Question

1) 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
-------------------------------------------------------------------------------------

Answer 1

I drew the compounds.

heptan-1-ol

----H---H--H---H---H---H--H
-----I----I----I----I----I----I----I
H - C - C - C - C - C - C - C - OH
-----I----I----I-----I----I----I----I
----H---H---H---H---H---H---H

heptanal

----H---H--H---H---H---H---O
-----I----I----I----I----I----I----II
H - C - C - C - C - C - C - C - H
-----I----I----I-----I----I----I
----H---H---H---H---H---H

heptane

----H---H--H---H---H---H--H
-----I----I----I----I----I----I----I
H - C - C - C - C - C - C - C - H
-----I----I----I-----I----I----I----I
----H---H---H---H---H---H---H

heptene

---------------H---H---H---H--H
----------------I----I----I----I----I
H - C = C - C - C - C - C - C - H
-----I----I----I-----I----I----I----I
----H---H---H---H---H---H---H

I do not know what to do.

Answer 2

First let me assure you that interpreting IR spectra is not an easy thing to do AND it is something that takes practice, practice, practice. I haven't done one of these in more than 50 years so take what I say with a grain of sand concerning identification of A, B, C and D. Look at what's common. ALL of them have something in the 750-1100. These are the C-C bonds and that makes sense since all are heptanes or heptanes. The heptane sample should have the C-C and C-H bonds ONLY; i.e., nothing in the C=O (aldehyde) or C=C (alkene) So you look for a hydrocarbon only and that would be heptane. Thank about heptane being sample B. Next you look at the C=C and C=O. The C=C is higher number than C=O with C=C being 1620-1680 and C=O being 1680-1750, On that basis I would pick C as heptene and D as heptanal. That leave A for the alcohol. Take all of this with a grain of sand since I've not done this in such a long time and I don't remember all of those absorption troughs and the nuances that go along with the interpretation of IR spectrograms. If I had to do this in practice, I would run samples of the four different compounds and compare them with the unknown spectra to help identify each unknown. The last time I did any of this for purposes of identification was 1952-1954. Count it up--that's abour 70 years ago. Remember, take my identifications with a grain of sand.