giving higher ppm value. And proton at the nearby distance from oxygen are shielded gives the lower ppm value .

Below the protons ate marked acording to their ppm region.

(3 H) A(w) 2.0

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please find the integration and tell what each peak represent H3C CH3 CH3 camphor
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