Question

For the molecule CH3CH(OH) CH(OH)CH(OH) CH3 total number of stereoisomers and meso form respectively are?
pls explain with structures

Answer 1

Answer:Trihydroxyglutaric

Explanation:

Trihydroxyglutaric acid (HO2C-CHOH-CHOH-CHOH-CO2H) exists in four stereoisomeric forms, two of which are optically acitve while the other are meso-forms. Reason: It contains two asymmeric and one pseudo-asymmeric carbon atom.

Answer 2

A total number of stereoisomers and meso form  are 6 and 4 respectively.

Explanation:

• For the molecule, $H3C-CH(OH)-CH(OH)-CH3$ the number of chiral carbon atoms is two and they are symmetrical. so $n=2$
• In this molecule the number of d and l isomers are $2^{n-1} =2^{2-1}=2$
• The number of meso forms in the given molecule is $2^{n/2 -1} =2^{2}=4$
• Total number of stereo isomers= number of d and l isomers plus the number of meso forms$= 2+4 =6$
• In the given molecule six different stereoisomers are found and four meso forms are also present in it.