(a) HIGHER BOILING POINT OF CARBOXYLIC ACIDS

Step 1 - Both classes form hydrogen bonds.
Both alcohols $(R-OH)$ and carboxylic acids $(R-COOH)$ have an $O-H$ bond and therefore
associate through intermolecular hydrogen bonding.

Step 2 - The difference lies in the NUMBER of hydrogen bonds.
An alcohol molecule can form essentially ONE hydrogen bond per pair, giving a LINEAR
(open chain) association.
A carboxylic acid has BOTH a hydrogen-bond donor $(O-H)$ and a strong hydrogen-bond
acceptor $(C=O)$ in the same functional group. Two molecules therefore lock together
through TWO hydrogen bonds, forming a stable CYCLIC DIMER.

Step 3 - Consequence.
Because of the dimer, the effective molecular mass in the liquid state is nearly DOUBLE,
and TWO hydrogen bonds (instead of one) have to be broken before a molecule can escape
into the vapour phase.
More energy is needed, so carboxylic acids have HIGHER BOILING POINTS than alcohols of
comparable molecular mass.

(b) ACIDITY OF ALPHA-HYDROGENS

Step 1 - Inductive effect.
The carbonyl group $>C=O$ is strongly ELECTRON-WITHDRAWING $(-I$ effect$)$. It pulls
electron density away from the alpha-carbon, weakening the $C_\alpha-H$ bond and making
the hydrogen easier to remove as $H^+$.

Step 2 - Resonance stabilisation of the conjugate base (the main reason).
When the alpha-hydrogen is removed, the carbanion left behind is stabilised by
RESONANCE: the negative charge is delocalised onto the highly ELECTRONEGATIVE OXYGEN
atom, giving the enolate ion.
$$-\overset{\ominus}{C}-C=O \; \longleftrightarrow \; -C=C-\overset{\ominus}{O}$$

Step 3 - Conclusion.
Because the conjugate base (enolate) is resonance stabilised, the loss of the
alpha-hydrogen is favoured. Hence alpha-hydrogens of aldehydes and ketones are ACIDIC.
This acidity is what makes the ALDOL condensation possible.

(c) NO NUCLEOPHILIC ADDITION OF $H_2N-Z$ IN STRONGLY ACIDIC MEDIUM

Step 1 - The role of acid (why some acid IS needed).
A little acid protonates the carbonyl oxygen. This makes the carbonyl carbon MORE
electrophilic (more positive), so it is attacked more readily by the nucleophile.
$$>C=O + H^+ \rightarrow >\overset{+}{C}-OH$$

Step 2 - What goes wrong if the medium is STRONGLY acidic.
Ammonia and its derivatives $(H_2N-Z$, e.g. $NH_2OH$, $NH_2NH_2$, $NH_2NHC_6H_5)$ are
BASIC, because of the lone pair on nitrogen. In a strongly acidic medium they are
completely PROTONATED:
$$H_2N-Z + H^+ \longrightarrow H_3\overset{+}{N}-Z$$

Step 3 - Consequence.
Once protonated, the lone pair on nitrogen is used up in bonding to $H^+$ and is NO
LONGER AVAILABLE. The reagent therefore LOSES ITS NUCLEOPHILIC CHARACTER and cannot
attack the carbonyl carbon.

Step 4 - Optimum condition.
These reactions are therefore carried out in a WEAKLY ACIDIC medium of about pH 4.5 -
acidic enough to activate the carbonyl group, but not acidic enough to destroy the
nucleophile.