In the structure of 3-bromo-2,2,3-trimethylpentane, what is the total number of sets of chemically n — Haloalkanes and Haloarenes Chemistry Question

💡 Solution & Explanation

We must systematically analyze the $\beta$ -hydrogens relative to the $\alpha$ -carbon bearing the bromine atom. Structure: $C^1(H_3) - C^2(CH_3)_2 - C^3(Br)(CH_3) - C^4(H_2) - C^5(H_3)$ The $\alpha$ -carbon is C3. The adjacent $\beta$ -carbons are C2, C4, and the methyl group directly attached to C3. 1) C2 is a quaternary carbon bonded to three methyls; it has **zero** $\beta$ -hydrogens. 2) The methyl group on C3 has 3 equivalent hydrogens. Due to free rotation around the C-C bond, these represent **one set** of equivalent $\beta$ -hydrogens. 3) C4 is a methylene ($CH_2$) group with 2 hydrogens. We must check the stereochemistry of C3. C3 is bonded to four different groups: $-Br$, $-CH_3$, $-C(CH_3)_3$, and $-CH_2CH_3$. Since C3 is a chiral center, the two protons on the adjacent C4 carbon are diastereotopic. Diastereotopic protons are chemically and magnetically non-equivalent. Therefore, the two protons on C4 represent **two distinct sets** of non-equivalent $\beta$ -hydrogens. Total sets of non-equivalent $\beta$ -hydrogens = $0 + 1 + 2 = 3$.