1) 2-iodopentane was passed through a mass spectrometer and the
most abundant fragment ion gave a peak with m/z = 183.
Draw the displayed formula of this fragment ion. Include its
charge.
Show your WORKING OUT for determining the abundent fragment ion.
2) NMR spectroscopy can be used to study the structures of organic compounds. Compound M was studied using 1H NMR spectroscopy:
---------------------CH3
-------a---------------I
Br - CH2 - CH2 - C - CH3
------------------------I
----------------------CH3
Compound M is shown above.
a) Explain why tetrachloromethane would be a suitable solvent for this type of spectroscopy.
b) Give the number of peaks in the 1H NMR spectrum of Compound M.
Take into consideration hydrogen species that are in the same environment.
c) Give the splitting pattern of the protons labelled a in Compound M.
Apply the n+1 rule.
d) Give the IUPAC name of Compound M.
3) A chemist discovered four bottles of liquid simply labelled A-D, each
of which contained a different pure organic compound. The compounds were known to be heptan-1-ol, heptanal, heptane and heptene. The chemist put a sample from each bottle through Infrared Spectroscopy and recorded the results below in Table 1.
a) Use the information in the data table (Table 2) to suggest which compound is in which bottle.
b) For each compound justify your conclusion
Table 1 Infrared Spectroscopic Results for unknown samples:
---------------------------------------------------------------------------------------
Bottle----- Infrared Spectroscopy main Troughs (wavelength cm-1)
------------------------------------- 3300
-- A ---------------------------------3000
--------------------------------------1100
------------------------------------750-1000
---------------------------------------------------------------------------------------
---------------------------------------3000
--- B --------------------------------750-1000
---------------------------------------------------------------------------------------
---------------------------------------3000
--- C --------------------------------1650
---------------------------------------750-1000
----------------------------------------------------------------------------------------
----------------------------------------3000
---- D --------------------------------1710
---------------------------------------750-1000
--------------------------------------------------------------------------------------
Table 2 Wavenumber data for different bonds between atoms:
---------------------------------------------------------------------------------------
---------------------------Infrared Absorption Data---------------------------
-----Bond--------------------------------- Wavenumber / cm-1--------------
N – H
(amines) -----------------------------------3300 - 3500
--------------------------------------------------------------------------------------
O – H
(alcohols) ----------------------------------3230 - 3550
-------------------------------------------------------------------------------------
C – H --------------------------------------- 2850 - 3300
-------------------------------------------------------------------------------------
O – H
(acids) --------------------------------------2500 - 3000
------------------------------------------------------------------------------------
C ≡ N --------------------------------------- 2220 - 2260
-----------------------------------------------------------------------------------
C = O --------------------------------------- 1680 - 1750
-----------------------------------------------------------------------------------
C = C --------------------------------------- 1620 - 1680
----------------------------------------------------------------------------------
C – O --------------------------------------- 1000 - 1300
----------------------------------------------------------------------------------
C – C ----------------------------------------- 750 - 1100
-------------------------------------------------------------------------------------
1) To determine the most abundant fragment ion, we need to consider the structure of 2-iodopentane and the fragmentation patterns observed in the mass spectrometer.
2-iodopentane has the following structure:
CH3-CH2-CH2-CH(I)-CH3
When passing through the mass spectrometer, the most common fragmentation occurs at the C-I bond, resulting in the formation of the fragment ion.
To determine the mass of the fragment ion, we need to calculate the molecular mass of the remaining portion of the molecule after the loss of iodine atom (I).
The molecular mass of pentane (CH3-CH2-CH2-CH3) is 72 g/mol.
Therefore, the molecular mass of the fragment ion = 72 - 127 (molar mass of iodine) = -55 g/mol.
Since the fragment ion has a positive charge, we need to include the charge in the displayed formula. So, the displayed formula of the fragment ion is CH3-CH2-CH2-CH3+ with m/z = 183.
2)
a) Tetrachloromethane (CCl4) is a suitable solvent for NMR spectroscopy because it is nonpolar and does not possess any protons that overlap with the protons in the compound being studied. Additionally, it has a high boiling point and is chemically inert, making it a good choice to dissolve a wide range of organic compounds.
b) The number of peaks in the 1H NMR spectrum of Compound M can be determined by considering the number of unique hydrogen environments.
Looking at the structure of Compound M, we can identify four unique hydrogen environments:
- The three hydrogen atoms attached to the methyl group will give a singlet at one chemical shift.
- The two hydrogen atoms attached to the methylene group adjacent to the iodine atom will give a multiplet at a different chemical shift.
- The two hydrogen atoms attached to the methylene group next to the bromine atom will give a multiplet at another chemical shift.
- The three hydrogen atoms attached to the methyl group on the far right will give a singlet at a different chemical shift.
Therefore, Compound M will have four peaks in the 1H NMR spectrum.
c) The splitting pattern of the protons labeled a in Compound M can be determined using the n+1 rule. According to this rule, if a proton has n equivalent neighboring protons, it will produce a signal that is split into n+1 peaks.
In the case of Compound M, the protons labeled a have two equivalent neighboring protons (the two hydrogen atoms in the neighboring methylene group), so they will produce a triplet signal. The triplet signal will appear as a center peak with two smaller peaks on either side in a 1:2:1 ratio.
d) The IUPAC name of Compound M is 3-bromo-4-iodoheptane.
3)
a) To determine which compound is in each bottle based on the infrared spectroscopy data, we need to compare the observed infrared absorption troughs with the known absorption wavelengths for different bonds between atoms.
- Bottle A shows a trough at 3000 cm-1, which matches the C-H absorption range.
- Bottle B shows a trough at 750-1000 cm-1, which matches the C-C absorption range.
- Bottle C shows a trough at 1650 cm-1, which matches the C = C absorption range.
- Bottle D shows a trough at 1710 cm-1, which matches the C = O absorption range.
b) Justification for each conclusion:
- The absorption at 3000 cm-1 in Bottle A indicates the presence of C-H bonds, suggesting that the compound is heptane.
- The absorption in the range of 750-1000 cm-1 in Bottle B indicates the presence of C-C bonds, suggesting that the compound is heptene.
- The absorption at 1650 cm-1 in Bottle C indicates the presence of C=C bonds, suggesting that the compound is heptene.
- The absorption at 1710 cm-1 in Bottle D indicates the presence of C=O bonds, suggesting that the compound is heptanal.