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Hello.
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In this question here we need to find out the correct option which will help in the elimination of substitution reaction of one bromopentine.
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So here we have pantin it is there.
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Ch3, ch2, ch2, ch2, ch2, ch2, and iodine it is there.
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1, 2, 3, 4, 5.
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Bromo it is there, bromine atom will be there.
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So this is one bromopentane.
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So the highest elimination reaction it will occur only then when it is when a most hindered base if we introduced here, which will difficult to act as a nucleophile.
00:54
So therefore among the following we can say that is this is the o negative and k positive it is.
01:03
There and tertiary alcohol at 55 degrees celsius it will lead to the formation of hair or this removal of this bromine and formation of this alken product that is more than 90 % it will form here this product will form so therefore here this elimination takes place hence here option c is the right answer now here for the second one the sn1 reaction s in one, that is substitution nucleophilic reaction, that is first it will convert into carbocatin and then the new formation of newborn formation will take place.
01:49
So here if we see that is among the option here, this ots, it is there on reacting with h2o, this ots group, it is a very good leaving group among the other leaving group here.
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So therefore it will lead to the formation of this carbocetine and hence on reacting with h2o it will form this alcohol, this oh group will get attached to this molecule...