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(A) The reaction of an alkene with $HBr$ proceeds via an electrophilic addition mechanism involving a carbocation intermediate.Step $1$: Protonation o

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$CH_3CH_2CH_2C(Br)(CH_3)_2$

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(A) The reaction of an alkene with $HBr$ proceeds via an electrophilic addition mechanism involving a carbocation intermediate.Step $1$: Protonation of the double bond can form two possible secondary carbocations:$(I)$ $CH_3-CH_2-CH^{+}-CH(CH_3)_2$$(II)$ $CH_3-CH^{+}-CH_2-CH(CH_3)_2$Step $2$: Carbocation $(I)$ undergoes a $1,2-$hydride shift from the adjacent tertiary carbon to form a much more stable tertiary carbocation:$CH_3-CH_2-CH^{+}-CH(CH_3)_2 \xrightarrow{1,2-H 	ext{ shift}} CH_3-CH_2-CH_2-C^{+}(CH_3)_2$Step $3$: The bromide ion $(Br^{-})$ attacks this highly stable tertiary carbocation to form the major product:$CH_3-CH_2-CH_2-C^{+}(CH_3)_2 + Br^{-} \longrightarrow CH_3-CH_2-CH_2-C(Br)(CH_3)_2$Thus,the major product is $2-$bromo$-2-$methylpentane.

The major product formed in the following reaction is :
$CH_3CH=CHCH(CH_3)_2 \xrightarrow{HBr}$

Similar Questions

For the following reactions,the major products are shown: $H_2C=CH-CH=CH_2$ $\xrightarrow[0^\circ C]{HBr} CH_2=CH-CH(Br)-CH_3$ $\xrightarrow{25^\circ C} Br-CH_2-CH=CH-CH_3$. These provide an example of $1$ control at low temperature and $2$ control at higher temperature.

Predict the major product for the following reaction:
$Cyclopentyl-ethene \xrightarrow[(ii) H_2O_2, NaOH]{(i) B_2H_6(THF)} ?$

Alkene $(x) \xrightarrow[\Delta]{KMnO_4/H^+} \text{Products}$. Identify the alkene $(x)$ from the given products:

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The decreasing order of reactivity of the following alkenes towards $HBr$ addition is:
$(I) \ CH_3-CH=CH_2$
$(II) \ CF_3-CH=CH_2$
$(III) \ MeOCH=CH_2$
$(IV) \ CH_3-CO-CH=CH_2$

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The major product formed in the following reaction is :$CH_3CH=CHCH(CH_3)_2 \xrightarrow{HBr}$