JEE Advanced 2016 · Paper 1 · Chemistry Q36

ChemistryInteger+3Question 36 of 54

In the following monobromination reaction, the number of possible chiral products is

(Enantiomerically pure 2-bromopentane, 1.0 mole, shown as a Fischer projection) $\xrightarrow[300\ ^\circ\mathrm C]{\ce{Br2}\ (1.0\text{ mole})}$

Your answer

Answer and worked solution

Official answer: 5

Start. Optically pure 2-bromopentane, $\ce{CH3-CHBr-CH2-CH2-CH3}$ . Free-radical monobromination can replace an H at each carbon.

Site	Product	Chiral products
C1	1,2-dibromopentane (C2 untouched)	1
C2	2,2-dibromopentane	0 (achiral)
C3	2,3-dibromopentane, two diastereomers (no meso: the ends differ)	2
C4	2,4-dibromopentane: one chiral and one meso	1
C5	1,4-dibromopentane	1

Total: 5 chiral products.

Answer: 5

Topics:Optical isomerism and chirality Alkanes

JEE Advanced 2016 · Paper 1, Chemistry question 36