00:01
Hi, in this question we are given with these three reactions and we are asked to give the name of the product formed here.
00:08
It is given that these are the sn2 reaction and we know that in sn2 reaction backside attack of nucleophile takes place and it is a concerted reaction.
00:29
Here we can say that in the first part we are having nac triple bond ch.
00:37
We can say we can write this as na plus and c triple bond ch and carbon will get a negative charge.
00:46
We are having this alkyne as a nucleophile.
00:51
This will attack on the carbon atom on which we are having a leaving group.
00:56
Br is the leaving group.
00:58
This will attack from the back side at this carbon atom as a result of which this br minus will be released and we will be getting the product here in this manner.
01:09
Now we have to name this.
01:11
We will start numbering from this carbon atom as 1, 2 and 3.
01:15
We can see that at third number position we are having cyclopentyl ring hence we can write 3 cyclopentyl propyne.
01:29
As for three carbon atoms we use the propyfix and in is the suffix for alkyne.
01:38
Here along with this product there is formation of nabr.
01:44
Hence we can say this is the answer for first part.
01:46
Now let us look for the second part.
01:49
Here we can see we are having this alkyl halide and nai.
01:54
Nai is having na plus ion or i minus ion.
02:00
I minus is the nucleophile.
02:02
It will attack from back side at this position where leaving group is attached.
02:07
As a result the cl minus will be released and we are getting this product along with this nacl is formed.
02:15
Let us name this compound...