The correct statement about the synthesis of erythritol $$(C(CH_2OH)_4)$$ used in the preparation of PETN is

We have to see how the four-carbon polyhydric alcohol erythritol, whose condensed formula is $$C(CH_2OH)_4$$ and whose open-chain structure is $$HOCH_2-CHOH-CHOH-CH_2OH$$, can be obtained from the simplest carbonyl compound formaldehyde $$HCHO$$ under strongly basic conditions. The classical sequence that operates in such a medium is popularly called the formose reaction. In this reaction two fundamental name reactions occur again and again:

(i) Aldol condensation: “An enolate ion of one aldehyde (or ketone) adds to the carbonyl carbon of another aldehyde (or ketone) to give a β-hydroxy-aldehyde (or β-hydroxy-ketone).”

(ii) Cannizzaro reaction: “An aldehyde that has no α-hydrogen atom undergoes disproportionation in strong base; one molecule is reduced to an alcohol and a second molecule is oxidised to the salt of the corresponding acid.”

Formaldehyde itself has no α-hydrogen, therefore every time it is present in base it is potentially able to suffer a Cannizzaro reaction; however, as soon as only one new carbon-carbon bond has been made, the product does possess α-hydrogen and can now participate as the enolate component in a fresh aldol step. Working this way the carbon chain grows by one carbon unit per aldol step until the desired tetrose (four-carbon aldehyde) is reached. At that stage the aldehyde function is finally removed (reduced) by a Cannizzaro step, giving the required tetra-alcohol erythritol.

Let us follow every individual transformation, showing all the algebraic details of the stoichiometry.

First aldol condensation

Two molecules of formaldehyde combine in base:

$$HCHO + HCHO \xrightarrow{\,OH^-\,} HOCH_2CHO$$

The product is glycolaldehyde $$HOCH_2CHO$$ (a C2 unit). This is the first aldol condensation.

Second aldol condensation

The new C2 aldehyde has an α-hydrogen at the CH2 group. Its enolate therefore adds to a fresh molecule of formaldehyde:

$$HOCH_2CHO + HCHO \xrightarrow{\,OH^-\,} HOCH_2CH(OH)CH_2CHO$$

The product is glyceraldehyde $$HOCH_2CH(OH)CH_2CHO$$ (C3). This is the second aldol condensation.

Third aldol condensation

Glyceraldehyde again possesses an α-hydrogen (on the central carbon). Its enolate can add to yet another molecule of formaldehyde:

$$HOCH_2CH(OH)CH_2CHO + HCHO \xrightarrow{\,OH^-\,} HOCH_2CH(OH)CH(OH)CH_2CHO$$

The product is the tetrose erythrose $$HOCH_2CH(OH)CH(OH)CHO$$ (C4). This is the third aldol condensation. At this point the required four-carbon skeleton and three of the four hydroxyl groups are already present; only the aldehydic carbon must still be reduced to an alcohol.

One Cannizzaro reaction

Erythrose by itself has an α-hydrogen at C-2, so a normal Cannizzaro reaction cannot occur intramolecularly. In practice a very concentrated strongly basic medium is used so that two molecules of erythrose participate in a “cross-Cannizzaro” process: one molecule is reduced to the alcohol while the other is oxidised to the corresponding acid salt. Symbolically we write

$$2\,HOCH_2CH(OH)CH(OH)CHO \xrightarrow{\,conc.\;OH^-\,} HOCH_2CH(OH)CH(OH)CH_2OH \;+\; HOCH_2CH(OH)CH(OH)COO^-$$

The reduced product on the right is precisely erythritol $$HOCH_2CH(OH)CH(OH)CH_2OH$$, while the oxidised product (erythronic acid salt) is discarded. This step is the single Cannizzaro reaction involved in the overall synthesis.

Collecting the evidence we see clearly that

• exactly three aldol condensations were required to build the C4 framework, and
• exactly one Cannizzaro reaction was required to convert the terminal aldehyde to the fourth hydroxyl group.

No α-hydrogen of ethanol or methanol is involved (they are alcohols, not aldehydes), and there is certainly no need for four aldol condensations or for two Cannizzaro steps. Therefore only statement A is true.

Hence, the correct answer is Option A.