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ANSWER ASAP !!!!

DISTINGUISHING TEST BETWEEN
1-C2H5Br AND C2H5Cl

2- phenol and cholobenzene

qs2) although chlorine is an electron withdrawing group yet it is ortho ,para, directing in electrophilic substitution reaction .why?

Asked by Ruchi Soni(student), on 10/1/13

Answers

EXPERT ANSWER

A-1)

(i) C2H5Br and C2H5Cl:

  • LASSAIGNE'S TEST: The sodium fusion extract, SFE can be used to detect the presence of chlorine, bromine. To detect their presence, the SFE is first acidified with HNO3 and then added with AgNO3 solution.

i) The formation of a curdy white precipitate that is soluble in NH4OH indicates the presence of chlorine in the organic compound.

detection of chlorine

ii) The formation of a pale yellow precipitate that is partially soluble in NH4OH confirms the presence of bromine.

detection of bromine

 

(ii) Phenol and cholobenzene:

Phenol reacts with sodium hydroxide solution to give a colourless solution containing sodium phenoxide.

Chlorobenzene does not react with NaOH.

A-2)

The answer to this question is based upon two factors. These are:

 

Reactivity towards an electrophile

The reactivity towards an electrophile depends upon the electron density on the benzene ring. If the electron density increases, electrophile attacks easily. As Cl withdraws electrons through inductive effect (-I effect), it tends to destabilize the intermediate carbocation towards electrophilic substitution.

 

Directive influence

Although Cl benzene is less reactive towards electrophilic aromatic substitution but if electrophile attacks and the product is to be formed, then it will be at o- and p- positions. This is because through resonance halogen tends to stabilize the carbocation and the effect is more pronounced at o- and p- positions.

Posted by Urvashi Yadav(MeritNation Expert), on 14/1/13

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