Question

$C_2H_5Cl \xrightarrow[2.anhyd. AlCl_3/HCl]{1.Na/dry ether} (P) \xrightarrow{Cl_2/hv} Product(s)$ $\xrightarrow{3.KMnO_4}$ $\xrightarrow{4.conc. H_2SO_4/\Delta}$ $\xrightarrow{5.H_2/Ni}$

The total number of monochlorinated product(s) including stereoisomer is/are____.

Answer 1

3

Answer 2

Step 1 (Wurtz reaction):

2 C₂H₅Cl treated with Na/dry ether couple to give n‐butane.

Step 2 (Monochlorination):

Upon chlorination (Cl₂/hν), n‐butane gives two different types of substitution:

• Replacement at a terminal (primary) C–H gives 1‑chlorobutane (achiral, one product).

• Replacement at an internal (secondary) C–H gives 2‑chlorobutane, which is chiral (thus two enantiomers).

Total products = 1 + 2 = 3.

n‑Butane has primary and secondary hydrogens. Chlorination gives 1‑chlorobutane and 2‑chlorobutane; the latter exists as a pair of enantiomers, so overall there are 3 monochlorinated products.