Q.Which of the carbon atoms present in the molecule given below are asymmetric?
$\mathrm{\overset{a}{\underset{\phantom{x}}{HOOC}}-\overset{b}{CH(OH)}-\overset{c}{CH(OH)}-\overset{d}{CHO}}$
(carbon atoms labelled a, b, c, d from the carboxylic-acid carbon to the aldehyde carbon; in the Exemplar the molecule is drawn expanded, with the OH/H pairs shown above and below carbons b and c)

The key idea is Optical Isomerism — specifically, identifying chiral (asymmetric) carbon atoms. An asymmetric carbon is one bonded to four different substituents.

Step 1: Label each carbon in the given molecule:

$\mathrm{\overset{a}{HOOC}-\overset{b}{CH(OH)}-\overset{c}{CH(OH)}-\overset{d}{CHO}}$

Step 2: Check each carbon:

• Carbon a (carboxylic acid carbon): bonded to two oxygens (one via double bond), an OH, and the rest of the chain — it is $sp^2$ hybridised, not tetrahedral, so not asymmetric.
• Carbon b: bonded to H, OH, COOH (on left), and CH(OH)CHO (on right) — all four groups are different → asymmetric. …

The molecule is $\mathrm{\overset{a}{HOOC}-\overset{b}{CH(OH)}-\overset{c}{CH(OH)}-\overset{d}{CHO}}$. An asymmetric carbon is an $sp^3$ carbon bonded to four different groups.

• Carbon a ($\mathrm{-COOH}$): $sp^2$ carbonyl carbon — not a chiral centre.
• Carbon b ($\mathrm{CH(OH)}$): bonded to $\mathrm{H}$, $\mathrm{OH}$, $\mathrm{-COOH}$ and $\mathrm{-CH(OH)CHO}$ — four different groups ⇒ asymmetric.
• Carbon c ($\mathrm{CH(OH)}$): bonded to $\mathrm{H}$, $\mathrm{OH}$, $\mathrm{-CHO}$ and $\mathrm{-CH(OH)COOH}$ — four different groups ⇒ asymmetric. …

Concept: Chirality and Asymmetric Carbon Atoms

An asymmetric carbon (chiral centre) is a carbon atom bonded to four different substituents. To identify them, check each carbon for four distinct groups attached.

Method: Substituent Comparison Method

Steps:

1. Draw the molecule with all bonds and atoms clearly shown

   The given molecule is:

   $\mathrm{HOOC - CH(OH) - CH(OH) - CHO}$

   Expanded form:

   • Carbon a: $\mathrm{-COOH}$ (carboxylic acid carbon)
   • Carbon b: $\mathrm{-CH(OH)-}$ (with H and OH)
   • Carbon c: $\mathrm{-CH(OH)-}$ (with H and OH)
   • Carbon d: $\mathrm{-CHO}$ (aldehyde carbon)

2. Check each carbon for four different groups

   • Carbon a (carboxylic acid carbon):

     Attached to $\mathrm{-OH}$, $\mathrm{=O}$ (double bond to oxygen), and $\mathrm{-C(b)}$. Because of the double bond to oxygen it is sp² hybridised, not tetrahedral.

     → Not asymmetric (cannot have four different groups in tetrahedral geometry)

   • Carbon b: …

Common Mistakes & How to Avoid Them

Mistake 1: Thinking the aldehyde carbon (d) is asymmetric

Why students make this mistake:

They see the aldehyde group ($-\mathrm{CHO}$) and think the carbon is bonded to four different groups because it has a double bond to oxygen.

The correct reasoning:

Carbon d is $\mathrm{sp^2}$ hybridised (trigonal planar, double-bonded to O). Asymmetric carbons must be $\mathrm{sp^3}$ hybridised (tetrahedral) with four different substituents. A carbon with a double bond cannot be a chiral centre.

How to avoid:

• Check hybridisation first: if the carbon has a double or triple bond → not asymmetric.
• Remember: chiral carbons are always $\mathrm{sp^3}$ with four single bonds.

Mistake 2: Thinking the carboxylic acid carbon (a) is asymmetric

Why students make this mistake:

They see four bonds (C–OH, C=O, C–C, and a lone pair or hydrogen) and assume four different groups.

The correct reasoning:

Carbon a is also $\mathrm{sp^2}$ hybridised (carbonyl carbon of $-\mathrm{COOH}$). It has a double bond to oxygen, so it is planar, not tetrahedral. No chiral centre possible.

How to avoid:

• Memorise: Carbonyl carbons (in $-\mathrm{CHO}$, $-\mathrm{COOH}$, $-\mathrm{COOR}$, $-\mathrm{COR}$, $-\mathrm{CONH_2}$) are never asymmetric.
• Draw the structure: if you see C=O, that carbon is out.

Mistake 3: Missing that both b and c are asymmetric

Why students make this mistake:

Some think only one of the two middle carbons is chiral, or that they are identical and therefore not chiral.

The correct reasoning:

• Carbon b: bonded to $-\mathrm{COOH}$, $-\mathrm{OH}$, $-\mathrm{H}$, and $-\mathrm{CH(OH)CHO}$ → four different groups → asymmetric.
• Carbon c: bonded to $-\mathrm{CH(OH)COOH}$, $-\mathrm{OH}$, $-\mathrm{H}$, and $-\mathrm{CHO}$ → four different groups → asymmetric.
• They are not identical because the groups on either side are different.

How to avoid:

• For each carbon, list all four substituents explicitly.
• If any two are the same (e.g., two H atoms, two CH$_3$ groups), it's not chiral. …