Q.Write a note on --
a. Cannizzaro reaction
b. Stephen reaction.

The Cannizzaro reaction occurs only with aldehydes that have NO alpha-hydrogen; formaldehyde (HCHO) qualifies.

In the Cannizzaro reaction, two molecules of an aldehyde lacking an alpha-hydrogen undergo self oxidation-reduction (disproportionation) with concentrated alkali to give an alcohol and a salt of a carboxylic acid:

2 HCHO + NaOH --> CH3OH + HCOONa

Acetaldehyde (CH3CHO) and acetone have alpha-hydrogens (and undergo aldol/other reactions instead), and formic acid is not an aldehyde. Only HCHO among the options has no alpha-hydrogen.

The Cannizzaro reaction (base-induced disproportionation into an alcohol and a carboxylate) only happens for aldehydes with NO alpha-hydrogen — because an alpha-H aldehyde instead undergoes the much faster aldol condensation.

Check each option for an alpha-hydrogen (an H on the carbon directly attached to -CHO):

1. cyclohexanecarbaldehyde — the ring carbon attached to CHO carries an alpha-H → undergoes aldol, not Cannizzaro.
2. CH3CHO (acetaldehyde) — the CH3 carbon has alpha-H's → aldol.
3. 2-methylpropanal, (CH3)2CH-CHO — the CH carbon has an alpha-H → aldol.
4. benzaldehyde, C6H5-CHO — the carbon attached to CHO is an aromatic ring carbon with NO sp3 C-H adjacent to the carbonyl, i.e. no alpha-hydrogen at all → Cannizzaro is the only path available: 2 C6H5CHO + conc. NaOH → C6H5CH2OH (benzyl alcohol) + C6H5COONa (sodium benzoate).

The Cannizzaro reaction is given only by aldehydes with NO $\alpha$-hydrogen; CH3CHO has $\alpha$-hydrogens (on its methyl group), so it is oxidised/reduced by the aldol route instead and does not undergo Cannizzaro.

The Cannizzaro reaction is a base-mediated disproportionation (self oxidation–reduction) of an aldehyde lacking $\alpha$-hydrogens: two molecules of the aldehyde react with concentrated $NaOH$, one being oxidised to the carboxylate and the other reduced to the alcohol. If $\alpha$-hydrogens ARE present, the base instead deprotonates the $\alpha$-carbon and the aldehyde undergoes base-catalysed aldol condensation, which is much faster.

• (B) benzaldehyde ($C_6H_5CHO$): the carbon bonded to CHO is an aromatic ring carbon with no removable $\alpha$-H for enolisation — undergoes Cannizzaro (classic example, gives benzyl alcohol + sodium benzoate).
• (C) $HCHO$ (formaldehyde): has no $\alpha$-carbon at all — undergoes Cannizzaro (gives methanol + sodium formate).
• (D) $CH_3CHO$ (acetaldehyde): its methyl group carries three $\alpha$-hydrogens, so it undergoes base-catalysed aldol condensation preferentially and is the textbook example of an aldehyde that does NOT give Cannizzaro's reaction.

(Note: strictly, a ring aldehyde such as (A) that also carries an $\alpha$-hydrogen on the ring carbon bonded to –CHO would likewise be excluded from Cannizzaro on the same reasoning; but the standard NCERT teaching example for "does not give Cannizzaro" is acetaldehyde, so (d) is the intended answer here.)

The Cannizzaro reaction is a self oxidation-reduction of aldehydes that have NO alpha-hydrogen; formaldehyde (HCHO) qualifies.

In the Cannizzaro reaction, two molecules of an aldehyde lacking alpha-hydrogen react in concentrated alkali — one is oxidised to the acid (salt) and the other reduced to the alcohol.

• HCHO has no alpha-H → gives Cannizzaro reaction:

  2 HCHO + NaOH → CH3OH + HCOONa

• CH3CHO and CH3COCH3 have alpha-hydrogens → undergo aldol condensation instead.

• HCOOH is a carboxylic acid, not an aldehyde.

Cannizzaro reaction (base-catalysed disproportionation of an aldehyde to an alcohol and a carboxylate) only happens for aldehydes that have NO alpha-hydrogen.

Cannizzaro's reaction requires an aldehyde lacking alpha-hydrogens, because if alpha-H were present, concentrated base would instead deprotonate it and trigger aldol condensation, which is much faster.

• (a) o-methylbenzaldehyde: the carbonyl carbon is attached directly to the aromatic ring — no alpha-carbon with H exists on that side, so it undergoes Cannizzaro.

• (c) HCHO (formaldehyde): the carbonyl carbon has only H atoms on it, no alpha-carbon at all — undergoes Cannizzaro (indeed reacts fastest of all).

• (d) Benzaldehyde: same reasoning as (a), no alpha-H — undergoes Cannizzaro (the classic textbook example, giving benzyl alcohol + benzoate).

• (b) CH3CHO (acetaldehyde): the methyl group IS an alpha-carbon bearing 3 hydrogens. With base, these alpha-H's are removed instead, and the resulting enolate attacks another acetaldehyde molecule — aldol condensation, not Cannizzaro.

Formaldehyde has no α-H → Cannizzaro disproportionation with NaOH giving methanol + sodium formate.

Formaldehyde (HCHO) has no alpha-hydrogen atom, so it cannot undergo aldol condensation. Instead, with concentrated NaOH it undergoes the Cannizzaro reaction — a self oxidation-reduction (disproportionation):

2 HCHO + NaOH → CH3OH + HCOONa

One molecule of formaldehyde is reduced to methyl alcohol (CH3OH) and the other is oxidised to sodium formate (the sodium salt of formic acid).

Since the reaction is run with NaOH, the acid appears as its sodium salt (sodium formate). The two products are methyl alcohol and formate; among the given options the redox partners are formic acid (as formate) and methyl alcohol.