00:01
Hello student in this question we have to draw the hnmr structure for the given compounds.
00:11
So let's start.
00:13
The compound is given as dichloromethane.
00:17
Here you can see we have to draw the hnmr structure so for these two hydrogen the value will become near 5 .30.
00:27
So somewhere between 5 and 6 the source.
00:31
Spectrum will come so the line will be like over here next one this is given one one dial iodohethane here you can see one hydrogen is present over here and another three is hydrogen present over here so for this ch3 the value will be 2 .90 and for this single hydrogen the value will be 6 .7 so somewhere near this three one long spike will come like this and somewhere over six to say between between six to seven here three small small spike will come as here you can see three hydrogen is present one it is given one one di bromone three fluoropropon two on here for this particular hydrogen, the value is 5 .25.
01:40
For this, again 5 .25.
01:43
For this, it is 6 .96.
01:48
So, two spike will come in between 5 and 6 .6.
01:55
So here, two spike will come like this.
01:59
And another will come near about 7...