TBR orgo Ch. 4 Practice exam #11 - UV peak

[plage noire]

Can anyone elaborate on the explanation for this question? I can't figure out why there would have been a UV peak. The last sentence in the chapter 2 information on UV-vis spectroscopy says "in organic chemistry, a compound must have a pi bond to be UV-visible active." There is no pi bond in the reagents in this passage, so I'm totally lost. Any thoughts?

Thanks!

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[amy_k]

We need to have a conjugated pi system for the compound to be UV-active, I would need to see the question to understand more.. do the reagents react to give a product which has conjugated pi bonds?

 

[Cawolf]

There is a UV peak around there for C-Br and C-I bonds.

As the reaction proceeds, this bond is broken - reducing the absorbance.

As an aside I did not know anything about UV spectroscopy, but just read a quick lecture about it after encountering it in chapter 1 of TBR.

 

[amy_k]

What are the reagents it is using?

 

[Cawolf]

It is a generic Sn2 reaction with a bromide leaving group and "nucleophile" - primary carbon.

 

[plage noire]

There is a UV peak around there for C-Br and C-I bonds.

As the reaction proceeds, this bond is broken - reducing the absorbance.

As an aside I did not know anything about UV spectroscopy, but just read a quick lecture about it after encountering it in chapter 1 of TBR.

Awesome, thanks!