Alcohols

(b) $CH_3CH_2\text{-}OH$

$LiAlH_4$ reduces ethanal (X) to ethanol (Y). When ethanol vapour is passed over hot copper at 300 °C, it is dehydrogenated back to ethanal:

$$CH_3CHO \xrightarrow{LiAlH_4} CH_3CH_2OH \xrightarrow{Cu,\ 300^\circ C} CH_3CHO + H_2$$

Functional isomers of $C_2H_6O$ [2]

Preparation of monohydric alcohol (ethanol as the example) [3]

From an alkyl halide: boiling with aqueous alkali replaces the halogen by –OH (nucleophilic substitution).

$$C_2H_5Cl + KOH(aq) \xrightarrow{\Delta} C_2H_5OH + KCl$$

From a primary amine: treating with nitrous acid ($NaNO_2 + HCl$, 0–5 °C) replaces $-NH_2$ by –OH and liberates nitrogen.

$$C_2H_5NH_2 + HNO_2 \rightarrow C_2H_5OH + N_2 + H_2O$$

From an ester: hydrolysis with dilute acid or alkali gives the alcohol and the acid (or its salt).

$$CH_3COOC_2H_5 + H_2O \underset{}{\overset{H^+}{\rightleftharpoons}} CH_3COOH + C_2H_5OH$$ $$CH_3COOC_2H_5 + NaOH \rightarrow CH_3COONa + C_2H_5OH$$

(a) acidic nature of alcohol

The O–H hydrogen of the alcohol is replaced by the active metal sodium and $H_2$ gas is set free, forming sodium alkoxide. This shows that alcohols are weakly acidic (a protic hydrogen behaves like an acid).

i) Fermentation: the slow decomposition of complex organic compounds into simpler compounds by the action of enzymes produced by micro-organisms such as yeast. It is usually accompanied by effervescence of $CO_2$. Example: sugar → ethanol + $CO_2$.

ii) Molasses: the dark brown, thick syrupy mother liquor left after crystallisation of cane sugar from sugarcane juice. It still contains about 50–60% sugar (sucrose, glucose and fructose) that cannot be crystallised. It is the cheapest raw material for ethyl alcohol.

iii) Function of yeast: yeast is the source of the enzymes that act as biochemical catalysts.

• Invertase hydrolyses sucrose into glucose and fructose.
• Zymase converts glucose and fructose into ethanol and $CO_2$.

iv) Reactions:

$$\underset{\text{sucrose}}{C_{12}H_{22}O_{11}} + H_2O \xrightarrow{\text{invertase}} \underset{\text{glucose}}{C_6H_{12}O_6} + \underset{\text{fructose}}{C_6H_{12}O_6}$$ $$C_6H_{12}O_6 \xrightarrow{\text{zymase}} 2C_2H_5OH + 2CO_2\uparrow$$

v) Absolute vs denatured alcohol:

Hydroboration–oxidation of propene: propene reacts with diborane to form tripropylborane. This is then oxidised with alkaline $H_2O_2$ to give propan-1-ol (anti-Markovnikov addition of water).

$$6CH_3CH=CH_2 + B_2H_6 \rightarrow 2(CH_3CH_2CH_2)_3B$$ $$(CH_3CH_2CH_2)_3B + 3H_2O_2 \xrightarrow{NaOH} 3CH_3CH_2CH_2OH + H_3BO_3$$

(a) $Al_2O_3$

When alcohol vapour is passed over heated alumina (about 350 °C), water is removed and an alkene forms, e.g. $C_2H_5OH \xrightarrow{Al_2O_3,\ 350^\circ C} CH_2=CH_2 + H_2O$. CaO, $CaCl_2$ and $Ca(OH)_2$ do not act as dehydration catalysts for alcohols.

i) Isomers A and B [1]

• (A) Propan-1-ol (1° alcohol): $CH_3\text{-}CH_2\text{-}CH_2\text{-}OH$
• (B) Propan-2-ol (2° alcohol):

        OH
        |
  CH3 - CH - CH3

ii) Nitrolic acid from A (Victor Meyer's method) [2]

$$CH_3CH_2CH_2OH \xrightarrow{red\ P/I_2} CH_3CH_2CH_2I \xrightarrow{AgNO_2} CH_3CH_2CH_2NO_2$$

The 1-nitropropane is treated with nitrous acid ($NaNO_2$ + dil. $H_2SO_4$). The $-CH_2-$ group next to $-NO_2$ reacts:

$$CH_3CH_2CH_2NO_2 + HO\text{-}N=O \rightarrow CH_3CH_2\text{-}C(=NOH)\text{-}NO_2 + H_2O$$

The product is propanenitrolic acid. It dissolves in NaOH and gives a blood-red colour.

iii) Conversion of A into B [2]

Step 1. Dehydrate propan-1-ol to propene:

$$CH_3CH_2CH_2OH \xrightarrow[443\ K]{conc.\ H_2SO_4} CH_3CH=CH_2 + H_2O$$

Step 2. Hydrate propene (Markovnikov addition) with conc. $H_2SO_4$, then boil with water:

$$CH_3CH=CH_2 + H_2SO_4 \rightarrow CH_3CH(OSO_3H)CH_3 \xrightarrow[\Delta]{H_2O} CH_3CH(OH)CH_3 + H_2SO_4$$

The product is propan-2-ol (B).

(a) Methyl alcohol

Illicit liquor is often mixed with methanol. In the body, methanol is oxidised to methanal (formaldehyde) and methanoic (formic) acid. These damage the optic nerve, causing blindness, and cause acidosis, which can lead to death.

Victor Meyer's test distinguishes primary, secondary and tertiary alcohols.

Principle: the alcohol is converted into a nitroalkane, which is treated with nitrous acid. The result depends on the number of H atoms on the carbon carrying the $-NO_2$ group (2 H for 1°, 1 H for 2°, none for 3°).

Steps (same for every alcohol):

1. Heat the alcohol with red phosphorus and iodine (P + $I_2$ forms $PI_3$). This gives the alkyl iodide.
2. Warm the alkyl iodide with silver nitrite ($AgNO_2$). This gives the nitroalkane.
3. Treat the nitroalkane with nitrous acid ($NaNO_2$ + dil. $H_2SO_4$), then make the solution alkaline with NaOH.

(a) Primary alcohol (e.g. ethanol)

$$CH_3CH_2OH \xrightarrow{P/I_2} CH_3CH_2I \xrightarrow{AgNO_2} CH_3CH_2NO_2$$ $$CH_3CH_2NO_2 + HONO \rightarrow \underset{\text{ethanenitrolic acid}}{CH_3\text{-}C(=NOH)\text{-}NO_2} + H_2O$$

Nitrolic acid dissolves in NaOH and gives a blood-red colour.

(b) Secondary alcohol (e.g. propan-2-ol)

$$(CH_3)_2CHOH \xrightarrow{P/I_2} (CH_3)_2CHI \xrightarrow{AgNO_2} (CH_3)_2CHNO_2$$ $$(CH_3)_2CHNO_2 + HONO \rightarrow \underset{\text{propylpseudonitrol}}{(CH_3)_2C(NO)NO_2} + H_2O$$

Pseudonitrol gives a blue colour with NaOH.

(c) Tertiary alcohol (e.g. 2-methylpropan-2-ol)

$$(CH_3)_3COH \xrightarrow{P/I_2} (CH_3)_3CI \xrightarrow{AgNO_2} (CH_3)_3CNO_2$$ $$(CH_3)_3CNO_2 + HONO \rightarrow \text{no reaction}$$

There is no H on the carbon carrying $-NO_2$, so the solution stays colourless with NaOH.

Summary

(c) $CH_3CH_2OH$

The iodoform test is given by alcohols containing the $CH_3\text{-}CH(OH)\text{-}$ group. Ethanol is first oxidised by $I_2$/NaOH to $CH_3CHO$, which then forms yellow $CHI_3$. Methanol, propan-1-ol and tert-butyl alcohol lack this group.

Principle

Ethyl alcohol is made from sugar-containing materials such as molasses by fermentation. The enzymes of yeast catalyse two reactions:

$$\underset{\text{sucrose}}{C_{12}H_{22}O_{11}} + H_2O \xrightarrow{\text{invertase}} \underset{\text{glucose}}{C_6H_{12}O_6} + \underset{\text{fructose}}{C_6H_{12}O_6}$$ $$C_6H_{12}O_6 \xrightarrow{\text{zymase}} 2C_2H_5OH + 2CO_2$$

Process (from molasses)

1. Dilution: molasses (about 50–60% sugar) is diluted with water to a sugar concentration of about 8–10%. Too much sugar or alcohol would kill the yeast.
2. Adding nutrients: a little ammonium sulphate or ammonium phosphate is added as food for the yeast.
3. Acidification: a small amount of dilute $H_2SO_4$ is added to keep the pH at about 4–5. This favours yeast and checks bacterial growth.
4. Fermentation: yeast is added and the mixture is kept at 25–30 °C for 2–3 days. Air is excluded, because oxygen would oxidise ethanol to acetic acid. $CO_2$ escapes as a by-product. The fermented liquid, called the wash, contains about 8–10% alcohol.
5. Fractional distillation: distilling the wash gives rectified spirit, which is 95.6% ethanol (an azeotrope with water).
6. Absolute alcohol: rectified spirit is distilled with benzene (azeotropic distillation), or kept over quicklime and then distilled. This removes the remaining water and gives 100% ethanol.

Molasses -> dilute (8-10% sugar) + (NH4)2SO4 + dil. H2SO4 (pH 4-5)
        -> + yeast, 25-30 °C, 2-3 days -> Wash (8-10% alcohol)
        -> fractional distillation -> Rectified spirit (95.6%)
        -> benzene / CaO -> Absolute alcohol (100%)

Propan-1-ol is a 1° alcohol and propan-2-ol is a 2° alcohol. Victor Meyer's method separates them by the colour of the final product.

Principle: the alcohol is converted into a nitroalkane, which is treated with nitrous acid. The result depends on the number of H atoms on the carbon carrying the $-NO_2$ group (2 H for 1°, 1 H for 2°, none for 3°).

Steps (same for every alcohol):

1. Heat the alcohol with red phosphorus and iodine (P + $I_2$ forms $PI_3$). This gives the alkyl iodide.
2. Warm the alkyl iodide with silver nitrite ($AgNO_2$). This gives the nitroalkane.
3. Treat the nitroalkane with nitrous acid ($NaNO_2$ + dil. $H_2SO_4$), then make the solution alkaline with NaOH.

Propan-1-ol

$$CH_3CH_2CH_2OH \xrightarrow{red\ P/I_2} \underset{\text{1-iodopropane}}{CH_3CH_2CH_2I} \xrightarrow{AgNO_2} \underset{\text{1-nitropropane}}{CH_3CH_2CH_2NO_2}$$ $$CH_3CH_2CH_2NO_2 + HONO \rightarrow \underset{\text{propanenitrolic acid}}{CH_3CH_2C(=NOH)NO_2} + H_2O$$

Propanenitrolic acid dissolves in NaOH and gives a blood-red colour.

Propan-2-ol

$$CH_3CH(OH)CH_3 \xrightarrow{red\ P/I_2} \underset{\text{2-iodopropane}}{CH_3CHICH_3} \xrightarrow{AgNO_2} \underset{\text{2-nitropropane}}{CH_3CH(NO_2)CH_3}$$ $$(CH_3)_2CHNO_2 + HONO \rightarrow \underset{\text{propylpseudonitrol}}{(CH_3)_2C(NO)NO_2} + H_2O$$

Propylpseudonitrol gives a blue colour with NaOH.

Conclusion: propan-1-ol gives a blood-red colour and propan-2-ol gives a blue colour.

(a) Ether

Alcohols $C_nH_{2n+2}O$ and ethers have the same molecular formula but different functional groups, for example $C_2H_5OH$ and $CH_3OCH_3$.

(d) Azeotropic distillation

Rectified spirit (95.6% ethanol) is an azeotrope with water, so it cannot be separated further by simple distillation. Benzene is added and the mixture is distilled; the ternary azeotrope of benzene, ethanol and water distils first and removes all the water, leaving absolute (100%) alcohol.

Ethyl ethanoate is hydrolysed by dilute acid (or heated with NaOH) to give ethanol:

$$CH_3COOC_2H_5 + H_2O \overset{H^+}{\rightleftharpoons} CH_3COOH + C_2H_5OH$$

(b) Ethylene glycol

Ethylene glycol, $HOCH_2CH_2OH$ (ethane-1,2-diol), has two $-OH$ groups. Ethyl alcohol is monohydric, glycerol is trihydric, and phenol is a monohydric aromatic compound.

(c) Ethanal

Aqueous KOH hydrolyses chloroethane to ethanol (A). Hot copper at $300^\circ C$ dehydrogenates a primary alcohol to an aldehyde (B):

$$CH_3CH_2Cl \xrightarrow{aq.\ KOH} CH_3CH_2OH \xrightarrow{Cu,\ 300^\circ C} CH_3CHO + H_2$$

Isomeric alcohols of $C_3H_8O$ (2 marks)

Victor Meyer's test (3 marks)

Principle: the alcohol is converted into a nitroalkane, which is treated with nitrous acid. The result depends on the number of H atoms on the carbon carrying the $-NO_2$ group (2 H for 1°, 1 H for 2°, none for 3°).

Steps (same for every alcohol):

1. Heat the alcohol with red phosphorus and iodine (P + $I_2$ forms $PI_3$). This gives the alkyl iodide.
2. Warm the alkyl iodide with silver nitrite ($AgNO_2$). This gives the nitroalkane.
3. Treat the nitroalkane with nitrous acid ($NaNO_2$ + dil. $H_2SO_4$), then make the solution alkaline with NaOH.

Propan-1-ol (1°):

$$CH_3CH_2CH_2OH \xrightarrow{P/I_2} CH_3CH_2CH_2I \xrightarrow{AgNO_2} CH_3CH_2CH_2NO_2$$ $$CH_3CH_2CH_2NO_2 + HONO \rightarrow \underset{\text{propanenitrolic acid}}{CH_3CH_2C(=NOH)NO_2} + H_2O$$

Propanenitrolic acid dissolves in NaOH and gives a blood-red colour.

Propan-2-ol (2°):

$$(CH_3)_2CHOH \xrightarrow{P/I_2} (CH_3)_2CHI \xrightarrow{AgNO_2} (CH_3)_2CHNO_2$$ $$(CH_3)_2CHNO_2 + HONO \rightarrow (CH_3)_2C(NO)NO_2 + H_2O$$

The product is propylpseudonitrol (2-nitro-2-nitrosopropane). With NaOH it gives a blue colour.

The blood-red colour shows propan-1-ol and the blue colour shows propan-2-ol.

A primary alcohol with molecular mass 46 is $C_2H_5OH$ (2 × 12 + 6 × 1 + 16 = 46), i.e. ethanol.

1. Boiling with NaOH and iodine (iodoform reaction): a yellow precipitate of iodoform forms.

$$C_2H_5OH + 4I_2 + 6NaOH \rightarrow CHI_3\downarrow + HCOONa + 5NaI + 5H_2O$$

2. Heating with ethanoic acid and conc. $H_2SO_4$ (esterification): the acid derivative formed is the ester ethyl ethanoate, which has a fruity smell.

$$CH_3COOH + C_2H_5OH \overset{conc.\ H_2SO_4}{\rightleftharpoons} CH_3COOC_2H_5 + H_2O$$

3. Heating with conc. $H_2SO_4$: the product depends on the temperature.

• With excess conc. $H_2SO_4$ at 170 °C (443 K), ethanol is dehydrated to ethene:

$$C_2H_5OH \xrightarrow[170^\circ C]{conc.\ H_2SO_4} CH_2=CH_2 + H_2O$$

• With excess alcohol at 140 °C (413 K), it gives ethoxyethane (diethyl ether):

$$2C_2H_5OH \xrightarrow[140^\circ C]{conc.\ H_2SO_4} C_2H_5OC_2H_5 + H_2O$$

4. Distinguishing ethanol from methanol (iodoform test): warm each alcohol with $I_2$ and NaOH. Ethanol gives a yellow precipitate of iodoform ($CHI_3$). Methanol has no $CH_3CH(OH)-$ group, so it gives no yellow precipitate.

(a) A primary alcohol

$$RCH_2OH \xrightarrow{[O]} RCHO \xrightarrow{[O]} RCOOH$$

In this oxidation, $-CH_2OH$ (mass 31) becomes $-COOH$ (mass 45), so the mass rises by 45 − 31 = 14. An aldehyde gains 16 units (one O atom). Ketones and secondary alcohols do not give a carboxylic acid with the same carbon chain.

Tertiary alcohol $C_4H_{10}O$ [1]

         CH3
         |
   CH3 - C - OH
         |
         CH3

$(CH_3)_3COH$ — IUPAC name: 2-methylpropan-2-ol (tert-butyl alcohol).

Victor Meyer's method for 1-propanol and 2-propanol [4]

Principle: the alcohol is converted into a nitroalkane, which is treated with nitrous acid. The result depends on the number of H atoms on the carbon carrying the $-NO_2$ group (2 H for 1°, 1 H for 2°, none for 3°).

Steps (same for every alcohol):

1. Heat the alcohol with red phosphorus and iodine (P + $I_2$ forms $PI_3$). This gives the alkyl iodide.
2. Warm the alkyl iodide with silver nitrite ($AgNO_2$). This gives the nitroalkane.
3. Treat the nitroalkane with nitrous acid ($NaNO_2$ + dil. $H_2SO_4$), then make the solution alkaline with NaOH.

1-Propanol (1°):

$$CH_3CH_2CH_2OH \xrightarrow{P/I_2} CH_3CH_2CH_2I \xrightarrow{AgNO_2} CH_3CH_2CH_2NO_2 \xrightarrow{HNO_2} \underset{\text{propanenitrolic acid}}{CH_3CH_2C(=NOH)NO_2}$$

Nitrolic acid gives a blood-red colour with NaOH.

2-Propanol (2°):

$$(CH_3)_2CHOH \xrightarrow{P/I_2} (CH_3)_2CHI \xrightarrow{AgNO_2} (CH_3)_2CHNO_2 \xrightarrow{HNO_2} \underset{\text{propylpseudonitrol}}{(CH_3)_2C(NO)NO_2}$$

Pseudonitrol gives a blue colour with NaOH.

The blood-red colour shows 1-propanol and the blue colour shows 2-propanol. (A 3° alcohol, such as the one above, would stay colourless.)

(a) 2

Propanol has the formula $C_3H_7OH$. The two possible alcohols are propan-1-ol ($CH_3CH_2CH_2OH$) and propan-2-ol ($CH_3CHOHCH_3$). They are position isomers, and no other alcohol structure fits $C_3H_8O$.

(a) Ethanol is converted to chloroethane by thionyl chloride, usually with a little pyridine. The by-products are gases:

$$CH_3CH_2OH + SOCl_2 \rightarrow CH_3CH_2Cl + SO_2 + HCl$$

Alternatively use $PCl_5$, or dry HCl with anhydrous $ZnCl_2$.

(b) Ethanal is obtained by controlled oxidation of ethanol. One method is to pass ethanol vapour over hot copper at 573 K, which dehydrogenates it:

$$CH_3CH_2OH \xrightarrow{Cu,\ 573\ K} CH_3CHO + H_2$$

Another method is to use acidified $K_2Cr_2O_7$ and distil the aldehyde out as it forms.

(c) Ethanol to ethanoic acid:

$$CH_3CH_2OH \xrightarrow{[O],\ K_2Cr_2O_7/H^+} CH_3CHO \xrightarrow{[O]} CH_3COOH$$

Use excess acidified $K_2Cr_2O_7$ (or alkaline $KMnO_4$ followed by acidification) and heat under reflux, so that the intermediate ethanal is oxidised further.

(d) In sunlight and air, diethyl ether slowly oxidises to give explosive peroxides (diethyl ether peroxide and hydroperoxide). This is why ether is stored in dark bottles and checked for peroxides before distillation.