Find out the number of 1-2-shifts during the | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) The starting material is $3,3-dimethylbutan-2-yl$ cation (a secondary carbocation). To reach the final product,which is $2,3-dimethylbut-2-ene$,th

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(A) The starting material is $3,3-dimethylbutan-2-yl$ cation (a secondary carbocation). To reach the final product,which is $2,3-dimethylbut-2-ene$,the carbocation must undergo rearrangements to reach a more stable tertiary form. $1$. First,a $1,2-methyl$ shift occurs to convert the secondary carbocation into a more stable tertiary carbocation ($2,3-dimethylbutan-2-yl$ cation). $2$. Then,a $1,2-hydride$ shift occurs to further stabilize or rearrange the system if necessary,or simply to facilitate the elimination. However,looking at the skeletal structure provided,the conversion from the initial $3,3-dimethylbutan-2-yl$ cation to the $2,3-dimethylbut-2-ene$ involves a total of $2$ shifts ($1,2-methyl$ shift followed by $1,2-hydride$ shift) to reach the most stable intermediate before elimination. Thus,the total number of $1-2-$shifts is $2$.

Find out the number of $1-2-$shifts during the conversion of the given carbocation to the final alkene product.

Similar Questions

Which of the following species is a radical?

The correct order of stability of the following carbocations is:

Which of the following carbocations is the most stable?

Using curved-arrow notation,show the formation of reactive intermediates when the following covalent bonds undergo heterolytic cleavage:
$(a)$ $CH_{3}-SCH_{3}$
$(b)$ $CH_{3}-CN$
$(c)$ $CH_{3}-Cu$

The stability of the following carbocations decreases in the order:

Vedclass Test Series

Exam Paper Generator

Online Exam Module