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896 mL vapour of a hydrocarbon 'A' having carbon 87.80%and hydrogen 12.19 % weighs 3.29 g at STP, Hydrogenation of 'A' gives 2-methylpentane. Also 'A' on hydration in the presence of H_2SO_4 and HgSO_4 gives a ketone 'B' having molecular formula C_6H_12O. The ketone 'B' gives a positive iodoform test. Find the structures of 'A' and give the reactions involved.

Answer» <html><body><p></p>Solution :<a href="https://interviewquestions.tuteehub.com/tag/step-25533" style="font-weight:bold;" target="_blank" title="Click to know more about STEP">STEP</a> 1 . To determine the molecular mass of hydrocarbon (A). <br/> 896 mL vapours of hydrocarbon (A) weigh at STP =3.28 g <br/> `therefore` 22700 mL vapours of A will weigh at STP `=(3.28xx22700)/896 "g mol"^(-1)=83.1 g` <br/> `therefore` Molecular mass of hydrocarbon (A) = `83.1 g mol^(-1)` <br/> Step 2. To determine the empirical formula of hydrocarbon (A). <br/> `{:("<a href="https://interviewquestions.tuteehub.com/tag/element-969236" style="font-weight:bold;" target="_blank" title="Click to know more about ELEMENT">ELEMENT</a>","% age","Atomic mass","<a href="https://interviewquestions.tuteehub.com/tag/relative-1183377" style="font-weight:bold;" target="_blank" title="Click to know more about RELATIVE">RELATIVE</a> ratio","Relative no. of atoms","Simplest ratio"),(C,87.8,12,7.31,1,3),(H,12.19,1,12.19,1.66,5):}` <br/> Thus, empirical formula of hydrocarbon (A) =`C_3H_5` and empirical formula mass = 12 x 3 + 5 x 1 = 41 u <br/> `therefore n="Molecular mass "/"Empirical formula mass "=83.1/41=2.02 approx 2` <br/> Thus, molecular formula of hydrocarbon (A)= 2 x Empirical formula = `2 xx C_3H_5 = C_6H_10` <br/> Step 3.To determine the structures of compounds (A) and (B) <br/> <img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/PR_CHE_02_XI_C13_E14_002_S01.png" width="80%"/> <br/> (i)Since hydrogenation of hydrocarbon (A) requires 2 <a href="https://interviewquestions.tuteehub.com/tag/moles-1100459" style="font-weight:bold;" target="_blank" title="Click to know more about MOLES">MOLES</a> of hydrogen to form 2-methylpentane, therefore, hydrocarbon (A) is an alkyne having five carbon atoms in a straigth chain and a methyl substitutent at <a href="https://interviewquestions.tuteehub.com/tag/position-1159826" style="font-weight:bold;" target="_blank" title="Click to know more about POSITION">POSITION</a> 2. Thus, the two possible structures for the alkyne (A) are I and II : <br/> <img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/PR_CHE_02_XI_C13_E14_002_S02.png" width="80%"/> <br/> (ii)Since addition of `H_2O` to alkyne (A) in presence of `Hg^(2+)` , gives a ketone which gives positive iodoform test, therefore, ketone (B) must be a methyl ketone , i.e., it must contain a `COCH_3` group. Now addition of `H_2O` to alkyne (II) should give a mixture of two ketone in which ketone (B) (Which shows +ve iodoform test ) predominates. <br/> <img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/PR_CHE_02_XI_C13_E14_002_S03.png" width="80%"/> <br/> In contrast, addition of `H_2O` 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/PR_CHE_02_XI_C13_E14_002_S04.png" width="80%"/> <br/>Thus, hydrocarbon (A) is 4-methylpent-1-yne.</body></html>


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