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

$${H_2}/Pd/BaS{O_4}$$    reduces an alkyne to cis - alkene, $${H_2}/Pt$$  reduces it to alkane, $$NaB{H_4}$$  does not reduce an alkyne. Reduction ofan alkyne by active metal in liq. $$N{H_3}$$  gives trans - alkene.

Bromine is more selective
∴ it will form \[{{3}^{\circ }}\] halide if there is presence of \[{{3}^{\circ }}\] hydrogen.
∴ $$Q, S, T,$$   form \[{{3}^{\circ }}\] halide as major and $$P, R, U$$   form \[{{2}^{\circ }}\] halide as major.

The Friedal-crafts acylation reaction will give propyl phenyl ketone which further on Clemmenson’s reduction will give butyl benzene
$${C_6}{H_6} + C{H_3}C{H_2}C{H_2}COCl$$       \[\xrightarrow{AlC{{l}_{3}}}{{C}_{6}}{{H}_{5}}COC{{H}_{2}}C{{H}_{2}}C{{H}_{3}}\]       \[\xrightarrow{Zn-Hg/HCl}\underset{\text{Butyl benzene}}{\mathop{{{C}_{6}}{{H}_{5}}C{{H}_{2}}C{{H}_{2}}C{{H}_{2}}C{{H}_{3}}}}\,\]

2, 3-dichloro butane will exhibit
optical isomerism due to the presence of two asymmetric carbon atom.

$$C{H_2} = CH - CH = CH - CH$$       $$ = CH - CH = C{H_2}$$
No. of $$\sigma $$ - bonds = 17, No. of $$\pi $$ - bonds = 4

Ethylene has restricted rotation [ due to $$C = C$$  ], acetylene no rotation [ due to $$C \equiv C$$  ], hexachloroethane has more rotation than ethylene but less than ethane because of greater size of the substituent (chlorine) than in ethane ( substituent is hydrogen ).

Reaction of $$HBr$$  with propene in the presence of peroxide gives $$n$$ - $$propyl$$  $$bromide.$$  This addition teaction is an example of anti-Markownikoff's addition reaction.
( i.e. it is completed in form of free radical addition )
\[C{{H}_{3}}-CH=C{{H}_{2}}+HBr\xrightarrow{\text{Peroxide}}\]        \[\underset{n\text{-propyl bromide}}{\mathop{C{{H}_{3}}-C{{H}_{2}}-C{{H}_{2}}BR}}\,\]
Mechanism of this reaction is represented as follows :
Step (i) Formation of free radical of peroxide by means of decomposition.

Step (ii) Benzoate free radical forms bromine free radical with $$HBr.$$
$${C_6}{H_5}CO\dot O + H - Br \to $$      $${C_6}{H_5}COOH + B\dot r$$
Step (iii) Bromine free radical attacks on $$C=C$$  of propene to form intermediate free radical.

Hence, $$C{H_3} - \dot CH - C{H_2}Br$$      is the major product of this step.
Step (iv) More stable free radical accept hydrogen free radical from benzoic acid and give final product of reaction. i.e. $$n$$ - $$propyl$$  $$bromide.$$
$$C{H_3} - \dot CH - C{H_2}Br + {C_6}{H_5}COOH$$        $$ \to \mathop {C{H_3} - C{H_2} - C{H_2}Br}\limits_{n{\text{ - propyl bromide}}} + {C_6}{H_5}CO\dot O$$
Step V Benzoate free radicals are changed into benzoyl peroxide for the termination of free radical chain.
$${C_6}{H_5}CO\dot O + {C_6}{H_5}CO\dot O$$      $$ \to {\left( {{C_6}{H_5}COO} \right)_2}$$

When alkyl benzene are oxidised with alkaline $$KMn{O_4},$$   ( strong oxidising agent ) the entire alkyl group is oxidised to $$ - COOH$$   group regardless of length of side chain.