Iodine Mapping

Chemistry Level 3

$\ce{Propyne}$ is reacted in the following way:

1) $\ce{NaNH_2}$
2) $\ce{BrCH_3}$
3) 2 moles of $\ce{HI}$

The major product will be $\ce{\underline{\ },\underline{\ }~-~diiodo~butane}$ . Fill in the blanked positions of the iodines.

Use IUPAC numbering protocol to formulate your answer.

David's Organic Chemistry Set

David's Physical Chemistry Set

1,1 3,3 2,2 3,4 1,2 2,3

1 solution

David Hontz
Jul 10, 2016

$NaNH_2$ will remove a hydrogen to form the acetylene anion
This anion will attack $BrCH3$ , by bonding the the methyl carbon and force the $Br$ to leave
Lastly, 2 moles of $HI$ will add two iodines to the same carbon in the triple bonded carbons.

$Answer \rightarrow \boxed{2,2}$

How the last step follows.When we add a single HI its quite apparent(I at carbon-2).But next Iodine has to be added at carbon-3.Because the carbonation at C-2 is unstable owing to -I effect of Iodine.Where as in the both the choices the hyper conjugation remains the same.?

Spandan Senapati - 3 years, 11 months ago

@David Hontz Yes, I think the other one should be at 3rd position.. @Md Zuhair

Ankit Kumar Jain - 3 years ago

I thought it to be 2,2. There is a reason. This is because the carbon attached to iodine will (after 1st Iodo addition) will have less -ve charge on it that the other C attached with it through the double bond. This is because I is electron withdrawing, Which means that the electron density given by CH3 is taken up by I in some extent, where as it doesnt takes place so actively at the other C.

That was my reason. I dont know if you have a counter reason.

Md Zuhair - 3 years ago

The carbocation that will be formed is more stable at the carbon other than the one to which Iodine is attached (Due to -I effect of Iodine) , so why won't that product be major?

Ankit Kumar Jain - 3 years ago

@Ankit Kumar Jain – Ha... TU bhi bhool nahi hai. But where is my concept going wrong?

Md Zuhair - 3 years ago

@Md Zuhair – I couldn't understand your point ...Can you please explain it again? I mean what happened if there is less -ve charge on that carbon?

Ankit Kumar Jain - 3 years ago

@Ankit Kumar Jain – Hangout pe aa

Md Zuhair - 3 years ago

@Ankit Kumar Jain – The carbocation will become stable when it forms below I due to Resonance effect of I ( by lone pair).

Aaron Jerry Ninan - 3 years ago

@Aaron Jerry Ninan – Aint my xplanation right?

Md Zuhair - 3 years ago

@Md Zuhair – See...may be ur logic wont work always....always look for the stability of intermediate formed in RDS( In this case it is formation of carbocation so stability of carbocation should be seen)

Aaron Jerry Ninan - 3 years ago

@Aaron Jerry Ninan – Ok. Ya you are true... You are top. I am 0

Md Zuhair - 3 years ago

@Aaron Jerry Ninan – But , do we consider the Resonance effect of I? ..The size difference between iodine and carbon is large.

Ankit Kumar Jain - 3 years ago

@Thomas Jacob @Aaron Jerry Ninan

Ankit Kumar Jain - 3 years ago

See the reason mentioned by @Aaron Jerry Ninan

Thomas Jacob - 3 years ago

Bhai tu repost kyu nahi karta Jo qs attempt karta hai.. hum log try karenge

Md Zuhair - 3 years ago

Iodine Mapping

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