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Highlight, by clicking on, the asymmetric carbons (if any) in each structure. A selected atom will turn green. H3C- CH3 CH3 CI CI Indicate whether each compound contains a plane of symmetry (internal mirror plane) or not. o yes yes no no no cannot cannot cannot determine determine determine There is additional feedback available! View this feedback by clicking on the bottom divider bar. Click on the divider bar again to hide the additional feedback. Close

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Answer #1
Concepts and reason

Chirality is the fundamental behind the optical activity of an organic molecule.

Chiral center: A‘C’ atom in a molecule becomes a chiral center when all the four valences of that atom are satisfied by chemically different groups. For a carbon to act as chiral center, it should be sp
hybridization

The absence of symmetry is known as asymmetry. The carbon atom bonded to four unlike groups is called the asymmetric carbon which is also known as the chiral carbon.

In a non-chiral compound, the plane of symmetry is present and it does not have a chiral center.

The plane of symmetry: The plane of the mirror that cuts the compound into two equal halves.

Fundamentals

Representative example for chiral center and optical isomers is given below:

Mirror
B-CD
A
D-C
-B
A
optical isomers

Here C is bonded to four different groups A, B C and D, Thus, it forms two optical isomers as the mirror images of each other

(1)

Assume the given structures as follows:

Cн
н
н
CCH
Н,с-
C
CH3
Structure 3
Structure 2
Structure

Structure 1:

н
C-
Нс-
-C-
-сCH,
C indicates Asymmetric carbons

Structure 2:

сн,
C indicates Asymmetric carbons

Structure 3:

CH3
No Asymmetric carbons

There is no asymmetric carbon in methylcyclohexane.

(2)

If a molecule is not chiral, then it has a plane of symmetry.

Structure 1:

н
н
, chiral center
Н,с—с.
-CH
two chiral center
four isomers possible (only two isomers contains the plane of symmetry)

Structure 2:

CH3
one chiral center
No symmetry plane

Structure 3:

CH
Plane of symmetry

Ans: Part 1

The asymmetric carbons in the given structures are indicated as follows:

Structure 2
Structure 3
Structure
Н
Н
Н
ШA
ш
-c-CH3
Hас—
C
CHз
CHз
CI
CI

Part 2

The plane of symmetry element is indicated for the given structures as follows:

Structure
Structure 2
Structure3
Yes
Yes
Yes
No
No
No
Cannot determine
Cannot determine
Cannot determine

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