Messages posted by adamabhilash

in 1st part

firstly H(+) attacks at OH grp ad H2O being good leaving grp leaves resulting in formation of carbocation.

ring(OH)-(+)ring

Now there are two possibilities

i)Arylic shift within the ring(ring contraction):- in this from the adjacent C-atom within the ring shift takes place forming-membered ring which is unstable.

so,this doesn't occur

ii)Arylic shift outside the molecule(ring expansion):-in this the bond from adjacent ring migrates and ring expands from 5 memered to 6 membered and then it loses H(+) from adjacent atom and the final product is

A cyclohexanone with five membered ring attached in ortho position including the ortho carbon.

don't know abt second ques sorry

If there is any mistake i first ques please tell me

yours is the correct answer because if we see the mechanism:-

i) Ph-OCH3+HI(H(+) source)----->Ph-OH(+)-CH3

now there are two possibilities either Ph-OH is removed and CH3(+) cation is formed or

CH3-OH is removed and Ph(+) cation is formed.

But we know phenyl carbocation is not formed anywhere therefore first possibility takes place and that CH3(+) reacts with I(-) and forms CH3-I

and more over O has partial double bond character(due to resonance with benzene ring) and is difficult to cleave.

I think answer is (a).
because rearrangement occurs y shifting or losing from adjacent atom but in this case adjacent atom can either lose H(+) or cyclobutane can shift,but shifting forms primary carbocation which is less stable than secondary. Therefore elimination occurs ad ot rearragement.

(cyclobutyl)CH2-CH(OH)-CH3 ----->{H2SO4}---->(cyclobutyl)CH2-CH(+)-CH3-----E1----> (cyclobutyl)CH--double ond--CH-CH3

is it correct
if there is some mistake please tell me

Psedohalides are those anions which resemle halide ions.
These are nitrogen containing anions whose:-
i)Hydrides are acids
ii)Sodium salts are water soluble
iii)Silver salts are water insoluble
eg. CN(-),CNO(-),SCN(-),SeCN(-),N3(-)
i)except N3 all these exist in dimer form
ii)HCN,HOCN,HSCN,HSeCN,HN3=all are acids
iii)NaCN,NaOCN,NaSCN,NaSeCN,NaN3=all are water soluble
iv)AgCN,AgOCN,AgSCN,AgSeCN,AgN3=all are water insoluble

is it ok,if somethig is wrong please tell me

It is not so imp.
but some questions(1 or 2) come from it.
So,i think just once go through it and remember some imp ones like of Fe and Al

hey (i)you may recollect that attack of nucleophile will be r.d.s.
moreover,(ii) we know that R(-) attacks on that carbon atom which has more delta(+) and yields the product.(iii)second step is just hydrolysis of OMgX to OH and it is done easily.
therefore, attack of CH3(-) is the r.d.s

NBS=N-Bromosuccinamide does allylic addition od Br with little HBr or Br an light heating.
therefore A=3-Bromohex-1-ene(Allylic bromide)
Any alkyl halide on treatment with Mg/ether forms grignard reagent,
therefore B=Cyclohexene magnesium bromide
In B cyclohexene has -ve charge and in CH3CN,C has delta+ve charge,therefore cyclohexe attacks C-triple bond-N and one of the triple bond goes over N.
therefore C= CH3-C(cyclohexene)-double bond-NH

is it ok
if something is wrong please tell me

some aldehydes primarily exist in gem-diol form in aq solution.
but it is not possible to isolate it.
basically
CH3CHO+H2O <---->CH3CH(OH)2
<----> = in equilibirium
but as one H2O evaporates due more randomness(delta{G}<0) the eqm shifts forward.
but there are some exceptios
like
Compounds with strong electron withdrawing grps attached to the carbonyl group can for stable gem-diols.
for eg. Chloral hydrate
(Cl)3C-CH-(OH)2
this is stable.

is it ok. if something is wrong tell me

i) hydrolise it to form methanol(CH3-OH)
ii)oxidise to form HCHO(using KMnO4 or K2Cr2O7)
iii)add CH3MgBr to form CH3CH2-OH
iv)again oxidise to form CH3CHO
v)again add CH3MgBr to form CH3COCH3

Is It ok?????????

see in chlorobenzene the carbon to which Cl is attached is sp2 hybridised in which s characteer is more and therefore it draws the bonded pair of electron more towards itself as compared to carbon of the other compound. hence the (delta -) on chlorine is less in chlorobenzene.

2) as chlorine in chlorobenzene has partial double bond character its bond length is little less and as dipole moment is proporional to length it is less in chlorobenzene..

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