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@Dexter
Note for OP
+solved @user1 @user2... to close the thread when your doubt is solved. Mention the users who helped you solve the doubt. This will be added to their stats.multi correct?
mujhe toh cd lag rha
yes same i also feel ki c and d hoga
reason kya ha?
HCN aldehyde aur ketone ke saath cynaide deta hai
to iske karan ur b is eliminated
and KCN reacts with halo
and in a we have alc to vo bhi possible nhi
c me we have alc to halo conversion and then kcn is added
to vo dega cynaide
and then same in d
bhai kaise kr rhe ho yar, ek baar pura thinking process batao
basically what she said above
c and d right? both will have better leaving group so CN- can do sn2 very smoooothly
ye Hcn ald and ket se react karta
ye sb classses mei krva rkha ha kya apko
how is that relevant here
pehle ethyl halide banao and then sn2 hogi
a and b nahi honge cuz oh- poor leaving group
yeah standard stuff
cuz they need quite an electrophilic site to attack and so this points to the fact that reaction is not occurring in all the 4 options
wtf ?
LOL i mean agar if it was any good nucleophile then toh sn2 chaaro me hota na
leaving grp bhi toh dekho
NCERT me hai
Yea was just gonna say. You need to revise halo alkanes and sn1 sn2
You need to replace an OH with a CN in a primary alcohol
You’re looking for an SN2 reaction because cyanide is a strong nucleophile and the centre is primary (trademark conditions for an SN2 reaction)
Now OH- can’t leave directly of course
It’s bad leaving group
So first gotta convert it into a good one and then do the usual KCN or NaCN
A) me you’re not forming a good leaving group
B) me bhi you aren’t (and on top of that HCN is a weak acid; won’t really give you very high amounts on CN- in the solution)
C) me you’re converting OH to OTs (an excellent leaving group) and then doing the SN2
D) me you’re first putting a Cl in the place of OH and then going for an SN2
So the answer should be C and D
+solved @hardcoreisdead @Sephrina
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