(a) (i) Draw the structures of the isomers of the alkene with molecular formurat C\(_4\)H\(_8\) (ii) State the class of alkanols to which each of the follow...

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(a)(i) Structures of the isomers of the alkene C\(_4\)H\(_8\)

The alkene of molecular formula C\(_4\)H\(_8\) exists as the following isomers (three structural isomers, of which but-2-ene shows an additional pair of geometric isomers):

Thus the structural isomers are but-1-ene, but-2-ene and 2-methylprop-1-ene; but-2-ene occurs as cis- and trans-forms.

(a)(ii) Class of alkanol

Compound	Structure	Class
I	CH\(_3\)C(CH\(_3\))\(_2\)OH (the \(-\)OH carbon is joined to three other carbons)	Tertiary (3°) alkanol
II	CH\(_3\)CH(CH\(_3\))CH\(_2\)OH (the \(-\)OH carbon is joined to one other carbon)	Primary (1°) alkanol
III	CH\(_3\)CH\(_2\)CH(CH\(_3\))OH (the \(-\)OH carbon is joined to two other carbons)	Secondary (2°) alkanol

(b)(i) Formulae of the products

I. Propanoic acid with potassium metal:

\[2CH_3CH_2COOH + 2K \rightarrow 2CH_3CH_2COOK + H_2\]

Products: CH\(_3\)CH\(_2\)COOK and H\(_2\).

II. Propanoic acid with butanol (C\(_4\)H\(_9\)OH) and concentrated H\(_2\)SO\(_4\), heat (esterification):

\[CH_3CH_2COOH + C_4H_9OH \xrightarrow[\Delta]{conc.\,H_2SO_4} CH_3CH_2COOC_4H_9 + H_2O\]

Products: CH\(_3\)CH\(_2\)COOC\(_4\)H\(_9\) and H\(_2\)O.

III. Butan-1-ol oxidised by acidified KMnO\(_4\) (excess) — the primary alkanol is oxidised fully to the carboxylic acid:

\[CH_3CH_2CH_2CH_2OH \xrightarrow{H^+/KMnO_4\,(excess)} CH_3CH_2CH_2COOH\]

Product: CH\(_3\)CH\(_2\)CH\(_2\)COOH.

(b)(ii) Names of the major product(s)

• I. Potassium propanoate (and hydrogen gas)
• II. Butyl propanoate (and water)
• III. Butanoic acid

(c) Empirical and molecular formula of hydrocarbon R

Given: mass of R = 7.0 g, volume = 2.24 dm\(^3\) at s.t.p., % H by mass = 14.3, molar volume = 22.4 dm\(^3\), H = 1.00, C = 12.0.

(i) Empirical formula

Since R is a hydrocarbon, % carbon = \(100 - 14.3 = 85.7\%\).

Element	C	H
% by mass	85.7	14.3
Moles = %÷A\(_r\)	\(\frac{85.7}{12.0}=7.14\)	\(\frac{14.3}{1.0}=14.3\)
Divide by smaller (7.14)	\(\frac{7.14}{7.14}=1\)	\(\frac{14.3}{7.14}=2\)

\[\therefore \text{Empirical formula} = CH_2\]

(ii) Molecular formula

Number of moles of R:

\[n = \frac{\text{volume at s.t.p.}}{\text{molar volume}} = \frac{2.24}{22.4} = 0.1\ \text{mol}\]

Molar mass:

\[M_r = \frac{\text{mass}}{\text{moles}} = \frac{7.0}{0.1} = 70\ \text{g mol}^{-1}\]

Let the molecular formula be \((CH_2)_n\). The empirical formula mass \(= 12.0 + 2(1.0) = 14\).

\[14n = 70 \quad\Rightarrow\quad n = \frac{70}{14} = 5\]\[\therefore \text{Molecular formula} = C_5H_{10}\]

(d) Structural isomerism

Structural isomerism is the phenomenon in which two or more compounds have the same molecular formula but different structural (arrangement of atoms) formulae, and hence different properties.