n propyl bromide on treatment with ethanolic potassium

$$n - $$ Propyl bromide on treatment with ethanolic potassium hydroxide produces

A. Propane

B. Propene

C. Propyne

D. Propanol

Answer : Propene

Solution :
\[\underset{n-\text{Propyl bromide}}{\mathop{C{{H}_{3}}-C{{H}_{2}}-C{{H}_{2}}Br}}\,\xrightarrow[-HBr]{\text{ethanolic}\,\text{KOH}}C{{H}_{3}}-\underset{\text{Propene}}{\mathop{CH}}\,=C{{H}_{2}}\]
Further dehydrohalogenation of \[C{{H}_{3}}CH=C{{H}_{2}}\] can be done only by strong base like \[NaN{{H}_{2}}.\]

Releted MCQ Question on
Organic Chemistry >> Alkyl and Aryl Halide

Releted Question 1

Chlorobenzene can be prepared by reacting aniline with :

A. hydrochloric acid

B. cuprous chloride

C. chlorine in presence of anhydrous aluminium chloride

D. nitrous acid followed by heating with cuprous chloride

View Answer

Releted Question 2

The reaction conditions leading to the best yields of $${C_2}{H_5}Cl$$ are :

A. \[{{C}_{2}}{{H}_{6}}\text{(excess)}+C{{l}_{2}}\xrightarrow{\text{uv}\,\text{light}}\]

B. \[{{C}_{2}}{{H}_{6}}+C{{l}_{2}}\xrightarrow[\text{room temperature}]{\text{dark}}\]

C. \[{{C}_{2}}{{H}_{6}}+C{{l}_{2}}\text{(excess)}\xrightarrow{\text{uv}\,\text{light}}\]

D. \[{{C}_{2}}{{H}_{6}}+C{{l}_{2}}\xrightarrow{\text{uv}\,\text{light}}\]

View Answer

Releted Question 3