IIT JEE , AIEEE Forum - Chemistry , Organic Chemistry

joyfrancis

please solve example 16 on pg338 (arihant organic chemistry)

magiclko

wats the question, pls post i dnt have that..

khinart

Plzz post the question

ramyadiamond

Well, i think they've given it quite clearly, but i'm jst trying to gve a brief explanation of it.

Ok, so as in alkenes, for hydrogentaion process, one mole of H₂ eliminates one double bond or one pi-bond. So, here as they say 5 moles of Hydrogen is consumed, hence either there are 5 double bonds, or might also contain triple bonds. This we understand by the information that it forms ppt. with AgNO₃. So, it is a terminal alkyne,ok?

Now, 3 more pi-bonds are left. For this to find out, u need to make the structure by joining the points of oznolysis from the products, and n the structure after locating the triple bond at the terminating position, u can draw the rest of the bonds. Also, there is no triple bond in the ring, as small rings cannot accomodate triple bonds bcoz they get unstable.So, u'll have to draw the double bonds accordingly.

Well, i hope this helps. Since, u didn;t specify what type of doubt u were having, so i've written what i thought might be helpful. So, if there are any other doubts, plz specify it. Thanks.

madhusudan.chavan

1.As the compound takes 5 mol. of Hydrogen, 5 pi bonds are present.

2.It gives white ppt with AgNO3, indicates terminal alkyne, i.e. carbon carbon triple bond at the end.

3.Ozonolysis products are given.

4.Hydrogenation product is also give, showing cyclohexane ring, triple bond cannot be present in ring.

5.Write down ozonolysis products facing C=O towards each other, then join the carbon atoms by = bonds.

(For example: H2C=O O=CH2

H2C=CH2)

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