Conjugation and IR

Anyone able to help on part b)? All I know is that the conjugation will reduce the stretching freq, but don't know how the c=c behaves when it's in the position in structure 2.

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Reply 1

alow

Do you know how to draw the conformation of the molecules?

Reply 2

TSR George

Original post by alow

Do you know how to draw the conformation of the molecules?

I don't think so!

Reply 3

alow

Here's a post with energy profiles for cyclohexene and cyclohexane: http://chemistry.stackexchange.com/questions/45197/energy-profile-of-cyclohexene-conformations

Now all you have to do is work out which ones will be conjugated (because of having a planar region with π electrons).

Reply 4

TSR George

Original post by alow

Definitely haven't seen this before! So the 2nd structure would require less energy to change conformation, and would therefore have a lower vibrational freq than the 1st structure? And structure 2 isn't conjugated (or do the lps on O have an effect?)

Reply 5

alow

That's nothing to do with it.

You need to know which of the molecules will have a conjugated π region, which requires planarity.

Reply 6

TSR George

Original post by alow

You need to know which of the molecules will have a conjugated π region, which requires planarity.

But only the 3rd structure does?

Reply 7

alow

Yeah, so that gives you the lowest stretching frequency.

Now you need to consider which will have the (uncharacteristically, especially for an ester) high absorption.

Reply 8

TSR George

Original post by alow

Now you need to consider which will have the (uncharacteristically, especially for an ester) high absorption.

That was the bit I wasn't sure about

Reply 9

alow

Okay.

What I think the question wants you to think about to explain this is how the ester stretching will bend the ring, and how energetically favourable this will be for compound 2.

Reply 10

TSR George

Original post by alow

Okay.

What I think the question wants you to think about to explain this is how the ester stretching will bend the ring, and how energetically favourable this will be for compound 2.

so stretching is more favourable in structure 2 because the ring contains a c=c, so lower freq is needed for structure 2?

Reply 11

alow

Other way around.

I think the double bond stretching during the vibration as the ring bends is less favourable than a single bond stretching, so requires more energy and therefore has a higher frequency absorption.

This definitely assigns the stretches properly (I looked up the spectra) but there may be an alternate explanation.

Reply 12

TSR George

Original post by alow

Thank you! It just might be that I haven't covered it yet, but that really helps!!

Reply 13

MexicanKeith

I would have used an alternative explanation so here goes!

saturated 6 membered rings are basically unstrained, so the carbonyl in structure 1 should stretch like a normal ester (1735cm^-1).

The carbonyl of an ester is strengthened by the inductive withdrawal of the alcoholic O, but weakened by donation of the lone pair on the same O into the carbonyl pi* (you can draw resonance structures for both).

In structure 2 the lone pair is less able to donate into the carbonyl because it is delocalised into the neighbouring c=c pi* this would give compound 2 the highest stretching frequency 1769cm^-1.

For structure 3, the c=c bond is now in such a position that, rather than interacting with the lone pair on oxygen, it too can donate into the carbonyl pi* so structure 3 has 2 weakening effects and hence the lowest stretching frequency of 1720cm^-1.