**53.** How many different stereoisomers are possible for the given molecule? CH₃–CH–CH=CH–CH₃ | OH Options: - (1) 3 - (2) 1 - (3) 2 - (4) 4

Key ideas you need

1. Chiral (asymmetric) carbon: A carbon attached to four different groups becomes a stereocentre and can exist as $R$ or $S$.
2. E/Z (cis/trans) at a double bond: If each carbon of a $C=C$ has two different substituents, the alkene can exist in two geometrical forms.
3. Total stereoisomers: When more than one independent stereogenic unit exists, the count is the product of possibilities unless internal symmetry makes some forms identical or meso.

Step-by-step thought process

1. Locate stereogenic carbon.
   The second carbon ($C_2$) in $CH_3\text{–}CH(OH)\text{–}...$ is attached to $\{H, OH, CH_3, CH=CH\text{–}CH_3\}$ — all different.
   ⇒ One chiral centre ⇒ $2$ configurations $(R\,/\,S)$.

2. Check the $C=C$ bond.
   $C_3=C_4$ has for $C_3$: $(H, CH(OH)\ldots)$ and for $C_4$: $(H, CH_3)$. Both sides unequal ⇒ $E/Z$ possible ⇒ $2$ configurations.

3. Combine.
   No internal mirror plane (only one chiral centre, so no meso). Thus total stereoisomers

   $2_{(R/S)} \times 2_{(E/Z)} = 4.$

4. Choose answer.
   Option (4) 4 is correct.

What is the question?

We have a tiny zig-zag carbon chain:

CH3—CH(OH)—CH = CH—CH3

We want to know how many different 3-D shapes (stereoisomers) this molecule can have—even though the atoms are joined in exactly the same order.

How to think about it like a 10-year old

1. Bendy joint (chiral centre): Think of the middle carbon with OH as a ball that has four different toys attached. Because every toy is different, the ball can be either a right-hand or a left-hand version (mirror images that can’t be matched). That gives 2 ways.
2. Double-link (double bond): The middle double line is like a stiff stick that can’t spin. The two ends can either be on the same side (cis/Z) or on opposite sides (trans/E). That gives 2 ways.
3. Total ways: Each bendy joint choice combines with each stiff stick choice, so 2 × 2 = 4 different 3-D pictures.

So there are 4 stereoisomers.

Detailed calculation

1. Identify stereogenic centres The carbon marked with $OH$ is attached to $OH$, $H$, $CH_3$, and $CH=CH\text{–}CH_3$.
   ⇒ One chiral centre ⇒ $2$ configurations ($R$ and $S$).

2. Identify geometrical isomerism For the alkene $C_3=C_4$:

   • At $C_3$: substituents are $H$ and $CH(OH)\text{–}CH_3$.
   • At $C_4$: substituents are $H$ and $CH_3$.

   Both ends have two distinct groups ⇒ $E$ and $Z$ possible ⇒ $2$ configurations.

3. Total stereoisomers

$\text{Total} = 2\,(R/S) \times 2\,(E/Z) = 4.$

4. Answer

Option (4) 4.