IIT JEE , AIEEE Forum - Chemistry , Organic Chemistry

alisha_gupta_27

explain the reason of the formation of the products of the following reaction:

CH ₃- CH ₂-CH(CH ₃)-CH ₃-O-CH₂-CH₃+ HI ------------->CH ₃- CH ₂-CH(CH ₃)-CH ₃-I + H-O-CH₂-CH₃

Why does CH ₃- CH ₂-CH(CH ₃)-CH ₃-OH + CH₂-CH₃-I is not formed?

rooney

Because CH ₃- CH ₂-CH(CH ₃)-CH ₃+ seems to be a much more stable carbocation than +CH₂-CH₃(which is highly unstable) to react with the nucleophile I-.This stability factor runs over the steric hindrance factor coz both carbocations are primary.

nishantsingh89

@rooney

yaar if carbocation is formed then

CH3-CH2-CH(CH3)-CH2+ SHOULD REARRANGE TO FORM A SECONDARY CARBOCATION

CH3-CH2-C(CH3)+-CH3 AND THEN PRODUCT SHOULD HAVE I- ATTACHED TO SECONDARY CARBON

sahilgupta_iit

Because oxygen is more electronegative and it will prefer to hold unit negative charge on it. Thus sigma bond between carbon and oxygen is broken resulting in formation of a carbocation. HI will produce H+ & I-. I- will combine with the carbocation and H+ will combine with oxygen containing group. Also charge is conserved.

rooney

@nishant

I know man but I couldnt find any other reason to satisfy the occurence. What I had studied was that if one of the Carbocations formed is not 3 degree and doesnt form a highly stable carbocation species, it is the smaller of the two R groups which give the Iodide.

polymath

After protonation of the ether at the oxygen the I^- will attack on the more postive site and in my opinion the CH ₃- CH₂-CH(CH ₃)-CH₃-OH + CH₂-CH₃-I is formed

nikss

BCOZ IT HAS MORE +I EFFECT AND WILL PRODUCE A BETTER C+

khinart

See as soon as the protonation of the ether occurs a stable carbocation is formed by the release of a proton which further attacks the ether molecule to form the halide.The other part of the ether is then protonated to give alcohol.

shubham_sachdeva

i have found the reason, OH - is better nucleophile then I - . so OH - find a more suitable alkyl group ( i.e. alkyl gp. having less size) (in this case it is CH3CH2)

THIS IS THE SAME REASON WHY BELOW IS FORMED ...

rooney

@shubham

Then why is the rule that when neither of the carbocation are tertiary , smaller alkyl group forms the iodide. What you say should happen in this case as well.

Also, what do you mean by "suitable alkyl group" ? less size or more stable carbocation ?

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