IHD Nitrogen rules and application of mass

IHD ,NITROGEN RULE AND APPLICATION OF MASS Spectrometry: 

IHD ,NITROGEN RULE AND APPLICATION OF MASS Spectrometry PREPARED BY- ANKIT.J.PARMAR, Pharmaceutics department, Roll no-3 GUIDED BY- Mr. NISHIT.S.PATEL m.Pharm in Quality assurance Dharmaj degree pharmacy college

HYDROGEN DEFICIENCY INDEX: 

HYDROGEN DEFICIENCY INDEX Definition : The index of hydrogen deficiency is the number of pairs of hydrogen atoms that must be removed from the corresponding “saturated” formula to produce the molecular formula of the compound of interest. The degree of unsaturation (also known as the index of hydrogen deficiency (IHD) or rings plus double bonds ) formula is used to help draw chemical structures. The formula lets the user determine how many rings, double bonds, and triple bonds are present in the compound to be drawn.

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It does not give the exact number of rings or double or triple bonds, but rather the sum of the number of rings and double bonds plus twice the number of triple bonds. The final structure is verified with use of NMR, mass spectrometry and IR spectroscopy as well as inspection.

FORMULA FOR IHD: 

FORMULA FOR IHD The formula for calculating the number of unsaturated site is: No. of site=C+1-{[H-N]/2}-halogen/2 For example- C 7 H 8 NO, the IHD is 5 or number of site is 5 and C 6 H 5 CHN 2 would be the possibility structure.

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Note that benzene itself account for 4 site of un saturation: 3 for double bond and one for ring The formula above for the index can be applied to fragments ions as well as the molecular ion. When it is applied to even electron ions, the result is always an odd multiple of 0.5. As an example consider C7H7O + with an index of 5.5. The reasonable structure is :

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In a hydrocarbon where all the C atoms have only single bonds and no rings are involved, the compound would have the maximum number of H atoms. If any of the bonds are replaced with double or triple bonds, or if rings are involved, there would be a “deficiency” of H atoms. By calculating the index of hydrogen deficiency (IHD), we can tell from the molecular formula whether and how many multiple bonds and rings are involved.

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For Hydrocarbons (CxHy): Example 1: IHD for C 2 H 4 is 1 This means it can have either one double bond or one ring. It cannot have a triple bond. Example 2: IHD for C 4 H 6 is 2 This means it can have either one double bond and a ring such as or two double bonds such as CH 2 =CH−CH=CH 2 or CH 2 =C=CH−CH 3 or two rings, or one triple bond, such as CH 3 C≡CCH 3 .

NITROGEN RULE: 

NITROGEN RULE Once the integral molecular weight is known, the number of nitrogen per molecule may be determined from the nitrogen rule. This rule is “ All organic compound having an even integral molecular weight must contain either none or an even number of nitrogen atom and all compound with an odd molecular weight must contain an odd number of nitrogen atom ”

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No attempt will be made here to prove the rule. It suffices to note that it is depend on the mass number and distribution of the isotopes of C, H, O, S, N and covalent bonding requirement of organic molecule. For example, the two compounds, C 3 H 5 NO 2 (M.W=87) and C 2 H 7 NS (M.W=77) which contain one nitrogen atom per molecule, each have odd molecular weight while that of C 6 H 7 BrN 2 (M.W=186) is even.

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The nitrogen rule is only true for neutral structures in which all of the atoms in the molecule have a number of covalent bonds equal to their standard valency Therefore, the rule is typically only applied to the molecular ion signal in the mass spectrum. The nitrogen rule can be applied to all compounds which have covalent bonds and contain any combination of C, H, O, N, S, Si, As, P, halogen and alkaline earth metals.

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Compound Number of nitrogen atoms MW CH4 0 16 EVEN HCN 1 27 ODD H2NNH2 2 32 EVEN C2H5OH 0 46 EVEN C6H6 0 78 EVEN C2H5NH2 1 45 ODD

APPLICATION OF MASS SPECTROMETRY: 

APPLICATION OF MASS SPECTROMETRY Quantitative application of mass spectroscopy (1)Isotopic abundance Most elements appear in nature as isotopic mixture. The natural carbon is a mixture of 98.9% of isotope C 12 and 1.10% of isotope C 13 . The table given below indicate abundance of few elements.

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Isotope Relative abundance% Mass Mean value calculated Mean value measured 1 H 99.985 1.007825 1.007976 1.000794 2 H 0.015 2.0140 12 C 98.90 12.000000 12.011036 12.011 13 C 1.10 13.003355 14 N 99.63 14.003074 14.006762 14.0674 15 N 0.37 15.000108

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What are the relative intensity of the isotopes peaks accompanying the molecular peak of CS 2 . Sulphur shows the mixture of three main isotope with nominal mass 32, 33, 34. There are three isotopes occupy two possible position in the CS 2 molecule. Total number of possible combination is (3) 2 , that is 9.

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32 S 32 S Total mass 64, one combination 32 S 33 S OR 33 S 32 S Total mass 65, two combination 32 S 34 S OR 34 S 32 S Total mass 66, two combination 33 S 33 S Total mass 66, one combination 33 S 34 S OR 34 S 33 S Total mass 67, two combination 34 S 34 S Total mass 68, one combinarion

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Consider the probability to observing each mass. Mass 64 result from single possible combination of two isotope 64. This isotope amounts to 95.02% in nature. So the probability having two such atom simultaneously is (0.9503) 2 that is 90.31%. Mass 65 result from the possible combination of isotopes 32 and 33. Which amount 95.03 and 0.75% respectively. Each combination probably equal to 0.9503*0.0075=0.72%. Which must be multiplied number of combination 0.72*2=1.42%.

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Mass 66 occur in either two way, either two isotope 33 or one isotope 32 and one 34. 33 S has probability (0.0075) 2 =0.00562 in one combination and 32 S 33 S has probability of (0.9503)*0.0422= 4.01% which must be multiplied by two for the number of possible combination is 8.02%. total probability of mass 66 is 8.02+0.00562=8.026%. Mass 67 has probability equal to (2*0.0075)*0.0422=0.063% and mass 68 has (0.0422) 2 =0.1781.

In summury:: 

In summury : Mass % % of predominant peak 64 90.37 100 65 1.42 1.5724 66 8.026 8.889 67 0.063 0.06975 68 0.1781 0.1972

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We must now combine with each of this with the carbon isotope. As there is only one carbon atom in CS 2 . We have 98.90% of isotope 12 and 1.10 % of isotope 13. We must have look for the possible combination of both of following isotopes series with their possibilities while observing the mass is the sum of the masses and the probability is the product of probabilities. Finally, we sum the probability of all the ion with equal mass.

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S 2 mass % 64 90.31 65 1.42 66 8.026 67 0.063 68 0.1781 C mass % 12 98.90 13 1.10

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Then with the mass S 2 +massC = total mass and probability S 2 * probability C is total probability. Mass S 2 Mass C Total mass Probability S 2 Probability C Total probability 64 12 76 90.31 98.90 89.317 64 13 77 90.31 1.10 0.99341 65 12 77 1.42 98.90 1.404 65 13 78 1.42 1.10 0.0156 66 12 78 8.026 98.90 7.938 66 13 79 8.026 1.10 0.088 67 12 79 0.063 98.90 0.062 67 13 80 0.063 1.10 0.00069 68 12 80 0.1781 98.90 0.176 68 13 81 0.1781 1.10 0.002 Total=99.996%

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Gathering the identical mass we obtain the abundance of CS 2 peaks: Mass Total% % of predominant peak 76 89.31 100 77 2.3974 2.684 78 7.9536 8.9056 79 0.152 0.17 80 0.177 0.198 81 0.0019 0.002

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2)Isotopic dilution method: Mass spectrometry is used to determine amount of a component of a complicated mixture from which it cannot be separated quantitatively. This determination is often made possible by the determination of isotopes ratios. For example, determining the amount of glycine in protein hydrolysate. Let the weight of glycine present in the protein hydrolysate is x gm and the natural 14 N to 15 N ratio in it is p. When y gm of a 15N enriched synthetic sample of glycine having 15N to 14N ratio q is added the mixed samples will posses a new isotopic ratio r which can be determine by mass spectrometer.

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We know that he two separate weights of 15N containing glycine combined together are equal to the weight in final mixture. X+b/r+1=(x/p+1)+(b/q+1) Where the value of p and r can be calculated by the mass spectrometer from the ratio of peak heights of 14N and 15N in mass spectra. As b, p, q, r are known the value of x can be calculated from the above equation

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Isotopic dilution method is also useful in determining the amount of caffeine in coffee. In this method, determination of isotopes ratio in caffeine present in the coffee by mass spectrometry require the sample to be converted to a suitable form, usually nitrogen or carbon dioxide.

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3) Quantitative analysis of mixture If the identities of the compound in the sample mixture are known, the quantitative analysis is feasible. The principle of this method is based upon the fact that contribution for each peak are additive for each compound present in the mixture. It means that mass spectrum of a mixture record the sum of spectra, i.e., the total current at each peak will be represented by sum of the currents resulting in form each compound present in mixture.

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If I 1 , I 2 , I 3 …….etc , are the total current at various peak present in the mass spectrum, one can write I 1 =(i 1 ) a p a +(i 1 ) b p b +(i 1 ) c p c +…. (1) I 2 =(i 2 ) a p a +(i 1 ) b p b +(i 1 ) c p c +..... (2) I 3 =(i 3 ) a p a +(i 1 ) b p b +(i 1 ) c p c +…. (3) Where the numerical (1, 2, 3…) represent the various mass number, letter(a, b, c..) represent the various and p a , p b , p c …represent the partial pressure of compound, a, b, c…

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The values of (i 1 ) a , (i 2 ) a , (i 3 ) a , (i 1 ) b , (i 2 ) b , (i 3 ) b ,(i 1 ) c , (i 2 ) c , (i 3 ) c ,….. can be obtained from spectra of pure compound at known pressure which is equal to the sum of partial pressure, i.e., p=p a +p b +p c + (4) On solving equation (1), (2), (3), (4), one can obtain the value of p a , p b , p c ... These partial pressure are proportion to the concentration of the respective compound in mixture. Thus one can know the concentration of various components present in mixture.

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Let us illustrate the above method by analysing a mixture of methane, ethane, propane, n-butane, and isobutene. The mass spectrum of mixture is recorded. Both n-butane and isobutane shows peak due to (C 4 H 8 ) + (m/e=58) and (C 4 H 9 ) + (m/e=57), resulting in two equation which on solving give the amount of this two components. If an allowance is made for the contribution of butane to propane(m/e=44), propane content can be found. The peak at m/e=44 does not receive any contribution from methane and ethane. Similarly, ethane and methane can be found out by peaks at m/e=30 and m/e=11.

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4)Studied involved in reaction mechanism: Mass spectrometry is one of the best method to study the mechanism of certain reaction by knowing about precise steps of reaction and as well as the reaction intermediates. An interesting example is the reaction of carboxylic acid with an alcohol which form ester by mechanism in which the acyl-oxygen bond of acid is broken or by a path in which O-H bond of acid is broken. When the product is analyzed with mass spectrometry it is found that the ester contain 18 O atom.

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5)Trace gas analysis Several techniques use ions created in a dedicated ion source injected into a flow tube or a drift tube: selected ion flow tube(SIFT-MS), and proton transfer reaction (PTR-MS), are variants of chemical ionization dedicated for trace gas analysis of air, breath or liquid headspace using well defined reaction time allowing calculations of analyte concentrations from the known reaction kinetics without the need for internal standard or calibration.

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6) Atom probe An atom probe is an instrument that combines time-of-flight mass spectrometry and field ion microscopy (FIM) to map the location of individual atoms.

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7) Pharmacokinetics Pharmacokinetics is often studied using mass spectrometry because of the complex nature of the matrix (often blood or urine) and the need for high sensitivity to observe low dose and long time point data. The most common instrumentation used in this application is LC-MS with a triple quadrupole mass spectrometer. Standard curves and internal standards are used for quantization of usually a single pharmaceutical in the samples. The samples represent different time points as a pharmaceutical is administered and then metabolized or cleared from the body. Blank or t=0 samples taken before administration are important in determining background and insuring data integrity with such complex sample matrices.

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8) Protein characterization Mass spectrometry is an important emerging method for the characterization of proteins. The two primary methods for ionization of whole proteins are electrospray ionization (ESI) and matrix-assisted laser desorption/ionization (MALDI). In keeping with the performance and mass range of available mass spectrometers, two approaches are used for characterizing proteins. In the first, intact proteins are ionized by either of the two techniques described above, and then introduced to a mass analyzer. This approach is referred to as "top-down" strategy of protein analysis.

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In the second, proteins are enzymatically digested into smaller peptides using proteases such as trypsin or pepsin, either in solution or in gel after electrophoretic separation. The collection of peptide products are then introduced to the mass analyzer. When the characteristic pattern of peptides is used for the identification of the protein the method is called peptide mass fingerprinting (PMF), if the identification is performed using the sequence data determined in tandem MS analysis it is called de novo sequencing. These procedures of protein analysis are also referred to as the "bottom-up" approach.

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9)Space exploration As a standard method for analysis, mass spectrometers have reached other planets and moons. Two were taken to Mars by the Viking program. In early 2005 the Cassini-Huygens mission delivered a specialized GC-MS instrument aboard the Huygens probe through the atmosphere of Titan, the largest moon of the planet Saturn.

QUALITATIVE APPLICATION OF MASS SPECTROMETRY : 

QUALITATIVE APPLICATION OF MASS SPECTROMETRY 1) Molecular mass determination: - Mass spectrometry is one of the best method to determine the molecular mass accurately. When the substance is bombarded with moving electron, and the mass spectrum is recorded, the mass of the peak at the highest m/e reveals the molecular mass accurately.

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2)Distinction between Cis and Trance isomer: -Both cis and trance isomer yield similar mass spectra but the molecular ion peak for the trance isomer is more intense than the cis isomer. -Another distinction is that the dehydration fragment for the cis isomer is much stronger than the trance isomer. -For example, cis and trance isomer of hex-2-ene-1-ol have been characterized on the basis of mass spectrometry. CH 3 -CH 2 -CH 2 -CH=CH-CH 2 OH<--- >CH 3 -CH 2 -CH=CH-CH=CH 2

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3)Evaluation of heat of sublimation: -If a subliming solid in a thermostat sample reservoir at the given temperature, the solid phase and vapour phase in the reservoir are in the equilibrium -Suppose a small quantity of the vapour is diffused from the reservoir into the ionization chamber. -If mass spectrum is run, there occurs a change in a peak intensity which is directly proportional to the vapour pressure of the sample in the ion source. -Similarly, the vapour pressures of the subliming substance are calculated from the peak heights in the mass spectrum at different temperature. -Then the heat of sublimation can be obtained from the slope of the plot of log p against I/T

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4)Determination of bond dissociation energy: Mass spectrometer can be used to determine the bond dissociation energies of molecule by employing the concept of appearance potential of fragmentation ion. The appearance potential of fragment ion may be defined as energy required to produce that ion in the mass spectrometer. As the production of any ion other than the molecular ion involves cleavage of bonds, the appearance potential of a fragment ion will furnish information on bond dissociation energy. The value of ionization potential can be determined by ionization efficiency curve as it related to energies of electron beam .

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When a molecule AB is bombarded with electron beam a positive molecular ion is produced. i)AB + +e -  AB + +2e - ii)AB +  A + +B . iii)AB +  A . +B + Appearance potential of A + for the step(ii) can be measured from the ionization efficiency curve of the ion A + .

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Appearance potential of A + for the step(ii) can be measured from the ionization efficiency curve of the ion A + . Then can write following equation. A(A + )=I(A . )+D(A-B) Where, A(A + )= appearance potential A + I(A . )= ionization potential of A D(A-B)=bond dissociation energy The value of A + (A . ) can be measured from the ionization efficiency curve of A + using the mass spectra. Ionization potential of A . can be obtained from photo ionization method. If one know the experimental value of A + , I(A . ) then one can calculate bond dissociation energy.

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5)Impurity detection: Mass spectrometry is one of the best method to detect the impurities. The detection of impurities in parts per million is only possible if their structures differ from those of major components. If the molecular weights of the impurities are much larger than major components their detection is more easy because their higher mass peaks are free from contribution by those of major components. On other hand if the molecular weight of the impurities are much lower than the major components, the detection is not easy because of formation of common fragments ion.

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6)Characterization of polymer : Mass spectrometry is used for the characterization of polymer. First of all, the polymer is pyrolized i.e. decomposed by heating and pyrolized product are fed into the inlet of mass spectrometer and identified. From the results, one may obtained much information concerning the structure of polymer. For example, the mass spectrometry will be able to distinguish chlorinated polymer to know whether the chlorine occurs randomly or occur in the block in the structure of polymer.

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7)Latent heat of vaporization liquids: Mass spectrometry can be used to determination of latent heat of vaporization of liquid by using clausius-clapeyron equation which express the saturated vapour pressure of a liquid as a function of absolute temperature. log p= -delta H/RT +constant From the above equation one can calculate delta H if the value of p at given temperature T is known. Similarly, the value of p are determined from the height of peaks in mass spectrum at different temperature. The value is obtained from the slop of the plot of log p against 1/T.

REFERENCE: 

REFERENCE Instrumental method of chemical analysis by G.R.Chatwal, page no.-2.296 Spectrometric identification of organic compound by Rbert.M.Silverstain and Webster, sixth edition, page no- 2 Principle and application of Mass spectrometry by Edmond de Hoffman and Vincent stroobent, third edition, page no-251 http://en.wikipedia.org/wiki/Mass_spectrometry

Thank you………: 

Thank you………