# 896 mL vapour of a hydrocarbon ‘A’ having carbon 87.80% and hydrogen 12.19% weighs 3.28g at STP. Hydrogenation of ‘A’ gives 2-methylpentane. Also ‘A’ on hydration in the presence of and  gives a ketone ‘B’ having molecular formula . The ketone ‘B’ gives a positive iodoform test. Find the structure of ‘A’ and give the reactions involved.

To calculate molar mass of hydrocarbon A it is given that 896 ml of Hydrocarbon ACxHy weighs 3.28 g at STP.

22400 ml of has ACxHy mass =

= 83.1 g/mol.

Thus, the molar mass of A = 83.1 g/mol

 Element C H Percentage 87.8% 12.19% Atomic mass 12 1 Relative ratio 7.31 12.19 Relative no. of atoms 1 1.66 Simplest ratio 3 4.98 = 5

We know that empirical formula =

Empirical formula with weight =

Molecular formula = (empirical formula) n, where n = mol mass/empirical mass =

Molecular formula =

Now, determining the structure of (A) and (B)-

is obtained when hydrogenation of happens with two moles of H2. The structure is methylpentane.

(which gives positive iodoform test) is obtained when hydration of (A) takes place in the presence of dil. H= and .

Thus, the structure of $A = (CH_{3})_{2}CH-CH-C\equiv CH$(4-methyl pent-yne) and (B) is 4-methyl pentanone-2.

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