00:01
Continuing to look at chemical transformations so what we're starting off with is pentan one of treat that way as pbr3 to generate one bromopentane pbrr three in the second part we then take this material in b we react it with tb u k tb u o h to generate pent one in so following this we take our pent one in that is treated with hgo a c2 th f h2o sodium boroh hydride oh h minus we then generate penton 2 oh so then we're taking pent one in again one pentine from part b directed with h2 nickel it's generate pentane then we can take two pentone from part c, reacted with p, br3.
01:59
Let's generate two bromopentane.
02:05
Next, we have our material from a, which is one bromopentane.
02:11
React that with mg, br, e t2o.
02:20
Following this reaction, we have h c, h -c -o, h -3o, plus, so it's running hexan -1ol.
02:34
Next we have pentile magnesium bromide formed in part f.
02:50
React that with our epoxide, proton source, peptan one -all is formed.
03:01
Next we take pentan one -all, react it with dmso, cocl2, pot -berry cold conditions, et23n.
03:29
We generate pentanel upon a swine oxidation.
03:32
Next we have two pentanol.
03:44
React that with h2, cro4 to generate two pentanone.
03:55
Next we take pentan one all.
04:01
React that with kmno4, oh h minus and heat and proton source in our second step.
04:13
Generate a carboxylic acid, pentanoic acid.
04:23
Next we can have pentan one o 'l...