1. The ion types formed in SID are similar with those fragmented in CID.Immonion ions,a,a*,a0,a++,b,b*,b0,b++, c,x,y,y*,y0,y++,z and some side-chain cleavage ions,like v and w ions.However, SID mainly yields y, b and a type backbone cleavage ions. It depends on the collision energy. Sometimes the ionization methods also play a role in it. For example, the authors said there were not significant abundance in the side-chain cleavage ions using their ESI-QTOF instrument while some people found many side-chain cleavage ions using the LSIMS ion source(Analytical Chemistry,1993,65,2859-2872).
Immonium ions are H2N=CHR.
2. The mass of the target surface is higher. Then the internal energy in the center of mass reference frame is higher too. The fragment efficiency depends on the relative masses of the target gas and the colliding ion. So the efficiency of SID is higher. The internal energy deposition is higher too. These leads to more conversion from translational energy to internal energy for SID.
3.Charge-directed cleavage is initiated through intramolecular proton transfers. One possibility is multiple proton dimers cause charge reduction. The target surface in SID can receive the proton from the projectile ion while the noble gas in CID does not allow the reaction happen. So single charged ions are more abundant in SID than in CID.
In regards to Chad's question on number 2, I believe the increase in fragmentation energy is a result of kinetic energy and momentum increases. The kinetic energy is 1/2*m*v^2 and the momentum is m*v. In both of these cases, the larger the mass of the colliding ions results in a larger momentum and larger kinetic energy being transferred. This results in more complete fragmentation.
I don't really understand the deposition process of the ions onto the surface in the first place. Could you explain? And how much time does a run with SID fragmentation take compared to CID?
The authors didn't compared the time of SID with the time of CID. I guess it's similar.Maybe the time of SID is less than that of CID, since SID is approximately a single collision method.
The ion collided with the surface. It didn't deposite on the surface. For example, AB changed to A and B after it colliding with the surface. Neither of them deposited on the surface.
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Comments
I have some questions for Yixiao and the rest of the class.
1. What ion types are formed in SID? What are immonium ions?
2. In SID, why more translational energy is converted into internal energy?
3. In SID, why singly charged ions are more abundant than in CID?
Posted by: Edgar Arriaga | March 27, 2008 04:29 PM
1. The ion types formed in SID are similar with those fragmented in CID.Immonion ions,a,a*,a0,a++,b,b*,b0,b++, c,x,y,y*,y0,y++,z and some side-chain cleavage ions,like v and w ions.However, SID mainly yields y, b and a type backbone cleavage ions. It depends on the collision energy. Sometimes the ionization methods also play a role in it. For example, the authors said there were not significant abundance in the side-chain cleavage ions using their ESI-QTOF instrument while some people found many side-chain cleavage ions using the LSIMS ion source(Analytical Chemistry,1993,65,2859-2872).
Immonium ions are H2N=CHR.
2. The mass of the target surface is higher. Then the internal energy in the center of mass reference frame is higher too. The fragment efficiency depends on the relative masses of the target gas and the colliding ion. So the efficiency of SID is higher. The internal energy deposition is higher too. These leads to more conversion from translational energy to internal energy for SID.
3.Charge-directed cleavage is initiated through intramolecular proton transfers. One possibility is multiple proton dimers cause charge reduction. The target surface in SID can receive the proton from the projectile ion while the noble gas in CID does not allow the reaction happen. So single charged ions are more abundant in SID than in CID.
Posted by: Yixiao Sheng | March 29, 2008 11:03 PM
I understand your statement about 2 but what is the science behind this? Why does the mass make a difference here?
Posted by: Chad Satori | March 30, 2008 10:28 AM
It's about momentum and kinetic energy. I don't know how to explain it in several sentences. There are some detailed equations in this paper:
http://www.sciencedirect.com/science?_ob=ArticleURL&_udi=B6TH2-44WK94T-67&_user=616288&_coverDate=09%2F30%2F1992&_rdoc=1&_fmt=&_orig=search&_sort=d&view=c&_acct=C000032378&_version=1&_urlVersion=0&_userid=616288&md5=3fe0afb46038dab61bf1f8a5526bb90e
I think the chapter "Frames of Reference" helps a lot. Thanks!
Posted by: Yixiao Sheng | March 30, 2008 01:30 PM
In regards to Chad's question on number 2, I believe the increase in fragmentation energy is a result of kinetic energy and momentum increases. The kinetic energy is 1/2*m*v^2 and the momentum is m*v. In both of these cases, the larger the mass of the colliding ions results in a larger momentum and larger kinetic energy being transferred. This results in more complete fragmentation.
Posted by: Josh Ochocki | March 30, 2008 06:37 PM
I don't really understand the deposition process of the ions onto the surface in the first place. Could you explain? And how much time does a run with SID fragmentation take compared to CID?
Posted by: Melissa Maurer-Jones | March 30, 2008 10:07 PM
The authors didn't compared the time of SID with the time of CID. I guess it's similar.Maybe the time of SID is less than that of CID, since SID is approximately a single collision method.
The ion collided with the surface. It didn't deposite on the surface. For example, AB changed to A and B after it colliding with the surface. Neither of them deposited on the surface.
Posted by: Yixiao Sheng | March 30, 2008 11:08 PM