00:01
Which of the alkali chlorides listed below undergoes dehydrogenation in the presence of aström based to give 2 pantine as the only alkyne product? so we have to prepare which will give only 1 alkene, which will give only 1 alkene, that is the 1 chlorotumethyl butane, 2 pantine, base to give only 2 pantine as the product.
00:27
So 2 panting will give 3 chloropentine, only 2 chloropentine, let us check 3 -kilropentine.
00:43
Yes, it is the correct answer.
00:46
When 3 -kilopentine is reacting with the base, then what will happen? cl will get eliminated, positive charge will come here.
00:54
Now it has two options to remove h from this carbon, from this carbon.
01:02
If this h will get eliminated, then the double bound will come here.
01:06
If this chance will eliminate, then the double bond will come here.
01:12
Right.
01:13
And both the double bond containing alkin are the 2 -pentine and both are the same.
01:20
Correct...