Which gives maximum yield of $C_{2} H_{5} C l$ ?

C_{2} H_{6} + {C l}_{2} h v, l i g h t - ------------- \to C_{2} H_{5} C l + H C l

C l_{2}

is in excess.

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C_{2} H_{6} + {C l}_{2} h v ⟶ l i g h t C_{2} H_{5} C l + H C l

C_{2} H_{6}

is in excess.

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C_{2} H_{6} + {C l}_{2} h v ⟶ l i g h t C_{2} H_{5} C l

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C_{2} H_{6} + {C l}_{2} D a r k - ---------- \to C_{2} H_{5} C l + H C l

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Solution

The correct option is B $C_{2} H_{6} + {C l}_{2} h v ⟶ l i g h t C_{2} H_{5} C l + H C l$
$C_{2} H_{6}$ is in excess.
$C_{2} H_{6} + {C l}_{2} h v ⟶ l i g h t C_{2} H_{5} C l + H C l$
excess
if halide will be in excess, then di- and tri- halide will form.

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Similar questions

A

B

C

C_{2} H_{5} O H P {C l}_{5} - ------ \to A K C N - ------- \to B H_{3} O^{+} - ------- \to C_{2} H_{5} C O O H N H_{3} - ------ \to Δ C

C_{6} H_{5} C O O H S O {C l}_{2} - ------- \to A N H_{3} - ------ \to B {B r}_{2} / K O H - ----------- \to C

C_{6} H_{5} N H_{2} + C H {C l}_{3} + 3 K O H A l c o h o l i c - ---------- \to S o l u t i o n C_{6} H_{5} N C + 3 K C l + 3 H_{2} O

(A) α - Amino acid (C_{3} H_{7} N O_{3}) M e O H - --- \to H C l (B) (C_{4} H_{10} N O_{3} C l) P C l_{5} - -- \to (C) (C_{4} H_{9} N O_{2} C l_{2})

(C) (C_{4} H_{9} N O_{2} C l_{2}) H_{3} O^{+} - -- \to (D) (C_{3} H_{6} N O_{2} C l) N a - H g - ---- \to H C l Alanine

Condition for maximum yield of $C_{2} H_{5} C l$ is

C_{6} H_{5} N_{2} C l C u C l / H C l - --------- \to C_{6} H_{5} C l + N_{2}

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