each of foll. reagents, consider normal charge polarisation to identify that part of the reagent that can act as an electrophile...
a) HBr
b) HOCl
c) Cl2
d) Br2
e) H2O
f) CH3COCl
g) CH3COOH
h) 
...n this is a benzene ring....sry for wrng drawing...
i)
some r obvious but others????
39b) Chlorine
c) Chlorine(One + other -)
d) Bromine(One + other -)
e) Proton
f) and g) are both CH3CO where + charge is on carbonyl carbon.
h) S(=O)2OH mein OH will leave and electrophile will be S(=O)2 with plus charge on sulphur.
i) Plus charge will be on carbonyl group (chlorine cannot leave).
1Well, c and d are cases in which a dipole moment is induced in the molecule, and one atom gets a partial + and one a partial -. The partial + atom then acts as electrophile
1http://www.avogadro.co.uk/definitions/electrophile.htm
very simply explained with diagrams
39HOCl is generated insitu by the reaction of NaNO2 and HCl(used in diazotization). It is used to form chlorohydrins when added to a double bond ( -C(Cl)-C(OH)- systems are chlorohydrins). Here Cl is in +1 oxidation state, and acts as the electrophile(because O is more electronegative than Cl).
In X-X(Only Br and Cl = X) type EAS reactions, since both atoms are of same electronegativity, the bond cannot be polarised without the help of a catalyst. Catalysts used are Lewis acids like FeBr3. They induce polarisation of the X-X bond, and then react with the more electronegative part of the molecule in an acid-base reaction. (Br - Br+ - FeBr3- complex forms, for example). Now this is a good leaving group(the + charged bromine, in this example), so a double bond from benzene(double bonds act as nucleophiles) attacks one X atom and frees the other. To regain aromaticity(last step in all EAS reactions), a hydrogen is given away to a base.
What more do you need to know, sir?
