Chapter 4 : Stereochemistry 1st Year
Page : 11 
Organic Chemistry

Typical problems with their Solutions

Stereochemistry Problem 1 :

Calculate the d- and l- isomers formed by the following compound and also give the number of meso-forms.


Solution :

Structure shows that the compounds can be divided into two equal halves and it consists five asymmetric carbon atoms i.e. odd number of asymmetric carbon atoms.

Hence, the number of d- and l- isomers( a) = 2(n-1) – 2(n/2-0.5) = 2(5-1) – 2(5/2–0.5)

= 24-2(2.5-0.5) = 24-22 = 12

and number of meso forms (m) = 2(n/2 – 0.5) = 2(5/2-0.5) = 2(2.5-0.5) = 22 = 4

Thus, d- and l-isomers = 12, meso forms = 4, and total optical isomers = 16

Stereochemistry Problem 2:

What is criterion of optical activity in organic compounds? Which of the following compounds show optical isomerism.

Solution :

Conditions for optical activity: An organic compounds of the type C*abcd, in which the atoms or groups attached to the carbon are all different, can show optical isomerism and therefore it is optical active. But as soon as two or more atoms or groups attached to carbon become identical, optical isomerism disappears. T) A molecule containing an asymmetric carbon atom exhibits optical activity. Now a days, a new term chiral carbon is used for asymmetric C-atom. (UI) A molecule that can not be superimposed on its mirror image is said to be chiral (Greek; heir = handedness or dissymmetric. Thus, a molecule is said to be dissymmetric when one or more of its carbon atoms are linked with four different atoms or groups and also it should not have any element of symmetry (like plane of symmetry, Centre of symmetry and alternative axis of symmetry).

Thus, the compounds (0) (CH3CH(Br)CH3 and (ii) (CH3)2CHCH2CH3 do not show optical isomerism because they do not have asymmetric carbon atom while the compound (iii) CH3CH(OH)CH3 shows optical isomerism.

Stereochemistry Problem 3:

Explain the difference between racemic and meso compounds giving suitable example.

Solution :

Differences between racemic and meso tartaric acid Racemic Tartaric acid.

Stereochemistry Problem 4:

Why maleic acid dehydrates at 130°C while fumaric acid does not ?

Solution :

Maleic acid on heating at 130°C dehydrates to give maleic anhydride because two bulky-COOH groups are present in this molecule on the same side which causes crowding and hence steric effect results cyclisation by removal of a water molecule. The fumeric acid on the other hand does not fort anhydride at this temperature because both of its -COOH groups lie on the opposite sides which results there is no interaction occur between them and therefore it provide more stability of this molecule.

Stereochemistry Problem 5 :

Assign reason the trans-isomers of alkenes are usually more stable than the corresponding cis-isomers.

Solution :

The trans-isomers of alkenes are most stable than their corresponding cis-isomers because in the cis- isomer two bulky alkyl groups are lying very closer to each other which interact one another and make this isomer less stable .Whereas in trans-isomer bulky groups are lying far apart and consiguent they exert less interaction to one another and therefore they provide more stability to this isomer.For example cis-butene-2 is less stable than trans-butene-2.

Problem 5 :

Write the type of isomerism exhibited by each of the following compounds. Give the structure and the name of each isomers. (DDUGU)


Solution :

In the above compounds, ethyl acetoacetate ,CH,COCH,COOC,H, exhibits keto-enol tautomeism. This somerism arises due to transfer of an acidic hydrogen atom from a-carbon atom to the carbonyl oxygen.

On the other hand compound (II) but-2-enedioic acid exhibits cis-trans isomerism because this compound consists a carbon -carbon double bond and on each carbon atom of carbon-carbon double, two different atom or group are attached.

Problem 6:

Which of the following compounds have greater value of dipole moments and why?

Solution :

In such alkenes which have one electron withdrawing and one electron releasing group on either side of the carbon-carbon double bond, have greater value of dipole moments because, the bond moments are fully additive. In such cases a trans isomer has a higher value of dipole moment as compared to the cisisomer. In cis isomer, bond moments developed by C-Cl. bond is decreased by the activating nature of electron releasing; (-CH,) group. Thus the trans 1-chloro propene has higher value of dipole moment than cis-1 chloropropene.

Problem 7:

What is the differences between d, l, dl and meso tartaric acids? Explain. (DDUGU)

Solution :

Differences between d, I, di and meso tartaric acids :

Problem 8:

why the dipole moment (u) of cis-1-2 dichloride ethane is 1.85D while the trans 1-2 dichloride ethane has u = 0.0D? (DDUGU)

 Solution :

The dipole moment of cis -isomer is higher than that of the trans -isomer because in trans-dichroic ethane the chlorine atoms being on opposite sides, cancel the bond moments whereas in the case of cis-dichloride ethane, the bond moments are additive rather than subtractive.

Problem 9 :

Write R or S name of the following configuration.(DDUGU) ·


Problem 10 :

Which of the following compounds will exhibit geometrical isomerism ? Draw the Z and E geometrical isomers where possible (DDUGU)

Solution :

Compound (i) 2-bromo-1-chloro-propene;CI.CH=C.BrCH, only exhibits geometrical isomerism. On the other hand, compounds (ii) and (iii) do not. Two possible geometrical Z and E isomers of compound (i) are :

According to sequence rule:

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