Alkyl Halides - Result Question 54
####54. In the following monobromination reaction, the number of possible chiral product(s) is (are)…
$$ H \xrightarrow[CH _3]{\stackrel{CH _2 CH _2 CH _3}{-} Br \quad \xrightarrow[300^{\circ} C]{Br _2(1.0 \text { mole })}} $$
(1.0 mole) (Enantiomerically pure)
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Solution:
- Given compound undergoes free-radical bromination under given conditions, replacing $H$ by $Br$.
$C^{*}$ is chiral carbon.
I. Chiral
Achiral
III. Chiral
Chiral
V. Chiral
(III) has two chiral centres and can have two structures.
(III) A
(III) B (IV) has also two chiral centres and can have two structures.
It has plane of symmetry thus, achiral.
Thus, chiral compounds are five. I, III A, III B, IV B and V.
