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1st reaction  :-                                              2nd reaction :-                                                     3rd reaction  :-                                           
In  reactions, we see the formation of carbocation and this is the rate determining step for these kinds of reactions.So the compound having more stable carbocation will react faster. Hence 2-Chloroheptane will react faster than 1- Chlorohexane.
In  reactions, we see the formation of carbocation and this is the rate determining step for this kind of reactions. So the compound having more stable carbocation will react faster. In the given case 2- Chloro, 2- Methylpropane we have  carbon whereas in 3- Chloropentane we have  carbon.
In these kinds of cases, we see where is the substituent is attached i.e., how far from the halide group. It can be clearly seen that the methyl group attached in 1-bromo-2-methylbutane is near than that attached in 1-bromo-3-methyl butane. Hence the rate of reaction will be faster in case of 1-bromo-3-methylbutane.
The rate of  reaction decreases with increase in hindrance to the attack of the nucleophile. So 2-bromobutane will react faster than 2-bromo-2-methylpropane in the nucleophilic attack.
In this case, the rate of    reaction will depend on the hindrance of the substrate. Since 1- Bromobutane is a  alkyl halide and 2- Bromobutane is a  alkyl halide hence 2- Bromobutane gives more hindrance to the nucleophile. Hence  1- Bromobutane reacts faster.
In the given compounds the halide groups are same. In these cases, the boiling point depends on the bulkiness of the alkyl group. The boiling point increases with an increase in the chain length. Also, the boiling point decreases with an increase in branching. So the order is :-    1- Chlorobutane  >  1- Chloropropane  >  Isopropyl Chloride

When it comes to group 17 elements being the substituents, the boiling point increases as we go down the group and thus the increasing order of boiling point is:

Chloromethane, Bromomethane, Dibromomethane, Bromoform

The final products are:-                                             
For four monochlorides we need four different hydrogens which can be replaced by chlorine. Hence the required isomer is :-                                              
For the given condition we must have three different hydrogens so that we can get three different monochlorides on the replacement. Thus the isomer is n-pentane.                                 
In this we have to find an isomer in which replacement of any hydrogen atom gives the singel compound for all replacements. So the isomer is Neopentane.                                                   
We obtain four dihalogen derivatives of propane  :- (i) 1,1 Dibromopropane                                                             (ii) 2, 2 Dibromopropane                                                       (iii) 1, 2 Dibromopropane                                                    (iv) 1, 3 Dibromopropane                                               
We don't use sulphuric acid because it acts as an oxidising agent and the required alkyl iodide is not produced. The reactions are given below :-                       2KI  +  H2SO4     →    2KHSO4  + 2HI                       2HI  + H2SO4    →  I2  +  SO2  +  H2O
The structure of  1-Bromo-4-sec. butyl-2-methylbenzene is shown below :-