MCQ HELP! Finding major product formed!

Hi, I dont remember exactly but read up on electrophilic addition. Major product is what is the most stable, so primary or secondary carbocation or tertiary. Tertiary/Secondary carbocations are the most stable and would form major product and the electrophile is the most electronegative halogen, which would bond at the carbocation. So the answer would be the most electonegative halogen bonded to the secondary/tertiary carbocation.

(edited 5 years ago)

Reply 2

So Ive just had a go and i think the answer is C. This is because there is either a secondary carbocation as an intermediate or a tertiary carbocation. Tertiary is the most stable, so forms the major product. Chlorine is the most electronegative, so when the halogen bond breaks, chlorine becomes the electrophile and accepts a pair electrons from the carbon of the teriiary carbocation. Iodine bonds to the other carbon of the double bond.

Reply 3

OP

Original post by kenj_

So Ive just had a go and i think the answer is C. This is because there is either a secondary carbocation as an intermediate or a tertiary carbocation. Tertiary is the most stable, so forms the major product. Chlorine is the most electronegative, so when the halogen bond breaks, chlorine becomes the electrophile and accepts a pair electrons from the carbon of the teriiary carbocation. Iodine bonds to the other carbon of the double bond.

I see where you are coming from now. I've tried drawing the mechanism out and this is what I got.

https://imgur.com/8OtmNi4

Knowing that there are two options between having a tertiary carbocation on the left forming or a secondary carbocation on the right forming by looking at the number of alkyl groups attached to it.

And stability of carbocations goes like 'teriary > secondary > primary', so then the chlorine electrophile attacks the tertiary carbocation which is by far the most stable so a major product.

This matches to option C, so you are most certainly correct. Thanks a lot for your help!

(edited 5 years ago)

Reply 4

OP

Original post by kenj_

So Ive just had a go and i think the answer is C. This is because there is either a secondary carbocation as an intermediate or a tertiary carbocation. Tertiary is the most stable, so forms the major product. Chlorine is the most electronegative, so when the halogen bond breaks, chlorine becomes the electrophile and accepts a pair electrons from the carbon of the teriiary carbocation. Iodine bonds to the other carbon of the double bond.

One question if you don't mind?

If we had a scenario where there was only secondary or tertiary carbocations available, it would be correct to say that the reaction would have a 50% chance of producing either isomeric products; in other words we have a racemic mixture, with two equal concentrations of RHS and LHS enantiomers present?

Reply 5

Original post by Yatayyat

One question if you don't mind?

If we had a scenario where there was only secondary or tertiary carbocations available, it would be correct to say that the reaction would have a 50% chance of producing either isomeric products; in other words we have a racemic mixture, with two equal concentrations of RHS and LHS enantiomers present?

hmm, I’m not sure about 50/50 because the major product is described as what is MOST LIKELY to form, and seeing as the tertiary carbocation is the most stable intermediate, this would have a higher conc ( remember molecules/atoms/compounds looove stability)

(edited 5 years ago)

Reply 6

OP

Original post by kenj_

hmm, I’m not sure about 50/50 because the major product is described as what is MOST LIKELY to form, and seeing as the tertiary carbocation is the most stable intermediate (and remember molecules/atoms/compounds looove stability)

Sorry I should have worded the question better. I meant if only one type of carbocation was seen. So let's suppose we only had secondary carbocations available for the electrophile to bond to...

Reply 7

Original post by Yatayyat

One question if you don't mind?

If we had a scenario where there was only secondary or tertiary carbocations available, it would be correct to say that the reaction would have a 50% chance of producing either isomeric products; in other words we have a racemic mixture, with two equal concentrations of RHS and LHS enantiomers present?

Are you thinking about something along the lines of 2,3-dimethylpent-2-ene + ICl?

Reply 8

Original post by kenj_

hmm, I’m not sure about 50/50 because the major product is described as what is MOST LIKELY to form, and seeing as the tertiary carbocation is the most stable intermediate, this would have a higher conc ( remember molecules/atoms/compounds looove stability)

OOPS, sorry I just got what you meant. Yes, I think you’re 100% correct. And in this case, there would be no major and minor product if the same carbocation type is formed either way. I’m sure there’s some complex stuff we dont need to know which could be another determiner though 🤔

Reply 9

OP

Original post by Pigster

Are you thinking about something along the lines of 2,3-dimethylpent-2-ene + ICl?

If 2,3-dimethylpent-2-ene can form only one type of carbocation, once the double bond attracts the iodine ion and breaks the polar covalent bond between ICl, then yes.

Reply 10

Original post by Yatayyat

If 2,3-dimethylpent-2-ene can form only one type of carbocation, once the double bond attracts the iodine ion and breaks the polar covalent bond between ICl, then yes.

The two intermediates would be (CH3)2C+CI(CH3)CH2CH3 OR (CH3)2CIC+(CH3)CH2CH3. Both carbocations are tertiary. BUT, due to the inductive effect, the second one would be the major product, as the carbocation is every so slightly more stable. But, only just the major product.

Reply 11

Original post by Pigster

The two intermediates would be (CH3)2C+CI(CH3)CH2CH3 OR (CH3)2CIC+(CH3)CH2CH3. Both carbocations are tertiary. BUT, due to the inductive effect, the second one would be the major product, as the carbocation is every so slightly more stable. But, only just the major product.

Could you please explain this inductive effect a little more?

Reply 12

OP

Original post by Pigster

The two intermediates would be (CH3)2C+CI(CH3)CH2CH3 OR (CH3)2CIC+(CH3)CH2CH3. Both carbocations are tertiary. BUT, due to the inductive effect, the second one would be the major product, as the carbocation is every so slightly more stable. But, only just the major product.

https://imgur.com/4huYZ1B

I've just drawn the displayed formula of the two possible carbocations just to make it easier to visualise. They are indeed both tertiary carbocations, but what specifically makes the second carbocation a bit more stable than the first carbocation, so it can be considered the major product.

Reply 13

The reason it goes 3o>2o>1o is due to the relative stabilities of the C+ atom. They are stabalised by the bonding to other alkyl groups, which are able to donate some e- density to the C+ atom. The more alkyl chains the more e- density donated and hence the smaller the + seems to be. Also the longer the alkyl chains are (up to propyl) the more e- density they can donate, so my left one has 2xCH3's attached (and the C which was across the C=C), whereas the right one has 1xCH3 and 1xCH2CH3. The extra chain length means there would be slightly more e- donated to the C+ atom and hence slightly more stability.

Reply 14

Original post by Pigster

The reason it goes 3o>2o>1o is due to the relative stabilities of the C+ atom. They are stabalised by the bonding to other alkyl groups, which are able to donate some e- density to the C+ atom. The more alkyl chains the more e- density donated and hence the smaller the + seems to be. Also the longer the alkyl chains are (up to propyl) the more e- density they can donate, so my left one has 2xCH3's attached (and the C which was across the C=C), whereas the right one has 1xCH3 and 1xCH2CH3. The extra chain length means there would be slightly more e- donated to the C+ atom and hence slightly more stability.

Ah okay, so the longer the alkyl chain, the more electron donating it is?

Reply 15

Original post by kenj_

ah okay, so the longer the alkyl chain, the more electron donating it is?

-ch3 < -ch2ch3 < -ch2ch2ch3 = -ch2ch2ch2ch3...

Reply 16

Original post by Pigster

-ch3 < -ch2ch3 < -ch2ch2ch3 = -ch2ch2ch2ch3...

are propyl and butyl equal due to the length of butyl’s chain weakening the effect of the extra carbon atoms? As in they’re too far away to have an effect?

Reply 17

OP

Original post by Pigster

The reason it goes 3o>2o>1o is due to the relative stabilities of the C+ atom. They are stabalised by the bonding to other alkyl groups, which are able to donate some e- density to the C+ atom. The more alkyl chains the more e- density donated and hence the smaller the + seems to be. Also the longer the alkyl chains are (up to propyl) the more e- density they can donate, so my left one has 2xCH3's attached (and the C which was across the C=C), whereas the right one has 1xCH3 and 1xCH2CH3. The extra chain length means there would be slightly more e- donated to the C+ atom and hence slightly more stability.

Thank you, this clears it up. Would there ever be any exceptions to the trend, when the larger the alkyl (more electron releasing) has the effect of stabilising the C+ more greatly?

Reply 18

Why don't you both just look up "the inductive effect" via google/youtube etc? I also checked and actually the effect increases up to butyl and pentyl etc. is roughly the same as butyl. Pesky memory.

Reply 19