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 H_{2}SO_{4}and HgSO_{4} gives a ketone ‘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.

Answers (1)

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 = \frac{3.28 \times 22400 ml}{896 ml}

                                                                = 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 = C_{3}H_{5}

Empirical formula with weight = 3 \times12 = 36 + 5 = 41

Molecular formula = (empirical formula) n, where n = mol mass/empirical mass = \frac{831}{41}=2.02

Molecular formula = [C_{3}H_{5}]_{2} = C_{6}H_{10}

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

C_{6}H_{12} is obtained when hydrogenation of C_{5}H_{10} happens with two moles of H2. The structure is methylpentane.

C_{6}H_{12}O (which gives positive iodoform test) is obtained when hydration of (A) takes place in the presence of dil. H=H_{2}SO_{4} and HgSO_{4}.

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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