Q.Describe how cold, dilute alkaline $\text{KMnO}_4$ (Baeyer's reagent) can be used as a test for unsaturation, and state what happens instead with hot, concentrated $\text{KMnO}_4$.

Ozonolysis of propene gives exactly acetaldehyde + formaldehyde, identifying C; both isomeric propyl bromides eliminate HBr to give that same propene, identifying A and B; the two named conversions follow standard elimination and electrophilic-substitution routes.

(i) Identifying A, B, and C:

Ozonolysis cleaves a C=C double bond and replaces it with two C=O groups. Since ozonolysis of C gives acetaldehyde ($CH_3CHO$) and formaldehyde ($HCHO$), C must be an alkene where cleaving the double bond regenerates exactly these two carbonyl fragments — that is propene:

$CH_3\text{-}CH=CH_2 \xrightarrow{O_3,\ Zn/H_2O} CH_3CHO + HCHO$

So C = propene, $CH_3CH=CH_2$.

A and B are the two isomers of $C_3H_7Br$ that both give propene on dehydrobromination (E2/elimination with alcoholic KOH):

• A = 1-bromopropane, $CH_3CH_2CH_2Br$ (n-propyl bromide) — eliminates HBr from C1–C2 to give $CH_3CH=CH_2$.
• B = 2-bromopropane, $CH_3CHBrCH_3$ (isopropyl bromide) — eliminates HBr symmetrically to also give $CH_3CH=CH_2$.

(ii)(a) $CH_2=CH_2 \rightarrow HC\equiv CH$:

Step 1 — add bromine across the double bond: $CH_2=CH_2 + Br_2 \rightarrow CH_2Br\text{-}CH_2Br$ (1,2-dibromoethane).

Step 2 — double dehydrohalogenation with excess alcoholic KOH: $CH_2Br\text{-}CH_2Br + 2KOH_{(alc,\ excess)} \rightarrow HC\equiv CH + 2KBr + 2H_2O$.

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