NEET 2020 Chemistry Question 114 Solution

When 2-bromopentane is heated with alcoholic potassium hydroxide ($$\text{alc. KOH}$$), the base abstracts a proton from a $$\beta$$-carbon while the bromide leaving group departs from the $$\alpha$$-carbon: $$\text{CH}_3-\text{CH(Br)}-\text{CH}_2-\text{CH}_2-\text{CH}_3 \xrightarrow{\text{alc. KOH, } \Delta} \text{CH}_3-\text{CH}=\text{CH}-\text{CH}_2-\text{CH}_3 + \text{KBr} + \text{H}_2\text{O}$$ Because hydrogen is abstracted from the $$\beta$$-position, this is universally termed a $$\beta$$-elimination reaction (Statement a).

2-Bromopentane has two different $$\beta$$-carbons: $$\beta_1$$ ($$\text{C}_1$$, methyl group) and $$\beta_2$$ ($$\text{C}_3$$, methylene group). According to Zaitsev's rule, in dehydrohalogenation reactions, the preferred product is that alkene which has the greater number of alkyl groups attached to the doubly bonded carbon atoms. Hence, pent-2-ene (disubstituted alkene, 81%) is formed as the major product over pent-1-ene (monosubstituted alkene, 19%). Thus, it follows Zaitsev's rule (Statement b).

- Dehydrohalogenation: Elimination of a hydrogen halide ($$\text{HX}$$, here $$\text{HBr}$$) from an alkyl halide. Statement (c) is TRUE. - Dehydration: Elimination of a water molecule ($$\text{H}_2\text{O}$$) from an alcohol to form an alkene. Here, $$\text{HBr}$$ is lost, not $$\text{H}_2\text{O}$$. Thus, Statement (d) is FALSE.

Statements (a), (b), and (c) are all correct descriptions, corresponding to Option 1.