foreign [Music] in this video we'll be looking at the M plus 2 Peak and learning about what the significance of recognizing the M plus 2 Peak is so the M plus two peak is a peak in the Mass Spectrum which is two Master charge ratios higher than the parent Island Peak it's important to note that it is higher because of the plus two the reason why we observe the M plus 2 Peak is because of the presence of two relatively abundant isotopes of a particular element let's quickly review what Isotopes are Isotopes are atoms of the same element but with different Mass numbers they must have the same number of protons because they have the same atomic identity while they have different numbers of neutrons hence they're different Mass numbers the molar mass of an isotope is going to be equal to the relative isotopic mass which is based on abundance so the significance of the N plus two peak is that it is a Telltale sign of the presence of chlorine or bromine and the reason why that is is because both atoms contain two common isotopes 75 of chlorine exists as the chlorine 35 isotope whereas 25 of chlorine exists as the chlorine 37 isotope the molar mass of Cl which we find in the NASA data sheet is equivalent to this relative abundance so 75 times 35 plus 25 times 37 which equals to 35.5 grams per mole if we look at bromine we can see that both bromine 1719 and bromine 81 exist in a relatively comparable abundance Romans 79 makes up 51 of the abundance of bromine while bromine 81 makes up 49 of the abundance of bromine to calculate the molar mass of bromine we take the abundances and multiply them by the mass numbers again to work out that the molar mass of BR according to the abundance of its Isotopes is equal to 79.98 grams per mole here we have an example of the Mass Spectrum for the molecule chloroethane the Mass Spectrum shows relative intensity versus The Master charge ratio here we can see the molecular ion Peak at this value of 64. and we can see that there's an M plus two peak and the value of 66. knowing that this is chloroethane what the M plus 2p therefore is going to correspond to is the molecule of chloroethane which contains the chlorine 37 isotope this means that the M plus P coincides with the chloroethane which contains the chlorine 35 isotope looking at the Mass Spectrum we can also notice that the relative intensity is equal to the relative abundance of the isotope if we remember from our previous slide around 25 exists as chlorine 37 while 75 exists as chlorine 35 we can see that the M plus 2p is approximately one-third of the size of the molecular ion Peak looking at primary ethane we see the same circumstances the existence of two isotopes with comparable abundances leads to two peaks which reflect their relative abundances bromine 81 if we remember makes up approximately 49 of the bromine isotope while bromine 79 makes up 51 percent we can see that the relative intensity of these Peaks reflects the same isotopic relative abundance so this means that this second Peak at this about 110 range is going to indicate the M plus 2 Peak or the bromoethane which is going to contain the isotope bromine 81. let's have a look at this example the question reads a series of preliminary tests on a small water-soluble organic molecule were conducted the infrared and Mass Spectrum of the unknown molecule are shown note two parent ion Peaks are shown draw the structural formula for this compound so it's important to note that the question hints to you that there are two parent ion Peaks and these are labeled 152 and 154 in the mass to charge ratios this means that the compound is suspected to be halogenated with either chlorine or bromine and as usual we'll Begin by analyzing the infrared spectrum and what we should immediately recognize is that there is a very defining functional group in this 1680 to 2000 Range which indicates a carbonyl carbon furthermore we also see that there is a broad signal in the study 200 kind of range which indicates to us that there is a carboxylic acid the fact that there's a carboxylic acid indicated by this peak is supported by the fact that we have a peak in the 1680 range as we mentioned previously indicating that there is a carbonyl carbon this is the peak that helps us distinguish whether or not this compound is an alcohol or carboxylic acid there's a Peak at 45 on the Mass Spectrum which indicates the presence of a carboxylic acid this is consistent without suspicion after analyzing the infrared Spectrum so that means our formula will be c n h2n minus 1 x where X is our halogen O2 what we can see next is on the Mass Spectrum there is a Peak at 107 and 109. the relevance of this is that these two numbers are exactly 45 away from a molecular and our n plus two peaks this is for the evidence that our compound is a carboxylic acid meaning that 107 and 109 are the masses of the chain that does not include the carboxylic acid e o two h 107 and 109 is going to be c n minus 1 h 2 N minus 2 x where X is our halogen we can substitute our suspected halogens into X in order to work out what n is for C and H so let's try Co first chlorine exists as either chlorine 35 or 37. if we subtract 35 and 37 from 107 and 109 respectively we get c n minus 1 h2n minus 2 equals 272. in this case there is no number which we can fit into the formula in order to satisfy this equation therefore we eliminate the chance that it could be CL and next we'll attempt for Bromine if the halogen was bromine Roman exists as either 79 or 81. if bromine is in the compound CN minus 1 h2n minus 2 is going to equal to 28. if n was equal to 3 our formula C2 h 4 is going to have a molar mass of 28 grams per mole that means that n equals to 3 is consistent with what we see in our Mass Spectrum so now we know what the formula is if n equals to 3 our formula will be C3 H5 O2 ER this formula indicates to us ribanoic acid however at this stage we still have not finished because it asks us to draw the structural formula of this compound unfortunately bromopropanolic acid can exist in two forms it can either exist as two bromine propanolic acid or three broken paranoic acid what we need to do is we need to search for what is the distinguishing feature of each of these isomers what we notice first is that if there's a fragmentation that occurs in this position for two bromine propionic acid we should expect that there is a peak for the methyl group and the methyl group will have a signal in the Mass Spectrum for 15 Master charge ratios although it is small we do see the presence of a methyl ion which occurs due to the fragmentation of the larger molecule this is evidence that the compound is two bromopropicnic acid as three bromiparbonic acid does not contain a methyl fragment therefore we're going to draw our final compound structure which is going to be two bromo propanolic acid and our answer is now complete hey everyone if you found this video helpful smash that like button and 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