Question

896 mL vapour of a hydrocarbon ‘A’ having carbon 87.80% and hydrogen 12.19% weighs 3.28 g at STP. Hydrogenation of ‘A’ gives 2-methylpentane. Also ‘A’ on hydration in the presence of H₂SO₄ and HgSO₄ gives a ketone ‘B’ having molecular formula C₆H₁₂O. The ketone ‘B’ gives a positive iodoform test. Find the structure of ‘A’ and give the reactions involved.

Answer 1

Step I

896 mL vapour of C_xH_y (A) weighs 3.28 g

22700 mL vapour of C_xH_y (A) weighs `(3.28 xx 22700)/896` g mol^–1 = 83.1 g mol^–1 

Step II

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

Empirical formula of 'A'C3H5

Empirical formula mass = 35 + 5 = 41 u

n = `"Molecular mass"/"Empirical formula mass" = 83.1/41` = 2.02 ≈ 2

⇒ Molecular mass is double of the empirical formula mass.

∴ Molecular Formula is C₆H₁₀

Step III

\[\ce{\underset{(A)}{C6H10} ->[2H2] 2-methylpentane}\]

Structure of 2-methylpentane is

\[\begin{array}{cc}
\ce{CH3}\phantom{...................}\\
\backslash\phantom{..............}\\
\phantom{............}\ce{CH - CH2 - CH2 - CH3}\\
/\phantom{.............}\\
\ce{CH3}\phantom{...................}
\end{array}\]

Hence, the molecule has a five carbon chain with a methyl group at the second carbon atom.

'A' adds a molecule of H₂O in the presence of Hg²⁺ and H⁺, it should be an alkyne. Two possible structures for 'A' are:

\[\begin{array}{cc} \ce{CH3}\phantom{...................}\\ \backslash\phantom{..............}\\ \phantom{.......}\ce{CH - C ≡ C - CH3}\\ /\phantom{..............}\\ \ce{CH3}\phantom{...................} \end{array}\]

I

or

\[\begin{array}{cc} \ce{CH3}\phantom{...................}\\ \backslash\phantom{..............}\\ \phantom{.........}\ce{CH - CH2 - C ≡ CH}\\ /\phantom{..............}\\ \ce{CH3}\phantom{...................} \end{array}\]

II

Since the ketone (B) gives a positive iodoform test, it should contain a – COCH₃ group. Hence the structure of ketone is as follows:

\[\begin{array}{cc}
\ce{CH3}\phantom{...................}\\
\backslash\phantom{..............}\\
\phantom{...........}\ce{CH - CH2 - CO - CH3}\\
/\phantom{..............}\\
\ce{CH3}\phantom{...................}
\end{array}\]

Therefore structure of alkyne is II.