Draw Dash and$/$or Bond$-$Line Formulas for all the Constitutional Isomers for all $17$ alkyl halides with the molecular formula $C_6{H}_{13}Cl$ : The $6,5$ & $4-$Carbon$-$Chain Templates are Drawn. Add one Chlorine Atom to each consecutively different C$-$Atom in the Template Bond$-$Line Structures $1-17$

$10\mathrm{.}1$ Name the Isomers;

$10\mathrm{.}2$ Classify the Isomers within the classes of Primary $(1),$ Secondary $(2),$ & Tertiary $(3)$ Alkyl Halid

$10\mathrm{.}3$ Identify the Number of Alkyl Halides of each class, namely:

$10\mathrm{.}3\mathrm{.}1\frac{?}{?}$# $of$ Primary $(1)$ Alkyl Halides $=$

$10\mathrm{.}3\mathrm{.}2\frac{?}{?}$# $of$ Secondary $(2)$ Alkyl Halides $=$

$10\mathrm{.}3\mathrm{.}3\frac{?}{?}$# $of$ Tertiary $(3)$ Alkyl Halides $=$

$10\mathrm{.}3\mathrm{.}4$ Total $\frac{?}{?}$# $of$ Alkyl Halides$\frac{}{b}=ar(17)$

Procedure: Complete using the C$-$C Templates $-$ the straight carbon and branched carbon chains are drawn.

A$.$ Consider a $6-$Carbon Chain and replace $1H$ for $1Cl$ atom for each set of$,$ Chemically Equivalent $H-$aton Bond$-$line$-$structure

Name Class $($e$.$g$.1,2.$

$1.1-$chlorohexane Primary $2-$chlorohexane Pripary

$2.$

$3.$ obrace$({)}^{Cl}3-$chlorohexane secondary Bond$-$line$-$structure

Name

Class e$.$g$.($:$1,2,3\}$

$9.$

$10.$

$11.$

$12.$

Consider a $4$ Carbon$-$Chain, add $5^{th}$ and $6^{th}C-$atoms, and replace $H-$atoms with $Cl-$atoms as above.

$13.$

$13.$

B$.$ Consider a $5C-$Chain, add a $6$ th $C-$atom, and replace $1H$ for $1Cl$ atom for each set of$,$ Chem$-$Eq$-$H$-$atom $13.$

$4.$

$5.$

$6.$

$7.$

F$.$ Total # of Primary: $,1_{},$ Alkyl Halides $=$

G$.$ Total # of Secondary: $\frac{2}{3},$ Alkyl Halides $=$

H$.$ Total # of Tertiary: $,3,$ Alkyl Halides $=$

$8.$

$7$