Assignment 3 hhwhwkwlwlqlqlp1l2jehegrkek

Well

1.  Altogether, the free radical chlorination of S-1-chloro-2-
methyl butane gave six fractions of formula C5H10Cl2. Four
fractions were found to be optically active, and two
fractions optically inactive. Draw structural formulas for
the compounds making up each fraction. Account in detail
for optical activity or inactivity in each case.

Solution
(1) Cl—CH2—CH—CH2—CH2—CH3 → Optically active
Cl
↓

Cl
H—C— H
Cl—C*— H
H—C— H
H—C— H
H—C— H
H

2. This compound is optically active because it has a chiral
center (C2) that enable it to rotate the plane polarized
light either towards left or towards right. Also there is no
plane of symmetry, so no meso-form in this compound.

(2) Cl—CH2—CH2—CH—CH2—CH3 → Optically active
Cl
↓

Cl
H—C— H
H—C— H
H—C*—Cl
H—C— H
H—C— H
H

This compound is optically active because it has a chiral
center (C3) that enable it to rotate the plane polarized
light either towards left or towards right. Also there is no
plane of symmetry, so no meso-form in this compound.

3. (3) Cl—CH2—CH2—CH2—CH—CH3 → Optically active
Cl
↓

Cl
H—C— H
H—C— H
H—C—H
H—C*—Cl
H—C— H
H

This compound is optically active because it has a chiral
center (C4) that enable it to rotate the plane polarized
light either towards left or towards right. Also there is no
plane of symmetry, so no meso-form in this compound.

4. (4) Cl—CH2—CH2—CH2—CH2—CH2—Cl → Optically inactive

↓

Cl
H—C— H
H—C— H
H—C—H Plane of symmetry
H—C—H
H—C— H
Cl

This compound is optically inactive because it has not any
chiral atom in its structure, so it can not rotate the plane
polarized light. Also it has plane of symmetry at C3, so
there is meso-form which makes the compound optically
inactive.

5. (5) Cl—CH—CH2—CH2—CH2—CH3 → Optically inactive
Cl
↓
Cl
H—C—Cl
H—C— H
H—C—H
H—C—H
H—C— H
H

This compound is optically inactive because it has not any
chiral atom in its structure, so it can not rotate the plane
polarized light.

6. (6) Cl—CH2—CH—CH—CH 3 → Optically active
Cl CH3
↓

Cl
H—C—H
H—C*—Cl
H—C—CH3
H—C—H
H

This compound is optically active because it has a chiral
center (C2) that enable it to rotate the plane polarized
light either towards left or towards right. Also there is no
plane of symmetry, so no meso-form in this compound.