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

Similar Questions

$A$ compound with the formula $C_4H_{10}O$ reacts with sodium and upon oxidation gives a carbonyl compound that does not reduce Tollen's reagent. What is the original substance?

$Ph-C(OH)(Ph)-C(OH)(CH_3)_2$ $(A)$ $+$ $Ph-C(OH)(Ph)-C(OH)(Et)_2$ $(B)$ $\xrightarrow{H_2SO_4}$
$(p)$ $Ph-C(Ph)(CH_3)-CO-CH_3$
$(q)$ $Ph-C(Ph)(Et)-CO-Et$
$(r)$ $Ph-C(Ph)(Et)-CO-CH_3$
$(s)$ $Ph-C(Ph)(CH_3)-CO-Et$
When $(A)$ and $(B)$ react with $H_2SO_4$,the products obtained are:

The compound which gives the most stable carbocation on dehydration is:

When $3, 3-$dimethyl$-2-$butanol is heated with ${H_2SO_4}$,the major product obtained is:

Distinction between primary,secondary and tertiary alcohol is done by

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