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Published on: 02/11/2019
Carbonyl Compounds
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1.
What are the Cartesian sign conventions for a spherical mirror?
2.
Derive the relation between f and R for a spherical mirror.
3.
Give the characteristics of image formed by a plane mirror.
4.
What is angle of deviation due to reflection?
5.
State the laws of reflection.
6.
A hydrocarbon A (molecular formula C8H10) on ozonolysis gives B (C4H6O2) only. Compound C (C3H5Br) on treatment with magnesium in dry ether gives (D) which on treatment with CO2 followed by acidification gives(C). Identify A, B and C.
7.
Identify X and Y.
\({ CH }_{ 3 }CO{ CH }_{ 2 }{ CH }_{ 2 }COO{ C }_{ 2 }{ H }_{ 5 }\overset { { CH }_{ 3 }MgBr }{ \longrightarrow } X\overset { { H }_{ 3 }{ O }^{ + } }{ \longrightarrow } Y\)
8.
How is propanoic acid is prepared starting from
(a) an alcohol
(b) an alkylhalide
(c) an alkene
9.
A Compound (A) with molecular formula C2H3N on acid hydrolysis gives (B) which reacts with thionylchloride to give compound(C). Benzene reacts with compound (C) in presence of anhydrous AlCl3 to give compound(D). Compound (D) on reduction with Zn/Hg and Conc.HCl gives (E). Identify (A), (B), (C) (D) and (E). Write the equations.
10.
Identify A, B and C
1.
(i) The Incident light is taken from left to right (i.e. object on the left of mirror).
(ii) All the distances are measured from the pole of the mirror (pole is taken as origin).
(iii) The distances measured to the right of pole along the principal axis are taken as positive.
(iv) The distances measured to the left of pole along the principal axis are taken as negative.
(v) Heights measured in the upward perpendicular direction to the principal axis are taken as positive.
(vi) Heights measured in the downward perpendicular direction to the principal axis, are taken as negative.
2.
(i) C be the centre of curvature of the mirror.
(ii) Consider a light ray parallel to the principal axis is incident on the mirror at M and passes through the principal focus F after reflection. An.
(iii) The line CM is the normal to the mirror at M. Let i be the angle of incidence and the same will be the angle of reflection.
(iv) If MP is the perpendicular from M on the principal axis, then from the geometry.
(v) The angles ∠MCP = i and ∠MFP = 2i
(vi) From right -angle triangles ΔMCP and ΔMFP,
\(tani=\cfrac { PM }{ PC } \) and \(tan2i=\cfrac { PM }{ PF } \)
(vii) As the angles are small, tan i≈i
\(i=\cfrac { PM }{ PC } \) and \(2i=\cfrac { PM }{ PF } \)
(viii) Simplifying
\(2\cfrac { PM }{ PC } =\cfrac { PM }{ PF } ;2PF=PC\)
3.
(i) The image formed by a plane mirror is virtual, erect, and laterally inverted.
(ii) The size of the image is equal to the size of the object.
(iii The image distance far behind the mirror is equal to the object distance in front of it.
(iv) If an object is placed between two plane mirrors inclined at an angle e, then the number of images n formed is as,
\(n=\left( \cfrac { 360 }{ \theta } -1 \right) \)
4.
The angle between the incident and deviated light ray is called angle of deviation of the light ray.
5.
According to law of reflection,
(i) The incident ray, reflected ray and normal to the reflecting surface all are coplanar (i.e: lie in the same plane).
(ii) The angle of incidence i is equal to the angle of reflection r.
i = r
6.
The compound with molecular formula C8H10
Compound (C) with molecular formula C3HSBr is
7.
8.
An alcohol:
An alkylhalide:
An alkene:
9.
10.
| Compound | Name |
| A | Benzoyl Chloride |
| B | Benzophenone |
| C | Ethylbenzoate |
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