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Question 3 Not yet answered Marked out of 3.00Flag questionWhich of the following compounds is likely to absorb the lower frequency of IR radiation? In brief; suppo...

Question

Question 3 Not yet answered Marked out of 3.00Flag questionWhich of the following compounds is likely to absorb the lower frequency of IR radiation? In brief; support your answer.

Question 3 Not yet answered Marked out of 3.00 Flag question Which of the following compounds is likely to absorb the lower frequency of IR radiation? In brief; support your answer.



Answers

Which compound, NO or $\mathrm{NO}_{2},$ absorbs IR radiation of a longer wavelength?

So an absorption in the 3000 region indicates that the structure does have hydrogen is present. Not all bonded to SP two hybridized carbons as well as SP three hybridized carbons. When considering all of the given compounds Benzel alcohol on Benzel aiming are the only structures that have SP three carbons. So there is also an observed peak at 3300 which corresponds to our O age group. We do not have an absorption at about 1600 so I'm ruling out the present often an H Therefore, from the data above, I have determined that our structure is in fact Ben Zyl alcohol So we can simply draw this out below. Start off with our benzene ring With all of the three of the conjugated bonds, we have a CH two and then an O. H

Absorption that we see at 1600 wave numbers on 15 hundreds of 14. 30. We've numbers indicate that the compound we're looking at toe identify maybe a benzene ring. So we also see an absorption of 3000. So this suggests that we have hydrogen attached to SP two hybridized carbons, not SP three hybridized carbons. So the absence off any sort of absorption of 1700 suggest that we do not have any carbon. Are groups present here? So for the above need reasoning, we may have a female so weaken simply just draw out our penal structure. Well, we can identify our functional groups where we have our benzene ring should R c six h six and we can draw on all functional group which is our alcohol on any given carbon because it's money substituted, it doesn't matter. We always assume that the carbon that are substitution is on. In a mono substituted structure is carbon one

So now we're just gonna be looking at the UV region a bit more and compounds that may be active in the region. So typically, in order to be active in this region, we need a conjugated system. So if we look our first example here, we have two double bonds. However, there's no conjugation, so this would not be active. Now, if you look our second example, we do have conjugation as we concious. If this set of electrons which would then push this serve electrons and then we will have residents structures. Now our residents structure will look almost identical, dependent on rotation because of the symmetry, however, that you still get the point. Now we have the localization off the electrons. Hence we have conjugated system and so its active now our last example here that the bottom might be a little bit less obvious. However, we do have conjugation here because but if we take a look at this first structure and might not be as obvious. But when you draw out and full like I have hit with the triple bond between the sea and end, then we take our double bond here shift those electrons. Then this will bump a set of electrons onto the nitrogen. And so then we will have this structure and again remember that in resident structures, but charges need to cancel. So we have a positive charge on the and then we'll have a negative charge on the nitrogen because we've shifted a pair of electrons onto it. So in our former residents structure, we have no formal charges. And then in our resident structure, after we've pushed a few arrows, we have a positive under negative charge which essentially cancel, and so are charges violence. So just to recap for a compound to be active in the UV region, we need a conjugated system.

And we're just going over infrared spectroscopy. So recall we have our definition at the top on what we will be doing here is looking at absorption is that we might expect in these molecules given in our examples structure, we have a seat. This is an AL the hide. If it is saturated, then we would see a absorption of 17 25 if it's conjugated 9 16 90 of its aromatic about 1700 all in Wade, Numbers said they're all in very close proximity. Secondly, we're looking at an ode to group. So in an acid metric molecule, this is between 15 and 1600. On in a symmetric molecule is 1300 to 13. 90. Lastly, we've got benzene at 16 82 7 to 5. So our second molecule now we can see an ESTA in areas of 17 35 17 20 17 26 on 17 35. Again, we've gotten our keen hair and nice scene and 1600 to 16 80 wave number on Lastly, our final molecule. Here we have a carbon ill on If it's in a saturated compound, we have an absorption of 17 50. If it's conjugated 16 80 aromatic, 16 90. And lastly, if it's in an autumn ring that 17 15 and just the last absorption will luck out here we have the alcohol functional groups, so that will appear at 606,500.


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