Question

1) If there were two bromine, what isotopic distribution (The ratio) would we see in Mass Spectrum?

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2) If there were three bromine, what isotopic distribution (The ratio) would we see in Mass Spectrum?

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3) If we had chlorinated the ring once (and not brominated), what would we expect to see in Mass Spectrum?

Answer 1

1) IF COMPOUND CONTAINS TWOBROMINE ATOMS WE WILLOBSERVE THREE CLUSTER OF PEAKS SEPARATED BY 2 UNITS M, M+2,M+4.THE RATIO OF THESE THREE PEAKS CAN BE KNOWN BY A FORMULA

(a+b)ⁿ where a= bromine - 79 isotope b= bromine -81 isotopewith their relative abundances.since Br ⁷⁹ and Br⁸¹ are available in 1;1ratio. a=1Br ⁷⁹ and b=1Br⁸¹ ,n= number of bromine atoms.=2

there fore intensity oflines equal to the coefficients of terms corresponding to m, m+2,m+4terms.

(a+b)ⁿ =   (1Br ⁷⁹ +1 Br⁸¹)² = 1( Br ⁷⁹)² +2.  Br ⁷⁹ . Br⁸¹+ 1( Br⁸¹)²  

m m+2 m+4

1 : 2 : 1

therefore the three peaks are observedwith 1:2:1 ratio.

2) if 3 bromine atoms are present then four cluster of peaks m, m+2, m+4, m+6 are observed and n= 3 in above formula

(a+b)ⁿ =   (1Br ⁷⁹ +1 Br⁸¹)³ = 1( Br ⁷⁹)³+3. (Br ⁷⁹)² . Br⁸¹+ 3Br ⁷⁹( Br⁸¹)² + 1( Br⁸¹)³ [a³+3a²b+3ab²+b^3]

m m+2 m+4 m+6

1 : 3 : 3 ; 1

therefore four peaksare observed in 1;3;3;1ratio.

3) if one chlorine atom is present then we will observe two peaks separated by 2 units are observed due to cl-35 and cl-37 isotopes .relative abundance is 3;1 so a= 3.cl³⁵ and b= 1cl³⁷ ,and n= 1since only one chlorine atom is present.

(a+b)ⁿ = (3.cl³⁵ + 1cl³⁷) ¹ =  3.cl³⁵ + 1cl³⁷

^{m m+2}

3 : 1

therefore m and m+2 peaks are observed in 3;1 ratio.