Preparation of haloalkanes - class-XII

Treatment with alcoholic KOH causes dehydrohalogenation to give an unsaturated hydrocarbon, while aqueous KOH substitutes halogens to form a carbonyl compound (ethanal). A geminal dihalide, specifically 1,1-dichloroethane (CH3CHCl2), yields ethanal upon hydrolysis with aqueous KOH.

Preparation of alkylhalides by the reaction of alcohols and phosporous halide is nucleophilic substitution Reaction.
Becuase the nucleophile halide ion from phosporous halide like $PX _3,PX _5$  substitutes the $OH^{-}$ ion in alcohol, which is called as nucleophilic substitution reaction
e.g. $3CH _3-CH _2-OH+PCl _3\rightarrow 3C _2H _5-Cl+H _3PO _3$
Hence option A is correct.

${C} _{2}{H} _{5}OH\xrightarrow [  ]{ SO{ Cl } _{ 2 } }  {C} _{2}{H} _{5}Cl+{SO} _{2}+HCl$
(a). $HCl$ bond is strong, no reaction.

$C _2H _5-Cl$ is the alkyl halide formed in the above reaction
$C _2H _5OH \ + \ PCl _5 \longrightarrow C _2H _5Cl \ + \ POCl _3 \ + \ HCl$
Hence option A is correct.

A,B,C are $R-Cl,POCl _3,HCl$
$R-OH+PCl _5\rightarrow R-Cl+POCl _3+HCl$
Hence option B is correct.

A is $C _2H _5Br$
$3CH _3-CH _2-OH+PBr _3\rightarrow 3C _2H _5-Br+H _3PO _3$

A is $H _3PO _3$
$3CH _3-CH _2-OH+PCl _3\rightarrow 3C _2H _5-Cl+H _3PO _3$
Hence option A is correct.

The products formed are $3C _2H _5-Cl+H _3PO _3$ when $3CH _3-CH _2-OH$ is reacted with $PCl _3$.

When ethanol reacts with phosphorous pentachloride, we get ethyl chloride as the main product whereas, phosphorous oxychloride and HCl as the byproducts.

A is $PCl _3$ in the above reaction.
$3CH _3-CH _2-OH+PCl _3\rightarrow 3C _2H _5-Cl+H _3PO _3$
Hence option A is corrcet.

By the action of $PCl _3$ or $PCl _5$ with alcohols alkyl halides and a
by product $HCl$ is formed.The byproduct $HCl$ is absorbed by using Triethyl amine($Et _3N$),pyridine($C _5H _5N$) which are bases and neutralise the acid like $HCl$ and $HBr$ etc.

Trimethyl amine and pyridine used for the absorption of hydrogen halide.in  the preparation of alkyl halide from alchohol and phosphorous halide.As these two are bases which can neutralize the acids like $HCl$ and $HBr$
Hence option A is correct.

When Alcohol reacts with $PCl _{3}$ in presence of Trimethyl Amine. It is a nucleophile substitution halogenation reaction replaces $OH^{-}$  with $Cl^{-}$  

The reagent used in the above reaction is $PBr _3$
$3CH _3-CH _2-OH+PBr _3\rightarrow 3C _2H _5-Cl+H _3PO _3$
Hence option A is correct.

$CH _3CH _2C(CH _3) _2CH _2-OH+PCl _3\rightarrow CH _3CH _2C(CH _3) _2CH _2-Cl+H _3PO _4$
So in the above reaction A is $CH _3CH _2C(CH _3) _2CH _2-Cl$
Hence option A is correct.

B is $ \displaystyle POX _3$.
For example, when ethyl alcohol is refluxed with $ \displaystyle PCl _5$, ethyl chloride, HCl and phosphorous oxy chloride is obtained.
$ \displaystyle CH _3-CH _2-OH +PCl _5 \to CH _3-CH _2-Cl+POCl _3+HCl$

Bases like tri-methyl amine and pyridine are used to mop up the acidic byproduct and make it neutral.
Hence option A is correct.

Alkyl Bromide is prepared by heating alcohol with Red phosphorous in the presence of Bromine gas.