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Question

(D) The photobromination of alkanes involves the formation of free radicals. The stability of free radicals follows the order: $3^\circ > 2^\circ > 1^

Vedclass · Accepted Answer

$2$-bromo-$2$-methylbutane

Vedclass · Answer

(D) The photobromination of alkanes involves the formation of free radicals. The stability of free radicals follows the order: $3^\circ > 2^\circ > 1^\circ$. In $2$-methylbutane $(CH_3-CH(CH_3)-CH_2-CH_3)$,there are four types of hydrogen atoms. Abstraction of the hydrogen atom from the $C-2$ position leads to a $3^\circ$ free radical,which is the most stable. Therefore,the reaction proceeds via the $3^\circ$ free radical intermediate to form $2$-bromo-$2$-methylbutane as the major product.

Which product is the major product in the photobromination of $2$-methylbutane?

Similar Questions

Give the methods of preparation of alkanes. Write the general reactions.

The complete combustion of $CH_4$ gives

$C_2H_6 \xrightarrow{450^{\circ}C} C_2H_4 + H_2$
The above reaction is called:

Different hydrogen atoms in the compound are represented by alphabets:
$CH_3(A)-CH=CH(B)-CH_2(C)-CH_2(D)-CH(E)(CH_3)_2(F)$
Arrange them in decreasing order of reactivity towards free radical substitution.

Complete the following reactions:
$1. CH_4 + I_2 \rightleftharpoons ?$
$2. CH_3-CH_3 + Cl_2 \xrightarrow{h\nu} ?$

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