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
-------------------------------------------------------------------------------------
2 answers