Chem P4 HELP !!

[ammarelahi]

Can anyone explain me N09 P42 Q5 part (a) ???

 

[ammarelahi]

anybody ??

 

[kahynickel]

I am writting the answer...........it's lengthy......... please wait........

 

[WellWIshER]

CAN ANY ONE PLEASE TELL ME NO9 41 Q1 PART D IV????

wts up with da ratio concept i neva did it be4

please help

thanx

 

[ammarelahi]

waiting....

 

[WellWIshER]

please someone ans my quesiton

 

[kahynickel]

Sorry I did not write due to Electricity problem. Here is the answer.

The important point here is to use the PLANE OF ATOMS and the rotation around the axes as the question requires.

compound A: This resembles benzene since it only has C=C bonds and C-H bonds. Here the rotation around the axes is restricted making all C-atoms to lie at the same plane. Therefore coplanar is the answer.

compound B: This partially resembles benzene the other ring does not have C=C bonds. C-C bonds are instead present in the hexagon condensed with benzene ring. Therefore the PLANE of all the Carbon atoms is not same. so all the C-atoms are not coplanar.

compound C: It has 3-dimensional shape so its coplanar. Compound D also has the 3-dimensional shape factor. so coplanar
and compound E are all coplanar C-atoms due to rotation factor.

 

[ammarelahi]

thanks but what does 3d shape has to do with rotation ??

 

[kahynickel]

Don't deal 3-D shape with rotation. Here Rotation apply for i) molecules having sigma bonds only ii) molecules having pi-bonds and sigma bonds.

When a sigma bond is formed the orbitals undergo maximum rotation around its axes thus producing a strong bond (single bond like H-H, C-C). Sigma bonds such as in C2H6 rotate around their axes since they are very strong they are not broken. Another example is H3CCH2CH2CH3 in which three C-C bonds are present. They are constantly rotating around the axes and it does not effect the chemical properties of this molecule.

While in pi-bonds there involves side-way overlapping orbitals do not undergo maximum rotation around the axes and a weak bond results. In ethene C2H4 there is thus a restricted rotation around the C=c bonds unlike a C-C bond in an alkane. Since pi-bonds are weak they may break during rotation. From here arises the CIS-TRANS isomerism if u recall.

Now bearing this in mind consider the molecules in the question.

There is another APPROACH to solve this question but that is not required by the question- the hybridization approach.
But unfortunately that was asked in the very next session i.e in June 10/42 Q.5(a) and most of the candidates had no answer because the most teachers did not cover this aspect. So solve this paper.

 

[ManishAdhikari]

Think about r0tating it. Benzene is itself planar and if carbon atoms are joined to it directly it must lie on the same plane as carbon atoms in the ring. Cyclic non aromatic can never be made to have all Carbon atoms in the same plane. But if there are straight chains the carbon atoms can be made to lie on same plane. Think about one plane and rotate bond freely until all are in it. for the one with oxygen one oxygen can be above planes of two rings and one can be below and because all the C-O bonds have equal length the planes must coincide.

 

[ammarelahi]

what is the other approach ?