Q24P

Question

Where would you expect each of the following compounds to absorb in the IR spectrum?

(a) 4-Penten-2-one (b) 3-Penten-2-one

(c) 2,2-Dimethylcyclopentanone (d) m-Chlorobenzaldehyde

(e) 3-Cyclohexenone (f) 2-Hexenal

Step-by-Step Solution

Verified
Answer

(a) 4-penten-2-one shows IR absorption at 1715 cm - 1

(b) 3-penten-2-one shows IR absorption at 1685 cm - 1 

(c) 2,2-Dimethylcyclopentanoneshows   IR absorption at 1750 cm - 1

(d) m-Chlorobenzaldehyde shows IR absorption at 1705, 2720 and 2820 cm - 1

(e) 3-Cyclohexenone shows IR absorption at 1715 cm - 1

(f) 2-Hexenal shows IR absorption at 1705 cm - 1

1Step 1: IR spectroscopy Regions

The infrared spectroscopy (IR) region lies under 50 cm - 1- 12500 cm - 1, which is further divided into three more regions which are as follows:

  • Near IR region,which is from 50 to 650 cm-1
  • Middle IR region, which is from 650 to 4000  cm-1
  • Far IR region, which is from 4000to 12500  cm-1

 

The Middle IR region is classified under two more regions:

  • Finger Print region (800 to 14,500 cm-1)
  • Functional group region (14,500to 4000 cm-1)
2Step 2: IR absorption of 4-Penten-2-one

4-penten-2-one is the ketone that is not alpha,beta-unsaturated as there is no double bond present between the alpha (carbon next to carbonyl carbon) and beta carbon of the 4-penten-2-one so it will show the infrared absorption at 1715 cm - 1.

 

CH2 = CHCH2COCH3

4-penten-2-one IR absorption at 1715  cm-1

3Step 3: IR absorption of 3-Penten-2-one

3-penten-2-one is the alpha, beta-unsaturated ketone (conjugate system); due to a double bond between alpha and beta carbon, it will show the infrared absorption at 1685 cm - 1.

The decrease in the absorption frequency in conjugated systems is because the compound 3-penten-2-one is involved in resonance. The Delocalization of electron density into the carbonyl reduces the bond order of the C=O group, which lowers the bond order, the bond’s force constant, and, in turn, lowers its vibrational frequency.


CH3 - CH = CHCOCH3

3-penten-2-one IR absorption at 1685  cm-1

 

4Step 4: IR absorption of 2,2-Dimethylcyclopentanone


2,2-Dimethylcyclopentanone is the cyclic ketone that is five-membered, and it will cause the angle strain in the carbonyl group as the ring size of the cyclic ketone is only five carbon long, which in turn increases the absorption position of the Infrared spectrum.

It will show the infrared absorption at 1750 cm - 1.




2,2-Dimethylcyclopentanone IR absorption at 1750 cm-1

5Step 5: IR absorption of m-Chlorobenzaldehyde


m-Chlorobenzaldehyde is the aromatic benzene having formyl substituent and chlorine substituent at the meta position.

In m-Chlorobenzaldehyde, the angle strain in the carbonyl group is less as the ring size is six carbon long, which decreases the absorption position of the Infrared spectrum. Therefore it shows the infrared absorption at 1705 cm - 1.

The compound is an aldehyde, showing two other absorptions at 2720 cm - 1 and 2820 cm - 1.

 



m-Chlorobenzaldehyde IR absorption at 1705, 2720 and 2820 cm-1

6Step 6 : IR absorption of 3-Cyclohexenone


3-Cyclohexenone is the cyclic ketone that is not analpha,beta-unsaturated ketone unsaturated as there is no double bond present between the alpha (carbon next to carbonyl carbon) and beta carbon of the 4-penten-2-one so it will show the infrared absorption at 1715 cm - 1.





3-Cyclohexenone IR absorption at 1715 cm-1

7Step 7 : IR absorption of 2-Hexenal

2-Hexenal is the alpha, beta-unsaturated aldehyde (conjugate system); due to a double bond between alpha and beta carbon, it will show the infrared absorption at 1705 cm - 1.

The decrease in the absorption frequency in conjugated systems is because the compound 3-penten-2-one is involved in resonance. The Delocalization of electron density into the carbonyl reduces the bond order of the C=O group, which lowers the bond order, the bond’s force constant, and, in turn, lowers its vibrational frequency.


CH3 - CH2 - CH2CH = CHCHO

2-Hexenal IR absorption at 1705 cm-1