Alcohol, Phenol & Ether flash cards

1. 1.How are alcohols and phenols classified based on the number of $-OH$ groups?

   By the number of hydroxyl groups: mono-, di-, tri- or polyhydric. E.g. methanol (mono), ethane-1,2-diol / glycol (di), propane-1,2,3-triol / glycerol (tri).

   Hint: Count the $-OH$ groups.

2. 2.Define primary $(1^\circ)$, secondary $(2^\circ)$ and tertiary $(3^\circ)$ alcohols.

   Based on the carbon bearing $-OH$: in $1^\circ$ the $-OH$ carbon is attached to one C, in $2^\circ$ to two C's, in $3^\circ$ to three C's. E.g. $\text{CH}_3\text{CH}_2\text{OH}$ ($1^\circ$), $(\text{CH}_3)_2\text{CHOH}$ ($2^\circ$), $(\text{CH}_3)_3\text{COH}$ ($3^\circ$).

   Hint: How many carbons on the carbinol carbon?

3. 3.Distinguish allylic, benzylic and vinylic alcohols.

   Allylic: $-OH$ on a carbon adjacent to $C=C$ ($\text{CH}_2=\text{CH-CH}_2\text{OH}$). Benzylic: $-OH$ on a carbon attached to a benzene ring. Vinylic: $-OH$ directly on a doubly-bonded (sp$^2$) carbon ($\text{CH}_2=\text{CH-OH}$), which is unstable (enol).

   Hint: Position of $-OH$ relative to double bond / ring.

4. 4.What is the difference between a phenol and an aromatic alcohol?

   In a phenol the $-OH$ is attached directly to a carbon of the aromatic ring. In an aromatic (benzylic) alcohol the $-OH$ is on a side-chain carbon, e.g. benzyl alcohol $\text{C}_6\text{H}_5\text{CH}_2\text{OH}$.

   Hint: Is $-OH$ on the ring or the side chain?

5. 5.How are ethers classified as simple and mixed?

   Simple (symmetrical): both groups on oxygen are identical, e.g. $\text{CH}_3\text{OCH}_3$. Mixed (unsymmetrical): the two groups differ, e.g. $\text{CH}_3\text{OC}_2\text{H}_5$.

   Hint: Are the two R groups the same or different?

6. 6.What is the general formula and functional group of alcohols and ethers?

   Both have general formula $\text{C}_n\text{H}_{2n+2}\text{O}$ (saturated). Alcohols contain the hydroxyl group $-OH$; ethers contain $-O-$ linking two carbons ($R-O-R'$). They are functional isomers of each other.

   Hint: Same molecular formula, different linkage.

7. 7.Why is the C–O–H bond angle in methanol ($108.9^\circ$) close to tetrahedral?

   The oxygen in alcohols is $sp^3$ hybridised. The slightly larger-than-water angle is due to lower repulsion between the bulky methyl group and H compared with lone-pair repulsion in water.

   Hint: Hybridisation of O and steric effect.

8. 8.What industrial method gives alcohols by acid-catalysed hydration of alkenes?

   Alkene + water in presence of acid (e.g. $\text{H}_2\text{SO}_4$) adds $\text{H-OH}$ across $C=C$ following Markovnikov's rule, giving the alcohol with $-OH$ on the more substituted carbon.

   Hint: Markovnikov addition of water.

9. 9.How does hydroboration–oxidation of an alkene give an alcohol?

   $\text{B}_2\text{H}_6$ adds across $C=C$ (anti-Markovnikov, syn) to give a trialkylborane, which on oxidation with alkaline $\text{H}_2\text{O}_2$ gives the alcohol with $-OH$ on the less substituted carbon (anti-Markovnikov product).

   Hint: Diborane then $\text{H}_2\text{O}_2/\text{OH}^-$; anti-Markovnikov.

10. 10.What alcohol forms when a primary alkyl halide is hydrolysed by aqueous $KOH$?

    A primary alcohol. $R-X + \text{OH}^- \rightarrow R-OH + X^-$ (nucleophilic substitution). Aqueous alkali favours substitution (alcohol) whereas alcoholic $KOH$ favours elimination (alkene).

    Hint: Aqueous vs alcoholic KOH.

11. 11.How are alcohols prepared from aldehydes and ketones by reduction?

    Aldehydes give primary alcohols; ketones give secondary alcohols. Reducing agents: $\text{H}_2/\text{Ni}$ (catalytic), or $\text{NaBH}_4$ / $\text{LiAlH}_4$.

    Hint: $>C=O \rightarrow >CH-OH$.

12. 12.What alcohols result from $LiAlH_4$ reduction of carboxylic acids and esters?

    Carboxylic acids and esters are reduced to primary alcohols. $\text{RCOOH} \xrightarrow{\text{LiAlH}_4} \text{RCH}_2\text{OH}$. $\text{LiAlH}_4$ is a strong reducing agent needed for these; $\text{NaBH}_4$ generally cannot reduce acids/esters.

    Hint: Strong hydride reduces the carbonyl fully.

13. 13.How do Grignard reagents give alcohols from carbonyl compounds?

    $R\text{MgX}$ adds to $>C=O$; the alkoxide intermediate is hydrolysed to the alcohol. Formaldehyde → $1^\circ$ alcohol, other aldehydes → $2^\circ$ alcohol, ketones → $3^\circ$ alcohol.

    Hint: HCHO, RCHO, RCOR' give $1^\circ,2^\circ,3^\circ$.

14. 14.QUESTION: What is the product when $\text{HCHO}$ reacts with $\text{CH}_3\text{MgBr}$ followed by hydrolysis?

    Ethanol ($\text{CH}_3\text{CH}_2\text{OH}$), a primary alcohol. Formaldehyde always gives a $1^\circ$ alcohol with a Grignard reagent.

    Hint: Formaldehyde → primary alcohol.

15. 15.QUESTION: Which alcohol forms from acetone $(\text{CH}_3)_2\text{CO}$ + $\text{CH}_3\text{MgBr}$ then hydrolysis?

    2-methylpropan-2-ol (tert-butyl alcohol), $(\text{CH}_3)_3\text{COH}$ — a tertiary alcohol, since ketones give $3^\circ$ alcohols with Grignard reagents.

    Hint: Ketone → tertiary alcohol.

16. 16.How is phenol prepared industrially from cumene?

    Cumene (isopropylbenzene) is oxidised by air to cumene hydroperoxide, which on treatment with dilute acid gives phenol and acetone. This is the major industrial route.

    Hint: Cumene → hydroperoxide → phenol + acetone.

17. 17.How is phenol prepared from chlorobenzene (Dow process)?

    Chlorobenzene is heated with $\text{NaOH}$ at ~$623\,\text{K}$ and $320\,\text{atm}$; acidification of the sodium phenoxide gives phenol.

    Hint: Harsh conditions: high T and P with NaOH.

18. 18.How is phenol prepared from benzenesulphonic acid?

    Benzene is sulphonated, then the sodium benzenesulphonate is fused with $\text{NaOH}$ (molten alkali). Acidification of the resulting sodium phenoxide gives phenol.

    Hint: Fuse the sulphonate with NaOH.

19. 19.How is phenol prepared from aniline via diazonium salt?

    Aniline is diazotised with $\text{NaNO}_2/\text{HCl}$ at $273$–$278\,\text{K}$ to give benzenediazonium chloride, which on warming with water gives phenol (with release of $\text{N}_2$).

    Hint: Diazonium salt + warm water.

20. 20.Why do lower alcohols have much higher boiling points than comparable hydrocarbons or ethers?

    Alcohols form intermolecular hydrogen bonds through the $-OH$ group, requiring extra energy to break. Ethers and hydrocarbons lack $-OH$ so cannot self-associate by H-bonding.

    Hint: Presence of $-OH$ enables association.

21. 21.Why does boiling point of alcohols increase with molecular mass but decrease with branching?

    Increasing chain length increases van der Waals surface area, raising boiling point. Branching makes the molecule more spherical, decreasing surface contact and lowering the boiling point.

    Hint: Surface area effect on van der Waals forces.

22. 22.Why are lower alcohols soluble in water while solubility decreases with chain length?

    The $-OH$ group forms hydrogen bonds with water. As the hydrophobic hydrocarbon (alkyl) part grows, it dominates and reduces solubility.

    Hint: Polar $-OH$ vs non-polar alkyl chain.

23. 23.Why are ethers only slightly soluble in water although they resemble alcohols?

    Ether oxygen can accept hydrogen bonds from water (its lone pairs), giving slight solubility comparable to alcohols of similar mass, but ethers cannot donate H-bonds among themselves, so solubility is limited.

    Hint: O accepts but cannot donate H-bonds.

24. 24.Why is the boiling point of ethers close to that of alkanes of comparable mass?

    Ethers cannot form intermolecular hydrogen bonds (no $-OH$); they are held only by weak dipole–dipole and van der Waals forces, like alkanes, so their boiling points are much lower than isomeric alcohols.

    Hint: No H-bonding in pure ether.

25. 25.Are alcohols and phenols acidic, basic, or amphoteric toward reactions at $-OH$?

    They are weakly acidic (can donate the $-OH$ proton) and also weakly basic (O lone pair can accept a proton). Reaction of $-OH$ can involve either C–O or O–H bond cleavage.

    Hint: $-OH$ can lose H$^+$ or accept H$^+$.

26. 26.Why are alcohols weaker acids than water?

    Alkyl groups are electron-donating ($+I$), increasing electron density on oxygen and destabilising the alkoxide ion ($RO^-$). This makes $RO^-$ a stronger base and $ROH$ a weaker acid than water.

    Hint: +I effect destabilises the conjugate base.

27. 27.Arrange the acidity order of $1^\circ$, $2^\circ$ and $3^\circ$ alcohols.

    Acidity: $1^\circ > 2^\circ > 3^\circ$ (in solution/toward acid strength of the alcohol). More alkyl groups increase $+I$ effect, destabilise the alkoxide, and decrease acidity.

    Hint: More alkyl groups → weaker acid.

28. 28.Why is phenol more acidic than ethanol?

    In phenol, the phenoxide ion is resonance-stabilised (negative charge delocalised onto the ring), and the O is attached to $sp^2$ carbon. In ethanol the ethoxide has no such stabilisation, so phenol is far more acidic (pKa ~10 vs ~16).

    Hint: Resonance stabilisation of phenoxide.

29. 29.Why is phenol a weaker acid than carboxylic acids?

    In carboxylate ion the negative charge is delocalised over two electronegative oxygen atoms (equivalent resonance), giving greater stabilisation than the phenoxide ion where charge is spread onto less electronegative ring carbons.

    Hint: Carboxylate: charge on two O atoms.

30. 30.How do electron-withdrawing groups (e.g. $-NO_2$) affect phenol's acidity?

    Electron-withdrawing groups, especially at ortho/para positions, stabilise the phenoxide ion and increase acidity. E.g. $2,4,6$-trinitrophenol (picric acid) is a very strong acid.

    Hint: EWG stabilise the negative charge → stronger acid.

Open the interactive deck for the other 62 cards, with self-grading so the ones you keep missing come back.