The number of primary (1^{circ}),secondary (2 | vedclass

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(C) For the molecular formula $C_5H_{12}O$,the possible alcohols are:$1^{\circ}$ Alcohols: $CH_3CH_2CH_2CH_2CH_2OH$ (pentan$-1-$ol),$CH_3CH_2CH_2CH(CH

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$4, 3, 1$

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(C) For the molecular formula $C_5H_{12}O$,the possible alcohols are:$1^{\circ}$ Alcohols: $CH_3CH_2CH_2CH_2CH_2OH$ (pentan$-1-$ol),$CH_3CH_2CH_2CH(CH_3)OH$ (is incorrect,this is $2^{\circ}$),$CH_3CH(CH_3)CH_2CH_2OH$ ($3$-methylbutan$-1-$ol),$CH_3CH_2CH(CH_3)CH_2OH$ ($2$-methylbutan$-1-$ol),and $(CH_3)_3CCH_2OH$ ($2$,$2$-dimethylpropan$-1-$ol). Total $1^{\circ} = 4$.$2^{\circ}$ Alcohols: $CH_3CH_2CH_2CH(OH)CH_3$ (pentan$-2-$ol),$CH_3CH_2CH(OH)CH_2CH_3$ (pentan$-3-$ol),and $CH_3CH(CH_3)CH(OH)CH_3$ ($3$-methylbutan$-2-$ol). Total $2^{\circ} = 3$.$3^{\circ}$ Alcohols: $CH_3CH_2C(CH_3)(OH)CH_3$ ($2$-methylbutan$-2-$ol). Total $3^{\circ} = 1$.Thus,the count is $4, 3, 1$.

The number of primary $(1^{\circ})$,secondary $(2^{\circ})$ and tertiary $(3^{\circ})$ alcohols possible for the formula $C_5H_{12}O$ respectively are

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