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2) 2) ID pea.6,1 E H 32 6' for thls correpte-answer H of Inner product Selecting dot product and 1 J024p8031 efvectocur 08500u penalty, the H end fsooree 1 &#...

Question

2) 2) ID pea.6,1 E H 32 6' for thls correpte-answer H of Inner product Selecting dot product and 1 J024p8031 efvectocur 08500u penalty, the H end fsooree 1 'molaq 1

2) 2) ID pea.6,1 E H 32 6' for thls correpte-answer H of Inner product Selecting dot product and 1 J024p8031 efvectocur 08500u penalty, the H end fsooree 1 'molaq 1



Answers

$\mathrm{SO}_{2}+\mathrm{H}_{2} \mathrm{~S} \longrightarrow$ Product. The final product is: (a) $\mathrm{H}_{2} \mathrm{SO}_{4}$ (b) $\mathrm{H}_{2} \mathrm{SO}_{3}$ (c) $\mathrm{H}_{2} \mathrm{~S}_{2} \mathrm{O}_{3}$ (d) $\mathrm{H}_{2} \mathrm{O}+\mathrm{S}$

In this problem, this compound, which I am writing here, just look at it carefully here. It is opec And in this I did teach CS three. So this compound in presence of as to as a four hit After two years of four. Yeah. As well as we are too CHCL three will produce the final product ID. This component, the educated CS terry group. In this side, it is beer therapy at heritage Beer. So according to the option in this problem, option B. It correct answer.

Okay, so I'm gonna go ahead and multiply these. The 16 times 16 is going to be won over 36 16 times. H is going to give me 168 negative. H times 16 is not going to give me negative 16 h and negative h Times eight is going to give me negative age squared by. Look, we got a positive 16 h and the negative 168 So you're going to cancel, so I'm going to be left with 1/36 minus h squared.

In this problem, the reaction will happen. Something like this. Just look at it carefully. This compound which I am writing here independence of access or three as well as as to to give the product ID this compound here it is double Bondo heritage. They will Bondo he edited or h he edited or H So according to the option option D. H. Correct answer. Obstinately it's correct answer for this problem.

Quite a complex mechanism. So we'll start off with our starting material where we have a four member ID rings. That's quite strained. We are alcohol group on DSO. What we do is we protein ate that with H two s 04 Wait two plus we then lose this now. It formed a cat ion, but is a secondary carbon cat ion. So what we have is a 12 hydride shift. So now that develops a five members ring where we now have a Tash Ricardo cotton which is more stable. So our final step here is the elimination. So we will eliminate a proton using any common base to form our final product.


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