Carbylamine reaction:
Chloroform reacts with aliphatic or aromatic primary amine and alcoholic caustic potash, to give foul smelling alkyl isocyanide (carbylamines).

This reaction is used to test primary amine.

i) DDT: DDT can be prepared by heating a mixture of chlorobenzene with chloral (Trichloro acetaldehyde) in the presence of con. H₂SO₄,

ii) Chloroform :The reaction of methane with excess of chlorine in the presence of sunlight will give carbon tetrachloride as the major product.
\(\underset { Methane }{ { CH }_{ 4 }+4C{ l }_{ 2 }\overset { h\gamma }{ \longrightarrow } } +\underset { Carbon \ tetrachloride }{ 4HCl } \)
iii) Bipheenyl :
\(\underset { Chlorobenzene }{ { C }_{ 6 }{ H }_{ 5 }Cl } +2Na+Cl-{ C }_{ 6 }{ H }_{ 5 }\overset { Ether }{ \longrightarrow } \underset { Biphenyl }{ { C }_{ 6 }{ H }_{ 5 }-{ C }_{ 6 }{ H }_{ 5 }+2NaCl } \)
iv) Chloropicrin : Chloroform reacts with nitric acid to form chloropicrin.(Trichloro nitro methane)
\(\underset { Chloroform }{ { CH }_{ 3 }+HN{ O }_{ 3 }\overset { \triangle }{ \longrightarrow } } \underset { Chloropicrin }{ C{ Cl }_{ 3 }N{ O }_{ 2 }+{ H }_{ 2 }O } \)
v) Freon-12 :Freon - 12 is prepared by the action of hydrogen fluoride on carbon tetrachloride in . the presence of catalylic amount of antimony pentachloride. is is called swartz reaction
\(\underset { Carbontetrachloridew }{ { CCl }_{ 3 }+2HF\overset { SbC{ l }_{ 5 } }{ \longrightarrow } } \underset { Freon-12 }{ 2HC{ l }+CCl_{ 2 }F_{ 2 } } \)

The Chlorination of methane is carried out by free radical mechanism. The initiation step to form free radical needs high energy which is supplied by light energy.
\(\mathrm{Cl}-\mathrm{Cl} \stackrel{h \nu}{\longrightarrow} 2 \mathrm{Cl}\)
So this reaction is not possible in dark.

(i) C₅H₁₀ can be either alkene or cyclo alkane.
(ii) Since, the hydrocarbon does not react with CI in dark, hence it is not an alkene, it may be cyclopentane.
(iii) Since, it forms only single monochloro derivative in bright sunlight, all the H-atoms should be identical. Hence it is cyclopentane

The boiling points of isomeric dihalobenzene are pearly the same. The melting point of para isomer is generally higher than the melting points of ortho and meta isomers. The higher melting point of p-isomer is due to its symmetry which leads to more close packing of its molecules in the crystal lattice and consequently strong intermolecular attractive force which requires more energy for melting. p-Dihalobenzene> o-Dichlorobenzene> m-Dichloro benzene.

True.
Some haloalkanes yield a mixture of olefins in different amounts. It is explained by Saytzeff's Rule, which states that 'In a dehydrohalogenation reaction, the preferred product is that alkene which has more number of alkyl groups attached to the doubly bonded carbon (more substituted double bond is formed).
\({ CH }_{ 3 }-\underset { \overset { | }{ Br } }{ CH } -{ CH }_{ 2 }-{ CH }_{ 3 }\)
2-Bromopropane

\(\overset { \underset { | }{ Cl } }{ \underset { \overset { | }{ Cl } }{ CH } } -{ CH }_{ 3 }\quad (A)\) 1, 1,-dichloro ethane
\(\underset { \overset { | }{ Cl } }{ { CH }_{ 2 } } -\underset { \overset { | }{ Cl } }{ { CH }_{ 2 } } \) (B) 1, 2, dichloro ethane
\(\overset { \underset { | }{ Cl } }{ \underset { \overset { | }{ Cl } }{ CH } } -{ CH }_{ 3 }\underset { aq\quad KOH }{ \longrightarrow } \left[ \overset { \underset { | }{ OH } }{ \underset { \overset { | }{ OH } }{ CH } } -{ CH }_{ 3 } \right] \underset { -{ H }_{ 2 }) }{ \longrightarrow } { CH }_{ 3 }CHO\)
(A)                                                                    (C)
1, 1, dichloro ethane                                      Acetaldhyde
\(A \ \ \overset { \underset { | }{ Cl } }{ \underset { \overset { | }{ Cl } }{ CH } } -{ CH }_{ 3 }\)  -1,1 dichloro ethane
B  \(\underset { \overset { | }{ Cl } }{ { CH }_{ 2 } } -\underset { \overset { | }{ Cl } }{ { CH }_{ 2 } } \)-1, 2 dichloro ethane
C CH₃CHO -Acetaldehyde 
D \(\underset { \overset { | }{ OH } }{ { CH }_{ 2 } } -\underset { \overset { | }{ OH } }{ { CH }_{ 2 } } \)-Ethyleneglycol

Reaction between iso-butylene and HCI are as follows:

(ii) When CH₃- Br is treated with KCN, methyl cyanide is formed.
\(\underset { Methyl\\ Bromide }{ { CH }_{ 3 } } -Br+\underset { Potcyanide }{ KCN } \longrightarrow \underset { Methyl\\ Cyanide }{ { CH }_{ 3 } } -CN+\underset { Pot\\ Bromide }{ KBr } \)
(iii)  is a chiral molecule, as four atoms attached to *C - atom are different
(iv) 
This order is due to steric reasons, the order of reactivity in S_N2 follows the order: 1°>2°> s 3°.

S_N1stands for unimolecular nucleophilic substitution
(i) Hence rate ofthe reaction = k(alkyl halide. This S_N1 reaction follows first order kinetics and occurs in two steps.
(ii) We understand ~l reaction-mechanism by taking a reaction between tertiary butyl bromide with aqueous KOH.
\({ CH }_{ 3 }-\overset { \underset { | }{ { CH }_{ 3 } } }{ \underset { \overset { | }{ { CH }_{ 3 } } }{ C } } -Br\overset { OH(aq) }{ \underset { -Br }{ \longrightarrow } } { CH }_{ 3 }-\overset { \underset { | }{ { CH }_{ 3 } } }{ \underset { \overset { | }{ { CH }_{ 3 } } }{ C } } -OH\)
Tert-Butyl bromide                      Tert-Butyl alcohol
This reaction takes place in two steps as shown below
Step - 1 Formation of carbocation :
(i) The polar C - Br bond breaks forming a carbocation and bromide ion. This step is slow and hence it is the rate determining step.

(ii) The carbocation has 2 equivalent lobes of the vacant 2p orbital, so it can react equally rapidly from either face.
Step - 2 :
(i) The nucleophile immediately reacts with the carbocation. This step is fast and hence does not affect the rate of the reactions.

(ii) As shown above, the nucleophilic reagent OHcan attack carbocation from both the sides.
(iii) In the above example the substrate tert-butyl bromide is not optically active, hence the obtained product is optically inactive. If halo alkane substrate is optically active then, the product obtained will be optically inactive racemic mixture. As nucleophilic reagent OR- can attack carbocation from both the sides, to form equal proportion of dextro and levorotatory optically active isomers which results in optically inactive racemic mixture

S_N2 Mechanism :
(i) The rate of S_N2 reaction depends upon the . concentration of both alkyl halide and the nucleophile.
(ii) Rate of reaction = is [alkylhalide] [nucleophile]. It follows second order kinetics and occurs in one step. 
(iii) This reaction involves the formation of a transition state in which both the reactant molecules are partially bonded to each other. The attack of nucleophile occurs from the back side (i.e opposite to the side in which the halogen is attacked).
(iv) The carbon at which substitution occurs has inverted configuration during the course of reaction just as an umbrella has tendency to invert in a wind stonri. is inversion of configuration is called Walden inversion; after paul walden who 1st discovered the inversion of configuration of a compound in S_N2reaction.
(v) S_N2reaction of an optically active haloalkane . is always accompanied by inversion of configuration at the asymmetric centre.
(vi) When 2 - Bromooctane is heated with sodium hydroxide, 2 - octanol is formed with invesion of configuration. (-) - 2 - Bromo octane is heated with sodium hydroxide (+) - 2 - Octanol is formed in which - OR group occupies a position opposite to what bromine had occupied,

(a) (-).2 - Bromo octane
(b) Transition State
(c) (+) 2 - Octanol (product)