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Write Brief Answer · Q32

Q.Identify the compound A, B, C and D in the following series of reactions
CH3-CH2-Br →alc. KOH\xrightarrow{\text{alc. KOH}} A →Cl2/CCl4\xrightarrow{Cl_2/CCl_4} B
A →ii) Zn/H2Oi) O3\xrightarrow[\text{ii) Zn/H}_2O]{\text{i) O}_3} C
B →NaNH2\xrightarrow{NaNH_2} D

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✓ Free question

Step 1. CH3-CH2-Br + alcoholic KOH is a dehydrohalogenation (elimination of HBr), giving A = CH2=CH2 (ethene).

Step 2. A (ethene) + Cl2/CCl4 is simple electrophilic halogen addition across the double bond, giving B = ClCH2-CH2Cl (1,2-dichloroethane, a vicinal dihalide).

Step 3. A (ethene) + i) O3 ii) Zn/H2O is reductive ozonolysis; since ethene is symmetric (=CH2 on both alkene carbons), BOTH fragments are identical, giving C = 2 HCHO (formaldehyde).

Step 4. B (1,2-dichloroethane) + NaNH2 is a double dehydrohalogenation (a strong base removing two successive HCl's from the vicinal dihalide, via a vinyl chloride intermediate), giving D = CH≡\equivCH (ethyne/acetylene).

✓Final answer

A = ethene; B = 1,2-dichloroethane; C = formaldehyde (2 molecules); D = ethyne.

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