00:01
Hello students, in this question we have to synthesize three products.
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In the first part we have to convert one butene to hexanol that will be this product.
00:13
This can be achieved by breaking this bond.
00:16
Hence we will get an alkene here and this acetyldehyde here.
00:24
So, these are going to be our main reagents.
00:27
So, we will initially have an alkene.
00:31
We will create a leaving group at this carbon.
00:34
Now, general acidic hydrolysis or other reagent will produce a leaving group on this carbon because this carbocation will be stable.
00:42
So, we will use a reagent like borane.
00:45
Now, in borane bh2 will be added on the less hindered carbon.
00:50
So, bh2 will be added on this terminal carbon and h will be added on this carbon.
00:57
Now, we can treat it with alkaline solution of hydrogen peroxide which will convert this bh2 to hydroxyl group.
01:09
So, we have created a hydroxyl group on the terminal carbon.
01:12
Now, this hydroxyl group is not a good leaving group.
01:15
So, we will treat it with tms chloride.
01:18
Now, tms chloride is trimethylsilane chloride.
01:24
So, this will give a simple nucleophilic substitution reaction on hence, this chlorine will be removed which will afterward abstract this proton from hydrogen as well.
01:37
So, hcl will be removed and we will finally get ao tms group here which will be o -sime3.
01:46
Now, this otms is a very good leaving group because oxygen and silicon bond is very stable.
01:52
Now, we will make some changes in our acetyl aldehyde.
01:58
So, this is our aldehyde and we will treat it with base.
02:02
Base will abstract the alpha hydrogen which will create and enolate like this.
02:09
Now, in the next step, we will treat it with our otms substrate that was created in the previous step.
02:17
So, this will attack on this carbon causing the nucleophilic substitution.
02:22
Hence, this otms will be removed and our final product that will be formed will be our hexanol that we had to synthesize.
02:32
So, this is how we will prepare our final product.
02:35
Now, let's look at the second product.
02:37
In second question, we have to synthesize cis -3 -hexene from but -1 -ene.
02:42
So, for obtaining this from but -1 -ene, we will have to cleave this bond.
02:47
Hence, we will now need one more bond that means triple bond between these carbons and here we will need a leaving group.
02:57
So, let's start with the question.
03:00
We will start with butene.
03:02
Now, this contains ch2 group.
03:06
If we treat it with br2, this will generate a dibromoalkene.
03:12
Now, we will treat it with nanh2.
03:17
This is a very strong base and we will use its two equivalents.
03:21
So, first nh2 negative will abstract a proton from here...