49 E - Organic Chemistry (Mcmurry)

Question

Question: 1,3,5-Hexatriene has ${\lambda }_{max}$=258 nm. In light of your answer to Problem

14-48, approximately where would you expect 2,3-dimethyl-1,3,5-hexatriene to absorb?

Step-by-Step Solution

Answer

From problem 14-48, we concluded that one alkyl group increases  ${\lambda }_{max}$of conjugated diene by approximately 5 nm. Since 2,3-dimethyl-1,3,5-hexatriene has two methyl substituents, So its UV${\lambda }_{max}$ should be about 10 nm longer than the of 1,3,5-hexatriene. 

Step-by-step Solution Step 1: Ultraviolet spectroscopy

Ultraviolet spectroscopy refers to absorption spectroscopy in part of the ultraviolet and the full adjacent visible region of the electromagnetic spectrum.

• The UV region of the electromagnetic spectrum extends from 100-400 nm.  
• The visible region extend from 400-800 nm.

 

Step 2: Factor affecting value of wavelength in UV spectroscopy

The absorption of light in UV spectroscopy promotes the$\pi -\pi *$ electron transition. In order for this transition occur the energy gap between HOMO and LUMO has to correspond to energies of the UV radiation.

Generally, the most probable transition is from the highest occupied molecular orbital (HOMO) to lowest unoccupied molecular orbital (LUMO).

• The most important factor that will affect the HOMO and LUMO energy gap, and, therefore, the energy and the wavelength of absorbed radiation is the extent of conjugation. The conjugation lowers the difference between HOMO and LUMO and the more extended it is, the lower the energy difference will be.
• The value of in the ${\lambda }_{max}$ultraviolet spectrum of dienes becomes larger with increasing alkyl substitution.

Step 3: Calculate λ m a x for 2,3-dimethyl-1,3,5-hexatriene

From problem 14-48, we concluded that one alkyl group ${\lambda }_{max}$ increases of conjugated diene by approximately 5 nm. Since 2,3-dimethyl-1,3,5-hexatriene has two methyl substituents, So its UV${\lambda }_{max}$ should be about 10 nm longer than the of 1,3,5-hexatriene .

 

${\lambda }_{max}=258nm$

                                           ${\lambda }_{max}=268nm$

The value of ${\lambda }_{max}$in the ultraviolet spectrum of dienes becomes larger with increasing alkyl substitution.