Introduction to Organic Chemistry: Question 10

Worked solution

Part (a): General formula of the monochloroalkanes

Replacing exactly one hydrogen atom of a saturated, non-cyclic alkane, $\text{C}_n\text{H}_{2n+2}$, with a chlorine atom removes one H and adds one Cl, giving the general formula:

$\text{C}_n\text{H}_{2n+1}\text{Cl}$

For $n=4$: $2n+1 = 9$, giving $\text{C}_4\text{H}_9\text{Cl}$. Exactly the molecular formula shared by J, K, L and M, confirming all four are consistent with a four-carbon, single-chlorine, saturated, non-cyclic chloroalkane.

Part (b): Classifying the isomerism between J & K, and J & M

J and K: $\text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{Cl} \ (\text{J}) \qquad \text{CH}_3\text{CH}_2\text{CH(Cl)CH}_3 \ (\text{K})$

Both compounds share the same unbranched, four-carbon skeleton. What differs is where the chlorine atom sits along that chain (at the end, C1, in J; at C2 in K). Because the carbon skeleton is identical and only the position of the substituent changes, this is positional isomerism.

J and M: $\text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{Cl} \ (\text{J}) \qquad (\text{CH}_3)_3\text{CCl} \ (\text{M})$

Here the carbon skeleton itself is different: J is an unbranched chain of four carbons, while M has a branched skeleton, a central carbon bonded to three methyl groups. Because the underlying skeleton differs (straight vs branched), rather than just the position of the chlorine on an identical skeleton, this is chain isomerism.

Part (c): Which isomer contains a chiral centre?

A carbon is a chiral centre only if it is bonded to four different atoms or groups. Testing the carbon bearing the chlorine atom (or, where relevant, the branch-point carbon) in each isomer:

• J, $\text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{Cl}$: the chlorine-bearing carbon (C1) is bonded to Cl, $\text{-CH}_2\text{CH}_2\text{CH}_3$, and two identical hydrogen atoms. Two of the four groups are the same, so C1 is not a chiral centre.
• K, $\text{CH}_3\text{CH}_2\text{CH(Cl)CH}_3$: the chlorine-bearing carbon (C2) is bonded to Cl, H, a methyl group ($\text{-CH}_3$, from C1) and an ethyl group ($\text{-CH}_2\text{CH}_3$, from C3-C4). All four groups (Cl, H, methyl, ethyl) are different, so C2 is a chiral centre.
• L, $(\text{CH}_3)_2\text{CHCH}_2\text{Cl}$: the branch-point carbon is bonded to two identical methyl groups, one hydrogen atom, and $\text{-CH}_2\text{Cl}$. Two of the four groups are the same, so this carbon is not a chiral centre (nor is the $\text{CH}_2\text{Cl}$ carbon, which carries two identical H atoms).
• M, $(\text{CH}_3)_3\text{CCl}$: the central carbon is bonded to three identical methyl groups and Cl. Three of the four groups are the same, so this carbon is not a chiral centre.

Only K (2-chlorobutane) contains a chiral centre.

Final answers

• (a) $\text{C}_n\text{H}_{2n+1}\text{Cl}$; consistent with $\text{C}_4\text{H}_9\text{Cl}$ since $2(4)+1=9$
• (b) J & K: positional isomerism; J & M: chain isomerism
• (c) Only K (2-chlorobutane) has a chiral centre (C2, bonded to Cl, H, methyl, ethyl, all different)