Q.Name the following: How many monochlorination products are possible for
a. 2-methylpropane? b. 2-methylbutane?
Draw their structures and write their IUPAC names.

Step 1. 2-Methylpropane, (CH3)3CH. This molecule has only two chemically distinct types of hydrogen: the 9 equivalent hydrogens on its three identical methyl groups (primary), and the 1 hydrogen on the central carbon (tertiary). Replacing a primary H gives 1-chloro-2-methylpropane, (CH3)2CH-CH2Cl; replacing the tertiary H gives 2-chloro-2-methylpropane, (CH3)3C-Cl. So exactly 2 distinct monochlorination products are possible.

Step 2. 2-Methylbutane, (CH3)2CH-CH2-CH3. Numbering the longest (butane) chain C1-C2-C3-C4 with the methyl branch on C2, there are FOUR distinct hydrogen environments: (i) the two equivalent methyl groups on C2 (C1 and the branch methyl are constitutionally equivalent to each other), giving 1-chloro-2-methylbutane on substitution;

(ii) the single tertiary H on C2 itself, giving 2-chloro-2-methylbutane;

(iii) the CH2 hydrogens on C3, giving 2-chloro-3-methylbutane (renumbered from the other end for the lowest locant set);

(iv) the terminal CH3 hydrogens on C4, giving 1-chloro-3-methylbutane. …

Identifying every chemically distinct (constitutionally non-equivalent) hydrogen environment in each starting alkan …

• Under-counting 2-methylbutane's products by treating C1 and the C2-branch methyl as giving two DIFFERENT products, rather than recognising they are equivalent and give the SAME 1-chloro-2-methylbutane. …