aromatic doubts

NCERT and CAREY say that with nitration and sulphonation of aniline, meta and para products are obtained in an almost 50-50 ratio.

But as temperature is not applied to the second step, i think the product should be

2,4,6 -tribromo, 3-sulphonyl aniline (sorry for wrong nomenclature)

ABhi look at answr for first one #7

isnt fluorine most deactivating among halogens

Ans 6) I think that must be bcoz of the fact that in aromatic substitution S _N reacn, the C- X bond is not cleaved at r.d.s. Hence it cannot affect the reactivity.
The carbon ion intermediate is stabilized by the -I effect of halogens and hence Cl is most reactive and I is least reactive.

Q6. answ is F>Cl>Br>I

tush pls explain

Q6 I>F>Br>cl

dunno if correct but acc to me :-

The groups which are deactivating in electrophilic substitution will be activating in nucleophilic substitution. E.W.G will be activating groups. The activating groups must be ortho and para w.r.t the leaving group.
ALso, the -M effect of NO₂.

yeah but y??

I thot since there is max -I and -R for ortho, the carbanion TS is most stabilised.. then -para and finaly meta..

but y ortho>para ??

4) Y should preferably be :

also , SO2 gas would be released
it would hv been better if heat would hv been mentioned.

bonne annee Q2 reason i hav given it earlier itself in #4

strangely ans 1. was given to be on the

Q2. solomons mentions fluorobenzene

Q3. debo correct.. an explanation pls

Q4. What is (Y) ?

ans 3 ) looks to be 5
ans 4) sulphanilic acid
ans1) correct answers on both occasions ! [1]

Ans 1)

Ans 2) The electron withdrawing effect of the halogens dec considerably form F to I.

Ans 3) 4

Ans 4) X → Sulphanilic acid

4) X will be tht Anilinium ion....n Br₂/H2O will form HBr + HOBr... so...thinking over Y...