Question

Classify the isomers of alcohols in $C_5{H}_{12}O$ as primary, secondary and teritiary alcohols.

Answer 1

$1.$ Organic compounds containing substituents from Group $C$ are named following this sequence of steps, as indicated on the examples below:

Step $1$: Find the longest continuous carbon chain. Determine the root name for this parent chain. In cyclic compounds, the ring is usually considered the parent chain, unless it is attached to a longer chain of carbons; indicate a ring with the prefix "cyclo" before the root name.

Step $2$: Number the chain in the direction such that the position number of the first substituent is the smaller number. If the first substituents from either end have the same number, then number so that the second substituent has the smaller number, etc.

Step $3$: Determine the name and position number of each substituent.

Step $4$: Indicate the number of identical groups by the prefixes di, tri, tetra, etc.

Step $5$: Place the position numbers and names of the substituents groups, in alphabetical order, before the root name. In alphabetizing ignore prefixes like sec-, tert-, di, tri, etc., but include iso and cyclo. Always include a position number for each substituents, regardless of redundancies.

Eight isomeric alcohols are possible:

$1.$ $pentan-1-ol\left(1^o\right)$ as shown in IMAGE $01$

$2.$ $2-Methylbutanol\left(2^o\right)$ as shown in IMAGE $02$

$3.$ $3-methylbutan-1-ol\left(1^o\right)$ as shown in IMAGE $03$

$4.$ $2,2-Dimethylpropan-1-ol\left(1^o\right)$ as shown in IMAGE $04$

$5.$ $pentan-2-ol\left(1^o\right)$ as shown in IMAGE $05$

$6.$ $3-methylbutan-2-ol\left(2^o\right)$ as shown in IMAGE $06$

$7.$ $pentan-3-ol\left(3^o\right)$ as shown in IMAGE $07$

$8.$ $2-methylbutan-2-ol\left(3^o\right)$ as shown in IMAGE $08$