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896 mL vapour of a hydrocarbon 'A' having carbon 87.80% and hydrogen 12.19% weighs 3.28 g at STP. Hydrotion of 'A' gives 2-methylpentane . Also 'A' on hydration in the presnce of H_(2)SO_(4) andHgSO_(4) gives a katone 'B' having molecular formula C_(6)H_(12)O. The ketone 'B' gives a positive iodoform test. Find the structure of 'A' and give the reactions involved.

Answer» <html><body><p></p>Solution :To <a href="https://interviewquestions.tuteehub.com/tag/determine-437910" style="font-weight:bold;" target="_blank" title="Click to know more about DETERMINE">DETERMINE</a> the molecular mass of hydrocarbon (A) 896 mL vapour of `C_(x)H_(y)` (A) weighs `3.28`g at STP <br/> 22700 mL vapour of `C_(x)H_(y)` (A) weighs `(328xx22700)/(896)g//`"mol at" STP <br/> = `83.1g//mol`<br/> Hence , molecular mass of `C_(x)H_(y)` (A) =`83.1g mol^(-)` To determine the empirical formula of hydrocarbon (A) <br/> `{:("Elament" ,%,"Atomic mass","Relative ratio","Relative no. of atoms","Somplest ratio"),(C,87.8,12,7.31,1,3),(H,12.19,1,12.19,1.66,4.98~~<a href="https://interviewquestions.tuteehub.com/tag/5-319454" style="font-weight:bold;" target="_blank" title="Click to know more about 5">5</a>):}` <br/> Thus, Empirical formula of A si `C_(3)H_(5)` . <br/> `:.` Empirical formul mass = 36 + 5 = 41. <br/> `n = ("Molecular mass")/("Emprirical formula mass")= 83.1/41 = 2.02~~ 2` <br/> Molecular mass is double of empirical formula mass. <br/> `:.` Molecular formula is `C_(6)H_(10)` <br/> To determine the structure of compounds (A) and (B) <br/><img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/ARH_NCERT_EXE_CHM_XI_C13_S01_049_S01.png" width="80%"/> <br/> Hence, hydrogenation of hydrocarbon (A) requires 2 moles of hydrogen to form 2-methylpentane. Therefor, hyrocarbon (A) is an alkyne having five carbon atoms in a staight chain and a methyl substituent at position 2. Thus the possible structures for the alkyne (A) are I and II. <br/> <img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/ARH_NCERT_EXE_CHM_XI_C13_S01_049_S02.png" width="80%"/> <br/> Since, addition of `H_(2)O` to alkyne (A) in presence of `Hg^(2+)` , give a ketong which gives positive iodoform test, therefore, therefore, ketone (B) must be a methyl katon , i.e., it must contain a `COOH_(3)` <a href="https://interviewquestions.tuteehub.com/tag/group-1013370" style="font-weight:bold;" target="_blank" title="Click to know more about GROUP">GROUP</a>. <br/> Now addition of `H_(2)O` to alkyne (II) should give a <a href="https://interviewquestions.tuteehub.com/tag/mixture-1098735" style="font-weight:bold;" target="_blank" title="Click to know more about MIXTURE">MIXTURE</a> of two ketones in which 2- methyl pentan -3 one (minor) and 4- methylpentan -2-one ketone (B) (which shows `+ve` iodoform test) <a href="https://interviewquestions.tuteehub.com/tag/predominates-7707028" style="font-weight:bold;" target="_blank" title="Click to know more about PREDOMINATES">PREDOMINATES</a>. <br/> <img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/ARH_NCERT_EXE_CHM_XI_C13_S01_049_S03.png" width="80%"/> <br/> In contrast, addition of `H_(2)O` to alkyne (I) will give only one ketone, i.e., 4-methylpentan-2- one which gives iodoform test. <br/> <img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/ARH_NCERT_EXE_CHM_XI_C13_S01_049_S04.png" width="80%"/> <br/> Thus, hydrocabon `C_(x)H_(y)` (A) is 4-methylpent -1-yne. 4-methylpentan -2 one (gives + ve iodoform test)</body></html>


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