[furaniki18]

if an organic compound contains both alkene and alkyne functional groups and 1 mole of bronsted acid or halogen is given as reagent then which of the two will undergo addition reaction alkene or alkyne?

[Dan]

I'd expect the major product to process via the lowest energy intermediate (carbocation or halonium ion).

Which do you think has a more stable intermediate?

[furaniki18]

i asked this question because i was having trouble in these 2 ques. in the image...in the first ques. if i shift one of the pi bonds of alkyne towards right it would result in formation of  a secondary vinylic carbocation which is unstable and if this carbocation shows resonance with the delocalisation of pi electrons on alkene it would result in formation of a cumulated diene which is again unstable.....so my guess is that it should be alkene which undergoes addition but the answer in my assignment is the one after addition on alkyne and my teacher seems to be very reluctant in addressing my queries.

[orgopete]

This would be my suggestion of how to approach this. If you had butadiene, protonation would occur such that the resultant carbocation can be resonance stabilized by the other double bond. I think you should try a similar analysis for this problem. Hint, it may be helpful to write out both possibilities, even if you think one may be wrong. Compare and analyze them. Hint #2, Q2 may be easier to recognize as one intermediate being more stable.

[furaniki18]

thank you for the hints...this is what i have worked out. i think intermediate I to be more stable because the positive charge is on sp2 hybridised carbon atom rather than on sp hybridised carbon atom in intermediate II. since sp2 hybridised carbon atom is less electronegative  than sp hybridised carbon atom, to have a positive charge on it will be more stable than to have it on sp hybridised carbon atom.
i just want to know if this is the right approach and can i proceed the same way for the second ques. as well...

[CaverKat]

I'm confused and just want to clarify.  Are your curly arrows a mechanism?

[furaniki18]

I'm confused and just want to clarify.  Are your curly arrows a mechanism?

i have drawn the curly arrows just to represent towards which side the pi electrons are shifting....if you're asking why exactly in the first step i am shifting the pi electrons towards a particular side and not the other then that's because in the first case shifting the pi electrons of alkyne towards left would result in formation of primary vinylic carbocation which is less stable than secondary vinylic and also cannot be further resonance stabilised....and similar explanation can be given for the second ques.

[Telamond]

Do you know how the two carbocations are stabilized?
The one below you drew is called an allene, it's p-orbital overlap looks a special way if you have never seen them before. How much more stable is that?

[furaniki18]

Do you know how the two carbocations are stabilized?
The one below you drew is called an allene, it's p-orbital overlap looks a special way if you have never seen them before. How much more stable is that?

i have heard of allene and know about its p orbital overlap..but in the second intermediate that i have drawn how does positive charge on the carbon atom change that p-orbital overlap which is due to formation of a pi bond with the adjacent carbon atom....i don't think this should be a reason for more stability of first intermediate

[Telamond]

Just saying that I'm pretty sure of the two structures you drew for the protonation of the alkene that the bottom one is the most stable carbocation. And what's the hybridization of the propargylic carbocation?

[furaniki18]

it is sp2 hybridised...

[orgopete]

thank you for the hints...this is what i have worked out. i think intermediate I to be more stable because the positive charge is on sp2 hybridised carbon atom rather than on sp hybridised carbon atom in intermediate II. since sp2 hybridised carbon atom is less electronegative  than sp hybridised carbon atom, to have a positive charge on it will be more stable than to have it on sp hybridised carbon atom.
i just want to know if this is the right approach and can i proceed the same way for the second ques. as well...

This was my thinking as well. The terminal sp carbocation is probably the least stable and therefore least likely to contribute. All others are sp2 and seemingly more stable. I think in the next example, the stabilities improve for protonation of the terminus.

By the way, the curved arrows are not standard usage (though my student surveys (enlargeable) suggests this is not unique). However, I do think that as we encounter questions about how to learn, I cite examples of curved arrow usage such as this to show a lack of connection between what the curved arrows and the result. The curved arrows are not contributing to an understanding in a clear and consistent manner. Because this isn't the worst example and I think the poster knew what they wanted to show, it may not have mattered greatly to the poster. But to anyone confused about the chemistry, these curved arrows would be confusing.