Addition reactions

Chemistry Level 2

Which of the following is the major product of the reaction shown?

A B C D

3 solutions

Deepanshu Gupta
Nov 18, 2014

Simple re-arrangement of carbo-cation after protonation of alkene ! { By Hydride 1-2 shift ) since 3 degree carbo-cation is more stable ! :)

Where on earth is option E ?

Vishnu Bhagyanath - 6 years, 4 months ago

Where is E seriously? how can you give me wrong? i clicked A?

Agastya Chandrakant - 6 years, 4 months ago

There's just one little mistake, bro! It'll be 1,2-Methyl shift, not Hydride shift, because a methyl group is being shifted to make the more stable tertiary $(3^{\circ})$ carbocation.

Prasun Biswas - 6 years, 6 months ago

Is this not Markonikov reaction?????

Rakib Hassan - 6 years, 5 months ago

Yeah, this is markovnikov addition but it is followed by 1,2-hydride shift to make the product stable.

Ravi Teja - 6 years, 4 months ago

it is but u have to first re arrange it to get a more stable product thus rearrangement takes place and three degree carbocation will be formed to gain more stability

ishan pradhan - 6 years, 3 months ago

"Simple", "of", "after", "since", "more", "stable". Would you please explain all other words. Thanks.

Satvik Golechha - 6 years, 6 months ago

Nothing much complicated here. It's just that this reaction will occur faster in Substitution Nucleophilic 1 process and in $S_N^1$ process, the more substituents there is on the central carbon atom of the carbocation to be formed, the more stable the carbocation will be. Since tertiary carbocation is more stable than secondary carbocation, there will be Whitmore 1,2-shift to create the more stable $(3^{\circ})$ carbocation rather than the expected reaction route.

Prasun Biswas - 6 years, 6 months ago

Asif Ali
Jan 18, 2015

because it follows the markonikhofs rule any attacking nucleophile will attack to that carbon atom which contains less number of hydrogen atoms

except that by markovnikov, they would go in either of the atoms of the double bond which is broken in the reaction

Vitor Kiguchi - 6 years, 4 months ago

Elle Chigf
May 12, 2015

As this might seem weird when you're used to witness Markonikov's reaction which would have given (CH3)2-CH-CH(Br)-CH3, it can be understood when you consider the formation of the carbocation. The carbocation formed by the breaking of the double bound on the second carbon is more stable when placed on a tertiary carbon by inductive effect. Therefore, molecular rearrangement happens after the addition of the hydrogen on the first carbon, so that the carbocation is placed on the tertiary carbon. Therefore, the Br- attacks the tertiary carbon, forming product A.

Addition reactions

3 solutions

0 pending reports