The total number of alcohols (including stetroisomers ) possible with the molecular formula $C_{5} H_{1} 2 O$ is:-

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Solution

There are $8$ alcohols possible for $C_{5} H_{12} O$
(1) $1 - p e n t a n o l$
$C H_{3} C H_{2} C H_{2} C H_{2} C H_{2} O H$
(2) $2 - p e n t a n o l$
$C H_{3} - C H_{2} - C H_{2} - C H | - C H_{3}$
$O H$
(3) $3 - p e n t a n o l$
$C H_{3} - C H_{2} - C H | - C H_{2} - C H_{3}$
$O H$
(4) $2 - m e t h y l b u t a n - 1 - o l$
(5) $2 - m e t h y l b u t a n - 2 - o l$
(6) $3 - m e t h y l b u t a n - 1 - o l$
(7) $3 - m e t h y l b u t a n - 2 - o l$
(8) $2, 2 - d i m e t h y l p r o p a n o l$

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