JEE QUES

The correct statement(s) about the compound given below is (are)

(A) The compound is optically active
(B) The compound possesses centre of symmetry
(C) The compound possesses plane of symmetry
(D) The compound possesses axis of symmetry

26 Answers

11
Subash ·

one ch3 on the other side

1
Grandmaster ·

in place of doing so much work whhy couldn't we go for r/s notation ,which will readily say whether its opticallty active or not..

if found optically in active now look for various symmetry.

e.g. its R for both carbons threrefore its opt. active

simple

one more thing do anyone know whether optically active compounds posses any

(B) The compound possesses centre of symmetry
(C) The compound possesses plane of symmetry
(D) The compound possesses axis of symmetry

33
Abhishek Priyam ·

yes.. thats true but this was not about the answer but the method convert 3d to fischer and to see the elements of symmetry... [1]

1
skygirl ·

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wenevr a molecule is optically active... jus forget

>> plane of symmatry

>> centre of symmetry

>> 'alternate' axis of symmetry

but..... there CAN BE 'AXIS OF SYMMETRY'...

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1
big looser ......... ·

thanksssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssss

33
Abhishek Priyam ·

first make any atom on both carbons opposite the point 'P'(central point..)( in this case i have made -H) by rotating at any one carbon..

now u can see (red line is coming out of paper.. from Cl to CH3..passing through P) that on moving from P one side you will get Cl other side CH3

33
Abhishek Priyam ·

no theres no point also.. wait a sec..

1
big looser ......... ·

thankssssssssssss............... but priyam .... kya centre mein point of symmetry hoga ?????????? becoz moving in opposite direction we get same molecule...............

33
Abhishek Priyam ·

just by rotating groups at both carbons so that... two are out of plane and one behind plane in between...(make this at both carbons..) and thats ur fischer projection....

33
Abhishek Priyam ·

33
Abhishek Priyam ·

kk.. just a sec

1
big looser ......... ·

can u tel how to convert it in fischer

11
Subash ·

same timing :p

1
Vivek ·

is it (A)?

33
Abhishek Priyam ·

33
Abhishek Priyam ·

u have converted wrongly....

33
Abhishek Priyam ·

yaar mere ko fischer nahi thik se dikhta hai... 3d is better..

1
big looser ......... ·

33
Abhishek Priyam ·

no.. plane of symmetry and point of symm is not there... just see once more..

11
Subash ·

better convert this to Fischer structure and then see

1
big looser ......... ·

and also ,,,, plane bisecting CH3 or plane parallel to the plane of page will be a plane of symmetry.
correct me if m wrong

1
big looser ......... ·


don't you think that point P is point of symmetry

33
Abhishek Priyam ·

is it clear or i should post some images... [7]

33
Abhishek Priyam ·

draw its mirror image and you will notice that they are not super imposable.. thats it..

1
big looser ......... ·

priyam can you post the solution

33
Abhishek Priyam ·

(A) (D)

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