How many monobrominated product(s) (including | vedclass

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(C) The free radical bromination of $n$-butane $(CH_3-CH_2-CH_2-CH_3)$ can occur at two different types of carbon atoms:$1$. At the terminal carbon ($

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$3$

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(C) The free radical bromination of $n$-butane $(CH_3-CH_2-CH_2-CH_3)$ can occur at two different types of carbon atoms:$1$. At the terminal carbon ($C_1$ or $C_4$),it forms $1$-bromobutane $(CH_3-CH_2-CH_2-CH_2Br)$. This product is achiral.$2$. At the internal carbon ($C_2$ or $C_3$),it forms $2$-bromobutane $(CH_3-CHBr-CH_2-CH_3)$. This product has a chiral center at $C_2$,resulting in a pair of enantiomers ($R$ and $S$ forms).Therefore,the total number of monobrominated products including stereoisomers is $1$ (from $1$-bromobutane) + $2$ (from $2$-bromobutane) = $3$ products.

How many monobrominated product$(s)$ (including stereoisomers) would form in the free radical bromination of $n$-butane?

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