1
Optimus Prime
·2009-03-24 08:36:25
answer for 2nd is C-C-(Cl)-C-C≡C
1
greatvishal swami
·2009-03-24 04:41:28
tapan it was to prov that double bonds reacts faster than triple as unstable vinyl cation is formed
( that i showed in the fig)
but also in Q 1 H+ still attacks triple bond such that overall stability is not disturbed
first the compound was stable due to conjugation hence for the product to remain in conjugation H+ attacks triple bond
always consider overall stability of compound first
1
gsns gannavarapu
·2009-03-24 04:44:00
sry if i m irritatin u i gt another doubt
can d 2nd compound b
C - C(Cl) - C - C ≡ C
21
tapanmast Vora
·2009-03-24 07:38:13
Vish : My dbt in Q1 is not werther duble or tripl bond is attacked........ that's triple 4 sure
Ther two Carbons mutually triply bonded......... extreme rite wala n the one next to it.....
Now Cl will go to wich C out of the two is ma DOUBT,.......
1
king_khan
·2009-03-24 08:28:54
hey dat is bcoz of greater stability of the carbocation !!!
1
greatvishal swami
·2009-03-24 08:32:53
hmm now i got u k k
next to it will be attacked as that carbocation is more stable due to conjugation with double bond
1
Optimus Prime
·2009-03-24 08:35:38
answer for 1st is C=C-C(Cl)=C
21
tapanmast Vora
·2009-03-24 04:34:14
Vish how does the abov diag. justify prodct 1?? [7]
in Q1 the terminal triply bonded C
1
vector
·2009-03-24 08:47:58
vishal cud u plz explain things a bit more clearly
1
greatvishal swami
·2009-03-24 08:53:37
after addition of H+
C
C-----C+
c----H ( charge is conjugated)
and
C
C-----C
c+----H ( charge is not conjugated)
r possible
clearly 1st carbocation is more stable than 2nd due to conjugation with double
bond hence Cl- will bond with 1st structure
1
vector
·2009-03-24 08:57:30
thanks wud u also explain 2nd one
1
greatvishal swami
·2009-03-24 09:08:26
richa 2nd one is easy simply on the facts that
1. double bonds are more reactive than triple
2. 20 carbocation is more stable than 10 carbocation
1
vector
·2009-03-24 09:15:54
bt wat s the significance of low temp
1
greatvishal swami
·2009-03-24 09:19:10
low temp hmm its exothermic reaction naa ( and i think highly isothermic maybee)
so low temp will provide slow rate of formation ( or kindof correct rate if its highly exothermic)
something like this [4][4]
21
tapanmast Vora
·2009-03-24 03:36:02
ANS :
1) C=C-C-(Cl)=C
2) C-C-(Cl)-C-C [TRIPLE BOND] C
I wanted to know the reasoning.......... I cudnt figure out one!! [2]
106
Asish Mahapatra
·2009-03-24 03:43:32
@tapan a possible explanation for 1:
a stable alkene forms as it becomes conjugated... but i dont think that applies for 2nd one then :(
1
greatvishal swami
·2009-03-24 03:52:38
at high both thebonds are attacked
1
greatvishal swami
·2009-03-24 03:54:56
well use two concepts here
1. in general duble bonds r more readily attacked than triple ( this explains for 2)
2. conjugated alkene is more stable so in 1 rather triple bond is attacked so such that the conjugation is not disturbed
13
deepanshu001 agarwal
·2009-03-24 03:58:18
how can both bond b attacked wen 1 equivalent is given
1
greatvishal swami
·2009-03-24 04:00:41
k for 1 eq u r asking
k then the same products will be formed [4]
106
Asish Mahapatra
·2009-03-24 03:16:13
i think .. double bonds are more reactive towards electrophilic addition reactions so i think with 1 eq. of HCl only the double bonds will get broken...
i.e.
1 will be Cl-C-C-C≡C
1
greatvishal swami
·2009-03-24 04:26:29
k
well in an electrophilic addition (add of H+ here)
less stable vinyl cation is formed
1
gsns gannavarapu
·2009-03-24 04:28:48
bt dis carbon z sp hybridised naaa