METASTABLE IONS, RING RULE, MC LAFFERY REARRANGMENT AND INTERPRETATION OF ORGANIC CHEMISTRY

THE TOPIC DESCRIBE ABOUT METASTABLE IONS, RING RULE, MC LAFFERY REARRANGMENT AND INTERPRETATION OF ORGANIC CHEMISTRY. AND THE TOPIC DESCRIBES IN PHARMACEUTICAL ORAGANIC CHEMISTRY.

1. CL BAID METHA COLLEGE OF PHARMACY, CHENNAI-97 DEPARTMENT OF PHARMACEUTICAL CHEMISTRY ADVANCED SPECTRAL ANALYSIS TO DR. N.RAMALAKSHMI MPHARM, Ph.D. BY INDRAKUMAR S 2 ND SEMESTER M PHARM PHARMACEUTICAL CHEMISTRY   In The Topic Of Metastable ion, Mc Lafferty Rearrangement, Ring Rule

2. THE RING RULE Used to calculate the Degree of Unsaturation. This rule is particularly analyzing the molecular formulas to understand the structure of the compound. Degree of Unsaturation Degree of Unsaturation= 2C+2−H+N−X C is no of carbon H is no of Hydrogen N is no of nitrogen X is no of halogen 2 Double bonds count as one unit of unsaturation. .

3. Interpretation of Degree of Unsaturation 0 DOU : No rings or double bonds (e.g., alkanes like propane, C3H8​). 1 DOU : One ring or one double bond (e.g., cyclohexane, C6H12​, or ethene, C2H4​) 2 DOU : Two double bonds, a triple bond, or a combination of a double bond and a ring (e.g., benzene, C6H6C_6H_6C6​H6​) 4 DOU : Indicates an aromatic ring system, such as benzene EXAMPLE Butene (C4H8​ ) : C=4C , H=8H DOU=2(4)+2−8/2 =8+2−8/2 =1 Interpretation: This could mean one double bond or one ring.

4. Benzene (C6H6​) : C=6C , H=6H DOU=2(6)+2−6/2 =12+2−6/2 =4 Interpretation: A degree of unsaturation of 4 indicates an aromatic ring. Caffeine (C8H10N4O2​) : C=8C, H=10, N=4, O=2 DOU=2(8)+2−10+4/2 =16+2−10+4/2 =6 Interpretation: This degree of unsaturation indicates a highly unsaturated structure with multiple rings and/or double bonds. APPLICATION The Ring Rule and degree of unsaturation calculations are useful tools for organic chemists to predict and verify structural information about a compound based on its molecular formula . They give valuable insights, particularly during the initial stages of structural elucidation and analysis.

5. METASTABLE IONS OR PEAKS

6. These ions are excited, unstable state, and typically break apart. They fragment in the field-free region in between the ion source and the mass analyzer. i.e. Second field-free region. These metastable ions are spontaneous decomposition of parent ions that are energetically unstable They also occur at the non-integral m/z value. These are much broader than the normal peaks. These are of relatively low abundance (o f ~ 10–2 or less than the parent or the daughter ions in the 70 eV spectrum). The molecular ions formed in the ionization chamber behave in one of the following ways. They decompose rapidly and completely in the ion source and never reach the collector (lifetime less than 10–5 sec)

7. REGIONS WHERE METASTABLE IONS ARE PRODUCED Interpretation : Analyzing the mass spectrum allows researchers to gain insights into the fragmentation pathways and structural characteristics of the original molecule. Detection of Product Ions : The resulting product ions are detected, often appearing as broad peaks at non-integer m/z values in the mass spectrum, indicating their metastable nature. Fragmentation Occurs : While traveling in the field-free region, metastable ions can spontaneously fragment into smaller product ions and neutral fragments due to their internal energy. Travel to Analyzer : Ions travel through the mass spectrometer towards the analyzer . During this transit, they may enter field-free regions where no external forces are acting on them. Excess Internal Energy : Some generated ions possess excess internal energy , making them unstable and prone to fragmentation . Ion Generation : Molecules are ionized in the ion source of the mass spectrometer, forming ions (e.g., molecular ions or fragment ions).

8. CALCULATION OF m/ z The apparent mass of a metatable ion A+ (m*) can be calculated fairly accurately from the masses of the parent ion (m1 ) and the normal daughter ion A+ (m2 ) using the formula For Example : Toluene has a strong peak at m/z 91 and m/z 65 and the broad metastable peak at 46.4. So, we interpret m/z 91 as decomposing by loss of 26 mass units to the daughter ion m/z 65, and that some of this fragmentation takes place in the second field-free region ( i.e between the two analyzers ) leading to metastable ion peak of m/z 46.4.

9. Mc LAFFERTY REARRANGEMENT STEP 1: Cleavage of α and β bond STEP 2: Forms an abstraction of γ molecule to the radical cation. The rearrangement eliminates neutral molecules from amines, aldehydes, ketones, unsaturated compounds, and substituent aromatics. Generally, a molecular ion with an even mass value cleaves to give fragment ions with odd mass values and vice-versa. But if the even mass molecular ion yields fragments of even mass and odd mass molecular ions yield fragments of odd mass, it indicates that a rearrangement has taken place

10. A rearrangement known as McLafferty rearrangement is commonly observed in compounds having the following structure where x and y can be C, O, S, N and P. Here X = C and Y = O Let us consider an aldehyde or a ketone with γ hydrogen. It can undergo the McLafferty rearrangement as shown below

11. where R = H, aldehyde R = alkyl, ketone The mass spectrum of butanal (CH3CH2CH2CHO) is shown as

12. Prominent peaks at m/z 72,57, 44 and 29. these peaks can be explained as follow The peak at m/z 44 (M-28) is an even mass peak from the even number z molecular ion and hence indicates a rearrangement. The McLafferty rearrangement explain this peak as follows

13. INTERPRETATION OF ORGANIC COMPOUNDS 1.Quercertin

14. Fragmentation pattern

15. 2. Isorhamnetin

16. Fragmentation pattern

17. 3.Kaempferol

18. Fragmentation process