5

OhOCH,CHg CH3 C-H OCHz( CHa'Hjo CHz CH OCHzCH;-C-CH; OCHaOHHO CHa-C-H OCHaOCHa CHa C-CHa OCHaCHz-C-CHzCHa OCHa...

Question

OhOCH,CHg CH3 C-H OCHz( CHa'Hjo CHz CH OCHzCH;-C-CH; OCHaOHHO CHa-C-H OCHaOCHa CHa C-CHa OCHaCHz-C-CHzCHa OCHa

Oh OCH,CHg CH3 C-H OCHz( CHa 'Hjo CHz CH OCHz CH;-C-CH; OCHa OH HO CHa-C-H OCHa OCHa CHa C-CHa OCHa CHz-C-CHzCHa OCHa



Answers

$\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COOH}$ B What is B? (a) $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CHO}$ (b) $\mathrm{ClCH}_{2} \mathrm{CH}_{2} \mathrm{COOH}$ (c) $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COCl}$ (d) $\mathrm{CH}_{2}=\mathrm{CHCOOH}$

In this problem I can write the reaction and CST CH two managed to in perchance of action or two 0-5° integrate will give this compound CS three CH 20 H. And this component to denso. PBL three will give CST CH two beyond and this compound in pageants of GCN we'll give CST CH two M. C. And this component presence of AL I am at full well finally give CST CH two NHCS three. This is compounded by this is component so according to the option, option C it correct here, option siege, correct answer.

In this problem, I can write DDX in it see http CH two C 00 H. This compound in pageants of Cl two A red P. That to be dad reaction. It will give the compound edge, see http C h C L C. O AT and this compound in perchance of alcoholic kO edge. The hydro chlorination happened here and the compound formed the CH two double 1 C h c o egg. Therefore, according to the option of indeed correct option, the is correct answer. I hope you understand the solution.

According to the given data here, two hydrogen every level Hollywood blood, they're dying. B dying. C h t v ch two C h double one ch, CH double one Ch. See http IT defense Clearly formed because it is a conjugated dying. It is a conjugated dying and it's more stable and it mode the stable, then an isolated dying. So absence. See it correct here.

In this problem, the reaction will happen. Something like this. Just look at it carefully. I'm writing the reaction here So he had CH two never one see it. CST in presence of NBS had to be on will give C b a. See, ohh, see be odd as the final product. So according to the options in this problem, option B is the correct answer. Option B is the correct answer for this problem. I hope you understand the solution of this problem.


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