is or is not favored at each of the three potential posilibns (oro, mc, 4. [12 p
ID: 876437 • Letter: I
Question
is or is not favored at each of the three potential posilibns (oro, mc, 4. [12 points] Consider the mass spectra of ketone A and ketone B shown to the right. One of these ketones gives a distinctive peak in its mass spectrum at me = 91; the other ketone gives a distinctive peak in its mass spectrum at me = 105. (a) Identify which one of these ketones gives the peak at m/e 91 and provide a CH-CHy CH CCHy ketone A ketone B Lewis structure that corresponds to the peak at m/e = 91. (b) Identify which one of these ketones gives the peak at m/e 105 and provide a Lewis structure that corresponds to the peak at mle = 105 Screen Shot 2015-06-23 at 9.45.09 PM each of the following reactions. ebeB kebme B 4b eine gf 8 f9Explanation / Answer
m/e is ratio of mass of ion to charge on it. Here m is relative formula mass of + ve ion & e is charge on it ((virtually all ions have charges of z = +1, so sorting by the mass to charge ratio is the same thing as sorting by mass).Both these ketone ionised as
C6H5COC2H5----->C6H5COC2H5.-----> C6H5CO+ + +C2H5
molecule (ionisation) moleculer ion fragmentation
C6H5CH2COCH3----->C6H5CH2COCH3.-----> C6H5CH2+ + +COCH3
molecule (ionisation) moleculerion--fragmentation
In mass spectroscopy first molecule is ionised and then moleculer ion and their fragment is subjected to mass analyser where they are separated according to their mass.
For ketone A
C6H5COC2H5.(m/z=134)------>C6H5CO+(m/z=105) + C2H5+(m/z= 29)
for ketone B
C6H5 CH2COCH3.(m/z= 134)-----> C6H5CH2+( m/z= 91) + COCH3(m/z=43)
m/z ratio is used to know the formula mass of given moleculerion and so the molecule.
these fragment can further breakes into smaller ions.this is apply for every mlecule.