Hydrocarbons (Alkane, Alkene and Alkyne) MCQ Questions & Answers in Organic Chemistry | Chemistry

Rate of reaction of alkanes with halogens is $${F_2} > C{l_2} > B{r_2} > {I_2}.$$
Reactivity decreases down the group as the electronegativity of the halogen decreases down the group.

All are aromatic compounds.
$$4n + 2 = 10 \Rightarrow n = 2$$
$$4n + 2 = 14 \Rightarrow n = 3$$
$$4n + 2 = 14 \Rightarrow n = 3$$

With ammoniacal $$C{u_2}C{l_2}$$  solution, terminal alkynes form red ppt. of copper acetylides.

Lower alkanes $$\left( {{C_1}\,{\text{to}}\,{C_4}} \right)$$   are gaseous at room temperature.

Key Idea In the presence of halogen carrier, electrophilic substitution occurs while in the presence of sunlight, substitution, occurs at the side chain.

$$FeC{l_3} + C{l_2} \to FeCl_4^ - + C{l^ + }$$
Electrophile attacking species
( $$\because \,\, - C{H_3}$$   is an $$o/p$$ - directing group. )
In presence of $$hv,$$  reaction is free radical substitution reaction.

Acetylene have shortest $$C-C$$  bond length because it have $$C \equiv C$$  triple bond. The bond length follows the following order
$$C - C > C \equiv C > C \equiv C$$

$$B{r^ \bullet }$$ is less reactive and more selective and so the most stable free radical $$\left( {{3^ \circ }} \right)$$  will be the major product.

Chlorination of methane is free radical substitution. To get chlorine free radical energy is required which is provided by sunlight to initiate the reaction.
\[C{{l}_{2}}\xrightarrow{h\upsilon }C\dot{l}+C\dot{l}\]

The reaction is hydroboration which is addition of $${H_2}O$$  to alkene ( anti Makownikoff's's rule ).

Conversion of $${C_4}{H_6}$$  to $${C_4}{H_6}B{r_2}$$   indicates that the compound is either butyne-1 or butyne-2. However, white precipitate with ammonical silver nitrate solution indicates that it is a terminal alkyne, i.e. butyne-1 and not butyne-2