[Kaladiscope]

Hi chemists, biochemist or whatever involved in science...

Let's guess that we have an alkyl chain like CH3-CH2-CH2-N3, that we can to use it as a substrate to perform on the azide a nucleophilic attack. And at the same time we perform the same reaction on an alkyl chain which have difluoromethyl instead of methylene, that is CH3-CF2-CH2-N3.

My question or better to say my assumption is:

Due to the presence of a difluoromethyl in the second case, and therefore making the azide more electrophilic (electronwithdrawing effect of the fluor atom), the nucleophilic attack on this azide would be faster than in the first case having an alkyl chain.

Is that right?

[opsomath]

What is the nature of your nucleophilic attack? Azide is not a leaving group in the sense of SNx.

[Kaladiscope]

is an attack of a phosphorous compound (III), like triphenylphosphine, to the azide to generate one phosphoazide (loss of N2 to form an iminophosphorane).

[discodermolide]

is an attack of a phosphorous compound (III), like triphenylphosphine, to the azide to generate one phosphoazide (loss of N2 to form an iminophosphorane).

I don't think the two F atoms will have much influence on that process.

[Kaladiscope]

Don't you think that the presence of the fluor could make the azide more electrophilic and therefore more prone to nucleophilic attack?

As far as I know the electrophilic behaviour of azides can change a lot depending on the R.

R-N3

[discodermolide]

Don't you think that the presence of the fluor could make the azide more electrophilic and therefore more prone to nucleophilic attack?

As far as I know the electrophilic behaviour of azides can change a lot depending on the R.

R-N3

No, not in the chemistry you described.

[Arctic-Nation]

I have no experience with azides whatsoever, but in the Staudinger reduction you not only have the nucleophilic addition of PPh3 to the azide, but also the conversion of the phosphazide into azaphosphetane, which will slow down if the phosphazide is more electron-poor.