Chemical Forums
Chemistry Forums for Students => Organic Chemistry Forum => Organic Spectroscopy => Topic started by: s30.93a on January 26, 2017, 12:53:47 PM
-
Mc Lafferty can occur for C=O or C=C, thus yielding 105 and 121 as two peaks, but which one will be base peak?
-
The base peak is that with the greatest relative intensity, so what determines rel int. Once an ion has been formed?
-
Obviously, the most stable cation will resemble base peak, if C=O will break its α-β bond -C-O+H will form, in other case -C-C+H will be formed. Please correct me if I'm wrong and thus the latter is preferred
-
Ionization of R1-CO-R2 will give R1-CO+ and R2-CO+
Which in your case will be more stable ?
-
Ionization of R1-CO-R2 will give R1-CO+ and R2-CO+
Which in your case will be more stable ?
I think it depends on the stability of the radical R2° or R1° as R-CO+ is stabilised by back bonding. please correct if i'm worng
I made these structures depending on Mc Lafferty, please tell which one will be base peak?
-
Ionization of R1-CO-R2 will give R1-CO+ and R2-CO+
Which in your case will be more stable ?
-
I don't believe that your drawn structure is 1-phenyl-1-butanone.
Please check !
-
Search NIST for MS of butyrophenon
-
Search NIST for MS of butyrophenon
AWK,
Why provide the answer on a platter, when our job is to help them think it through for themselves?
-
s30.93a discusses other compound (2-propyl-3,4-dihydro-2H-naphthalen-1-one). Two main peaks of butyrophenon are 105 and 77. Title of thread is incorrect.
-
No, i want to know base peak for both the compounds.
The latter which I posted has more possibilities for Mc Lafferty.
I want to know which of the fragmentation is preferred, thus I'll subsequently apply that fact to 1-phenyl-1-butanone
-
Search NIST for MS of butyrophenon
AWK,
Why provide the answer on a platter, when our job is to help them think it through for themselves?
I already knew such databases, but looking for reasons.
-
Then start from easier compound - butyrophenone. Show us McLafferty rearrangement
and two α-cleavages for this compound.
-
Then start from easier compound - butyrophenone. Show us McLafferty rearrangement
and two α-cleavages for this compound.
These are the two possible rearrangements.
-
These not give ions 105 and 77 which are most abundant in the spectrum of butyrophenone. Check α-cleavages.
-
The second rearrangement (if any) gives the same ion mass as one of α-cleavage (the more probable one)
Thank you, but I want to know why the second rearrangement would give base peak ion and not the first one?
-
Base peak shows odd number and cannot come from rearrangement.
-
I'm sorry but I'm not able to understand you.
There are two possibilities or two fragment according to Mc Lafferty then why one one is base peak ?
-
These not give ions 105 and 77 which are most abundant in the spectrum of butyrophenone. Check α-cleavages.
α cleavages ? what should I look for when I check it, I don't know how to look for it, sorry
-
Just broke bond left or right to CO and leave charge on oxygen (in the case of bond to phenyl group charge may be also left at phenyl group.
Elimination of molecule (ethane or propane) from butyrophenone should give even number for ion. MS spectrum of this compound show only abundant odd mass numbers. It means that McLafferty rearrangement is unimpotrant for this compound.
-
Simply Put:
If I ask you to tell me Base peak for 1-phenyl-1-butanone what would you say ? and why ? please help it's a question asked in my exam
-
The most intensive one - 105
Peaks 106 (partially) and 120 may come from McLafferty rearangement.
-
and why it's 105 ? not any other cation ?
-
This ion is the most stable C6H5CO+. this does not surprise me.
-
Please correct me if I'm wrong:
It's not imperative that ion formed after Mc Lafferty rearrangement will always show base peak.
Hence stability is preferred over Mc Lafferty.
-
Search google for mass spectra of some ketones (aldehydes) starting from 5 carbon atoms (2-pentanone, 2- and 3-hexanone ...). Compare base peaks (the strongest ones) to that of M-28, M-42... and so on (all with even m/e)... and find the answer on your own.