1, 2-dibromoethane reacts with alcoholic KOH to yield a product X. The hybridization state of the carbons present in X respectively are:

Question

1, 2-dibromoethane reacts with alcoholic KOH to yield a product X. The hybridization state of the carbons present in X respectively are: (A)  sp,sp  (B)  sp3,sp3  (C)  sp2,sp2  (D)  sp3,sp2

Tardigrade Admin · Accepted Answer

Alkyl halides give elimination reaction with alcoholic KOH and yield an alkene or alkyne (From dihalides) e.g., underset1,-2 textdibromo textethane mathopBr-CH2-CH2-Br xrightarrow[Δ ]alc.KOH undersetacetylene mathop oversetsp mathopCH ≡ C oversetsp mathopH Hence, product has both carbon sp- hybrid.

Q. 1, 2-dibromoethane reacts with alcoholic KOH to yield a product X. The hybridization state of the carbons present in X respectively are:

$s p, s p$

$s p^{3}, s p^{3}$

$s p^{2}, s p^{2}$

$s p^{3}, s p^{2}$

Alkyl halides give elimination reaction with alcoholic KOH and yield an alkene or alkyne (From dihalides) e.g., $1, - 2 dibromo ethane B r - C H_{2} - C H_{2} - B r a l c . K O H Δ a ce t y l e n e C H s p \equiv C H s p$ Hence, product has both carbon sp- hybrid.

Q. 1, 2-dibromoethane reacts with alcoholic KOH to yield a product X. The hybridization state of the carbons present in X respectively are:

sp,sp

sp3,sp3

sp2,sp2

sp3,sp2

Alkyl halides give elimination reaction with alcoholic KOH and yield an alkene or alkyne (From dihalides) e.g., 1,−2dibromoethaneBr−CH2​−CH2​−Br​alc.KOHΔ​acetyleneCHsp≡CHsp​ Hence, product has both carbon sp- hybrid.

$s p, s p$

$s p^{3}, s p^{3}$

$s p^{2}, s p^{2}$

$s p^{3}, s p^{2}$

Alkyl halides give elimination reaction with alcoholic KOH and yield an alkene or alkyne (From dihalides) e.g., $1, - 2 dibromo ethane B r - C H_{2} - C H_{2} - B r a l c . K O H Δ a ce t y l e n e C H s p \equiv C H s p$ Hence, product has both carbon sp- hybrid.